Question

A satellite moves in a circle around the earth. The radius of this circle is equal to one-half of the radius of the moon's orbit. The satellite completes one revolution is

• A. $ \frac{1}{2} $ lunar month
• B. $ \frac{2}{3} $ lunar month
• C. $ 2^{-3/2} $ lunar month
• D. $ 2^{3/2} $ lunar month

Answer 1

Answer: (C)

The period of revolution of a satellite at a height $h$ from the surface of earth is given by
$ T=2\pi\sqrt{\frac{(R_e+h)^3}{gR_e^2}} $
Given $ T_m = 1$ lunar month,
$= T_{\text{sat}}=2\pi\sqrt{\frac{\left(R+\frac{h}{2}\right)^2}{gR^2}}$
$\Rightarrow T_{\text{sat}}\approx\frac{1}{2^{3/2}} $
$ T_{\text{moon}}=2^{-3/2} $ lunar month