Solution:
The centripetal force is provided by the gravitational force: \(m \omega^2 R = \frac{k}{R^n} ⇒ \omega^2 \propto \frac{1}{R^{n+1}}\). Since \(T = \frac{2\pi}{\omega}\), we get \(T^2 \propto R^{n+1} ⇒ T \propto R^{\frac{n+1}{2}}\).
Suppose the gravitational force varies inversely as the \(n^{\text{th}}\) 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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