If the time period of revolution of a satellite is $T$. then its kinetic energy is proportional to
If the time period of revolution of a satellite is $T$. then its kinetic energy is proportional to
$\mathrm{T}^{-1}$
$\mathrm{T}^{-2}$
$\mathrm{T}^{-3}$
$\mathrm{T}^{-2 / 3}$
Solution
The time period of satellite is
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}} \Rightarrow \mathrm{R} \propto \mathrm{~T}^{\frac{2}{3}}$
$\therefore$ Kinetic energy,
$K=\frac{G M_m}{R} \Rightarrow k \propto \frac{1}{R} \propto \frac{1}{T^{\frac{2}{3}}} \propto T^{-\frac{2}{3}}$