In an orbit,if the time of revolution of a sa | vedclass

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(C) The potential energy $(PE)$ of a satellite of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by $PE = -\frac{GMm}{r}$

Vedclass · Accepted Answer

$T^{-2/3}$

Vedclass · Answer

(C) The potential energy $(PE)$ of a satellite of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by $PE = -\frac{GMm}{r}$.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 $(r)$: $T^2 \propto r^3$,which implies $r \propto T^{2/3}$.Substituting the expression for $r$ into the potential energy formula:$PE \propto -\frac{1}{r}$Since $r \propto T^{2/3}$,we have $PE \propto -\frac{1}{T^{2/3}}$.Therefore,the magnitude of the potential energy is proportional to $T^{-2/3}$.

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

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