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(D) The time period $T$ of a satellite revolving in a circular orbit of radius $R$ is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{R^3}{GM}}$.Sq

Vedclass · Accepted Answer

$T^{-2/3}$

Vedclass · Answer

(D) The time period $T$ of a satellite revolving in a circular orbit of radius $R$ is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{R^3}{GM}}$.Squaring both sides,we get $T^2 = \frac{4\pi^2 R^3}{GM}$,which implies $R^3 \propto T^2$,or $R \propto T^{2/3}$.The kinetic energy $K$ of a satellite in a circular orbit is given by $K = \frac{GMm}{2R}$.Since $G$,$M$,and $m$ are constants,$K \propto \frac{1}{R}$.Substituting the relation $R \propto T^{2/3}$,we get $K \propto \frac{1}{T^{2/3}}$.Therefore,$K \propto T^{-2/3}$.

If the time period of revolution of a satellite is $T$,then its kinetic energy is proportional to

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