Period of a Planet

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

(A) \(R^{\frac{n+1}{2}}\)


(C) \(R^n\)

(D) \(R^{n/2}\)

The gravitational force can be given as $$ \frac{GMm}{R^n}=mR\omega^2$$

Now, we know \(\omega=\frac{2\pi}{T}\),


$$\frac{GMm}{R^n}= mR(\frac{2\pi}{T})^2$$ $$ T^2= \frac{4\pi^2R^{n+1}}{GM}$$


Leave a Reply

Your email address will not be published. Required fields are marked *