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(B) For a planet of mass $m$ revolving in a circular orbit of radius $R$ with angular velocity $\omega$,the centripetal force is provided by the gravi

Vedclass · Accepted Answer

$T^2 \propto R^{7/2}$

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(B) For a planet of mass $m$ revolving in a circular orbit of radius $R$ with angular velocity $\omega$,the centripetal force is provided by the gravitational force.Given that the gravitational force $F \propto R^{-5/2}$,we can write $F = k R^{-5/2}$ for some constant $k$.The centripetal force is given by $F = m R \omega^2$.Equating the two,we have $m R \omega^2 \propto R^{-5/2}$.Since $\omega = \frac{2\pi}{T}$,we substitute this into the equation: $m R (\frac{2\pi}{T})^2 \propto R^{-5/2}$.This simplifies to $\frac{R}{T^2} \propto R^{-5/2}$.Rearranging for $T^2$,we get $T^2 \propto R \cdot R^{5/2} = R^{1 + 5/2} = R^{7/2}$.Thus,$T^2 \propto R^{7/2}$.

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

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