Question

Two satellites of earth, $ S_1 $ and $ S_2 $ , are moving in the same orbit. The mass of $ S_1 $ is four times the mass of $ S_2 $ . Which one of the following statements is true ?

• A. The time period of $ S_1 $ is four times that of $ S_2 $
• B. The potential energies of earth and satellite in the two cases are equal
• C. $ S_1 $ and $ S_2 $ are moving with the same speed
• D. The kinetic energies of the two satellites are equal

Answer 1

Answer: (C)

The satellite of mass $m$ is moving in a circular orbit of radius $r$.
$\therefore$ Kinetic energy of the satellite, $K=\frac{G M m}{2 r} \ldots$ (i)
Potential energy of the satellite, $U=\frac{-G M m}{r} \ldots$ (ii)
Orbital speed of satellite, $v=\sqrt{\frac{G M}{r}} \ldots$ (iii)
Time-period of satellite, $T=\left[\left(\frac{4 \pi^{2}}{G M}\right) r^{3}\right]^{1 / 2} \ldots$(iv)
Given $m_{S_{1}}=4 m_{S_{2}}$
Since $M, r$ is same for both the satellites $S_{1}$ and $S _{2}$
$\therefore$ From equation (ii), we get $U \propto m$
$\therefore \frac{U_{S_{1}}}{U_{S_{2}}}=\frac{m_{S_{1}}}{m_{S_{2}}}=4 $
or, $ U_{S_{1}}=4 U_{S_{2}} .$
From (iii), since $v$ is independent of the mass of a satellite, the orbital speed is same for both satellites $S_{1}$ and $S_{2}$.
Hence option (b) is correct.
From (i), we get $K \propto m$
$\therefore \frac{K_{S_{1}}}{K_{S_{2}}}=\frac{m_{S_{1}}}{m_{S_{2}}}=4 $
or, $ K_{S_{1}}=4 K_{S_{2}}$
Hence option (c) is wrong.
From (iv), since $T$ is independent of the mass of a satellite, time period is same for both the satellites $S_{1}$ and $S_{2}$.