2015IOAA理论短问题第2题-行星质量

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英文题目

One satellite of a planet has an orbital period of 7 days, 3 hours, 43 minutes, and the semi major axis is 15.3 times the mean radius of the planet. The Moon has an orbital period of 27 days, 7 hours, 43 minutes and the semi major axis is 60.3 times the Earth's mean radius. Assume that the mass of the moon and the satellite is negligible compared to the mass of the planet. Calculate the ratio of the planet’s mean density to that of the Earth.

中文翻译

AI翻译

一颗行星的卫星轨道周期为7天3小时43分钟，其半长轴为行星平均半径的15.3倍。月球轨道周期为27天7小时43分钟，其半长轴为地球平均半径的60.3倍。假设卫星和月球的质量相对于其中心行星的质量可忽略不计。计算该行星的平均密度与地球平均密度之比。

解答

官方答案

解法一

(30%) Kepler third Law for satellite and Moon \[\frac{P_S^2}{a_S^3} = \frac{P_S^2}{(15.3 R_P)^3} = \frac{4\pi^2}{GM_P};\] \[\frac{P_M^2}{a_M^3} = \frac{P_M^2}{(60.3 R_E)^3} = \frac{4\pi^2}{GM_E}\]

(30%) Express mass in term of mass density: \[\frac{P_S^2}{(15.3 R_P)^3} = \frac{4\pi^2}{G M_P} = \frac{4\pi^2}{G \rho_P \frac{4}{3} \pi R_P^3}\] \[\frac{P_M^2}{(60.3 R_E)^3} = \frac{4\pi^2}{G M_E} = \frac{4\pi^2}{G \rho_E \frac{4}{3} \pi R_E^3}\]

(30%) Evaluate ratio of mass-density: \[\frac{\rho_E}{\rho_P} = \frac{60.3^3 \times P_S^2}{15.3^3 \times P_M^2}\] (10%) \[\frac{\rho_E}{\rho_P} = 0.238 \approx 0.24\]

翻译
AI翻译

(30%) 针对卫星和月球运用开普勒第三定律 \[\frac{P_S^2}{a_S^3} = \frac{P_S^2}{(15.3 R_P)^3} = \frac{4\pi^2}{GM_P};\] \[\frac{P_M^2}{a_M^3} = \frac{P_M^2}{(60.3 R_E)^3} = \frac{4\pi^2}{GM_E}\]

(30%) 用质量密度表示质量： \[\frac{P_S^2}{(15.3 R_P)^3} = \frac{4\pi^2}{G M_P} = \frac{4\pi^2}{G \rho_P \frac{4}{3} \pi R_P^3}\] \[\frac{P_M^2}{(60.3 R_E)^3} = \frac{4\pi^2}{G M_E} = \frac{4\pi^2}{G \rho_E \frac{4}{3} \pi R_E^3}\]

(30%) 计算质量密度的比值： \[\frac{\rho_E}{\rho_P} = \frac{60.3^3 \times P_S^2}{15.3^3 \times P_M^2}\] (10%) \[\frac{\rho_E}{\rho_P} = 0.238 \approx 0.24\]

解法二

The period of the Satellite: $$P_S = 7 days, 3 hours, 43 minutes = 0.0196 years$$,

Mean orbital radius of the Satellite : $$a_S = 15.3 R_P$$; $$R_P$$: radius of the Planet.

The period of the Moon:$$ P_M = 27 days, 7 hours, 43 minutes = 0.0749 years$$,

Mean orbital radius of the Moon :$$ a_M = 60.3 R_E$$ ;$$ R_E$$: radius of the Earth.

Mass of the Sun ($$M$$)

Kepler's Third Law :$$ G(M + m) = 4\pi^2 \frac{a^3}{T^2} \rightarrow$$

  • Mass of the Planet: $$m_P$$
  • Radius of the Planet : $$R_P$$
  • Semi major axis of the Planet = Mean orbital radius of the Plant: $$a_P$$
  • Semi major axis of the Satellite = Mean orbital radius of the Satellite: $$a_S$$
  • Semi major axis of the Moon = Mean orbital radius of the Moon:$$a_M$$
  • Mass of the Earth :$$m_E$$
  • Radius of the Earth:$$R_E$$
  • Semi major axis of the Earth = Mean orbital radius of the Earth:$$a_E$$

(5%) Planet - Satellite: \begin{equation}G(m_P + m_S) = 4\pi^2 \frac{a_S^3}{P_S^2}, m_P \gg m_S \rightarrow G(m_P) = 4\pi^2 \frac{a_S^3}{P_S^2}\end{equation} (5%) The Earth-the Moon: \begin{equation}G(m_E + m_M) = 4\pi^2 \frac{a_M^3}{P_M^2}, m_E \gg m_M \rightarrow G(m_E) = 4\pi^2 \frac{a_M^3}{P_M^2}\end{equation} (5%） Planet-Sun: \begin{equation} G(M + m_P) = 4\pi^2 \frac{a_P^3}{P_P^2}, M \gg m_P \rightarrow G(M) = 4\pi^2 \frac{a_P^3}{P_P^2}\end{equation} (5%) Earth-Sun: \begin{equation} G(M + m_E) = 4\pi^2 \frac{a_E^3}{P_E^2}, M \gg m_E \rightarrow G(M) = 4\pi^2 \frac{a_E^3}{P_E^2}\end{equation}

(20%) From equation (1) and (3): \[\frac{m_P}{M} = \frac{4\pi^2 \frac{a_S^3}{P_S^2}}{4\pi^2 \frac{a_P^3}{P_P^2}} = \frac{(a_S)^3 (P_P)^2}{(a_P)^3 (P_S)^2} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (P_S)^2} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (0.0196)^2}\]

(20%) From equation (2) and (4): \[\frac{m_E}{M} = \frac{4\pi^2 \frac{a_M^3}{P_M^2}}{4\pi^2 \frac{a_E^3}{P_E^2}} = \frac{(a_M)^3 (P_E)^2}{(a_E)^3 (P_M)^2} = \frac{(60.3 R_E)^3 (P_E)^2}{(a_E)^3 (P_M)^2} = \frac{(60.3 R_E)^3 (P_E)^2}{(a_E)^3 (0.0749)^2} = \frac{(60.3 R_E)^3 (1)^2}{(1)^3 (0.0749)^2}\] $$\rightarrow$$ \[\frac{m_E}{m_P} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (0.0196)^2} = \frac{(15.3 R_P)^3 / (0.0196)^2}{(60.3 R_E)^3 / (0.0749)^2} = \frac{(60.3 R_E)^3 (1)^2}{(1)^3 (0.0749)^2}\] $$\rightarrow$$

(20%) \[\frac{\rho_P}{\rho_E} = \frac{m_P / Vol_P}{m_E / Vol_E} = \frac{m_P / \frac{4}{3} \pi (R_P)^3}{m_E / \frac{4}{3} \pi (R_E)^3} = \frac{(15.3 R_P)^3 / (0.0196)^2}{(60.3 R_E)^3 / (0.0749)^2} = \frac{4}{3} \pi (R_P)^3 / \frac{4}{3} \pi (R_E)^3 = \frac{(15.3)^3}{(60.3)^3} \frac{(0.0749)^2}{(0.0196)^2} = 0.24\]

翻译
AI翻译

卫星的周期： $$P_S = 7 日, 3 小时, 43 分钟 = 0.0196 年$$,

卫星的平均轨道半径：$$a_S = 15.3 R_P$$; $$R_P$$: 行星的半径。

月球的周期：$$ P_M = 27 日, 7 小时, 43 分钟 = 0.0749 年$$,

月球的平均轨道半径：$$ a_M = 60.3 R_E$$ ;$$ R_E$$: 地球的半径。

太阳的质量 ($$M$$)

开普勒第三定律：$$ G(M + m) = 4\pi^2 \frac{a^3}{T^2} \rightarrow$$

  • 行星的质量： $$m_P$$
  • 行星的半径： $$R_P$$
  • 行星的半长轴 = 行星的平均轨道半径： $$a_P$$
  • 卫星的半长轴 = 卫星的平均轨道半径： $$a_S$$
  • 月球的半长轴 = 月球的平均轨道半径：$$a_M$$
  • 地球的质量：$$m_E$$
  • 地球的半径：$$R_E$$
  • 地球的半长轴 = 地球的平均轨道半径：$$a_E$$

(5%) 行星 - 卫星： \begin{equation}G(m_P + m_S) = 4\pi^2 \frac{a_S^3}{P_S^2}, m_P \gg m_S \rightarrow G(m_P) = 4\pi^2 \frac{a_S^3}{P_S^2}\end{equation} (5%) 地球 - 月球： \begin{equation}G(m_E + m_M) = 4\pi^2 \frac{a_M^3}{P_M^2}, m_E \gg m_M \rightarrow G(m_E) = 4\pi^2 \frac{a_M^3}{P_M^2}\end{equation} (5%） 行星 - 太阳： \begin{equation} G(M + m_P) = 4\pi^2 \frac{a_P^3}{P_P^2}, M \gg m_P \rightarrow G(M) = 4\pi^2 \frac{a_P^3}{P_P^2}\end{equation} (5%) 地球 - 太阳： \begin{equation} G(M + m_E) = 4\pi^2 \frac{a_E^3}{P_E^2}, M \gg m_E \rightarrow G(M) = 4\pi^2 \frac{a_E^3}{P_E^2}\end{equation}

(20%) 由方程(1)和(3)得出： \[\frac{m_P}{M} = \frac{4\pi^2 \frac{a_S^3}{P_S^2}}{4\pi^2 \frac{a_P^3}{P_P^2}} = \frac{(a_S)^3 (P_P)^2}{(a_P)^3 (P_S)^2} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (P_S)^2} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (0.0196)^2}\]

(20%) 由方程(2)和(4)得出： \[\frac{m_E}{M} = \frac{4\pi^2 \frac{a_M^3}{P_M^2}}{4\pi^2 \frac{a_E^3}{P_E^2}} = \frac{(a_M)^3 (P_E)^2}{(a_E)^3 (P_M)^2} = \frac{(60.3 R_E)^3 (P_E)^2}{(a_E)^3 (P_M)^2} = \frac{(60.3 R_E)^3 (P_E)^2}{(a_E)^3 (0.0749)^2} = \frac{(60.3 R_E)^3 (1)^2}{(1)^3 (0.0749)^2}\] $$\rightarrow$$ \[\frac{m_E}{m_P} = \frac{(15.3 R_P)^3 (P_P)^2}{(a_P)^3 (0.0196)^2} = \frac{(15.3 R_P)^3 / (0.0196)^2}{(60.3 R_E)^3 / (0.0749)^2} = \frac{(60.3 R_E)^3 (1)^2}{(1)^3 (0.0749)^2}\] $$\rightarrow$$

(20%) \[\frac{\rho_P}{\rho_E} = \frac{m_P / Vol_P}{m_E / Vol_E} = \frac{m_P / \frac{4}{3} \pi (R_P)^3}{m_E / \frac{4}{3} \pi (R_E)^3} = \frac{(15.3 R_P)^3 / (0.0196)^2}{(60.3 R_E)^3 / (0.0749)^2} = \frac{4}{3} \pi (R_P)^3 / \frac{4}{3} \pi (R_E)^3 = \frac{(15.3)^3}{(60.3)^3} \frac{(0.0749)^2}{(0.0196)^2} = 0.24\]

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