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(D) The gravitational force provides the necessary centripetal force for circular motion.Given,$F \propto r^{-3/2}$,so $F = \frac{k}{r^{3/2}}$ where $

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$r^{5/2}$

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(D) The gravitational force provides the necessary centripetal force for circular motion.Given,$F \propto r^{-3/2}$,so $F = \frac{k}{r^{3/2}}$ where $k$ is a constant.The centripetal force is given by $F = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r = \frac{4\pi^2 m r}{T^2}$.Equating the two expressions for force:$\frac{k}{r^{3/2}} = \frac{4\pi^2 m r}{T^2}$.Rearranging for $T^2$:$T^2 = \frac{4\pi^2 m}{k} \cdot r \cdot r^{3/2} = \frac{4\pi^2 m}{k} \cdot r^{5/2}$.Since $\frac{4\pi^2 m}{k}$ is a constant,we have $T^2 \propto r^{5/2}$.

$A$ planet is revolving around the sun in a circular orbit with a radius $r$. The time period is $T$. If the force between the planet and the star is proportional to $r^{-3/2}$,then the square of the time period is proportional to

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