In a satellite,if the time of revolution is T | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$:$T^2 \

Vedclass · Accepted Answer

$T^{-2/3}$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$:$T^2 \propto r^3$This implies $r \propto T^{2/3}$.The potential energy $(PE)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by:$PE = -\frac{GMm}{r}$Since $G, M,$ and $m$ are constants,we have $PE \propto \frac{1}{r}$.Substituting $r \propto T^{2/3}$ into the expression for potential energy:$PE \propto \frac{1}{T^{2/3}}$$PE \propto T^{-2/3}$Therefore,the potential energy is proportional to $T^{-2/3}$.

In a satellite,if the time of revolution is $T$,then the potential energy $(PE)$ is proportional to ..........

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