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 $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,an

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$T^{-2/3}$

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(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $r$ is the orbital radius.Kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$,which implies $K.E. \propto v^2$.Substituting the expression for $v$,we get $K.E. \propto \frac{1}{r}$.According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius,i.e.,$T^2 \propto r^3$,which means $r \propto T^{2/3}$.Substituting $r \propto T^{2/3}$ into the relation $K.E. \propto \frac{1}{r}$,we get $K.E. \propto \frac{1}{T^{2/3}}$ or $K.E. \propto T^{-2/3}$.

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

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