Q117 / 150 · MHT CET 2009 · Physics
In a satellite, if the time of revolution is $T$, then $\mathrm{KE}$ is proportional to
- A$\frac{1}{T}$
- B$\frac{1}{T^{2}}$
- C$\frac{1}{T^{3}}$
- Correct answerD$T^{-2 / 3}$
Solution
Velocity of satellite $v=\sqrt{\frac{G M}{r}}$
$\therefore$
$$
\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}
$$
$$
\therefore \quad \mathrm{KE} \propto T^{-2 / 3}
$$
Hence, the correct answer is option: 'D'.