The period of oscillation of a near-Earth satellite (neglecting atmospheric effects) is 84.3 min. What is the period of a near-Moon satellite? (RE = 6.37 * 106 m; RM = 1.74 *106 m; ME = 5.98 *1024 kg; MM = 7.36 *1022 kg.)
A)6.03* 10-3 min.
B)0.713 min.
C)84.3 min.
D)108 min.
E)140 min.

Question

Quizplus · Accepted Answer

The period of a satellite is given by $T=2\pi\sqrt{\frac{a^3}{GM}}$, where $a$ is the average distance from the center of the body the satellite is orbiting, $G$ is the gravitational constant, and $M$ is the mass of the body the satellite is orbiting. We can use the fact that $\frac{T_M}{T_E}=\sqrt{\frac{a_M^3}{a_E^3}}$ for satellites orbiting two different bodies to find the period of the near-Moon satellite. Rearranging the equation, we get $\frac{a_M}{a_E}=\left(\frac{T_M}{T_E}ight)^{2/3}$. Plugging in the given values, we get $\frac{a_M}{a_E}=\left(\frac{84.3}{	ext{1 day}}ight)^{2/3}$ (since we are given the period of the near-Earth satellite, which is approximately the period of one day). Using the values $a_E=6.37	imes10^6	ext{ m}$ and $a_M=1.74	imes10^6	ext{ m}$, we can solve for the period of the near-Moon satellite: $$T_M=2\pi\sqrt{\frac{(1.74	imes10^6	ext{ m})^3}{GM_M}\left(\frac{a_M}{a_E}ight)^3}=108	ext{ min}.$$ Therefore, the answer is D.

The Period of Oscillation of a Near-Earth Satellite (Neglecting Atmospheric