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In a satellite, if the time of revolution is $T$, then $\mathrm{KE}$ is proportional to
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1681 Upvotes
Verified Answer
The correct answer is:
$T^{-2 / 3}$
Velocity of satellite $v=\sqrt{\frac{G M}{r}}$
$\therefore$
$$
\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}
$$
$$
\therefore \quad \mathrm{KE} \propto T^{-2 / 3}
$$
$\therefore$
$$
\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}
$$
$$
\therefore \quad \mathrm{KE} \propto T^{-2 / 3}
$$
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