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Question: Answered & Verified by Expert
In a satellite, if the time of revolution is $T$, then $\mathrm{KE}$ is proportional to
PhysicsGravitationMHT CETMHT CET 2009
Options:
  • A $\frac{1}{T}$
  • B $\frac{1}{T^{2}}$
  • C $\frac{1}{T^{3}}$
  • D $T^{-2 / 3}$
Solution:
1681 Upvotes Verified Answer
The correct answer is: $T^{-2 / 3}$
Velocity of satellite $v=\sqrt{\frac{G M}{r}}$
$\therefore$
$$
\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}
$$
$$
\therefore \quad \mathrm{KE} \propto T^{-2 / 3}
$$

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