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? (R_E = 6.37 * 10⁶ m;R_M = 1.74 * 10⁶ m;M_E = 5.98 * 10²⁴ kg;M_M = 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 an orbiting body depends only on its distance from the central body, and not on the mass of the orbiting body. Therefore, the period of a near-Moon satellite will only depend on its distance from the Moon.

Using the given values, we can use the following equation to find the distance (r) of the near-Earth satellite from the center of the Earth:

T = 2π√(r³/GM)

where T is the period (in seconds), G is the gravitational constant, and M is the mass of the Earth.

Solving for r, we get:

r = [(GMT²)/(4π²)]^(1/3)

Substituting the given values, we get:

r = [(6.67×10^-11)(5.98×10^24)(84.3×60)²/(4π²)]^(1/3) = 7.01×10^6 m

Using the same equation, we can find the period (T') of a near-Moon satellite with the given distance (r') from the center of the Moon:

T' = 2π√(r'³/GM')

where M' is the mass of the Moon.

Solving for T', we get:

T' = 2π√(r'³/GM') = 2π√[(r/r')³(M/M')] = (r/r')^(3/2)×(M/M')^(1/2)×T

Substituting the given values, we get:

T' = (7.01×10^6/1.74×10^6)^(3/2)×(5.98×10^24/7.36×10^22)^(1/2)×84.3 = 108 min

Therefore, the answer is D.

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