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(A) The gravitational force is given by $F \propto \frac{1}{R^n} = R^{-n}$.For a planet in a circular orbit of radius $R$,the gravitational force prov

Vedclass · Accepted Answer

$R^{\left( \frac{n+1}{2} \right)}$

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(A) The gravitational force is given by $F \propto \frac{1}{R^n} = R^{-n}$.For a planet in a circular orbit of radius $R$,the gravitational force provides the necessary centripetal force:$F = \frac{M v^2}{R} = M R \omega^2$,where $\omega = \frac{2\pi}{T}$.Equating the forces: $M R \omega^2 \propto R^{-n}$.$\omega^2 \propto R^{-n-1} \Rightarrow \omega \propto R^{-\frac{n+1}{2}}$.Since $\omega = \frac{2\pi}{T}$,we have $\frac{1}{T} \propto R^{-\frac{n+1}{2}}$.Therefore,$T \propto R^{\frac{n+1}{2}}$.

Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

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