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Astronomy_1135

The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$e. Taking the mass of $\mathrm{M}_{8} 7^{*}$ as $6.5 \times 10^{9} \mathrm{M}_{\odot}$ : i. Determine the period of a particle in the ISCO of $\mathrm{M} 87^{*}$ for the $\mathrm{a}=1, \mathrm{a}=-1$ and $\mathrm{a}=0$ (i.e. non-spinning) cases. Give your answer in days.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$ problem: e. Taking the mass of $\mathrm{M}_{8} 7^{*}$ as $6.5 \times 10^{9} \mathrm{M}_{\odot}$ : i. Determine the period of a particle in the ISCO of $\mathrm{M} 87^{*}$ for the $\mathrm{a}=1, \mathrm{a}=-1$ and $\mathrm{a}=0$ (i.e. non-spinning) cases. Give your answer in days. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of days, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_306

已知质量分布均匀的球壳对壳内物体的万有引力为零, 假设地球是质量分布均匀的球体, 如图若在地球内挖一球形内切空腔, 有一小球自切点 $A$ 自由释放, 则小球在球形空腔内将做（） [图1] A: 自由落体运动 B: 加速度越来越大的直线运动 C: 匀加速直线运动 D: 加速度越来越小的直线运动

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 已知质量分布均匀的球壳对壳内物体的万有引力为零, 假设地球是质量分布均匀的球体, 如图若在地球内挖一球形内切空腔, 有一小球自切点 $A$ 自由释放, 则小球在球形空腔内将做（） [图1] A: 自由落体运动 B: 加速度越来越大的直线运动 C: 匀加速直线运动 D: 加速度越来越小的直线运动 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_39

火卫一是太阳系最暗的天体之一, 假设火卫一围绕火星做匀速圆周运动的轨道半径为 $r_{1}$, 运行周期为 $T_{1}$, 火星半径为 $R$ 。已知行星与卫星间引力势能的表达式为 $E_{\mathrm{p}}=-\frac{G M_{0} m_{0}}{r}, r$ 为行星与卫星的中心距离, 则火星的第二宇宙速度为 $(\quad)$ A: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ B: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R}}$ C: $\frac{2 \pi r_{1}}{R T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ D: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 火卫一是太阳系最暗的天体之一, 假设火卫一围绕火星做匀速圆周运动的轨道半径为 $r_{1}$, 运行周期为 $T_{1}$, 火星半径为 $R$ 。已知行星与卫星间引力势能的表达式为 $E_{\mathrm{p}}=-\frac{G M_{0} m_{0}}{r}, r$ 为行星与卫星的中心距离, 则火星的第二宇宙速度为 $(\quad)$ A: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ B: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R}}$ C: $\frac{2 \pi r_{1}}{R T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ D: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_1212

Hanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy. [figure1] Figure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it. Credit: Keel et al. (2012) \& Galaxy Zoo. Subsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\left(3600\right.$ arcseconds $\left.=1^{\circ}\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \mathrm{kpc}$ and a mass of $10^{11} \mathrm{M}_{\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy. In this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar.c. The gravitational potential energy of the material falling to radius, which in this case is a black hole with radius equal to the Schwarzschild radius, $R_{S}=2 G M / c^{2}$, at a mass accretion rate $\dot{m} \equiv \delta m / \delta t$, is converted into radiation with an efficiency of. Show that the power (luminosity) output of the SMBH is given by $L=\frac{1}{2} \eta \dot{m} c^{2}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: Hanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy. [figure1] Figure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it. Credit: Keel et al. (2012) \& Galaxy Zoo. Subsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\left(3600\right.$ arcseconds $\left.=1^{\circ}\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \mathrm{kpc}$ and a mass of $10^{11} \mathrm{M}_{\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy. In this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar. problem: c. The gravitational potential energy of the material falling to radius, which in this case is a black hole with radius equal to the Schwarzschild radius, $R_{S}=2 G M / c^{2}$, at a mass accretion rate $\dot{m} \equiv \delta m / \delta t$, is converted into radiation with an efficiency of. Show that the power (luminosity) output of the SMBH is given by $L=\frac{1}{2} \eta \dot{m} c^{2}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_113

宇宙中组成双星系统的甲、乙两颗恒星的质量分别为 $m 、 k m(k>1)$, 甲绕两恒星连线上一点做圆周运动的半径为 $r$, 周期为 $T$, 根据宇宙大爆炸理论, 两恒星间的距离会缓慢增大, 若干年后, 甲做圆周运动的半径增大为 $n r(n>k)$, 设甲、乙两恒星的质量保持不变, 引力常量为 $G$, 则若干年后下列说法正确的是 A: 恒星甲做圆周运动的向心力为 $G \frac{\mathrm{km}^{2}}{(\mathrm{nr})^{2}}$ B: 恒星甲做圆周运动周期变大 C: 恒星乙做圆周运动的半径为 $k n r$ D: 恒星乙做圆周运动的线速度为恒星甲做圆周运动线速度的 $\frac{1}{k}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 宇宙中组成双星系统的甲、乙两颗恒星的质量分别为 $m 、 k m(k>1)$, 甲绕两恒星连线上一点做圆周运动的半径为 $r$, 周期为 $T$, 根据宇宙大爆炸理论, 两恒星间的距离会缓慢增大, 若干年后, 甲做圆周运动的半径增大为 $n r(n>k)$, 设甲、乙两恒星的质量保持不变, 引力常量为 $G$, 则若干年后下列说法正确的是 A: 恒星甲做圆周运动的向心力为 $G \frac{\mathrm{km}^{2}}{(\mathrm{nr})^{2}}$ B: 恒星甲做圆周运动周期变大 C: 恒星乙做圆周运动的半径为 $k n r$ D: 恒星乙做圆周运动的线速度为恒星甲做圆周运动线速度的 $\frac{1}{k}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_575

2022 年 9 月 27 日的“木星冲日”是观测木星的一次好机会。木星冲日就是指木星、 地球和太阳依次排列大致形成一条直线时的天象。已知木星与地球几乎在同一平面内沿同一方向绕太阳近似做匀速圆周运动, 木星质量约为地球质量的 318 倍, 木星半径约为地球半径的 11 倍, 木星到太阳的距离大约是地球到太阳距离的 5 倍, 则下列说法正确的是 ( ) A: 木星运行的加速度比地球运行的加速度大 B: 木星表面的重力加速度比地球表面的重力加速度小 C: 在木星表面附近发射飞行器的速度至少为 $7.9 \mathrm{~km} / \mathrm{s}$ D: 上一次“木星冲日”的时间大约在 2021 年 8 月份

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2022 年 9 月 27 日的“木星冲日”是观测木星的一次好机会。木星冲日就是指木星、 地球和太阳依次排列大致形成一条直线时的天象。已知木星与地球几乎在同一平面内沿同一方向绕太阳近似做匀速圆周运动, 木星质量约为地球质量的 318 倍, 木星半径约为地球半径的 11 倍, 木星到太阳的距离大约是地球到太阳距离的 5 倍, 则下列说法正确的是 ( ) A: 木星运行的加速度比地球运行的加速度大 B: 木星表面的重力加速度比地球表面的重力加速度小 C: 在木星表面附近发射飞行器的速度至少为 $7.9 \mathrm{~km} / \mathrm{s}$ D: 上一次“木星冲日”的时间大约在 2021 年 8 月份 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_992

Given the mass and radius of an exoplanet we can determine its likely composition, since the denser the material the smaller the planet will be for a given mass (see Figure 3). At a minimum orbital distance, called the Roche limit, tidal forces will cause a planet to break up, so very short period exoplanets place meaningful constraints on their composition as they must be very dense to survive in that orbit. The most extreme variant is called an 'iron planet', as it is made of iron with little or no rocky mantle ( $\gtrsim 80 \%$ iron by mass). The exoplanet KOI 1843.03 has the shortest known orbital period of a little over 4 hours, so is a potential candidate to be an iron planet. [figure1] Figure 3: Left: Comparisons of sizes of planets with different compositions [Marc Kuchner / NASA GSFC]. Right: An artist's impression of what an extreme iron planet (almost $\sim 100 \%$ iron) might look like. The Roche limiting distance, $a_{\min }$, for a body comprised of an incompressible fluid with negligible bulk tensile strength in a circular orbit about its parent star is $$ a_{\min }=2.44 R_{\star}\left(\frac{\rho_{\star}}{\rho_{p}}\right)^{1 / 3} $$ where $R_{\star}$ is the radius of the star, $\rho_{\star}$ is the density of the star and $\rho_{p}$ is the density of the planet. Over the mass range $0.1 M_{E}-1.0 M_{E}$ the mass-radius relation for pure silicate and pure iron planets is approximately a power law and can be described as $$ \log _{10}\left(\frac{R}{R_{E}}\right)=0.295 \log _{10}\left(\frac{M}{M_{E}}\right)+\alpha $$ where $\alpha=0.0286$ in the pure silicate case and $\alpha=-0.1090$ in the pure iron case. The exoplanet KOI 1843.03 is measured (from transit lightcurves) to have a radius of $0.61 R_{E}$. Calculate the minimum period for the exoplanet in the pure silicate and pure iron cases. Give your answer in hours.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Given the mass and radius of an exoplanet we can determine its likely composition, since the denser the material the smaller the planet will be for a given mass (see Figure 3). At a minimum orbital distance, called the Roche limit, tidal forces will cause a planet to break up, so very short period exoplanets place meaningful constraints on their composition as they must be very dense to survive in that orbit. The most extreme variant is called an 'iron planet', as it is made of iron with little or no rocky mantle ( $\gtrsim 80 \%$ iron by mass). The exoplanet KOI 1843.03 has the shortest known orbital period of a little over 4 hours, so is a potential candidate to be an iron planet. [figure1] Figure 3: Left: Comparisons of sizes of planets with different compositions [Marc Kuchner / NASA GSFC]. Right: An artist's impression of what an extreme iron planet (almost $\sim 100 \%$ iron) might look like. The Roche limiting distance, $a_{\min }$, for a body comprised of an incompressible fluid with negligible bulk tensile strength in a circular orbit about its parent star is $$ a_{\min }=2.44 R_{\star}\left(\frac{\rho_{\star}}{\rho_{p}}\right)^{1 / 3} $$ where $R_{\star}$ is the radius of the star, $\rho_{\star}$ is the density of the star and $\rho_{p}$ is the density of the planet. Over the mass range $0.1 M_{E}-1.0 M_{E}$ the mass-radius relation for pure silicate and pure iron planets is approximately a power law and can be described as $$ \log _{10}\left(\frac{R}{R_{E}}\right)=0.295 \log _{10}\left(\frac{M}{M_{E}}\right)+\alpha $$ where $\alpha=0.0286$ in the pure silicate case and $\alpha=-0.1090$ in the pure iron case. The exoplanet KOI 1843.03 is measured (from transit lightcurves) to have a radius of $0.61 R_{E}$. Calculate the minimum period for the exoplanet in the pure silicate and pure iron cases. Give your answer in hours. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of hrs, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy_51

天文观测发现, 天狼星 $\mathrm{A}$ 与其伴星 $\mathrm{B}$ 是一个双星系统。它们始终绕着 $O$ 点在两个不同椭圆轨道上运动, 如图所示, 实线为天狼星 $\mathrm{A}$ 的运行轨迹, 虚线为其伴星 $\mathrm{B}$ 的轨迹，则（ ) [图1] A: A 的运行周期小于 B 的运行周期 B: $\mathrm{A}$ 的质量小于 $\mathrm{B}$ 的质量 C: A 的加速度总是小于 $\mathrm{B}$ 的加速度 D: $\mathrm{A}$ 与 $\mathrm{B}$ 绕 $O$ 点的旋转方向可能相同, 可能相反

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 天文观测发现, 天狼星 $\mathrm{A}$ 与其伴星 $\mathrm{B}$ 是一个双星系统。它们始终绕着 $O$ 点在两个不同椭圆轨道上运动, 如图所示, 实线为天狼星 $\mathrm{A}$ 的运行轨迹, 虚线为其伴星 $\mathrm{B}$ 的轨迹，则（ ) [图1] A: A 的运行周期小于 B 的运行周期 B: $\mathrm{A}$ 的质量小于 $\mathrm{B}$ 的质量 C: A 的加速度总是小于 $\mathrm{B}$ 的加速度 D: $\mathrm{A}$ 与 $\mathrm{B}$ 绕 $O$ 点的旋转方向可能相同, 可能相反 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_1002

When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ One of the most famous examples of a system with the barycentre lying outside the larger object is the Pluto-Charon system (see Fig 4), which is why many considered it to be a double-planet long before it was reclassified as a dwarf planet. Given the barycentre of the system is at $1.83 \mathrm{R}_{\text {Pluto }}$ (where $\mathrm{R}_{\text {Pluto }}=1187 \mathrm{~km}$ ) and the average separation between Pluto and Charon (as measured from centre to centre) is $19570 \mathrm{~km}$, calculate the ratio of their masses (i.e. $m_{\text {Pluto }} / m_{\text {Charon }}$ ).

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ One of the most famous examples of a system with the barycentre lying outside the larger object is the Pluto-Charon system (see Fig 4), which is why many considered it to be a double-planet long before it was reclassified as a dwarf planet. Given the barycentre of the system is at $1.83 \mathrm{R}_{\text {Pluto }}$ (where $\mathrm{R}_{\text {Pluto }}=1187 \mathrm{~km}$ ) and the average separation between Pluto and Charon (as measured from centre to centre) is $19570 \mathrm{~km}$, calculate the ratio of their masses (i.e. $m_{\text {Pluto }} / m_{\text {Charon }}$ ). All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

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Astronomy_312

建立物理模型是解决实际问题的重要方法。 如图 1 所示, 圆和椭圆是分析卫星运动时常用的模型。已知, 地球质量为 $M$, 半径为 $R$, 万有引力常量为 $G$ 。 卫星在近地轨道I上围绕地球的运动, 可视做匀速圆周运动, 轨道半径近似等于地球半径。求卫星在近地轨道I上的运行速度大小 $v$ 。 图 1 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 建立物理模型是解决实际问题的重要方法。 如图 1 所示, 圆和椭圆是分析卫星运动时常用的模型。已知, 地球质量为 $M$, 半径为 $R$, 万有引力常量为 $G$ 。 卫星在近地轨道I上围绕地球的运动, 可视做匀速圆周运动, 轨道半径近似等于地球半径。求卫星在近地轨道I上的运行速度大小 $v$ 。 图 1 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy_632

深空中, 某行星 $\mathrm{X}$ 绕恒星 $\mathrm{Y}$ 逆时针方向公转, 卫星 $\mathrm{Z}$ 绕 $\mathrm{X}$ 逆时针方向运行, $\mathrm{X}$ 轨道与 $\mathrm{Z}$ 轨道在同一平面内。如图, 某时刻 $\mathrm{Z} 、 \mathrm{X}$ 和 $\mathrm{Y}$ 在同一直线上, 经过时间 $t, \mathrm{Z} 、 \mathrm{X}$和 $\mathrm{Y}$ 再次在同一直线上 (相对位置的顺序不变)。已知 $\mathrm{Z}$ 绕 $\mathrm{X}$ 做匀速圆周运动的周期为 $T, \mathrm{X}$ 绕 $\mathrm{Y}$ 做匀速圆周运动的周期大于 $T, \mathrm{X}$ 与 $\mathrm{Y}$ 间的距离为 $r$, 则 $\mathrm{Y}$ 的质量为 ( ) [图1] A: $\frac{4 \pi^{2} r^{3}(t-T)^{2}}{G t^{2} T^{2}}$ B: $\frac{4 \pi^{2} r^{3}(t+T)^{2}}{G t^{2} T^{2}}$ C: $\frac{4 \pi^{2} r^{3}}{G t^{2}}$ D: $\frac{4 \pi^{2} r^{3}}{G T^{2}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 深空中, 某行星 $\mathrm{X}$ 绕恒星 $\mathrm{Y}$ 逆时针方向公转, 卫星 $\mathrm{Z}$ 绕 $\mathrm{X}$ 逆时针方向运行, $\mathrm{X}$ 轨道与 $\mathrm{Z}$ 轨道在同一平面内。如图, 某时刻 $\mathrm{Z} 、 \mathrm{X}$ 和 $\mathrm{Y}$ 在同一直线上, 经过时间 $t, \mathrm{Z} 、 \mathrm{X}$和 $\mathrm{Y}$ 再次在同一直线上 (相对位置的顺序不变)。已知 $\mathrm{Z}$ 绕 $\mathrm{X}$ 做匀速圆周运动的周期为 $T, \mathrm{X}$ 绕 $\mathrm{Y}$ 做匀速圆周运动的周期大于 $T, \mathrm{X}$ 与 $\mathrm{Y}$ 间的距离为 $r$, 则 $\mathrm{Y}$ 的质量为 ( ) [图1] A: $\frac{4 \pi^{2} r^{3}(t-T)^{2}}{G t^{2} T^{2}}$ B: $\frac{4 \pi^{2} r^{3}(t+T)^{2}}{G t^{2} T^{2}}$ C: $\frac{4 \pi^{2} r^{3}}{G t^{2}}$ D: $\frac{4 \pi^{2} r^{3}}{G T^{2}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_3

理论研究表明地球上的物体速度达到第二宇宙速度 $11.2 \mathrm{~km} / \mathrm{s}$ 时, 物体就能脱离地球, 又知第二宇宙速度是第一宇宙速度的 $\sqrt{2}$ 倍. 现有某探测器完成了对某未知星球的探测任务悬停在该星球表面. 通过探测到的数据得到该星球的有关参量（1)其密度基本与地球密度一致. (2)其半径约为地球半径的 2 倍. 若不考虑该星球自转的影响, 欲使探测器脱离该星球, 则探测器从该星球表面的起飞速度至少约为 ( ) A: $7.9 \mathrm{~km} / \mathrm{s}$ B: $11.2 \mathrm{~km} / \mathrm{s}$ C: $15.8 \mathrm{~km} / \mathrm{s}$ D: $22.4 \mathrm{~km} / \mathrm{s}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 理论研究表明地球上的物体速度达到第二宇宙速度 $11.2 \mathrm{~km} / \mathrm{s}$ 时, 物体就能脱离地球, 又知第二宇宙速度是第一宇宙速度的 $\sqrt{2}$ 倍. 现有某探测器完成了对某未知星球的探测任务悬停在该星球表面. 通过探测到的数据得到该星球的有关参量（1)其密度基本与地球密度一致. (2)其半径约为地球半径的 2 倍. 若不考虑该星球自转的影响, 欲使探测器脱离该星球, 则探测器从该星球表面的起飞速度至少约为 ( ) A: $7.9 \mathrm{~km} / \mathrm{s}$ B: $11.2 \mathrm{~km} / \mathrm{s}$ C: $15.8 \mathrm{~km} / \mathrm{s}$ D: $22.4 \mathrm{~km} / \mathrm{s}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_702

航天技术的发展是当今各国综合国力的直接体现，近年来，我国的航天技术取得了让世界瞩目的成绩，也引领科技爱好者思索航天技术的发展，有人就提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 如图所示, 在地球上距地心 $h$ 处沿一条弦挖一光滑通道, 在通道的两个出口处 $A$ 和 $B$ 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, $M \gg m$, 在中点 $O^{\prime}$ 弹性正撞后，质量为 $m$ 的物体，即待发射的卫星就会从通道口 $B$ 冲出通道, 设置一个装置, 卫星从 $B$ 冲出就把速度变为沿地球切线方向, 但不改变速度大小, 这样就有可能成功发射卫星。已知地球可视为质量分布均匀 的球体, 且质量分布均匀的球壳对壳内物体的引力为零, 地球半径为 $R_{0}$, 表面的重力加速度为 $g$ 。 求地球的第一宇宙速度 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 航天技术的发展是当今各国综合国力的直接体现，近年来，我国的航天技术取得了让世界瞩目的成绩，也引领科技爱好者思索航天技术的发展，有人就提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 如图所示, 在地球上距地心 $h$ 处沿一条弦挖一光滑通道, 在通道的两个出口处 $A$ 和 $B$ 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, $M \gg m$, 在中点 $O^{\prime}$ 弹性正撞后，质量为 $m$ 的物体，即待发射的卫星就会从通道口 $B$ 冲出通道, 设置一个装置, 卫星从 $B$ 冲出就把速度变为沿地球切线方向, 但不改变速度大小, 这样就有可能成功发射卫星。已知地球可视为质量分布均匀 的球体, 且质量分布均匀的球壳对壳内物体的引力为零, 地球半径为 $R_{0}$, 表面的重力加速度为 $g$ 。 求地球的第一宇宙速度 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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multi-modal

Astronomy_956

On $24^{\text {th }}$ August 2016, astronomers discovered a planet orbiting the closest star to the Sun, Proxima Centauri, situated 4.22 light years away, which fulfils a long-standing dream of science-fiction writers: a world that is close enough for humans to send their first interstellar spacecraft. Astronomers have noted how the motion of Proxima Centauri changed in the first months of 2016, with the star moving towards and away from the Earth, as seen in the figure below. Sometimes Proxima Centauri is approaching Earth at $5 \mathrm{~km} \mathrm{hour}^{-1}-$ normal human walking pace - and at times receding at the same speed. This regular pattern of changing radial velocities caused by an unseen planet, which they named Proxima Centauri B, repeats and results in tiny Doppler shifts in the star's light, making the light appear slightly redder, then bluer. [figure1] From the radial velocity curve above, determine the period of the planet around Proxima Centauri.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: On $24^{\text {th }}$ August 2016, astronomers discovered a planet orbiting the closest star to the Sun, Proxima Centauri, situated 4.22 light years away, which fulfils a long-standing dream of science-fiction writers: a world that is close enough for humans to send their first interstellar spacecraft. Astronomers have noted how the motion of Proxima Centauri changed in the first months of 2016, with the star moving towards and away from the Earth, as seen in the figure below. Sometimes Proxima Centauri is approaching Earth at $5 \mathrm{~km} \mathrm{hour}^{-1}-$ normal human walking pace - and at times receding at the same speed. This regular pattern of changing radial velocities caused by an unseen planet, which they named Proxima Centauri B, repeats and results in tiny Doppler shifts in the star's light, making the light appear slightly redder, then bluer. [figure1] From the radial velocity curve above, determine the period of the planet around Proxima Centauri. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of days, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy_1006

At a latitude of $52^{\circ} \mathrm{N}$ what is the altitude of Polaris above the horizon? A: $38^{\circ}$ B: $48^{\circ}$ C: $52^{\circ}$ D: $90^{\circ}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: At a latitude of $52^{\circ} \mathrm{N}$ what is the altitude of Polaris above the horizon? A: $38^{\circ}$ B: $48^{\circ}$ C: $52^{\circ}$ D: $90^{\circ}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy_254

如图所示, 假设在太空中有恒星 A、B 双星系统绕点 $O$ 做顺时针匀速圆周运动, 运动周期为 $T_{l}$, 它们的轨道半径分别为 $r_{A} 、 r_{B}, r_{A}<r_{B}, \mathrm{C}$ 为 $\mathrm{B}$ 的卫星, 绕 $\mathrm{B}$ 做逆时针匀速圆周运动, 周期为 $T_{2}$ 。忽略 $\mathrm{A}$ 与 $\mathrm{C}$ 之间的引力, $\mathrm{A}$ 与 $\mathrm{B}$ 之间的引力远大于 $\mathrm{C}$ 与 $\mathrm{B}$ 之间的引力。引力常量为 $G$, 下列说法正确的是 ( ) [图1] A: 若知道 $\mathrm{C}$ 的轨道半径, 则可求出 $\mathrm{C}$ 的质量 B: 若 $\mathrm{A}$ 也有一颗轨道半径与 $\mathrm{C}$ 相同的卫星, 则其运动周期也一定为 $T_{2}$ C: 恒星 $\mathrm{A}$ 的质量为 $M_{\mathrm{A}}=\frac{4 \pi^{2} r_{\mathrm{A}}\left(r_{\mathrm{A}}+r_{\mathrm{B}}\right)^{2}}{G T_{1}^{2}}$ D: 设 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三星由图示位置到再次共线的时间为 $t$, 则 $t=\frac{T_{1} T_{2}}{2\left(T_{1}+T_{2}\right)}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, 假设在太空中有恒星 A、B 双星系统绕点 $O$ 做顺时针匀速圆周运动, 运动周期为 $T_{l}$, 它们的轨道半径分别为 $r_{A} 、 r_{B}, r_{A}<r_{B}, \mathrm{C}$ 为 $\mathrm{B}$ 的卫星, 绕 $\mathrm{B}$ 做逆时针匀速圆周运动, 周期为 $T_{2}$ 。忽略 $\mathrm{A}$ 与 $\mathrm{C}$ 之间的引力, $\mathrm{A}$ 与 $\mathrm{B}$ 之间的引力远大于 $\mathrm{C}$ 与 $\mathrm{B}$ 之间的引力。引力常量为 $G$, 下列说法正确的是 ( ) [图1] A: 若知道 $\mathrm{C}$ 的轨道半径, 则可求出 $\mathrm{C}$ 的质量 B: 若 $\mathrm{A}$ 也有一颗轨道半径与 $\mathrm{C}$ 相同的卫星, 则其运动周期也一定为 $T_{2}$ C: 恒星 $\mathrm{A}$ 的质量为 $M_{\mathrm{A}}=\frac{4 \pi^{2} r_{\mathrm{A}}\left(r_{\mathrm{A}}+r_{\mathrm{B}}\right)^{2}}{G T_{1}^{2}}$ D: 设 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三星由图示位置到再次共线的时间为 $t$, 则 $t=\frac{T_{1} T_{2}}{2\left(T_{1}+T_{2}\right)}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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multi-modal

Astronomy_388

如图所示, $\mathrm{A}$ 为静止于地球赤道上的物体, $B$ 为绕地球沿陏圆轨道运行的卫星, $C$为绕地球做圆周运动的卫星， $P$ 为 $B 、 C$ 两卫星轨道的交点。在 $\Delta t$ 时间内, 卫星 $B$ 、卫星 $C$ 与地心连线扫过的面积分别为 $S_{B} 、 S_{C}$ 。已知 $\mathrm{A} 、 B 、 C$ 绕地心运动的周期相同,下列说法中正确的是（） [图1] A: 卫星 $B$ 的运行速度可能会大于第一宇宙速度 B: 物体 $\mathrm{A}$ 的速度大于卫星 $C$ 的运行速度 C: $S_{B}<S_{C}$ D: 卫星 $B$ 在 $P$ 点的向心加速度与卫星 $C$ 在 $P$ 点的向心加速度大小相等

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, $\mathrm{A}$ 为静止于地球赤道上的物体, $B$ 为绕地球沿陏圆轨道运行的卫星, $C$为绕地球做圆周运动的卫星， $P$ 为 $B 、 C$ 两卫星轨道的交点。在 $\Delta t$ 时间内, 卫星 $B$ 、卫星 $C$ 与地心连线扫过的面积分别为 $S_{B} 、 S_{C}$ 。已知 $\mathrm{A} 、 B 、 C$ 绕地心运动的周期相同,下列说法中正确的是（） [图1] A: 卫星 $B$ 的运行速度可能会大于第一宇宙速度 B: 物体 $\mathrm{A}$ 的速度大于卫星 $C$ 的运行速度 C: $S_{B}<S_{C}$ D: 卫星 $B$ 在 $P$ 点的向心加速度与卫星 $C$ 在 $P$ 点的向心加速度大小相等 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1071

On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps.a. Calculate the width of the path of totality at GE (in $\mathrm{km}$ ). You may use the approximation that the Moon is directly overhead so that the shadow is circular, and small enough that the curvature of the Earth can be neglected.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps. problem: a. Calculate the width of the path of totality at GE (in $\mathrm{km}$ ). You may use the approximation that the Moon is directly overhead so that the shadow is circular, and small enough that the curvature of the Earth can be neglected. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~km}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_27

把地球看做一个质量均匀分布的标准球体, 已知质量均匀分布的球壳对其内部的物体的万有引力为零, 地球半径为 $R$, 地球表面的重力加速度为 $g_{0}$, 不考虑地球的自转。 求沿地球半径方向深度为 $h$ 的矿井底部质量为 $m$ 的小物体受到的万有引力大小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 把地球看做一个质量均匀分布的标准球体, 已知质量均匀分布的球壳对其内部的物体的万有引力为零, 地球半径为 $R$, 地球表面的重力加速度为 $g_{0}$, 不考虑地球的自转。 求沿地球半径方向深度为 $h$ 的矿井底部质量为 $m$ 的小物体受到的万有引力大小 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

text-only

Astronomy_508

2022 年 11 月 29 日 23 时 08 分, 搭载神舟十五号载人飞船的长征二号遥十五运载火箭在酒泉卫星发射中心点火发射, 11 月 30 日 7 时 33 分, 神舟十五号 3 名航天员顺利进驻中国空间站，与神舟十四号航天员乘组首次实现“太空会师”。中国空间站在距地面约 $390 \mathrm{~km}$ 的近圆形轨道上运行, 运行轨道面与赤道面夹角约为 $42^{\circ}$, 其运行方向和地球自转方向如图所示。已知地球表面的重力加速度大小约为 $9.8 \mathrm{~m} / \mathrm{s}^{2}$, 地球的半径为 $6400 \mathrm{~km}$, 地球的自转周期约为 $24 \mathrm{~h}$, 假设地球为质量分布均匀的球体。某次空间站依次经过赤道上的 $A$ 地和 $B$ 地（图中未画出）上空, 两地相距约为（） [图1] A: $1300 \mathrm{~km}$ B: $2600 \mathrm{~km}$ C: $6400 \mathrm{~km}$ D: $18800 \mathrm{~km}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2022 年 11 月 29 日 23 时 08 分, 搭载神舟十五号载人飞船的长征二号遥十五运载火箭在酒泉卫星发射中心点火发射, 11 月 30 日 7 时 33 分, 神舟十五号 3 名航天员顺利进驻中国空间站，与神舟十四号航天员乘组首次实现“太空会师”。中国空间站在距地面约 $390 \mathrm{~km}$ 的近圆形轨道上运行, 运行轨道面与赤道面夹角约为 $42^{\circ}$, 其运行方向和地球自转方向如图所示。已知地球表面的重力加速度大小约为 $9.8 \mathrm{~m} / \mathrm{s}^{2}$, 地球的半径为 $6400 \mathrm{~km}$, 地球的自转周期约为 $24 \mathrm{~h}$, 假设地球为质量分布均匀的球体。某次空间站依次经过赤道上的 $A$ 地和 $B$ 地（图中未画出）上空, 两地相距约为（） [图1] A: $1300 \mathrm{~km}$ B: $2600 \mathrm{~km}$ C: $6400 \mathrm{~km}$ D: $18800 \mathrm{~km}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_738

地球和木星绕太阳运行的轨道可以看作是圆形的, 它们各自的卫星轨道也可看作是圆形的。已知木星的公转轨道半径约为地球公转轨道半径的 5 倍, 木星半径约为地球半径的 11 倍, 木星质量大于地球质量。如图所示是地球和木星的不同卫星做圆周运动的半径 $r$ 的立方与周期 $T$ 的平方的关系图象, 已知万有引力常量为 $G$, 地球的半径为 $R$,下列说法正确的是（ ） [图1] A: 木星与地球的质量之比为 $\frac{b d}{11 a c}$ B: 木星与地球的线速度之比为 1:5 C: 地球密度为 $\frac{4 \pi a}{G d R^{3}}$ D: 木星密度为 $\frac{3 \pi_{b}}{1331 G c R^{3}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 地球和木星绕太阳运行的轨道可以看作是圆形的, 它们各自的卫星轨道也可看作是圆形的。已知木星的公转轨道半径约为地球公转轨道半径的 5 倍, 木星半径约为地球半径的 11 倍, 木星质量大于地球质量。如图所示是地球和木星的不同卫星做圆周运动的半径 $r$ 的立方与周期 $T$ 的平方的关系图象, 已知万有引力常量为 $G$, 地球的半径为 $R$,下列说法正确的是（ ） [图1] A: 木星与地球的质量之比为 $\frac{b d}{11 a c}$ B: 木星与地球的线速度之比为 1:5 C: 地球密度为 $\frac{4 \pi a}{G d R^{3}}$ D: 木星密度为 $\frac{3 \pi_{b}}{1331 G c R^{3}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_321

在地球上将质量为 $m$ 的物体 $\mathrm{A}$ 水平抛出, 物体的动能 $E_{k}$ 与运动时间 $t^{2}$ 的关系如图中实线甲所示。在某一星球 $\mathrm{X}$ 上将质量为 $2 m$ 的物体 $\mathrm{B}$ 水平抛出, 其动能 $E_{k}$ 与运动时间 $t^{2}$ 的关系如图中虚线乙所示。假设地球和星球 $\mathrm{X}$ 均为质量均匀分布的球体。已知地球的半径是星球 $\mathrm{X}$ 的 3 倍, 地球表面的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~A} 、 \mathrm{~B}$ 在运动过程中空气阻力均忽略不计, 下列说法正确的是（） [图1] A: 地球的质量是星球 $\mathrm{X}$ 的 18 倍 B: 地球的近地卫星的周期与星球 $\mathrm{X}$ 的近地卫星的周期的平方之比为 $3: \sqrt{2}$ C: 若 A、B 下落过程的水平位移大小相等, A 下降的高度是 B 的 2 倍 D: 若 $A 、 B$ 下落过程的动量变化量大小相等, 则 B下降的高度是 $\mathrm{A}$ 的 2 倍

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 在地球上将质量为 $m$ 的物体 $\mathrm{A}$ 水平抛出, 物体的动能 $E_{k}$ 与运动时间 $t^{2}$ 的关系如图中实线甲所示。在某一星球 $\mathrm{X}$ 上将质量为 $2 m$ 的物体 $\mathrm{B}$ 水平抛出, 其动能 $E_{k}$ 与运动时间 $t^{2}$ 的关系如图中虚线乙所示。假设地球和星球 $\mathrm{X}$ 均为质量均匀分布的球体。已知地球的半径是星球 $\mathrm{X}$ 的 3 倍, 地球表面的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~A} 、 \mathrm{~B}$ 在运动过程中空气阻力均忽略不计, 下列说法正确的是（） [图1] A: 地球的质量是星球 $\mathrm{X}$ 的 18 倍 B: 地球的近地卫星的周期与星球 $\mathrm{X}$ 的近地卫星的周期的平方之比为 $3: \sqrt{2}$ C: 若 A、B 下落过程的水平位移大小相等, A 下降的高度是 B 的 2 倍 D: 若 $A 、 B$ 下落过程的动量变化量大小相等, 则 B下降的高度是 $\mathrm{A}$ 的 2 倍 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1073

A "supermoon" is a new or full moon that occurs with the Moon at or near its closest approach to Earth in a given orbit (perigee). The media commonly associates supermoons with extreme brightness and size, sometimes implying that the Moon itself will become larger and have an impact on human behaviour, but just how different is a supermoon compared to the 'normal' Moon we see each month? Lunar Data: Synodic Period Anomalistic Period Semi-major axis Orbit eccentricity $$ \begin{aligned} & =29.530589 \text { days (time between same phases e.g. full moon to full moon) } \\ & =27.554550 \text { days (time between perigees i.e. perigee to perigee) } \\ & =3.844 \times 10^{5} \mathrm{~km} \\ & =0.0549 \\ & =1738.1 \mathrm{~km} \end{aligned} $$ $$ \begin{array}{ll} \text { Radius of the Moon } & =1738.1 \mathrm{~km} \\ \text { Mass of the Moon } & =7.342 \times 10^{22} \mathrm{~kg} \end{array} $$ In this question, we will only consider a full moon that is at perigee to be a supermoon.d). What change in magnitude does this brightness difference correspond to?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: A "supermoon" is a new or full moon that occurs with the Moon at or near its closest approach to Earth in a given orbit (perigee). The media commonly associates supermoons with extreme brightness and size, sometimes implying that the Moon itself will become larger and have an impact on human behaviour, but just how different is a supermoon compared to the 'normal' Moon we see each month? Lunar Data: Synodic Period Anomalistic Period Semi-major axis Orbit eccentricity $$ \begin{aligned} & =29.530589 \text { days (time between same phases e.g. full moon to full moon) } \\ & =27.554550 \text { days (time between perigees i.e. perigee to perigee) } \\ & =3.844 \times 10^{5} \mathrm{~km} \\ & =0.0549 \\ & =1738.1 \mathrm{~km} \end{aligned} $$ $$ \begin{array}{ll} \text { Radius of the Moon } & =1738.1 \mathrm{~km} \\ \text { Mass of the Moon } & =7.342 \times 10^{22} \mathrm{~kg} \end{array} $$ In this question, we will only consider a full moon that is at perigee to be a supermoon. problem: d). What change in magnitude does this brightness difference correspond to? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

NV

Astronomy

EN

text-only

Astronomy_493

如图所示，我国自行设计、制造的第一颗人造地球卫星“东方红一号”卫星，运行轨道为粗圆轨道，其近地点 $M$ 和远地点 $N$ 的高度分别为 $439 \mathrm{~km}$ 和 $2384 \mathrm{~km}$. 关于“东方红一号”卫星, 下列说法正确的是（ ） [图1] A: 在 $M$ 点的速度小于在 $N$ 点的速度 B: 在 $M$ 点的加速度小于在 $N$ 点的加速度 C: 在 $M$ 点受到的地球引力大于在 $N$ 点受到的地球引力 D: 从 $M$ 点运动到 $N$ 点的过程中角速度逐渐增大

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示，我国自行设计、制造的第一颗人造地球卫星“东方红一号”卫星，运行轨道为粗圆轨道，其近地点 $M$ 和远地点 $N$ 的高度分别为 $439 \mathrm{~km}$ 和 $2384 \mathrm{~km}$. 关于“东方红一号”卫星, 下列说法正确的是（ ） [图1] A: 在 $M$ 点的速度小于在 $N$ 点的速度 B: 在 $M$ 点的加速度小于在 $N$ 点的加速度 C: 在 $M$ 点受到的地球引力大于在 $N$ 点受到的地球引力 D: 从 $M$ 点运动到 $N$ 点的过程中角速度逐渐增大 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_754

The so-called dark energy is a model to explain ... A: the radiation of black holes. B: the mass distribution of galaxies. C: the acceleration of the universe. D: the microwave background of the universe.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The so-called dark energy is a model to explain ... A: the radiation of black holes. B: the mass distribution of galaxies. C: the acceleration of the universe. D: the microwave background of the universe. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_946

the ratio of Neptune's mass to the Sun's, $M_{\mathrm{Nep}} / M_{\odot}$ = $5.028 \times 10^{-5}$Neso is the outermost moon of Neptune, and is notable for being the moon with the greatest known orbital distance from its planet, corresponding to a period of over 26 Earth years. At its furthest point (apoapsis), it is further from Neptune than Mercury gets from the Sun, meaning it is close to the limit of being held by Neptune's gravity (see Figure 1). [figure1] Figure 1: Plan view of Neptune's outermost satellite orbits. Neptune is represented by the black dot at the centre. The green orbit corresponds to Nereid, the two in blue are the prograde orbits of Sao and Laomedeia, and the red are the retrograde orbits of Halimede, Psamathe and Neso (bottom right). The dotted circle shows the theoretical outer limit of stability for Neptune satellites. Taken from Shepperd et al. (2006). Orbital data about Neso and Mercury are summarised below: | | Mercury | Neso | | :--- | :---: | :---: | | Semi-major axis, $a\left(\times 10^{9} \mathrm{~m}\right)$ | 57.909 | 49.285 | | Eccentricity, $e$ | 0.2056 | 0.5714 | | Orbital period, $T$ (Earth days) | 87.97 | 9740.73 | Hence work out the mass of Neptune, $M_{\mathrm{Nep}}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: the ratio of Neptune's mass to the Sun's, $M_{\mathrm{Nep}} / M_{\odot}$ = $5.028 \times 10^{-5}$ problem: Neso is the outermost moon of Neptune, and is notable for being the moon with the greatest known orbital distance from its planet, corresponding to a period of over 26 Earth years. At its furthest point (apoapsis), it is further from Neptune than Mercury gets from the Sun, meaning it is close to the limit of being held by Neptune's gravity (see Figure 1). [figure1] Figure 1: Plan view of Neptune's outermost satellite orbits. Neptune is represented by the black dot at the centre. The green orbit corresponds to Nereid, the two in blue are the prograde orbits of Sao and Laomedeia, and the red are the retrograde orbits of Halimede, Psamathe and Neso (bottom right). The dotted circle shows the theoretical outer limit of stability for Neptune satellites. Taken from Shepperd et al. (2006). Orbital data about Neso and Mercury are summarised below: | | Mercury | Neso | | :--- | :---: | :---: | | Semi-major axis, $a\left(\times 10^{9} \mathrm{~m}\right)$ | 57.909 | 49.285 | | Eccentricity, $e$ | 0.2056 | 0.5714 | | Orbital period, $T$ (Earth days) | 87.97 | 9740.73 | Hence work out the mass of Neptune, $M_{\mathrm{Nep}}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of kg, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

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Astronomy_834

Given that the apparent magnitude of the system is 8.2 , what is the apparent magnitude of the brighter star? A: 8.4 B: 8.7 C: 8.9 D: 9.0 E: 9.2

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Given that the apparent magnitude of the system is 8.2 , what is the apparent magnitude of the brighter star? A: 8.4 B: 8.7 C: 8.9 D: 9.0 E: 9.2 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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Astronomy_289

如图所示, 探月卫星的发射过程可简化如下：首先进入绕地球运行的“停泊轨道”,在该轨道的 $P$ 处通过变速再进入“地月转移轨道”, 在快要到达月球时, 对卫星再次变速,卫星被月球引力“俘获”后, 成为环月卫星, 最终在环绕月球的“工作轨道”绕月飞行（视为圆周运动), 对月球进行探测. “工作轨道”周期为 $T$ 、距月球表面的高度为 $h$, 月球半径为 $R$, 引力常量为 $G$, 忽略其他天体对探月卫星在“工作轨道”上环绕运动的影响。 求月球的质量 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 探月卫星的发射过程可简化如下：首先进入绕地球运行的“停泊轨道”,在该轨道的 $P$ 处通过变速再进入“地月转移轨道”, 在快要到达月球时, 对卫星再次变速,卫星被月球引力“俘获”后, 成为环月卫星, 最终在环绕月球的“工作轨道”绕月飞行（视为圆周运动), 对月球进行探测. “工作轨道”周期为 $T$ 、距月球表面的高度为 $h$, 月球半径为 $R$, 引力常量为 $G$, 忽略其他天体对探月卫星在“工作轨道”上环绕运动的影响。 求月球的质量 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

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multi-modal

Astronomy_198

中国行星探测任务名称为“天问系列”，首次火星探测任务被命名为“天问一号”。若已知“天问一号”探测器在距离火星中心为 $r_{1}$ 的轨道上做匀速圆周运动, 其周期为 $T_{1}$ 。火星半径为 $R_{0}$, 自转周期为 $\mathrm{T}_{0}$ 。引力常量为 $G$, 若火星与地球运动情况相似, 下列说法正确的是（） A: 火星的质量为 $\frac{4 \pi^{2} R_{0}^{2}}{G T_{1}^{2}}$ B: 火星表面两极的重力加速度为 $\frac{4 \pi^{2} r_{1}^{3}}{T_{1}^{2} R_{0}^{2}}$ C: 火星的第一宇宙速度为 $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R_{0}}}$ D: 火星的同步卫星距离星球表面高度为 $r_{1} \sqrt{\frac{T_{0}^{2}}{T_{1}^{2}}}-R_{0}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 中国行星探测任务名称为“天问系列”，首次火星探测任务被命名为“天问一号”。若已知“天问一号”探测器在距离火星中心为 $r_{1}$ 的轨道上做匀速圆周运动, 其周期为 $T_{1}$ 。火星半径为 $R_{0}$, 自转周期为 $\mathrm{T}_{0}$ 。引力常量为 $G$, 若火星与地球运动情况相似, 下列说法正确的是（） A: 火星的质量为 $\frac{4 \pi^{2} R_{0}^{2}}{G T_{1}^{2}}$ B: 火星表面两极的重力加速度为 $\frac{4 \pi^{2} r_{1}^{3}}{T_{1}^{2} R_{0}^{2}}$ C: 火星的第一宇宙速度为 $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R_{0}}}$ D: 火星的同步卫星距离星球表面高度为 $r_{1} \sqrt{\frac{T_{0}^{2}}{T_{1}^{2}}}-R_{0}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy_483

如图所示, “嫦娥四号”飞船绕月球在圆轨道III上运动, 在 $B$ 位置变轨进入粗圆轨道 II, 在 $A$ 位置再次变轨进入圆轨道I, 下列判断正确的是（ ） [图1] A: 飞船在 $A$ 位置变轨时, 动能变大 B: 飞船在轨道I上的速度大于在轨道III上的速度 C: 飞船在轨道I上的加速度大于在轨道III上的加速度 D: 飞船在轨道I上的周期小于在轨道II的周期

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, “嫦娥四号”飞船绕月球在圆轨道III上运动, 在 $B$ 位置变轨进入粗圆轨道 II, 在 $A$ 位置再次变轨进入圆轨道I, 下列判断正确的是（ ） [图1] A: 飞船在 $A$ 位置变轨时, 动能变大 B: 飞船在轨道I上的速度大于在轨道III上的速度 C: 飞船在轨道I上的加速度大于在轨道III上的加速度 D: 飞船在轨道I上的周期小于在轨道II的周期 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_843

Two (spherical) asteroids, Ek and Do, are orbiting in free space around their stationary center of mass. Ek has mass 7M and Do has mass 1.4M, where M is the mass of moon. What is the ratio of the angular momentum of the whole system to the angular momentum of Do about the center of mass of the system? A: 26 B: 6 C: 1.2 D: 1.04 E: 0.1667

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Two (spherical) asteroids, Ek and Do, are orbiting in free space around their stationary center of mass. Ek has mass 7M and Do has mass 1.4M, where M is the mass of moon. What is the ratio of the angular momentum of the whole system to the angular momentum of Do about the center of mass of the system? A: 26 B: 6 C: 1.2 D: 1.04 E: 0.1667 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy_167

如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 求飞行器在轨道半径为 $R$ 的高空绕地球做圆周运动的周期; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 求飞行器在轨道半径为 $R$ 的高空绕地球做圆周运动的周期; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

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multi-modal

Astronomy_791

Calculate the speed of the sun around the center of mass due to the presence of Jupiter. A: $6 \mathrm{~m} / \mathrm{s}$ B: $12 \mathrm{~m} / \mathrm{s}$ C: $600 \mathrm{~m} / \mathrm{s}$ D: $1200 \mathrm{~m} / \mathrm{s}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Calculate the speed of the sun around the center of mass due to the presence of Jupiter. A: $6 \mathrm{~m} / \mathrm{s}$ B: $12 \mathrm{~m} / \mathrm{s}$ C: $600 \mathrm{~m} / \mathrm{s}$ D: $1200 \mathrm{~m} / \mathrm{s}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

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Astronomy_589

地月拉格朗日 $L_{2}$ 点, 始终位于地月连线上的如图所示位置, 该点距离地球 40 多万公里, 距离月球约 6.5 万公里. 若飞行器 $P$ 通过 $L_{2}$ 点时, 是在地球和月球的引力共同作用下沿圆轨道I绕地心做匀速圆周运动, 其周期与月球沿圆轨道II绕地心做匀速圆周运动的周期相等. 已知飞行器 $P$ 线速度为 $v$, 周期为 $T$, 受 地球和月球对它的万有引力大小之比为 $k$. 若飞行器 $P$ 只在地球万有引力作用下沿圆轨道I绕地心做匀速圆周运动的线速度为 $v^{\prime}$, 周期为 $T^{\prime}$, 则 [图1] A: $v^{\prime}>v$ B: $v^{\prime}<v$ C: $T^{\prime}=T \sqrt{1+\frac{1}{k}}$ D: $T^{\prime}=T \sqrt{1+k}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 地月拉格朗日 $L_{2}$ 点, 始终位于地月连线上的如图所示位置, 该点距离地球 40 多万公里, 距离月球约 6.5 万公里. 若飞行器 $P$ 通过 $L_{2}$ 点时, 是在地球和月球的引力共同作用下沿圆轨道I绕地心做匀速圆周运动, 其周期与月球沿圆轨道II绕地心做匀速圆周运动的周期相等. 已知飞行器 $P$ 线速度为 $v$, 周期为 $T$, 受 地球和月球对它的万有引力大小之比为 $k$. 若飞行器 $P$ 只在地球万有引力作用下沿圆轨道I绕地心做匀速圆周运动的线速度为 $v^{\prime}$, 周期为 $T^{\prime}$, 则 [图1] A: $v^{\prime}>v$ B: $v^{\prime}<v$ C: $T^{\prime}=T \sqrt{1+\frac{1}{k}}$ D: $T^{\prime}=T \sqrt{1+k}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

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multi-modal

Astronomy_517

火星周围笼罩着空气层, 与地球空气层相似。若宇航员在火星表面附近由静止释放一个质量为 $m$ 的小球, 小球在下落过程中由于受到空气阻力, 其加速度 $a$ 随位移 $x$ 变化的关系图像如图所示。已知火星的半径为 $R$ 、自转周期为 $T$, 引力常量为 $G$, 忽略火星自转对重力的影响, 则下列说法正确的是（） [图1] A: 火星的“第一宇宙速度”为 $\sqrt{a_{0} R}$ B: 火星的质量为 $\frac{a_{0} R}{G}$ C: 火星静止轨道卫星的轨道半径为 $T \sqrt{a_{0} R}$ D: 小球下落过程中的最大速度为 $\sqrt{a_{0} x_{0}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 火星周围笼罩着空气层, 与地球空气层相似。若宇航员在火星表面附近由静止释放一个质量为 $m$ 的小球, 小球在下落过程中由于受到空气阻力, 其加速度 $a$ 随位移 $x$ 变化的关系图像如图所示。已知火星的半径为 $R$ 、自转周期为 $T$, 引力常量为 $G$, 忽略火星自转对重力的影响, 则下列说法正确的是（） [图1] A: 火星的“第一宇宙速度”为 $\sqrt{a_{0} R}$ B: 火星的质量为 $\frac{a_{0} R}{G}$ C: 火星静止轨道卫星的轨道半径为 $T \sqrt{a_{0} R}$ D: 小球下落过程中的最大速度为 $\sqrt{a_{0} x_{0}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

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multi-modal

Astronomy_221

如图, 卫星在 $P$ 点由粗圆轨道II变轨到近椭圆形轨道I (可认为半径等于火星半径 $R)$ 。已知椭圆轨道II上的远火点 $Q$ 到火星表面高度为 $6 R$, 火星表面重力加速度为 $g_{0}$,引力常量为 $G$, 下列说法错误的是 ( ) [图1] A: 火星的平均密度为 $\frac{3 g_{0}}{4 \pi G R}$ B: 卫星在轨道II上由 $Q$ 点运动到 $P$ 点所用时间为 $16 \pi \sqrt{\frac{R}{g_{0}}}$ C: 卫星在轨道II上运动到 $Q$ 点时的速度小于 $\sqrt{\frac{g_{0} R}{7}}$ D: 卫星在轨道II上经过 $P$ 点时的加速度等于在轨道I上经过 $P$ 点时的加速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图, 卫星在 $P$ 点由粗圆轨道II变轨到近椭圆形轨道I (可认为半径等于火星半径 $R)$ 。已知椭圆轨道II上的远火点 $Q$ 到火星表面高度为 $6 R$, 火星表面重力加速度为 $g_{0}$,引力常量为 $G$, 下列说法错误的是 ( ) [图1] A: 火星的平均密度为 $\frac{3 g_{0}}{4 \pi G R}$ B: 卫星在轨道II上由 $Q$ 点运动到 $P$ 点所用时间为 $16 \pi \sqrt{\frac{R}{g_{0}}}$ C: 卫星在轨道II上运动到 $Q$ 点时的速度小于 $\sqrt{\frac{g_{0} R}{7}}$ D: 卫星在轨道II上经过 $P$ 点时的加速度等于在轨道I上经过 $P$ 点时的加速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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multi-modal

Astronomy_690

下列说法正确的是 A: 行星的运动和地球上物体的运动遵循不同的规律 B: 物体在转弯时一定受到力的作用 C: 月球绕地球运动时受到地球的引力和向心力的作用 D: 物体沿光滑斜面下滑时受到重力、斜面的支持力和下滑力的作用

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 下列说法正确的是 A: 行星的运动和地球上物体的运动遵循不同的规律 B: 物体在转弯时一定受到力的作用 C: 月球绕地球运动时受到地球的引力和向心力的作用 D: 物体沿光滑斜面下滑时受到重力、斜面的支持力和下滑力的作用 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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text-only

Astronomy_1097

It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$.a. Consider an observer in Oxford $\left(\phi=+51.8^{\circ}\right)$ on the June solstice. ii. By calculating the angle the solar path makes with the horizon, $\eta$, at sunrise for both the solstice and the equinox, estimate the duration of sunrise on the solstice if sunrise takes 3 mins 26 secs on the equinox. Assume the same solar angular velocity in both cases.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$. problem: a. Consider an observer in Oxford $\left(\phi=+51.8^{\circ}\right)$ on the June solstice. ii. By calculating the angle the solar path makes with the horizon, $\eta$, at sunrise for both the solstice and the equinox, estimate the duration of sunrise on the solstice if sunrise takes 3 mins 26 secs on the equinox. Assume the same solar angular velocity in both cases. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of min, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_133

如图, 虚线 I、II、III分别表示地球卫星的三条轨道, 其中轨道 I 为与第一宇宙速度 $7.9 \mathrm{~km} / \mathrm{s}$ 对应的近地环绕圆轨道, 轨道II为椭圆轨道, 轨道III为与第二宇宙速度 $11.2 \mathrm{~km} / \mathrm{s}$对应的脱离轨道, $a 、 b 、 c$ 三点分别位于三条轨道上, $b$ 点为轨道II的远地点, $b 、 c$ 点与地心的距离均为轨道 I 半径的 2 倍, 则 ( ) [图1] A: 卫星在轨道II的运行周期为轨道 I 的 2 倍 B: 卫星经过 $a$ 点的速率为经过 $b$ 点的 $\sqrt{2}$ 倍 C: 卫星在 $a$ 点的加速度大小为在 $c$ 点的 4 倍 D: 质量相同的卫星在 $b$ 点的机械能小于在 $c$ 点的机械能

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图, 虚线 I、II、III分别表示地球卫星的三条轨道, 其中轨道 I 为与第一宇宙速度 $7.9 \mathrm{~km} / \mathrm{s}$ 对应的近地环绕圆轨道, 轨道II为椭圆轨道, 轨道III为与第二宇宙速度 $11.2 \mathrm{~km} / \mathrm{s}$对应的脱离轨道, $a 、 b 、 c$ 三点分别位于三条轨道上, $b$ 点为轨道II的远地点, $b 、 c$ 点与地心的距离均为轨道 I 半径的 2 倍, 则 ( ) [图1] A: 卫星在轨道II的运行周期为轨道 I 的 2 倍 B: 卫星经过 $a$ 点的速率为经过 $b$ 点的 $\sqrt{2}$ 倍 C: 卫星在 $a$ 点的加速度大小为在 $c$ 点的 4 倍 D: 质量相同的卫星在 $b$ 点的机械能小于在 $c$ 点的机械能 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_857

The resolution of a space telescope is theoretically limited by diffraction from its primary mirror. In this problem, we will compare the diffraction limit of the Hubble Space Telescope (HST) (primary mirror diameter $d=2.4 \mathrm{~m})$ and the James Webb Space Telescope (JWST) $(d=6.5 \mathrm{~m})$. The operating wavelengths for the two telescopes are $500 \mathrm{~nm}$ and $10 \mu \mathrm{m}$ respectively. Calculate the ratio of the diffraction limited angular resolution $\frac{\theta(\mathrm{HST})}{\theta(\mathrm{JWST})}$. Which telescope can resolve smaller angular features if limited only by diffraction? A: 0.014, JWST B: $0.14, \mathrm{HST}$ C: 1.4, JWST D: 14, HST E: 140, JWST

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The resolution of a space telescope is theoretically limited by diffraction from its primary mirror. In this problem, we will compare the diffraction limit of the Hubble Space Telescope (HST) (primary mirror diameter $d=2.4 \mathrm{~m})$ and the James Webb Space Telescope (JWST) $(d=6.5 \mathrm{~m})$. The operating wavelengths for the two telescopes are $500 \mathrm{~nm}$ and $10 \mu \mathrm{m}$ respectively. Calculate the ratio of the diffraction limited angular resolution $\frac{\theta(\mathrm{HST})}{\theta(\mathrm{JWST})}$. Which telescope can resolve smaller angular features if limited only by diffraction? A: 0.014, JWST B: $0.14, \mathrm{HST}$ C: 1.4, JWST D: 14, HST E: 140, JWST You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_171

已知某卫星在赤道上空的圆形轨道运行, 轨道半径为 $r_{1}$, 运行周期为 $T$, 卫星运动方向与地球自转方向相同, 不计空气阻力, 万有引力常量为 $G \circ$ 。求: 卫星在赤道上空轨道半径为 $r_{I}$ 的圆形轨道上运行, 小明住在赤道上某城市, 某时刻，该卫星正处于小明的正上方，在后面的一段时间里，小明观察到每两天恰好三次看到卫星掠过其正上方, 求地球自转周期 $T_{0}$ 。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 已知某卫星在赤道上空的圆形轨道运行, 轨道半径为 $r_{1}$, 运行周期为 $T$, 卫星运动方向与地球自转方向相同, 不计空气阻力, 万有引力常量为 $G \circ$ 。求: 卫星在赤道上空轨道半径为 $r_{I}$ 的圆形轨道上运行, 小明住在赤道上某城市, 某时刻，该卫星正处于小明的正上方，在后面的一段时间里，小明观察到每两天恰好三次看到卫星掠过其正上方, 求地球自转周期 $T_{0}$ 。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_369

天体在引力场中具有的能叫做引力势能, 物理学中经常把无穷远处定为引力势能的零势能点, 引力势能表达式是 $E_{\mathrm{p} r}=-G \frac{M m}{r}$, 其中 $G$ 为引力常量, $M$ 为产生引力场物体 (中心天体) 的质量, $m$ 为研究对象的质量, $r$ 为两者质心之间的距离。已知海王星绕太阳做椭圆运动, 远日点和近日点的距离分别为 $r_{1}$ 和 $r_{2}$ 。另外已知地球绕太阳做圆周运动, 其轨道半径为 $R$ 。如果你还知道引力常量 $G$ 和地球公转周期 $T$, 结合已知数据和你掌握的物理规律，下列各选项中的两个物理量均可以推算出的是（） A: 海王星质量和地球质量 B: 太阳质量和海王星质量 C: 地球质量和海王星近日点速度大小 D: 太阳质量和海王星远日点速度大小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 天体在引力场中具有的能叫做引力势能, 物理学中经常把无穷远处定为引力势能的零势能点, 引力势能表达式是 $E_{\mathrm{p} r}=-G \frac{M m}{r}$, 其中 $G$ 为引力常量, $M$ 为产生引力场物体 (中心天体) 的质量, $m$ 为研究对象的质量, $r$ 为两者质心之间的距离。已知海王星绕太阳做椭圆运动, 远日点和近日点的距离分别为 $r_{1}$ 和 $r_{2}$ 。另外已知地球绕太阳做圆周运动, 其轨道半径为 $R$ 。如果你还知道引力常量 $G$ 和地球公转周期 $T$, 结合已知数据和你掌握的物理规律，下列各选项中的两个物理量均可以推算出的是（） A: 海王星质量和地球质量 B: 太阳质量和海王星质量 C: 地球质量和海王星近日点速度大小 D: 太阳质量和海王星远日点速度大小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_523

小型登月器连接在航天站上, 一起绕月球做圆周运动, 其轨道半径为月球半径的 3 倍, 某时刻, 航天站使登月器减速分离, 登月器沿如图所示的椭圆轨道登月, 在月球表面逗留一段时间完成科考工作后，经快速启动仍沿原椭圆轨道返回，当第一次回到分离点时恰与航天站对接, 登月器快速启动时间可以忽略不计, 整个过程中航天站保持原轨道绕月运行. 已知月球表面的重力加速度为 $\mathrm{g}$, 月球半径为 $\mathrm{R}$, 不考虑月球自转的影响, 则登月器可以在月球上停留的最短时间约为 ( ) [图1] A: $4.7 \pi \sqrt{\frac{R}{g}}$ B: $3.6 \pi \sqrt{\frac{R}{g}}$ C: $1.7 \pi \sqrt{\frac{R}{g}}$ D: $1.4 \pi \sqrt{\frac{R}{g}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 小型登月器连接在航天站上, 一起绕月球做圆周运动, 其轨道半径为月球半径的 3 倍, 某时刻, 航天站使登月器减速分离, 登月器沿如图所示的椭圆轨道登月, 在月球表面逗留一段时间完成科考工作后，经快速启动仍沿原椭圆轨道返回，当第一次回到分离点时恰与航天站对接, 登月器快速启动时间可以忽略不计, 整个过程中航天站保持原轨道绕月运行. 已知月球表面的重力加速度为 $\mathrm{g}$, 月球半径为 $\mathrm{R}$, 不考虑月球自转的影响, 则登月器可以在月球上停留的最短时间约为 ( ) [图1] A: $4.7 \pi \sqrt{\frac{R}{g}}$ B: $3.6 \pi \sqrt{\frac{R}{g}}$ C: $1.7 \pi \sqrt{\frac{R}{g}}$ D: $1.4 \pi \sqrt{\frac{R}{g}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_1121

It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$.d. This exam is being taken on $24^{\text {th }}$ January and is 3 hours long. ii. Hence, calculate the latitude where today's day length is equal to the exam length, taking the Sun to be a point source.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$. problem: d. This exam is being taken on $24^{\text {th }}$ January and is 3 hours long. ii. Hence, calculate the latitude where today's day length is equal to the exam length, taking the Sun to be a point source. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \circ, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_897

In 2015, physicists at the Laser Interferometer Gravitational-Wave Observatory (LIGO) announced the first-ever observation of gravitational waves. What triggered the pulse? A: Two neutron stars colliding B: Two black holes colliding C: A star going supernova D: A gamma ray burst

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: In 2015, physicists at the Laser Interferometer Gravitational-Wave Observatory (LIGO) announced the first-ever observation of gravitational waves. What triggered the pulse? A: Two neutron stars colliding B: Two black holes colliding C: A star going supernova D: A gamma ray burst You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_1043

The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically.c. Computer models suggest the first galaxies formed around $z \sim 10-20$. One of the best ways to look for high-redshift galaxies is to try and detect the emission from the Lyman alpha (Lya) emission line at $\lambda_{\text {emit }}=121.6 \mathrm{~nm}$ as it is a relatively bright line. Some of the brightest galaxies in that initial era of galaxy formation would have an absolute magnitude of $\mathcal{M} \sim 20$. In this question, you are given that $\Omega_{0, \mathrm{~m}}=0.3, \Omega_{0, \Lambda}=0.7, \Omega_{0, \mathrm{r}}=0$ and $\mathrm{H}_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. iv. If the minimum flux detectable decreases proportionally to $t_{\text {exp }}^{1 / 2}$ where $t_{\text {exp }}$ is the length of the exposure, estimate the minimum exposure time necessary for JWST to image this galaxy with $S / N=10$. Give your answer in hours.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically. problem: c. Computer models suggest the first galaxies formed around $z \sim 10-20$. One of the best ways to look for high-redshift galaxies is to try and detect the emission from the Lyman alpha (Lya) emission line at $\lambda_{\text {emit }}=121.6 \mathrm{~nm}$ as it is a relatively bright line. Some of the brightest galaxies in that initial era of galaxy formation would have an absolute magnitude of $\mathcal{M} \sim 20$. In this question, you are given that $\Omega_{0, \mathrm{~m}}=0.3, \Omega_{0, \Lambda}=0.7, \Omega_{0, \mathrm{r}}=0$ and $\mathrm{H}_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. iv. If the minimum flux detectable decreases proportionally to $t_{\text {exp }}^{1 / 2}$ where $t_{\text {exp }}$ is the length of the exposure, estimate the minimum exposure time necessary for JWST to image this galaxy with $S / N=10$. Give your answer in hours. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{hrs}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_978

After many delays, the James Webb Space Telescope (JWST) is due to be launched in December 2021. This will be the successor to the Hubble Space Telescope (HST) and has a much bigger primary mirror ( $6.5 \mathrm{~m}$ in diameter against $2.4 \mathrm{~m}$ for the HST) so will be able to study the Universe in unprecedented detail. Given the faintest objects the HST can see have a magnitude of $\sim 31$ (known as the limiting magnitude), what limiting magnitude might we expect for JWST? A: $\sim 29$ B: $\sim 31$ C: $\sim 33$ D: $\sim 36$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: After many delays, the James Webb Space Telescope (JWST) is due to be launched in December 2021. This will be the successor to the Hubble Space Telescope (HST) and has a much bigger primary mirror ( $6.5 \mathrm{~m}$ in diameter against $2.4 \mathrm{~m}$ for the HST) so will be able to study the Universe in unprecedented detail. Given the faintest objects the HST can see have a magnitude of $\sim 31$ (known as the limiting magnitude), what limiting magnitude might we expect for JWST? A: $\sim 29$ B: $\sim 31$ C: $\sim 33$ D: $\sim 36$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_323

2016 年 12 月 17 口是我国发射“悟空”探测卫星一周年的日子.“悟空”探测卫星的发射为人类对暗物质的研究,做出了重大贡献.假设两颗质量相等的星球绕其球心连线中心转动,理论计算的周期与实际观测周期有出入,且 $\frac{T_{\text {理论 }}}{T_{\text {观测 }}}=\frac{\sqrt{n}}{1}(n>1)$ 。科学家推测在以两星球球心连线为直径的球体空间中均匀分布着暗物质, 设两星球球心连线长度为 $L$, 质量均为 $m$,据此推测,暗物质的质量为( ) A: $(n-1) m$ B: $(2 n-1) m$ C: $\frac{n-1}{4} m$ D: $\frac{n-2}{8} m$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2016 年 12 月 17 口是我国发射“悟空”探测卫星一周年的日子.“悟空”探测卫星的发射为人类对暗物质的研究,做出了重大贡献.假设两颗质量相等的星球绕其球心连线中心转动,理论计算的周期与实际观测周期有出入,且 $\frac{T_{\text {理论 }}}{T_{\text {观测 }}}=\frac{\sqrt{n}}{1}(n>1)$ 。科学家推测在以两星球球心连线为直径的球体空间中均匀分布着暗物质, 设两星球球心连线长度为 $L$, 质量均为 $m$,据此推测,暗物质的质量为( ) A: $(n-1) m$ B: $(2 n-1) m$ C: $\frac{n-1}{4} m$ D: $\frac{n-2}{8} m$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_98

2020 年 11 月 24 日, 长征五号遥五运载火箭搭载嫦娥五号探测器成功发射升空并将其送入轨道, 11 月 28 日, 嫦娥五号进入环月轨道飞行, 12 月 17 日凌晨, 嫦娥五号返回器携带月壤着陆地球。假设嫦娥五号环绕月球飞行时, 在距月球表面高度为 $h$ 处,绕月球做匀速圆周运动 (不计周围其他天体的影响), 测出其飞行周期 $T$, 已知引力常量 $G$ 和月球半径 $R$, 则下列说法正确的是 ( ) A: 嫦娥五号绕月球飞行的线速度为 $\frac{2 \pi(R+h)}{T}$ B: 月球的质量为 $\frac{4 \pi^{2}(R+h)^{2}}{G T^{2}}$ C: 月球的第一宇宙速度为 $\frac{2 \pi(R+h)}{T} \sqrt{\frac{R+h}{R}}$ D: 月球表面的重力加速度为 $\frac{4 \pi^{2}(R+h)^{3}}{R^{2} T^{2}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2020 年 11 月 24 日, 长征五号遥五运载火箭搭载嫦娥五号探测器成功发射升空并将其送入轨道, 11 月 28 日, 嫦娥五号进入环月轨道飞行, 12 月 17 日凌晨, 嫦娥五号返回器携带月壤着陆地球。假设嫦娥五号环绕月球飞行时, 在距月球表面高度为 $h$ 处,绕月球做匀速圆周运动 (不计周围其他天体的影响), 测出其飞行周期 $T$, 已知引力常量 $G$ 和月球半径 $R$, 则下列说法正确的是 ( ) A: 嫦娥五号绕月球飞行的线速度为 $\frac{2 \pi(R+h)}{T}$ B: 月球的质量为 $\frac{4 \pi^{2}(R+h)^{2}}{G T^{2}}$ C: 月球的第一宇宙速度为 $\frac{2 \pi(R+h)}{T} \sqrt{\frac{R+h}{R}}$ D: 月球表面的重力加速度为 $\frac{4 \pi^{2}(R+h)^{3}}{R^{2} T^{2}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_1051

In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$.a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only iii. Sketch an acceleration-time graph of the journey from lift-off to reaching the parking orbit. Give accelerations in units of $g_{0}$. Assume the time between one stage finishing, detaching, and ignition of the next stage is negligible (i.e. you will have discontinuities in the graph at the end of each stage).

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$. problem: a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only iii. Sketch an acceleration-time graph of the journey from lift-off to reaching the parking orbit. Give accelerations in units of $g_{0}$. Assume the time between one stage finishing, detaching, and ignition of the next stage is negligible (i.e. you will have discontinuities in the graph at the end of each stage). All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_314

某地面卫星接收站的纬度为 $\theta(\theta>0)$. 已知地球半径为 $R$, 重力加速度为 $g$, 自转周期为 $T$. 光速为 $c$, 则地球同步卫星发射的电磁波到该接收站的时间不小于 ( ) A: $\frac{\sqrt[3]{\frac{R^{2} T^{2} g}{4 \pi^{2}}}}{c}$ B: $\sqrt[3]{\frac{R^{2} T^{2} g}{4 \pi^{2}}}-R$ C: $\frac{\sqrt{R^{2}+r^{2}+2 R r \cos \theta}}{c}\left(\right.$ 其中 $r=\sqrt[3]{\frac{g R^{2} T^{2}}{4 \pi^{2}}}$ ) D: $\frac{\sqrt{R^{2}+r^{2}-2 R r \cos \theta}}{c}\left(\right.$ 其中 $r=\sqrt[3]{\frac{g R^{2} T^{2}}{4 \pi^{2}}}$ )

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 某地面卫星接收站的纬度为 $\theta(\theta>0)$. 已知地球半径为 $R$, 重力加速度为 $g$, 自转周期为 $T$. 光速为 $c$, 则地球同步卫星发射的电磁波到该接收站的时间不小于 ( ) A: $\frac{\sqrt[3]{\frac{R^{2} T^{2} g}{4 \pi^{2}}}}{c}$ B: $\sqrt[3]{\frac{R^{2} T^{2} g}{4 \pi^{2}}}-R$ C: $\frac{\sqrt{R^{2}+r^{2}+2 R r \cos \theta}}{c}\left(\right.$ 其中 $r=\sqrt[3]{\frac{g R^{2} T^{2}}{4 \pi^{2}}}$ ) D: $\frac{\sqrt{R^{2}+r^{2}-2 R r \cos \theta}}{c}\left(\right.$ 其中 $r=\sqrt[3]{\frac{g R^{2} T^{2}}{4 \pi^{2}}}$ ) 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_534

2016 年 10 月 19 日 3 时 31 分, 神舟十一号载人飞船与天宫二号空间实验室成功实 现自动交会对接. 若对接前神舟十一号和天宫二号分别在图示轨道上绕地球同向 (天宫二号在前）做匀速圆周运动, 则 [图1] A: 神舟十一号的线速度小于天宫二号的线速度 B: 神舟十一号的周期大于天宫二号的周期 C: 神舟十一号的角速度小于天宫二号的角速度 D: 神舟十一号可在轨道上加速后实现与天宫二号的对接

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2016 年 10 月 19 日 3 时 31 分, 神舟十一号载人飞船与天宫二号空间实验室成功实 现自动交会对接. 若对接前神舟十一号和天宫二号分别在图示轨道上绕地球同向 (天宫二号在前）做匀速圆周运动, 则 [图1] A: 神舟十一号的线速度小于天宫二号的线速度 B: 神舟十一号的周期大于天宫二号的周期 C: 神舟十一号的角速度小于天宫二号的角速度 D: 神舟十一号可在轨道上加速后实现与天宫二号的对接 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_735

“神舟十三号”飞船开始在半径为 $r_{1}$ 的圆轨道 $\mathrm{I}$ 上运行, 运行周期为 $T_{1}$, 在 $A$ 点通过变轨操作后进入粗圆轨道II运动, 沿轨道II运动到远地点 $B$ 时正好与处于半径为 $r_{3}$ 的圆轨道III上的核心舱对接, $A$ 为粗圆轨道II的近地点。假设飞船质量始终不变, 关于飞船的运动, 下列说法正确的是（ ） [图1] A: 沿轨道I运动到 $A$ 时的速率大于沿轨道II运动到 $B$ 时的速率 B: 沿轨道I运行时的机械能等于沿轨道II运行时的机械能 C: 沿轨道II运行的周期为 $T_{1} \sqrt{\left(\frac{r_{1}+r_{3}}{2 r_{1}}\right)^{3}}$ D: 沿轨道I运动到 $A$ 点时的加速度小于沿轨道II运动到 $B$ 点时的加速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: “神舟十三号”飞船开始在半径为 $r_{1}$ 的圆轨道 $\mathrm{I}$ 上运行, 运行周期为 $T_{1}$, 在 $A$ 点通过变轨操作后进入粗圆轨道II运动, 沿轨道II运动到远地点 $B$ 时正好与处于半径为 $r_{3}$ 的圆轨道III上的核心舱对接, $A$ 为粗圆轨道II的近地点。假设飞船质量始终不变, 关于飞船的运动, 下列说法正确的是（ ） [图1] A: 沿轨道I运动到 $A$ 时的速率大于沿轨道II运动到 $B$ 时的速率 B: 沿轨道I运行时的机械能等于沿轨道II运行时的机械能 C: 沿轨道II运行的周期为 $T_{1} \sqrt{\left(\frac{r_{1}+r_{3}}{2 r_{1}}\right)^{3}}$ D: 沿轨道I运动到 $A$ 点时的加速度小于沿轨道II运动到 $B$ 点时的加速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1203

The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\text {peri }}=9.86 R_{\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft. [figure1] Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\mathrm{km} \mathrm{s}^{-1}$. Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2. [figure2] Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\angle y c z$ in the figure, such that the shaded areas are the same). [figure3] Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia. The eccentric anomaly can be related to the mean anomaly through Kepler's Equation, $$ M=E-e \sin E \text {. } $$c. After the first flyby of Venus on 3rd October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6th November 2018 was at a distance of $35.7 R_{\odot}$. Given its mass at launch was $685 \mathrm{~kg}$, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\text {peri }}=9.86 R_{\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft. [figure1] Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\mathrm{km} \mathrm{s}^{-1}$. Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2. [figure2] Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\angle y c z$ in the figure, such that the shaded areas are the same). [figure3] Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia. The eccentric anomaly can be related to the mean anomaly through Kepler's Equation, $$ M=E-e \sin E \text {. } $$ problem: c. After the first flyby of Venus on 3rd October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6th November 2018 was at a distance of $35.7 R_{\odot}$. Given its mass at launch was $685 \mathrm{~kg}$, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~J}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

EN

multi-modal

Astronomy_828

Consider Galaxies $\mathrm{A}$ and $\mathrm{B}$, both of which have radius $R$. At a distance $R$ from its center, Galaxy A's rotational velocity is equal to $v$. Meanwhile, Galaxy B's radial velocity dispersion is also equal to $v$. However, galaxy A is spiral while galaxy B is spherical elliptical and composed of uniform, evenly-spaced stars. Calculate the masses of both galaxies. (Answer choices are listed as $\left.m_{A} ; m_{B}\right)$. A: $v^{2} R / G ; v^{2} R / G$ B: $v^{2} R / G ; 5 / 6 v^{2} R / G$ C: $v^{2} R / G ; 5 / 4 v^{2} R / G$ D: $v^{2} R / G ; 5 v^{2} R / G$ E: $5 / 2 v^{2} R / G ; v^{2} R / G$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Consider Galaxies $\mathrm{A}$ and $\mathrm{B}$, both of which have radius $R$. At a distance $R$ from its center, Galaxy A's rotational velocity is equal to $v$. Meanwhile, Galaxy B's radial velocity dispersion is also equal to $v$. However, galaxy A is spiral while galaxy B is spherical elliptical and composed of uniform, evenly-spaced stars. Calculate the masses of both galaxies. (Answer choices are listed as $\left.m_{A} ; m_{B}\right)$. A: $v^{2} R / G ; v^{2} R / G$ B: $v^{2} R / G ; 5 / 6 v^{2} R / G$ C: $v^{2} R / G ; 5 / 4 v^{2} R / G$ D: $v^{2} R / G ; 5 v^{2} R / G$ E: $5 / 2 v^{2} R / G ; v^{2} R / G$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

EN

text-only

Astronomy_963

Currently, Polaris is very close to the north celestial pole (the projection of the Earth's rotational axis on the sky) and so all other stars appear to rotate around it. However, this axis is drawing out a large circle in the sky with an angular radius of $23.44^{\circ}$ so Polaris will only temporarily be the pole star (see Figure 2). This precession of the rotational axis is mainly driven by the gravitational pull of the Moon and the Sun. [figure1] Figure 2: Top left: The Earth's rotational axis itself rotates slowly (white circle), in what is known as axial precession. Credit: David Battisti / University of Washington. Top right: Due to precession, the pole star has changed over time. About 5000 years ago, the star Thuban in the constellation of Draco was the pole star. Credit: Richard W. Pogge / Ohio State University. Bottom: The position of the Sun at the spring equinox (where the celestial equator meets the ecliptic) has also changed over the same period, moving from Aries to Pisces. Credit: Guy Ottewell / Universal Workshop. Another consequence is that the position of the Sun at the equinoxes varies slightly, moving slowly westwards. This gives rise to two definitions of a year: - a sidereal year (the time taken for the Earth to orbit the Sun once with respect to the background stars) $=365.256363$ days - a tropical year (the time taken for the Sun to return to the same position in the cycle of the seasons) $=365.242190$ days The Gregorian calendar is a 400-year cycle with a system of leap years. The rule is: "every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100 , but these centurial years are leap years if they are exactly divisible by 400. ." By working out the average length of a year in the Gregorian calendar, is it closer to the sidereal or the tropical year?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Currently, Polaris is very close to the north celestial pole (the projection of the Earth's rotational axis on the sky) and so all other stars appear to rotate around it. However, this axis is drawing out a large circle in the sky with an angular radius of $23.44^{\circ}$ so Polaris will only temporarily be the pole star (see Figure 2). This precession of the rotational axis is mainly driven by the gravitational pull of the Moon and the Sun. [figure1] Figure 2: Top left: The Earth's rotational axis itself rotates slowly (white circle), in what is known as axial precession. Credit: David Battisti / University of Washington. Top right: Due to precession, the pole star has changed over time. About 5000 years ago, the star Thuban in the constellation of Draco was the pole star. Credit: Richard W. Pogge / Ohio State University. Bottom: The position of the Sun at the spring equinox (where the celestial equator meets the ecliptic) has also changed over the same period, moving from Aries to Pisces. Credit: Guy Ottewell / Universal Workshop. Another consequence is that the position of the Sun at the equinoxes varies slightly, moving slowly westwards. This gives rise to two definitions of a year: - a sidereal year (the time taken for the Earth to orbit the Sun once with respect to the background stars) $=365.256363$ days - a tropical year (the time taken for the Sun to return to the same position in the cycle of the seasons) $=365.242190$ days The Gregorian calendar is a 400-year cycle with a system of leap years. The rule is: "every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100 , but these centurial years are leap years if they are exactly divisible by 400. ." By working out the average length of a year in the Gregorian calendar, is it closer to the sidereal or the tropical year? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

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Astronomy

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multi-modal

Astronomy_353

“亚洲一号”地球同步通讯卫星的质量是 $1.2 \mathrm{t}$, 下列有关它的说法正确的是 ( ) A: 若将它的质量增加为 $2.4 \mathrm{t}$, 其同步轨道半径变为原来的 2 倍 B: 它的运行速度小于 $7.9 \mathrm{~km} / \mathrm{s}$ C: 它可以绕过北京的正上方, 所以我国能利用其进行电视转播 D: 它的周期是 $24 \mathrm{~h}$, 其轨道平面与赤道平面重合且距地面高度一定

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: “亚洲一号”地球同步通讯卫星的质量是 $1.2 \mathrm{t}$, 下列有关它的说法正确的是 ( ) A: 若将它的质量增加为 $2.4 \mathrm{t}$, 其同步轨道半径变为原来的 2 倍 B: 它的运行速度小于 $7.9 \mathrm{~km} / \mathrm{s}$ C: 它可以绕过北京的正上方, 所以我国能利用其进行电视转播 D: 它的周期是 $24 \mathrm{~h}$, 其轨道平面与赤道平面重合且距地面高度一定 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

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text-only

Astronomy_796

Edwin Hubble published in 1929 the paper "A Relation Between Distance and Extragalactic Nebulae" explaining his find that there exists a linear relationship between the radial velocities and distances for extragalactic objects. In the following graph you can see his data used in the paper. The units for the velocity are in $\mathrm{km} / \mathrm{s}$. (notice the "/ $\mathrm{s}$ " is missing from his original graph). What is the value for the Hubble constant that he derived from this data? Use the linear fit to the data marked with the continuous line to derive your estimation. [figure1] A: $500 \mathrm{~km} / \mathrm{s} / \mathrm{pc}$ B: $72 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ C: $500 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ D: $50 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ E: $67 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Edwin Hubble published in 1929 the paper "A Relation Between Distance and Extragalactic Nebulae" explaining his find that there exists a linear relationship between the radial velocities and distances for extragalactic objects. In the following graph you can see his data used in the paper. The units for the velocity are in $\mathrm{km} / \mathrm{s}$. (notice the "/ $\mathrm{s}$ " is missing from his original graph). What is the value for the Hubble constant that he derived from this data? Use the linear fit to the data marked with the continuous line to derive your estimation. [figure1] A: $500 \mathrm{~km} / \mathrm{s} / \mathrm{pc}$ B: $72 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ C: $500 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ D: $50 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ E: $67 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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multi-modal

Astronomy_972

Which of the following is not a zodiacal constellation, according to the astronomical definition? [figure1] A: Aquila B: Aquarius C: Ophiucus D: Pisces

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Which of the following is not a zodiacal constellation, according to the astronomical definition? [figure1] A: Aquila B: Aquarius C: Ophiucus D: Pisces You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

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multi-modal

Astronomy_588

2021 年 7 月和 10 月, SpaceX 公司星链卫星两次侵入中国天宫空间站轨道, 为保证空间站内工作的三名航天员生命安全, 中方不得不调整轨道高度, 紧急避碰。其中星链 -2305 号卫星, 采取连续变轨模式接近中国空间站, 中国发现且规避后, 该卫星轨道又重新回到正常轨道。已知中国空间站在高度 390 千米附近的近圆轨道，轨道倾角 $41.58^{\circ}$, 而星链卫星在高度为 550 千米附近的近圆轨道, 倾角为 $53^{\circ}$ 。已知地球半径为 $6400 \mathrm{~km}$, 引力常量为 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$ 。下列说法正确的是（） A: 中国空间站的运行速度约为 11.68 千米/秒 B: 由题目所给数据可以计算出地球的密度 C: 星链卫星在正常圆轨道需减速才能降到中国空间站所在高度 D: 星链卫星保持轨道平面不变, 降至 390 公里附近圆轨道时, 其速度与中国空间站速度接近, 不会相撞（轨道倾角是指卫星轨道面与地球赤道面之间的夹角）

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2021 年 7 月和 10 月, SpaceX 公司星链卫星两次侵入中国天宫空间站轨道, 为保证空间站内工作的三名航天员生命安全, 中方不得不调整轨道高度, 紧急避碰。其中星链 -2305 号卫星, 采取连续变轨模式接近中国空间站, 中国发现且规避后, 该卫星轨道又重新回到正常轨道。已知中国空间站在高度 390 千米附近的近圆轨道，轨道倾角 $41.58^{\circ}$, 而星链卫星在高度为 550 千米附近的近圆轨道, 倾角为 $53^{\circ}$ 。已知地球半径为 $6400 \mathrm{~km}$, 引力常量为 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$ 。下列说法正确的是（） A: 中国空间站的运行速度约为 11.68 千米/秒 B: 由题目所给数据可以计算出地球的密度 C: 星链卫星在正常圆轨道需减速才能降到中国空间站所在高度 D: 星链卫星保持轨道平面不变, 降至 390 公里附近圆轨道时, 其速度与中国空间站速度接近, 不会相撞（轨道倾角是指卫星轨道面与地球赤道面之间的夹角） 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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Astronomy_878

Consider that the absolute magnitude of a star is $m_{0}$. Imagine that the first star gets split into $\mathrm{N}$ smaller identical stars with the same temperature and average densities as the initial star, and that the sum of the masses of all $\mathrm{N}$ smaller stars is equal to the initial star's mass (i.e., total mass is conserved). What is the total combined absolute magnitude $(\mathrm{m})$ of all the $\mathrm{N}$ stars assuming that none of the stars obstruct each other's light (i.e. their luminosities add linearly)? A: $m=m_{0}-\log (N)$ B: $m=m_{0}-2.5 \log (N)$ C: $m=m_{0}-\frac{2.5}{3} \log (N)$ D: $m=m_{0}-\frac{2.5}{N}$ E: $m=m_{0}-2.5 \mathrm{~N}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Consider that the absolute magnitude of a star is $m_{0}$. Imagine that the first star gets split into $\mathrm{N}$ smaller identical stars with the same temperature and average densities as the initial star, and that the sum of the masses of all $\mathrm{N}$ smaller stars is equal to the initial star's mass (i.e., total mass is conserved). What is the total combined absolute magnitude $(\mathrm{m})$ of all the $\mathrm{N}$ stars assuming that none of the stars obstruct each other's light (i.e. their luminosities add linearly)? A: $m=m_{0}-\log (N)$ B: $m=m_{0}-2.5 \log (N)$ C: $m=m_{0}-\frac{2.5}{3} \log (N)$ D: $m=m_{0}-\frac{2.5}{N}$ E: $m=m_{0}-2.5 \mathrm{~N}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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Astronomy_863

A comet is approaching our solar system from the depths of space with a velocity of $10000 \mathrm{~m} / \mathrm{s}$, and if it continues moving in a straight line on its current trajectory, it will just barely graze the surface of the Sun! What is the eccentricity of the comet's orbit? A: 1.00014 B: 1.000014 C: 1.0000014 D: 1.00000014 E: 1.000000014

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: A comet is approaching our solar system from the depths of space with a velocity of $10000 \mathrm{~m} / \mathrm{s}$, and if it continues moving in a straight line on its current trajectory, it will just barely graze the surface of the Sun! What is the eccentricity of the comet's orbit? A: 1.00014 B: 1.000014 C: 1.0000014 D: 1.00000014 E: 1.000000014 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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Astronomy_467

利用金星凌日现象, 我们可以估算出地球与太阳之间的平均距离。日地平均距离也被定义为 1 个天文单位 (1A.U.), 是天文学中常用的距离单位。 金星轨道在地球轨道内侧, 某些特殊时刻, 地球、金星、太阳恰在一条直线上, 这时从地球上可以看到金星就像一个小黑点一样在太阳表面缓慢移动, 如图甲所示, 天文学称之为“金星凌日”。在地球上的不同地点, 比如图乙中的 $A 、 B$ 两点, 它们在同一时刻观察到的金星在日面上的位置是不同的，我们分别记为 $A^{\prime} 、 B^{\prime}$ 。 设金星与太阳的距离为 $k$ 倍日地距离, 即 $k \mathrm{~A}$. U. 可测得地球上 $A 、 B$ 之间的距离为 $l$, 估算 $A^{\prime} 、 B^{\prime}$ 在太阳表面的真实距离 [图1] 甲 [图2] 丙 [图3] 乙

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 利用金星凌日现象, 我们可以估算出地球与太阳之间的平均距离。日地平均距离也被定义为 1 个天文单位 (1A.U.), 是天文学中常用的距离单位。 金星轨道在地球轨道内侧, 某些特殊时刻, 地球、金星、太阳恰在一条直线上, 这时从地球上可以看到金星就像一个小黑点一样在太阳表面缓慢移动, 如图甲所示, 天文学称之为“金星凌日”。在地球上的不同地点, 比如图乙中的 $A 、 B$ 两点, 它们在同一时刻观察到的金星在日面上的位置是不同的，我们分别记为 $A^{\prime} 、 B^{\prime}$ 。 设金星与太阳的距离为 $k$ 倍日地距离, 即 $k \mathrm{~A}$. U. 可测得地球上 $A 、 B$ 之间的距离为 $l$, 估算 $A^{\prime} 、 B^{\prime}$ 在太阳表面的真实距离 [图1] 甲 [图2] 丙 [图3] 乙 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

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multi-modal

Astronomy_67

银河系处于某超星系团的边缘, 已知银河系距离星系团中心约 2 亿光年, 绕星系团中心运行的公转周期约为 1000 亿年, 引力常量 $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$, 根据上述数据可估算 A: 银河系绕超星系团中心运动的线速度 B: 银河系绕超星系团中心运动的加速度 C: 银河系的质量 D: 超星系团的质量

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 银河系处于某超星系团的边缘, 已知银河系距离星系团中心约 2 亿光年, 绕星系团中心运行的公转周期约为 1000 亿年, 引力常量 $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$, 根据上述数据可估算 A: 银河系绕超星系团中心运动的线速度 B: 银河系绕超星系团中心运动的加速度 C: 银河系的质量 D: 超星系团的质量 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

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text-only

Astronomy_391

2020 年 6 月 23 日上午, 北斗三号全球卫星导航系统的“收官之星”成功发射, 标志着北斗三号全球卫星导航系统全球星座组网部署最后一步完成。中国北斗, 将点亮世界卫星导航的天空。“收官之星”最后静止在地面上空（与地面保持相对静止）。该卫星距地面的高度为 $h$, 已知地球的半径为 $R$, 地球表面的重力加速度为 $g$, 万有引力常量为 $G$ 。由此可知（） A: “收官之星”运动的周期为 $2 \pi \sqrt{\frac{R}{g}}$ B: “收官之星”运动的加速度为 $\frac{g R}{R+h}$ C: “收官之星”运动的轨道一定与赤道不共面 D: 地球的平均密度为 $\frac{3 g}{4 \pi G R}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2020 年 6 月 23 日上午, 北斗三号全球卫星导航系统的“收官之星”成功发射, 标志着北斗三号全球卫星导航系统全球星座组网部署最后一步完成。中国北斗, 将点亮世界卫星导航的天空。“收官之星”最后静止在地面上空（与地面保持相对静止）。该卫星距地面的高度为 $h$, 已知地球的半径为 $R$, 地球表面的重力加速度为 $g$, 万有引力常量为 $G$ 。由此可知（） A: “收官之星”运动的周期为 $2 \pi \sqrt{\frac{R}{g}}$ B: “收官之星”运动的加速度为 $\frac{g R}{R+h}$ C: “收官之星”运动的轨道一定与赤道不共面 D: 地球的平均密度为 $\frac{3 g}{4 \pi G R}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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Astronomy_1141

In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$.a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only i. Show that the thrust generated by the S-IC is about $3.3 \times 10^{7} \mathrm{~N}$ and hence calculate the acceleration experienced by the astronauts firstly at lift-off and secondly when the S-IC finishes its burn (ignore that the S-IC ignites a few seconds before lift off). Give your answer in units of $g_{0}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$. problem: a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only i. Show that the thrust generated by the S-IC is about $3.3 \times 10^{7} \mathrm{~N}$ and hence calculate the acceleration experienced by the astronauts firstly at lift-off and secondly when the S-IC finishes its burn (ignore that the S-IC ignites a few seconds before lift off). Give your answer in units of $g_{0}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_9

在如图所示的双星系统中, $A 、 B$ 两个恒星靠着相互之间的引力正在做匀速圆周运 动, 已知恒星 $A$ 的质量为太阳质量的 29 倍, 恒星 $B$ 的质量为太阳质量的 36 倍, 两星之间的距离 $L=2 \times 10^{5} \mathrm{~m}$, 太阳质量 $M=2 \times 10^{30} \mathrm{~kg}$, 引力常量 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \pi^{2}$ $=10$. 若两星在环绕过程中会辐射出引力波, 该引力波的频率与两星做圆周运动的频率具有相同的数量级, 则根据题目所给信息估算该引力波频率的数量级是( ) [图1] A: $10^{2} \mathrm{~Hz}$ B: $10^{4} \mathrm{~Hz}$ C: $10^{6} \mathrm{~Hz}$ D: $10^{8} \mathrm{~Hz}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 在如图所示的双星系统中, $A 、 B$ 两个恒星靠着相互之间的引力正在做匀速圆周运 动, 已知恒星 $A$ 的质量为太阳质量的 29 倍, 恒星 $B$ 的质量为太阳质量的 36 倍, 两星之间的距离 $L=2 \times 10^{5} \mathrm{~m}$, 太阳质量 $M=2 \times 10^{30} \mathrm{~kg}$, 引力常量 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \pi^{2}$ $=10$. 若两星在环绕过程中会辐射出引力波, 该引力波的频率与两星做圆周运动的频率具有相同的数量级, 则根据题目所给信息估算该引力波频率的数量级是( ) [图1] A: $10^{2} \mathrm{~Hz}$ B: $10^{4} \mathrm{~Hz}$ C: $10^{6} \mathrm{~Hz}$ D: $10^{8} \mathrm{~Hz}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_932

When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ Verify that the barycentre of the Sun-Jupiter system lies outside the Sun (i.e. $r>\mathrm{R}_{\odot}$ ), given that the mass of Jupiter is $1.90 \times 10^{27} \mathrm{~kg}$ and it is on average $5.20 \mathrm{AU}$ from the Sun. [figure1] Figure 4: The relative positions of Pluto, Charon and the system's barycentre.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. problem: When two objects of unequal mass orbit around each other, they both orbit around a barycentre - this is the name given to the location of the centre of mass of the system. The masses of both objects, and the distance between their centres, affects the position of their barycentre. Imagine two objects, Object 1 and Object 2, with masses $m_{1}$ and $m_{2}$ respectively, and the average distance between the centre of both objects is $a$, then the average distance from the centre of Object 1 to the barycentre, $r$, is given by the formula: $$ r=a \frac{m_{2}}{m_{1}+m_{2}} $$ Verify that the barycentre of the Sun-Jupiter system lies outside the Sun (i.e. $r>\mathrm{R}_{\odot}$ ), given that the mass of Jupiter is $1.90 \times 10^{27} \mathrm{~kg}$ and it is on average $5.20 \mathrm{AU}$ from the Sun. [figure1] Figure 4: The relative positions of Pluto, Charon and the system's barycentre. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

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Astronomy

EN

multi-modal

Astronomy_711

如图甲, “星下点” 是指卫星和地心连线与地球表面的交点。图乙是航天控制中心大屏幕上显示卫星 FZ01 的“星下点”在一段时间内的轨迹, 已知地球同步卫星的轨道半径为 $r, \mathrm{FZ} 01$ 绕行方向与地球自转方向一致, 则下列说法正确的是 ( ) [图1] 甲 [图2] 乙 A: 卫星 FZ01 的轨道半径约为 $\frac{r}{2}$ B: 卫星 FZ01 的轨道半径约为 $\frac{r}{5}$ C: 卫星 FZ01 可以记录到北极点的气候变化 D: 卫星 FZ01 不可以记录到北极点的气候变化

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图甲, “星下点” 是指卫星和地心连线与地球表面的交点。图乙是航天控制中心大屏幕上显示卫星 FZ01 的“星下点”在一段时间内的轨迹, 已知地球同步卫星的轨道半径为 $r, \mathrm{FZ} 01$ 绕行方向与地球自转方向一致, 则下列说法正确的是 ( ) [图1] 甲 [图2] 乙 A: 卫星 FZ01 的轨道半径约为 $\frac{r}{2}$ B: 卫星 FZ01 的轨道半径约为 $\frac{r}{5}$ C: 卫星 FZ01 可以记录到北极点的气候变化 D: 卫星 FZ01 不可以记录到北极点的气候变化 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1154

It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$.d. This exam is being taken on $24^{\text {th }}$ January and is 3 hours long. i. Estimate the solar declination on this date.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: It is often said that the Sun rises in the East and sets in the West, however this is only true twice a year at the equinoxes. In the Northern hemisphere, the Sun will rise northwards of East on the June solstice, and southwards of East on the December solstice; this is directly tied in with the varying length of day too, since the Sun either has a greater or shorter distance to travel across the sky (see Figure 1). [figure1] Figure 1: Left: The path of the Sun across the sky during the equinoxes and solstices, as viewed by an observer in the Northern hemisphere at a latitude of $\sim 40^{\circ}$. Credit: Daniel V. Schroeder / Weber State University. Right: The same idea but viewed from Iceland at a latitude of $65^{\circ}$, where by being so close to the Artic circle the day length can get close to 24 hours in June and almost no daylight in December. Credit: Kristn Bjarnadttir / University of Iceland. During the equinox, the Sun travels along the projection of the Earth's equator. In this question, we will assume a circular orbit for the Earth, and all angles will be calculated in degrees. A simple model for the vertical angle between the Sun and the horizon (known at the altitude), $h$, as a function of the bearing on the horizon, $A$ (measured clockwise from North, also called the azimuth), the latitude of the observer, $\phi$ (positive in Northern hemisphere, negative in Southern hemisphere), and the vertical angle of the Sun relative to the celestial equator (known as the solar declination), $\delta$, is given as: $$ h=-\left(90^{\circ}-\phi\right) \cos (A)+\delta $$ The solar declination can be considered to vary sinusoidally over the year, going from a maximum of $\delta=+23.44^{\circ}$ at the June solstice (roughly $21^{\text {st }}$ June) to a minimum of $\delta=-23.44^{\circ}$ on the December solstice (roughly $21^{\text {st }}$ December). It can be shown using spherical trigonometry that the precise model connecting $\delta, h, \phi$ and $A$ is: $$ \sin (\delta)=\sin (h) \sin (\phi)+\cos (h) \cos (\phi) \cos (A) . $$ Using the precise model, the path of the Sun across the sky forms a shape that is not quite the cosine shape of the simple model, and is shown in Figure 2. [figure2] Figure 2: The altitude of the Sun as a function of bearing during the equinoxes and solstices, as viewed by an observer at a latitude of $+56^{\circ}$. Whilst it resembles the cosine shape of the simple model well at this latitude, there are small deviations. Credit: Wikipedia. By using further spherical trigonometry, we can derive a second helpful equation in the precise model: $$ \sin (h)=\sin (\phi) \sin (\delta)+\cos (\phi) \cos (\delta) \cos (H) $$ Here, $H$ is the solar hour angle, which measures the angle between the Sun and solar noon as measured along the projection of the Earth's equator on the sky. Conventionally, $H=0^{\circ}$ at solar noon, is negative before solar noon, and is positive afterwards. Since the sun's hour angle increases at an approximately constant rate due to the rotation of the Earth, we can convert this angle into a time using the conversion $360^{\circ}=24^{\mathrm{h}}$. problem: d. This exam is being taken on $24^{\text {th }}$ January and is 3 hours long. i. Estimate the solar declination on this date. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \circ, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_288

经长期观测发现, $A$ 行星运行的轨道半径为 $\mathrm{R}_{0}$, 周期为 $\mathrm{T}_{0}$, 但其实际运行的轨道与圆轨道总存在一些偏离, 且周期性地每隔 $\mathrm{t}_{0}$ 时间发生一次最大的偏离. 如图所示, 天文学家认为形成这种现象的原因可能是 $\mathrm{A}$ 行星外侧还存在着一颗未知行星 $\mathrm{B}$, 则行星 $\mathrm{B}$运动轨道半径表示错误的是( [图1] A: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}^{2}}{t_{0}-T_{0}}}$ B: $\mathrm{R}=\mathrm{R}_{0} \frac{t_{0}}{t_{0}-T}$ C: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}}{\left(t_{0}-T_{0}\right)^{2}}}$ D: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}^{2}}{t_{0}-T_{0}}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 经长期观测发现, $A$ 行星运行的轨道半径为 $\mathrm{R}_{0}$, 周期为 $\mathrm{T}_{0}$, 但其实际运行的轨道与圆轨道总存在一些偏离, 且周期性地每隔 $\mathrm{t}_{0}$ 时间发生一次最大的偏离. 如图所示, 天文学家认为形成这种现象的原因可能是 $\mathrm{A}$ 行星外侧还存在着一颗未知行星 $\mathrm{B}$, 则行星 $\mathrm{B}$运动轨道半径表示错误的是( [图1] A: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}^{2}}{t_{0}-T_{0}}}$ B: $\mathrm{R}=\mathrm{R}_{0} \frac{t_{0}}{t_{0}-T}$ C: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}}{\left(t_{0}-T_{0}\right)^{2}}}$ D: $\mathrm{R}=\mathrm{R}_{0} \sqrt[3]{\frac{t_{0}^{2}}{t_{0}-T_{0}}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_687

有人设想: 如果在地球的赤道上坚直向上建一座非常高的高楼, 是否可以在楼上直接释放人造卫星呢? 高楼设想图如图所示。已知地球自转周期为 $T$, 同步卫星轨道半径为 $r$; 高楼上有 $A 、 B$ 两个可视为点的小房间, $A$ 到地心的距离为 $\frac{r}{2}, B$ 到地心的距离为 $\frac{3 r}{2}$ 。 $A$ 房间天花板上连一坚直轻弹簧, 弹簧下端连一质量为 $m$ 的小球, 求弹簧给小球的作用力大小和方向;[图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 有人设想: 如果在地球的赤道上坚直向上建一座非常高的高楼, 是否可以在楼上直接释放人造卫星呢? 高楼设想图如图所示。已知地球自转周期为 $T$, 同步卫星轨道半径为 $r$; 高楼上有 $A 、 B$ 两个可视为点的小房间, $A$ 到地心的距离为 $\frac{r}{2}, B$ 到地心的距离为 $\frac{3 r}{2}$ 。 $A$ 房间天花板上连一坚直轻弹簧, 弹簧下端连一质量为 $m$ 的小球, 求弹簧给小球的作用力大小和方向;[图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_1171

In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966).b. Averaging over a series of pulses from August 1965, after correcting for the relative motion of the Earth and Mercury and the rotation rate of the Earth during the observations, the difference between the frequencies of photons from the extreme left and right parts of an annulus received $500 \mu$ s after the initial echo was $\Delta f_{\text {total }}=4.27 \mathrm{~Hz}$. ii. Mercury has a semi-major axis of 0.387 au. Rounding slightly if necessary, express the ratio orbital period : rotational period in a simple integer form (the integers should be $<10$ ).

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966). problem: b. Averaging over a series of pulses from August 1965, after correcting for the relative motion of the Earth and Mercury and the rotation rate of the Earth during the observations, the difference between the frequencies of photons from the extreme left and right parts of an annulus received $500 \mu$ s after the initial echo was $\Delta f_{\text {total }}=4.27 \mathrm{~Hz}$. ii. Mercury has a semi-major axis of 0.387 au. Rounding slightly if necessary, express the ratio orbital period : rotational period in a simple integer form (the integers should be $<10$ ). All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \text { days }, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1185

The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times.a. When it reached first perihelion, radio signals from the probe took $446.58 \mathrm{~s}$ to reach Earth. iii. Find the values of $\mathrm{D}$ and $\mathrm{E}$, given as fractions in their simplest terms, and hence calculate a new value for the distance travelled by the probe (also to 4 s.f.). Compare this to the approximation in the previous part and comment on your answer.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times. problem: a. When it reached first perihelion, radio signals from the probe took $446.58 \mathrm{~s}$ to reach Earth. iii. Find the values of $\mathrm{D}$ and $\mathrm{E}$, given as fractions in their simplest terms, and hence calculate a new value for the distance travelled by the probe (also to 4 s.f.). Compare this to the approximation in the previous part and comment on your answer. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of au, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_103

火星表面特征非常接近地球, 适合人类居住, 近期, 我国宇航员王跃与俄罗斯宇航员一起进行“模拟登火星”实验活动, 已知火星半径是地球半径的 $\frac{1}{2}$, 质量是地球质量的 $\frac{1}{9}$, 自转周期与地球的基本相同, 地球表面重力加速度为 $g$, 王跃在地面上能向上跳起的最大高度是 $h$, 在忽略自转影响的条件下, 下列分析不正确的是 ( ) A: 火星表面的重力加速度是 $\frac{4 g}{9}$ B: 火星的第一宇宙速度是地球第一宇宙速度的 $\frac{\sqrt{2}}{3}$ C: 王跃在火星表面受的万有引力是在地球表面受万有引力的 $\frac{2}{9}$ 倍 D: 王跃以相同的初速度在火星上起跳时, 可跳的最大高度是 $\frac{9 h}{4}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 火星表面特征非常接近地球, 适合人类居住, 近期, 我国宇航员王跃与俄罗斯宇航员一起进行“模拟登火星”实验活动, 已知火星半径是地球半径的 $\frac{1}{2}$, 质量是地球质量的 $\frac{1}{9}$, 自转周期与地球的基本相同, 地球表面重力加速度为 $g$, 王跃在地面上能向上跳起的最大高度是 $h$, 在忽略自转影响的条件下, 下列分析不正确的是 ( ) A: 火星表面的重力加速度是 $\frac{4 g}{9}$ B: 火星的第一宇宙速度是地球第一宇宙速度的 $\frac{\sqrt{2}}{3}$ C: 王跃在火星表面受的万有引力是在地球表面受万有引力的 $\frac{2}{9}$ 倍 D: 王跃以相同的初速度在火星上起跳时, 可跳的最大高度是 $\frac{9 h}{4}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_252

华为 Mate 60 Pro 通过中国自主研制的天通一号卫星通信系统实现了卫星电话功能,天通一号卫星是地球同步卫星。天通一号卫星发射首先利用火箭将卫星运载至地球附近圆形轨道 1, 通过多次变轨最终进入地球同步轨道 3。其变轨简化示意图如图所示, 轨道 1 离地面高度为 $h$ 。已知天通一号卫星质量为 $m$, 地球自转周期为 $T_{0}$, 地球半径为 $R$,地球表面的重力加速度为 $g, O$ 为地球中心，引力常量为 $G$ 。如果规定距地球无限远处为地球引力零势能点, 地球附近物体的引力势能可表示为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$ (未知)为地球质量, $m$ 为物体质量, $r$ 为物体到地心距离。求: 假设在变轨点 $P$ 和 $Q$ 通过两次发动机加速, “天通一号”卫星正好进入地球同步轨道 3，则发动机至少做多少功? [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 华为 Mate 60 Pro 通过中国自主研制的天通一号卫星通信系统实现了卫星电话功能,天通一号卫星是地球同步卫星。天通一号卫星发射首先利用火箭将卫星运载至地球附近圆形轨道 1, 通过多次变轨最终进入地球同步轨道 3。其变轨简化示意图如图所示, 轨道 1 离地面高度为 $h$ 。已知天通一号卫星质量为 $m$, 地球自转周期为 $T_{0}$, 地球半径为 $R$,地球表面的重力加速度为 $g, O$ 为地球中心，引力常量为 $G$ 。如果规定距地球无限远处为地球引力零势能点, 地球附近物体的引力势能可表示为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$ (未知)为地球质量, $m$ 为物体质量, $r$ 为物体到地心距离。求: 假设在变轨点 $P$ 和 $Q$ 通过两次发动机加速, “天通一号”卫星正好进入地球同步轨道 3，则发动机至少做多少功? [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_1221

GW170817 was the first gravitational wave event arising from a binary neutron star merger to have been detected by the LIGO \& Virgo experiments, and careful localization of the source meant that the electromagnetic counterpart was quickly found in galaxy NGC 4993 (see Figure 2). Such a combination of two completely separate branches of astronomical observation begins a new era of 'multi-messenger astronomy'. Since the gravitational waves allow an independent measurement of the distance to the host galaxy and the light allows an independent measurement of the recessional speed, this observation allows us to determine a new, independent value of the Hubble constant $H_{0}$. Another way of measuring distances to galaxies is to use the Fundamental Plane (FP) relation, which relies on the assumption there is a fairly tight relation between radius, surface brightness, and velocity dispersion for bulge-dominated galaxies, and is widely used for galaxies like NGC 4993. It can be described by the relation $$ \log \left(\frac{D}{(1+z)^{2}}\right)=-\log R_{e}+\alpha \log \sigma-\beta \log \left\langle I_{r}\right\rangle_{e}+\gamma $$ where $D$ is the distance in Mpc, $R_{e}$ is the effective radius measured in arcseconds, $\sigma$ is the velocity dispersion in $\mathrm{km} \mathrm{s}^{-1},\left\langle I_{r}\right\rangle_{e}$ is the mean intensity inside the effective radius measured in $L_{\odot} \mathrm{pc}^{-2}$, and $\gamma$ is the distance-dependent zero point of the relation. Calibrating the zero point to the Leo I galaxy group, the constants in the FP relation become $\alpha=1.24, \beta=0.82$, and $\gamma=2.194$. Figure 2: Left: The GW170817 signal as measured by the LIGO and Virgo gravitational wave detectors, taken from Abbott $e t$ al. (2017). The normalized amplitude (or strain) is in units of $10^{-21}$. The signal is not visible in the Virgo data due to the direction of the source with respect to the detector's antenna pattern. Right: The optical counterpart of GW170817 in host galaxy NGC 4993, taken from Hjorth et al. (2017). By measuring the amplitude (called strain, $h$ ) and the frequency of the gravitational waves, $f_{\mathrm{GW}}$, one can determine the distance to the source without having to rely on 'standard candles' like Cepheid variables or Type Ia supernovae. For two masses, $m_{1}$ and $m_{2}$, orbiting the centre of mass with separation $a$ with orbital angular velocity $\omega$, then the dimensionless strain parameter $h$ is $$ h \simeq \frac{G}{c^{4}} \frac{1}{r} \mu a^{2} \omega^{2} $$ where $r$ is the luminosity distance, $c$ is the speed of light, $\mu=m_{1} m_{2} / M_{\text {tot }}$ is the reduced mass and $M_{\text {tot }}=m_{1}+m_{2}$ is the total mass. The rate of change of frequency of the gravitational waves (called the 'chirp') from a merging binary can be written as $$ \dot{f}_{\mathrm{GW}}=\frac{96}{5} \pi^{8 / 3}\left(\frac{G \mathcal{M}}{c^{3}}\right)^{5 / 3} f_{\mathrm{GW}}^{11 / 3} $$e. Typically, you measure $\tau \equiv f_{G W} / \dot{f}_{G W}$, rather than $\dot{f}_{G W}$ directly. Given that just as the merger began the detectors measured $\tau=0.0023 \mathrm{~s}, f_{G W}=300 \mathrm{~Hz}$ and $h=6.0 \times 10^{-21}$, estimate the distance to GW170817 (in Mpc) and its absolute uncertainty (assuming a percentage uncertainty of $\pm 10 \%$ ). How does this compare with your answers in parts $a$. and $b$.?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: GW170817 was the first gravitational wave event arising from a binary neutron star merger to have been detected by the LIGO \& Virgo experiments, and careful localization of the source meant that the electromagnetic counterpart was quickly found in galaxy NGC 4993 (see Figure 2). Such a combination of two completely separate branches of astronomical observation begins a new era of 'multi-messenger astronomy'. Since the gravitational waves allow an independent measurement of the distance to the host galaxy and the light allows an independent measurement of the recessional speed, this observation allows us to determine a new, independent value of the Hubble constant $H_{0}$. Another way of measuring distances to galaxies is to use the Fundamental Plane (FP) relation, which relies on the assumption there is a fairly tight relation between radius, surface brightness, and velocity dispersion for bulge-dominated galaxies, and is widely used for galaxies like NGC 4993. It can be described by the relation $$ \log \left(\frac{D}{(1+z)^{2}}\right)=-\log R_{e}+\alpha \log \sigma-\beta \log \left\langle I_{r}\right\rangle_{e}+\gamma $$ where $D$ is the distance in Mpc, $R_{e}$ is the effective radius measured in arcseconds, $\sigma$ is the velocity dispersion in $\mathrm{km} \mathrm{s}^{-1},\left\langle I_{r}\right\rangle_{e}$ is the mean intensity inside the effective radius measured in $L_{\odot} \mathrm{pc}^{-2}$, and $\gamma$ is the distance-dependent zero point of the relation. Calibrating the zero point to the Leo I galaxy group, the constants in the FP relation become $\alpha=1.24, \beta=0.82$, and $\gamma=2.194$. Figure 2: Left: The GW170817 signal as measured by the LIGO and Virgo gravitational wave detectors, taken from Abbott $e t$ al. (2017). The normalized amplitude (or strain) is in units of $10^{-21}$. The signal is not visible in the Virgo data due to the direction of the source with respect to the detector's antenna pattern. Right: The optical counterpart of GW170817 in host galaxy NGC 4993, taken from Hjorth et al. (2017). By measuring the amplitude (called strain, $h$ ) and the frequency of the gravitational waves, $f_{\mathrm{GW}}$, one can determine the distance to the source without having to rely on 'standard candles' like Cepheid variables or Type Ia supernovae. For two masses, $m_{1}$ and $m_{2}$, orbiting the centre of mass with separation $a$ with orbital angular velocity $\omega$, then the dimensionless strain parameter $h$ is $$ h \simeq \frac{G}{c^{4}} \frac{1}{r} \mu a^{2} \omega^{2} $$ where $r$ is the luminosity distance, $c$ is the speed of light, $\mu=m_{1} m_{2} / M_{\text {tot }}$ is the reduced mass and $M_{\text {tot }}=m_{1}+m_{2}$ is the total mass. The rate of change of frequency of the gravitational waves (called the 'chirp') from a merging binary can be written as $$ \dot{f}_{\mathrm{GW}}=\frac{96}{5} \pi^{8 / 3}\left(\frac{G \mathcal{M}}{c^{3}}\right)^{5 / 3} f_{\mathrm{GW}}^{11 / 3} $$ problem: e. Typically, you measure $\tau \equiv f_{G W} / \dot{f}_{G W}$, rather than $\dot{f}_{G W}$ directly. Given that just as the merger began the detectors measured $\tau=0.0023 \mathrm{~s}, f_{G W}=300 \mathrm{~Hz}$ and $h=6.0 \times 10^{-21}$, estimate the distance to GW170817 (in Mpc) and its absolute uncertainty (assuming a percentage uncertainty of $\pm 10 \%$ ). How does this compare with your answers in parts $a$. and $b$.? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{Mpc}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

text-only

Astronomy_135

光电效应和康普顿效应深入地揭示了光的粒子性的一面。前者表明光子具有能量,后者表明光子除了具有能量之外还具有动量。由狭义相对论可知, 一定的质量 $m$ 与一 定的能量 $E$ 相对应: $E=m c^{2}$, 其中 $c$ 为真空中光速。 光照射到物体表面时, 光子被物体吸收或反射时, 光都会对物体产生压强, 这就是“光压”。已知太阳半径为 $R$, 单位时间辐射的总能量为 $P_{0}$, 光速为 $c$ 。求:科幻作品中经常以文明能够利用的能量程度来对文明进行分级。比如, 一级文明能够利用行星的全部能量, 人类处于 0.73 级; 二级文明能够利用恒星的全部能量; 三级文明能够利用整个星系的能量。科幻作家们设想了一种被称为“戴森球”的装置来收集整个恒星的能量。“戴森球”是一种包围整个恒星的球状膜, 它的内表面可以吸收恒星发出的所有光的能量。如果给太阳安装一个“戴森球”, 求: 戴森球的半径 $r$ 多大时, 戴森球可以靠着光压来抵抗太阳的万有引力, 避免其向太阳塌缩。已知引力常量为 $G$, 太阳的质量为 $M$ ，戴森球的总质量记为 $m$ 。

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 光电效应和康普顿效应深入地揭示了光的粒子性的一面。前者表明光子具有能量,后者表明光子除了具有能量之外还具有动量。由狭义相对论可知, 一定的质量 $m$ 与一 定的能量 $E$ 相对应: $E=m c^{2}$, 其中 $c$ 为真空中光速。 光照射到物体表面时, 光子被物体吸收或反射时, 光都会对物体产生压强, 这就是“光压”。已知太阳半径为 $R$, 单位时间辐射的总能量为 $P_{0}$, 光速为 $c$ 。求:科幻作品中经常以文明能够利用的能量程度来对文明进行分级。比如, 一级文明能够利用行星的全部能量, 人类处于 0.73 级; 二级文明能够利用恒星的全部能量; 三级文明能够利用整个星系的能量。科幻作家们设想了一种被称为“戴森球”的装置来收集整个恒星的能量。“戴森球”是一种包围整个恒星的球状膜, 它的内表面可以吸收恒星发出的所有光的能量。如果给太阳安装一个“戴森球”, 求: 戴森球的半径 $r$ 多大时, 戴森球可以靠着光压来抵抗太阳的万有引力, 避免其向太阳塌缩。已知引力常量为 $G$, 太阳的质量为 $M$ ，戴森球的总质量记为 $m$ 。 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

text-only

Astronomy_476

小型登月器连接在航天站上, 一起绕月球做圆周运动, 其轨道半径为月球半径的 3 倍, 某时刻, 航天站使登月器减速分离, 登月器沿如图所示的椭圆轨道登月, 在月球表面逗留一段时间完成科考工作后，经快速启动仍沿原椭圆轨道返回，当第一次回到分离点时恰与航天站对接, 登月器快速启动所用的时间可以忽略不计, 整个过程中航天站保持原轨道绕月运行, 不考虑月球自转的影响, 则下列说法正确的是 ( ) [图1] A: 从登月器与航天站分离到对接, 航天站至少转过半个周期 B: 从登月器与航天站分离到对接, 航天站至少转过 2 个周期 C: 航天站做圆周运动的周期与登月器在椭圆轨道上运动的周期之比为 $\sqrt{\frac{27}{8}}$ D: 航天站做圆周运动的周期与登月器在椭圆轨道上运动的周期之比为 $\frac{27}{8}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 小型登月器连接在航天站上, 一起绕月球做圆周运动, 其轨道半径为月球半径的 3 倍, 某时刻, 航天站使登月器减速分离, 登月器沿如图所示的椭圆轨道登月, 在月球表面逗留一段时间完成科考工作后，经快速启动仍沿原椭圆轨道返回，当第一次回到分离点时恰与航天站对接, 登月器快速启动所用的时间可以忽略不计, 整个过程中航天站保持原轨道绕月运行, 不考虑月球自转的影响, 则下列说法正确的是 ( ) [图1] A: 从登月器与航天站分离到对接, 航天站至少转过半个周期 B: 从登月器与航天站分离到对接, 航天站至少转过 2 个周期 C: 航天站做圆周运动的周期与登月器在椭圆轨道上运动的周期之比为 $\sqrt{\frac{27}{8}}$ D: 航天站做圆周运动的周期与登月器在椭圆轨道上运动的周期之比为 $\frac{27}{8}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_870

What is the time difference between the longest day of the year and the shortest day of the year in San Francisco $\left(37.7^{\circ} \mathrm{N}, 122.4^{\circ} \mathrm{W}\right)$ ? Neglect atmospheric refraction. A: $2 \mathrm{~h} 30 \mathrm{~min}$ B: $3 \mathrm{~h} 32 \mathrm{~min}$ C: $4 \mathrm{~h} 08 \mathrm{~min}$ D: $5 \mathrm{~h} 12 \mathrm{~min}$ E: 6 h25min

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What is the time difference between the longest day of the year and the shortest day of the year in San Francisco $\left(37.7^{\circ} \mathrm{N}, 122.4^{\circ} \mathrm{W}\right)$ ? Neglect atmospheric refraction. A: $2 \mathrm{~h} 30 \mathrm{~min}$ B: $3 \mathrm{~h} 32 \mathrm{~min}$ C: $4 \mathrm{~h} 08 \mathrm{~min}$ D: $5 \mathrm{~h} 12 \mathrm{~min}$ E: 6 h25min You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_526

$\mathrm{A} 、 \mathrm{~B}$ 是两颗人造地球卫星, 已知 $\mathrm{A}$ 的环绕半径大于 $\mathrm{B}$ 的环绕半径 $r_{A}>r_{B}$, 设 $\mathrm{A}$ 和 $\mathrm{B}$ 的运行速度分别为 $v_{A}$ 和 $v_{B}$, 最初的发射速度分别为 $v_{0 A}$ 和 $v_{0 B}$, 那么以下说法中正确的是( ) A: $v_{A}>v_{B}, v_{0 A}>v_{0 B}$ B: $v_{A}>v_{B}, v_{0 A}<v_{0 B}$ C: $v_{A}<v_{B}, v_{0 A}<v_{0 B}$ D: $v_{A}<v_{B}, \quad v_{0 A}>v_{0 B}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: $\mathrm{A} 、 \mathrm{~B}$ 是两颗人造地球卫星, 已知 $\mathrm{A}$ 的环绕半径大于 $\mathrm{B}$ 的环绕半径 $r_{A}>r_{B}$, 设 $\mathrm{A}$ 和 $\mathrm{B}$ 的运行速度分别为 $v_{A}$ 和 $v_{B}$, 最初的发射速度分别为 $v_{0 A}$ 和 $v_{0 B}$, 那么以下说法中正确的是( ) A: $v_{A}>v_{B}, v_{0 A}>v_{0 B}$ B: $v_{A}>v_{B}, v_{0 A}<v_{0 B}$ C: $v_{A}<v_{B}, v_{0 A}<v_{0 B}$ D: $v_{A}<v_{B}, \quad v_{0 A}>v_{0 B}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_1070

The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times.a. When it reached first perihelion, radio signals from the probe took $446.58 \mathrm{~s}$ to reach Earth. ii. Use the Ramanujan approximation to work out the distance travelled by the probe between launch and perihelion to 4 s.f.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times. problem: a. When it reached first perihelion, radio signals from the probe took $446.58 \mathrm{~s}$ to reach Earth. ii. Use the Ramanujan approximation to work out the distance travelled by the probe between launch and perihelion to 4 s.f. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of au, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_50

“天宫一号”的运行圆轨道离地高度为 $350 \mathrm{~km}$, “神舟十号”需要追赶“天宫一号”并成功与之对接, 对接开始前它们在同一平面绕地球做匀速圆周运动且运行方向相同, 要成功对接则对接前“神舟十号”应该（） A: 从离地高度等于 $350 \mathrm{~km}$ 的圆轨道上加速且对接成功后运行速度比开始对接前大 B: 从离地高度大于 $350 \mathrm{~km}$ 的圆轨道上减速且对接成功后运行速度比开始对接前小 C: 从离地高度小于 $350 \mathrm{~km}$ 圆轨道上加速且对接成功后运行速度比开始对接前小 D: 从离地高度小于 $350 \mathrm{~km}$ 圆轨道上加速且对接成功后运行速度比开始对接前大

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: “天宫一号”的运行圆轨道离地高度为 $350 \mathrm{~km}$, “神舟十号”需要追赶“天宫一号”并成功与之对接, 对接开始前它们在同一平面绕地球做匀速圆周运动且运行方向相同, 要成功对接则对接前“神舟十号”应该（） A: 从离地高度等于 $350 \mathrm{~km}$ 的圆轨道上加速且对接成功后运行速度比开始对接前大 B: 从离地高度大于 $350 \mathrm{~km}$ 的圆轨道上减速且对接成功后运行速度比开始对接前小 C: 从离地高度小于 $350 \mathrm{~km}$ 圆轨道上加速且对接成功后运行速度比开始对接前小 D: 从离地高度小于 $350 \mathrm{~km}$ 圆轨道上加速且对接成功后运行速度比开始对接前大 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_907

Star Wars Rogue One The new Star Wars film Rogue One concentrates on the creation of the first Death Star, which in one scene causes a total eclipse on the planet Scarif, where it is being built. [figure1] Assume the Death Star is being built in orbit around the Earth instead, but still causes a very brief total solar eclipse when it passes in front of the Sun. The Death Star has a diameter of $120 \mathrm{~km}$, and so by comparing it to the size and distance to the Sun, calculate the altitude it is being built at. How does that compare to the altitude of the International Space Station $(400 \mathrm{~km})$ ?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Star Wars Rogue One The new Star Wars film Rogue One concentrates on the creation of the first Death Star, which in one scene causes a total eclipse on the planet Scarif, where it is being built. [figure1] Assume the Death Star is being built in orbit around the Earth instead, but still causes a very brief total solar eclipse when it passes in front of the Sun. The Death Star has a diameter of $120 \mathrm{~km}$, and so by comparing it to the size and distance to the Sun, calculate the altitude it is being built at. How does that compare to the altitude of the International Space Station $(400 \mathrm{~km})$ ? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of km, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

EN

multi-modal

Astronomy_925

The cosmic microwave background (CMB) is measured today to have an almost perfectly uniform temperature of $T_{0}=2.725 \mathrm{~K}$. The temperature of the CMB at any redshift can be calculated using $T=T_{0}(1+z)$ where $z$ is the redshift. During the early expansion of the Universe, the temperature was high enough to ionize the hydrogen atoms filling the universe and make the universe opaque. Once the temperature dropped below $3000 \mathrm{~K}$ the universe became transparent and the CMB was emitted in a process known as photon decoupling. [figure1] Figure 2: Left: The CMB has been measured to have an almost perfect black-body spectrum. Credit: NASA/WMAP Science Team. Right: The Planck satellite was launched to measure the tiny deviations from a perfect black-body. This is the map of the sky from their final data release from July 2018. Credit: ESA/Planck Collaboration. The relative size of the Universe, called the scale factor $a$, varies with redshift as $a=a_{0}(1+z)^{-1}$, whilst in a matter dominated universe (a valid assumption after the CMB was emitted) the scale factor varies with time $t$ as $a \propto t^{2 / 3}$. If the scale factor today is defined to be $a_{0} \equiv 1$, calculate the age of the Universe when the CMB was emitted. [Current age of the Universe $t_{0}=13.8$ Gyr.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: The cosmic microwave background (CMB) is measured today to have an almost perfectly uniform temperature of $T_{0}=2.725 \mathrm{~K}$. The temperature of the CMB at any redshift can be calculated using $T=T_{0}(1+z)$ where $z$ is the redshift. During the early expansion of the Universe, the temperature was high enough to ionize the hydrogen atoms filling the universe and make the universe opaque. Once the temperature dropped below $3000 \mathrm{~K}$ the universe became transparent and the CMB was emitted in a process known as photon decoupling. [figure1] Figure 2: Left: The CMB has been measured to have an almost perfect black-body spectrum. Credit: NASA/WMAP Science Team. Right: The Planck satellite was launched to measure the tiny deviations from a perfect black-body. This is the map of the sky from their final data release from July 2018. Credit: ESA/Planck Collaboration. The relative size of the Universe, called the scale factor $a$, varies with redshift as $a=a_{0}(1+z)^{-1}$, whilst in a matter dominated universe (a valid assumption after the CMB was emitted) the scale factor varies with time $t$ as $a \propto t^{2 / 3}$. If the scale factor today is defined to be $a_{0} \equiv 1$, calculate the age of the Universe when the CMB was emitted. [Current age of the Universe $t_{0}=13.8$ Gyr.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value.

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Astronomy

EN

multi-modal

Astronomy_692

设想在地面上通过火箭将质量为 $m$ 的人造小飞船送入预定轨道, 至少需要做功 $W$ 。若预定轨道半径为 $r$, 地球半径为 $R$, 地球表面处的重力加速度为 $g$, 忽略空气阻力, 不考虑地球自转的影响。取地面为零势能面, 则下列说法错误的是 ( ) A: 地球的质量为 $\frac{g r^{2}}{G}$ B: 小飞船在预定轨道的周期为 $2 \pi \sqrt{\frac{R^{3}}{g r^{2}}}$ C: 小飞船在预定轨道的动能为 $\frac{m g r^{2}}{2 R}$ D: 小飞船在预定轨道的势能为 $W-\frac{m g R^{2}}{2 r}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 设想在地面上通过火箭将质量为 $m$ 的人造小飞船送入预定轨道, 至少需要做功 $W$ 。若预定轨道半径为 $r$, 地球半径为 $R$, 地球表面处的重力加速度为 $g$, 忽略空气阻力, 不考虑地球自转的影响。取地面为零势能面, 则下列说法错误的是 ( ) A: 地球的质量为 $\frac{g r^{2}}{G}$ B: 小飞船在预定轨道的周期为 $2 \pi \sqrt{\frac{R^{3}}{g r^{2}}}$ C: 小飞船在预定轨道的动能为 $\frac{m g r^{2}}{2 R}$ D: 小飞船在预定轨道的势能为 $W-\frac{m g R^{2}}{2 r}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_806

The problem of magnetic monopoles - that is, the apparent absence of magnetic monopoles in the universe - arises from the fact that some modern physical theories (such as string theory) predict that the number density of magnetic monopoles at the time of their creation was $n_{M}\left(t_{G U T}\right) \approx 10^{82} \mathrm{~m}^{-3}$. The inflation theory provides a possible solution to this problem, as the exponential expansion of the primordial universe would "dilute" the monopoles. Calculate, approximately, how much the universe expanded during the inflationary period so that today the probability of a single magnetic monopole existing in the observational universe is $1 \%$. Consider that the beginning of inflation coincides with the time of the creation of magnetic monopoles, and that the universe is flat (Euclidean geometry can be used on large scales). You can use that the diameter of the observational universe is $28.5 \mathrm{Gpc}$, and that between the end of inflation and today, the universe has linearly expanded by a factor of $5 \times 10^{26}$. A: $e^{40}$ B: $e^{50}$ C: $e^{55}$ D: $e^{65}$ E: $e^{85}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The problem of magnetic monopoles - that is, the apparent absence of magnetic monopoles in the universe - arises from the fact that some modern physical theories (such as string theory) predict that the number density of magnetic monopoles at the time of their creation was $n_{M}\left(t_{G U T}\right) \approx 10^{82} \mathrm{~m}^{-3}$. The inflation theory provides a possible solution to this problem, as the exponential expansion of the primordial universe would "dilute" the monopoles. Calculate, approximately, how much the universe expanded during the inflationary period so that today the probability of a single magnetic monopole existing in the observational universe is $1 \%$. Consider that the beginning of inflation coincides with the time of the creation of magnetic monopoles, and that the universe is flat (Euclidean geometry can be used on large scales). You can use that the diameter of the observational universe is $28.5 \mathrm{Gpc}$, and that between the end of inflation and today, the universe has linearly expanded by a factor of $5 \times 10^{26}$. A: $e^{40}$ B: $e^{50}$ C: $e^{55}$ D: $e^{65}$ E: $e^{85}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

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text-only

Astronomy_629

如图所示, 有一质量为 $M$, 半径为 $R$, 密度均匀的球体, 在距离球心 $O$ 为 $2 R$ 的地方有一质量为 $m$ 的质点, 现从 $M$ 中挖去一半径为 $\frac{R}{2}$ 的球体, 试求: 剩余部分对质点 $m$ 的引力大小; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 有一质量为 $M$, 半径为 $R$, 密度均匀的球体, 在距离球心 $O$ 为 $2 R$ 的地方有一质量为 $m$ 的质点, 现从 $M$ 中挖去一半径为 $\frac{R}{2}$ 的球体, 试求: 剩余部分对质点 $m$ 的引力大小; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_204

放置在水平平台上的物体, 其表观重力在数值上等于物体对平台的压力, 方向与压力的方向相同。微重力环境是指系统内物体的表观重力远小于其实际重力（万有引力）的环境。此环境下，物体的表观重力与其质量之比称为微重力加速度。 如图所示, 中国科学院力学研究所微重力实验室落塔是我国微重力实验的主要设施之一, 实验中落舱可采用单舱和双舱两种模式进行。已知地球表面的重力加速度为 $g$; 如图所示, 双舱模式是采用内外双舱结构, 实验平台固定在内舱中, 实验时让双舱同时下落。落体下落时受到的空气阻力可表示为 $f=k \rho v^{2}$, 式中 $k$ 为由落体形状决定的常数, $\rho$ 为空气密度, $v$ 为落体相对于周围空气的速率。若某次实验中, 内舱与舱内物体总质量为 $m_{1}$, 外舱与舱内物体总质量为 $m_{2}$ (不含内舱)。某时刻, 外舱相对于地面的速度为 $v_{1}$, 内舱相对于地面的速度为 $v_{2}$, 它们所受空气阻力的常数 $k$ 相同, 外舱中与外部环境的空气密度相同, 不考虑外舱内空气对外舱自身运动的影响。求此时内舱与外舱中的微重力加速度之比 $g_{2}: g_{3} ;$ [图1] 落塔 落舱

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 放置在水平平台上的物体, 其表观重力在数值上等于物体对平台的压力, 方向与压力的方向相同。微重力环境是指系统内物体的表观重力远小于其实际重力（万有引力）的环境。此环境下，物体的表观重力与其质量之比称为微重力加速度。 如图所示, 中国科学院力学研究所微重力实验室落塔是我国微重力实验的主要设施之一, 实验中落舱可采用单舱和双舱两种模式进行。已知地球表面的重力加速度为 $g$; 如图所示, 双舱模式是采用内外双舱结构, 实验平台固定在内舱中, 实验时让双舱同时下落。落体下落时受到的空气阻力可表示为 $f=k \rho v^{2}$, 式中 $k$ 为由落体形状决定的常数, $\rho$ 为空气密度, $v$ 为落体相对于周围空气的速率。若某次实验中, 内舱与舱内物体总质量为 $m_{1}$, 外舱与舱内物体总质量为 $m_{2}$ (不含内舱)。某时刻, 外舱相对于地面的速度为 $v_{1}$, 内舱相对于地面的速度为 $v_{2}$, 它们所受空气阻力的常数 $k$ 相同, 外舱中与外部环境的空气密度相同, 不考虑外舱内空气对外舱自身运动的影响。求此时内舱与外舱中的微重力加速度之比 $g_{2}: g_{3} ;$ [图1] 落塔 落舱 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_1178

A day on Earth can be defined in two ways: relative to the Sun (called solar or synodic time) or relative to the background stars (called sidereal time). The mean solar day is 24 hours (within a few milliseconds), whilst the mean sidereal day is shorter at 23 hours 56 minutes 4 seconds (to the nearest second). The solar day is longer as over the course of a sidereal day the Earth has moved slightly in its orbit around the Sun and so has to rotate slightly further for the Sun to be back in the same direction (see Figure 4). [figure1] Figure 4: A solar day is defined as the time between two consecutive passages of the Sun through the meridian, corresponding to local midday (which in the Northern hemisphere is in the South), whilst a sidereal day is the time for a distant star to do the same. The difference between the two is due to the Earth having moved slightly in its orbit around the Sun. Credit: Wikipedia. The length of a year on Earth is 365.25 solar days (to 2 d.p.), however some ancient civilizations used to believe that there were once exactly 360 solar days in a year, with various myths explaining where the extra days came from. In this question you will look at how to return the Earth to this time. [Note that this question is very sensitive to the precision of the fundamental constants used, so throughout please take $G=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}, R_{\oplus}=6371 \mathrm{~km}, M_{\oplus}=5.972 \times 10^{24} \mathrm{~kg}, M_{\odot}=$ $1.989 \times 10^{30} \mathrm{~kg}$ and $1 \mathrm{au}=1.496 \times 10^{11} \mathrm{~m}$.] ## Helpful equations: The moment of inertia, $I$, of a sphere of mass $M$ and radius $R$ is $I=\frac{2}{5} M R^{2}$. The angular momentum, $L$, of a spinning object with an angular velocity of $\omega$ is $L=I \omega=r \times p$, where $p$ is the linear momentum of a point particle a distance $r$ from the axis of rotation. The speed, $v$, of an object in an elliptical orbit of semi-major axis $a$ around an object of mass $M$ when a distance $r$ away can be calculated as $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$d. Imagine creating an incredibly powerful rocket, positioned on the Earth's equator, that when fired once can apply a huge force to the Earth in a very short time period, delivering a total impulse of $\Delta p$. Assuming the Earth's orbit is initially circular, calculate: ii. The total impulse required to change the orbit to give a year of 360 solar days, but with no change in the length of a solar day, also explaining how the rocket needs to be fired.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: A day on Earth can be defined in two ways: relative to the Sun (called solar or synodic time) or relative to the background stars (called sidereal time). The mean solar day is 24 hours (within a few milliseconds), whilst the mean sidereal day is shorter at 23 hours 56 minutes 4 seconds (to the nearest second). The solar day is longer as over the course of a sidereal day the Earth has moved slightly in its orbit around the Sun and so has to rotate slightly further for the Sun to be back in the same direction (see Figure 4). [figure1] Figure 4: A solar day is defined as the time between two consecutive passages of the Sun through the meridian, corresponding to local midday (which in the Northern hemisphere is in the South), whilst a sidereal day is the time for a distant star to do the same. The difference between the two is due to the Earth having moved slightly in its orbit around the Sun. Credit: Wikipedia. The length of a year on Earth is 365.25 solar days (to 2 d.p.), however some ancient civilizations used to believe that there were once exactly 360 solar days in a year, with various myths explaining where the extra days came from. In this question you will look at how to return the Earth to this time. [Note that this question is very sensitive to the precision of the fundamental constants used, so throughout please take $G=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}, R_{\oplus}=6371 \mathrm{~km}, M_{\oplus}=5.972 \times 10^{24} \mathrm{~kg}, M_{\odot}=$ $1.989 \times 10^{30} \mathrm{~kg}$ and $1 \mathrm{au}=1.496 \times 10^{11} \mathrm{~m}$.] ## Helpful equations: The moment of inertia, $I$, of a sphere of mass $M$ and radius $R$ is $I=\frac{2}{5} M R^{2}$. The angular momentum, $L$, of a spinning object with an angular velocity of $\omega$ is $L=I \omega=r \times p$, where $p$ is the linear momentum of a point particle a distance $r$ from the axis of rotation. The speed, $v$, of an object in an elliptical orbit of semi-major axis $a$ around an object of mass $M$ when a distance $r$ away can be calculated as $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ problem: d. Imagine creating an incredibly powerful rocket, positioned on the Earth's equator, that when fired once can apply a huge force to the Earth in a very short time period, delivering a total impulse of $\Delta p$. Assuming the Earth's orbit is initially circular, calculate: ii. The total impulse required to change the orbit to give a year of 360 solar days, but with no change in the length of a solar day, also explaining how the rocket needs to be fired. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

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multi-modal

Astronomy_582

以两天体 $\mathrm{A} 、 \mathrm{~B}$ 中心连线为底的等边三角形的第三个顶点被称为“三角拉格朗日点”, 如果在该点有一颗质量远小于 $\mathrm{A} 、 \mathrm{~B}$ 的卫星 $\mathrm{C}$, 则三者可以组成一个稳定的三星系统, 如图所示。由于 $\mathrm{C}$ 对 $\mathrm{A} 、 \mathrm{~B}$ 的影响很小, 故 $\mathrm{A} 、 \mathrm{~B}$ 又可视作双星系统绕连线上某定点 $P$ (未画出) 做匀速圆周运动。已知天体 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 的质量分布均匀, 且分别为 $m_{1} 、 m_{2} 、 m_{3}$, 已知 $m_{1}=2 m_{2}$, 两天体 $\mathrm{A} 、 \mathrm{~B}$ 中心间距为 $L$, 万有引力常量为 $G$, 则下列说法正确的是（ ） [图1] A: 天体 $\mathrm{A}$ 做匀速圆周运动的轨道半径为 $\frac{L}{2}$ B: 天体 $A 、 B$ 所需要的向心力大小之比为 2: 1 C: 卫星 $\mathrm{C}$ 所受合力恰好指向 $P$ 点 D: 卫星 $\mathrm{C}$ 的周期为 $2 \pi \sqrt{\frac{L^{3}}{G\left(m_{1}+m_{2}\right)}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 以两天体 $\mathrm{A} 、 \mathrm{~B}$ 中心连线为底的等边三角形的第三个顶点被称为“三角拉格朗日点”, 如果在该点有一颗质量远小于 $\mathrm{A} 、 \mathrm{~B}$ 的卫星 $\mathrm{C}$, 则三者可以组成一个稳定的三星系统, 如图所示。由于 $\mathrm{C}$ 对 $\mathrm{A} 、 \mathrm{~B}$ 的影响很小, 故 $\mathrm{A} 、 \mathrm{~B}$ 又可视作双星系统绕连线上某定点 $P$ (未画出) 做匀速圆周运动。已知天体 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 的质量分布均匀, 且分别为 $m_{1} 、 m_{2} 、 m_{3}$, 已知 $m_{1}=2 m_{2}$, 两天体 $\mathrm{A} 、 \mathrm{~B}$ 中心间距为 $L$, 万有引力常量为 $G$, 则下列说法正确的是（ ） [图1] A: 天体 $\mathrm{A}$ 做匀速圆周运动的轨道半径为 $\frac{L}{2}$ B: 天体 $A 、 B$ 所需要的向心力大小之比为 2: 1 C: 卫星 $\mathrm{C}$ 所受合力恰好指向 $P$ 点 D: 卫星 $\mathrm{C}$ 的周期为 $2 \pi \sqrt{\frac{L^{3}}{G\left(m_{1}+m_{2}\right)}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_1023

Star Wars Rogue One The new Star Wars film Rogue One concentrates on the creation of the first Death Star, which in one scene causes a total eclipse on the planet Scarif, where it is being built. [figure1] What will be its orbital period, assuming it moves in a circular orbit?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: Star Wars Rogue One The new Star Wars film Rogue One concentrates on the creation of the first Death Star, which in one scene causes a total eclipse on the planet Scarif, where it is being built. [figure1] What will be its orbital period, assuming it moves in a circular orbit? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of hours, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

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Astronomy_131

2019 年 2 月 15 日，一群中国学生拍摄的地月同框照, 被外媒评价为迄今为止最好的地月合影之一。如图所示, 把地球和月球看做绕同一圆心做匀速圆周运动的双星系统,质量分别为 $M 、 m$, 相距为 $L$, 周期为 $T$, 若有间距也为 $L$ 的双星 $\mathrm{P} 、 \mathrm{Q}, \mathrm{P} 、 \mathrm{Q}$ 的质量分别为 $2 M 、 2 m$ ，则 $(\quad)$ [图1] A: 地、月运动的轨道半径之比为 $\frac{M}{m}$ B: 地、月运动的加速度之比为 $\frac{M}{m}$ C: P 运动的速率与地球的相等 D: $\mathrm{P} 、 \mathrm{Q}$ 运动的周期均为 $\frac{\sqrt{2}}{2} T$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2019 年 2 月 15 日，一群中国学生拍摄的地月同框照, 被外媒评价为迄今为止最好的地月合影之一。如图所示, 把地球和月球看做绕同一圆心做匀速圆周运动的双星系统,质量分别为 $M 、 m$, 相距为 $L$, 周期为 $T$, 若有间距也为 $L$ 的双星 $\mathrm{P} 、 \mathrm{Q}, \mathrm{P} 、 \mathrm{Q}$ 的质量分别为 $2 M 、 2 m$ ，则 $(\quad)$ [图1] A: 地、月运动的轨道半径之比为 $\frac{M}{m}$ B: 地、月运动的加速度之比为 $\frac{M}{m}$ C: P 运动的速率与地球的相等 D: $\mathrm{P} 、 \mathrm{Q}$ 运动的周期均为 $\frac{\sqrt{2}}{2} T$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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multi-modal

Astronomy_937

A telescope with a focal length of $750 \mathrm{~mm}$ is used with an eyepiece. Which eyepiece focal length would give the greatest overall magnification? A: $25 \mathrm{~mm}$ B: $20 \mathrm{~mm}$ C: $15 \mathrm{~mm}$ D: $10 \mathrm{~mm}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: A telescope with a focal length of $750 \mathrm{~mm}$ is used with an eyepiece. Which eyepiece focal length would give the greatest overall magnification? A: $25 \mathrm{~mm}$ B: $20 \mathrm{~mm}$ C: $15 \mathrm{~mm}$ D: $10 \mathrm{~mm}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

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Astronomy_772

What is true for a type-la ("type one-a") supernova? A: This type occurs in binary systems. B: This type occurs often in young galaxies. C: This type produces gamma-ray bursts. D: This type produces high amounts of X-rays.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What is true for a type-la ("type one-a") supernova? A: This type occurs in binary systems. B: This type occurs often in young galaxies. C: This type produces gamma-ray bursts. D: This type produces high amounts of X-rays. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy_168

一颗人造卫星在地球赤道平面内做匀速圆周运动, 运动方向跟地球自转方向相同,每 6 天经过赤道上同一地点上空一次, 已知地球同步卫星轨道半径为 $\mathrm{r}$, 下列有关该卫星的说法正确的有 A: 周期可能为 6 天, 轨道半径为 $\sqrt[3]{36} r$ B: 周期可能为 1.2 天, 轨道半径为 $\sqrt[3]{\frac{36}{25}} r$ C: 周期可能为 $\frac{6}{7}$ 天, 轨道半径为 $\sqrt[3]{\frac{36}{49} r}$ D: 周期可能为 5 天, 轨道半径为 $\sqrt[3]{25} r$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 一颗人造卫星在地球赤道平面内做匀速圆周运动, 运动方向跟地球自转方向相同,每 6 天经过赤道上同一地点上空一次, 已知地球同步卫星轨道半径为 $\mathrm{r}$, 下列有关该卫星的说法正确的有 A: 周期可能为 6 天, 轨道半径为 $\sqrt[3]{36} r$ B: 周期可能为 1.2 天, 轨道半径为 $\sqrt[3]{\frac{36}{25}} r$ C: 周期可能为 $\frac{6}{7}$ 天, 轨道半径为 $\sqrt[3]{\frac{36}{49} r}$ D: 周期可能为 5 天, 轨道半径为 $\sqrt[3]{25} r$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy_529

地球赤道上的重力加速度为 $g$, 物体在赤道上随地球自转的向心加速度为 $a$, 卫星甲、乙、丙在如图所示三个粗圆轨道上绕地球运行, 卫星甲和乙的运行轨道在 $P$ 点相切,以下说法中正确的是（ ）[图1] A: 如果地球自转的角速度突然变为原来的 $\frac{g+a}{a}$ 倍, 则赤道上的物体刚好“飘”起来 B: 卫星甲、乙经过 $P$ 点时的加速度大小相等 C: 卫星甲的周期最小 D: 三个卫星在远地点的速度可能大于第一宇宙速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 地球赤道上的重力加速度为 $g$, 物体在赤道上随地球自转的向心加速度为 $a$, 卫星甲、乙、丙在如图所示三个粗圆轨道上绕地球运行, 卫星甲和乙的运行轨道在 $P$ 点相切,以下说法中正确的是（ ）[图1] A: 如果地球自转的角速度突然变为原来的 $\frac{g+a}{a}$ 倍, 则赤道上的物体刚好“飘”起来 B: 卫星甲、乙经过 $P$ 点时的加速度大小相等 C: 卫星甲的周期最小 D: 三个卫星在远地点的速度可能大于第一宇宙速度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_767

What does the astronomical term barycentre describe? A: The centre of mass of multiple each other orbiting objects. B: The central axis of rotation of a spinning stellar object. C: The central axis of the precession of a rotating stellar object.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What does the astronomical term barycentre describe? A: The centre of mass of multiple each other orbiting objects. B: The central axis of rotation of a spinning stellar object. C: The central axis of the precession of a rotating stellar object. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C].

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Astronomy

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text-only

Astronomy_546

星体 $\mathrm{P}$ (行星或彗星) 绕太阳运动的轨迹为圆锥曲线 $r=\frac{k}{1+\varepsilon \cos \theta}$ 式中, $r$ 是 $\mathrm{P}$ 到太阳 $\mathrm{S}$ 的距离, $\theta$ 是矢径 $\mathrm{SP}$ 相对于极轴 $\mathrm{SA}$ 的夹角 (以逆时针方向为正), $k=\frac{L^{2}}{G M m^{2}}, L$是 $\mathrm{P}$ 相对于太阳的角动量, $G=6.67 \times 10^{-11} \mathrm{~m} 3 \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}$ 为引力常量, $M \approx 1.99 \times 10^{30} \mathrm{~kg}$ 为太阳的质量, $\varepsilon=\sqrt{1+\frac{2 E L^{2}}{G^{2} M^{2} m^{3}}}$ 为偏心率, $m$ 和 $E$ 分别为 $\mathrm{P}$ 的质量和机械能。假设有一颗彗星绕太阳运动的轨道为抛物线, 地球绕太阳运动的轨道可近似为圆, 两轨道相交于 $C 、 D$两点, 如图所示。已知地球轨道半径 $R_{E}=1.49 \times 10^{11} \mathrm{~m}$, 彗星轨道近日点 $A$ 到太阳的距离为地球轨道半径的三分之一, 不考虑地球和彗星之间的相互影响。求彗星先后两次穿过地球轨道所用的时间; 已知积分公式 $\int \frac{x d x}{\sqrt{x+a}}=\frac{2}{3}(x+a)^{\frac{3}{2}}-2 a(x+a)^{\frac{1}{2}}+C$, 式中 $\mathrm{C}$ 是任意常数。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 星体 $\mathrm{P}$ (行星或彗星) 绕太阳运动的轨迹为圆锥曲线 $r=\frac{k}{1+\varepsilon \cos \theta}$ 式中, $r$ 是 $\mathrm{P}$ 到太阳 $\mathrm{S}$ 的距离, $\theta$ 是矢径 $\mathrm{SP}$ 相对于极轴 $\mathrm{SA}$ 的夹角 (以逆时针方向为正), $k=\frac{L^{2}}{G M m^{2}}, L$是 $\mathrm{P}$ 相对于太阳的角动量, $G=6.67 \times 10^{-11} \mathrm{~m} 3 \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}$ 为引力常量, $M \approx 1.99 \times 10^{30} \mathrm{~kg}$ 为太阳的质量, $\varepsilon=\sqrt{1+\frac{2 E L^{2}}{G^{2} M^{2} m^{3}}}$ 为偏心率, $m$ 和 $E$ 分别为 $\mathrm{P}$ 的质量和机械能。假设有一颗彗星绕太阳运动的轨道为抛物线, 地球绕太阳运动的轨道可近似为圆, 两轨道相交于 $C 、 D$两点, 如图所示。已知地球轨道半径 $R_{E}=1.49 \times 10^{11} \mathrm{~m}$, 彗星轨道近日点 $A$ 到太阳的距离为地球轨道半径的三分之一, 不考虑地球和彗星之间的相互影响。求彗星先后两次穿过地球轨道所用的时间; 已知积分公式 $\int \frac{x d x}{\sqrt{x+a}}=\frac{2}{3}(x+a)^{\frac{3}{2}}-2 a(x+a)^{\frac{1}{2}}+C$, 式中 $\mathrm{C}$ 是任意常数。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 请记住，你的答案应以s为单位计算，但在给出最终答案时，请不要包含单位。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。

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Astronomy

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multi-modal

Astronomy_590

暗物质是二十世纪物理学之谜, 对该问题的研究可能带来一场物理学的革命。为了探测暗物质，我国在 2015 年 12 月 17 日成功发射了一颗被命名为“悟空”的暗物质探测卫星, 已知“悟空”在低于同步卫星的轨道上绕地球做匀速圆运动, 经过时间 $t$ ( $t$ 小于其 运动周期), 运动的弧长为 $s$, 与地球中心连扫过的角度为 $\beta$ (弧度), 引力常量为 $G$,则下列说法中正确的是（） A: “悟空”的线速度大于第一宇宙速度 B: “悟空”的向心加速度小于地球同步卫星的向心加速度 C: “悟空”的环绕周期为 $\frac{2 \pi t}{\beta}$ D: “悟空”的质量为 $\frac{s^{3}}{\left(G t^{2} \beta\right)}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 暗物质是二十世纪物理学之谜, 对该问题的研究可能带来一场物理学的革命。为了探测暗物质，我国在 2015 年 12 月 17 日成功发射了一颗被命名为“悟空”的暗物质探测卫星, 已知“悟空”在低于同步卫星的轨道上绕地球做匀速圆运动, 经过时间 $t$ ( $t$ 小于其 运动周期), 运动的弧长为 $s$, 与地球中心连扫过的角度为 $\beta$ (弧度), 引力常量为 $G$,则下列说法中正确的是（） A: “悟空”的线速度大于第一宇宙速度 B: “悟空”的向心加速度小于地球同步卫星的向心加速度 C: “悟空”的环绕周期为 $\frac{2 \pi t}{\beta}$ D: “悟空”的质量为 $\frac{s^{3}}{\left(G t^{2} \beta\right)}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

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text-only