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(A) The gravitational force $F$ is given by $F \propto \frac{1}{R^n}$.For a planet in a circular orbit,this force provides the necessary centripetal f

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

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(A) The gravitational force $F$ is given by $F \propto \frac{1}{R^n}$.For a planet in a circular orbit,this force provides the necessary centripetal force:$F = m\omega^2 R = m\left( \frac{4\pi^2}{T^2} \right) R$.Equating the two expressions:$m\left( \frac{4\pi^2}{T^2} \right) R \propto \frac{1}{R^n}$.Since $m$ and $4\pi^2$ are constants,we have:$\frac{R}{T^2} \propto \frac{1}{R^n}$.Rearranging for $T^2$:$T^2 \propto R \cdot R^n = R^{n+1}$.Taking the square root on both sides:$T \propto R^{\left( \frac{n+1}{2} \right)}$.

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

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