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

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(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.Since kinetic energy $K.E. = \frac{1}{2}mv^2$,we have $K.E. \propto v^2

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$T^{-2/3}$

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(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.Since kinetic energy $K.E. = \frac{1}{2}mv^2$,we have $K.E. \propto v^2 \propto \frac{1}{r}$.According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius: $T^2 \propto r^3$,which implies $r \propto T^{2/3}$.Substituting this into the proportionality for kinetic energy: $K.E. \propto \frac{1}{r} \propto \frac{1}{T^{2/3}} = T^{-2/3}$.

In a satellite,if the time of revolution is $T$,then the kinetic energy $(K.E.)$ is proportional to:

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