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⊕= 6.37 × 10^6 m; RMoon= 1.74 × 10^6 m; m⊕ = 5.98 × 10^24 kg; mMoon = 7.36 × 10^22 kg.)

Answer 1

Recall that T^2 = 4π^2a^3/GM
That is, T^2M/a^3 = 4π^2/G, a constant
So, you want T such that

(7.36*10^22)T^2/(1.74*10^6)^3 = (5.98*10^24)(84.3)^2/(6.37*10^6)^3

Or, you know that

T^2 is proportional to M/a^3
so, the new (') values bear the relation to the old values:

(T'/T)^2 = (M'/M)(a'/a)^3

This kind of relation is useful if they say something like, what is the new period if you double the radius and triple the mass?