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(D) For a planet to orbit a star in a circular path,the gravitational force provides the necessary centripetal force.Given that the gravitational forc

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$R^{7/4}$

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(D) For a planet to orbit a star in a circular path,the gravitational force provides the necessary centripetal force.Given that the gravitational force $F_G \propto R^{-5/2}$.The centripetal force is given by $F_c = m \omega^2 R$,where $\omega = \frac{2\pi}{T}$.Thus,$F_c = m \left(\frac{4\pi^2}{T^2}\right) R$.Equating the forces: $m \left(\frac{4\pi^2}{T^2}\right) R \propto R^{-5/2}$.Since $m$ and $4\pi^2$ are constants,we have $\frac{R}{T^2} \propto R^{-5/2}$.Rearranging for $T^2$: $T^2 \propto R \cdot R^{5/2} = R^{7/2}$.Taking the square root on both sides: $T \propto R^{7/4}$.

$A$ small planet is revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force between the planet and the star were proportional to $R^{-5/2}$,then $T$ would be proportional to:

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