A light planet is revolving around a massive star in a circular orbit of radius $$R$$ with a period of revolution $$T$$. If the force of attraction between planet and star is proportional to $$R^{-3/2}$$ then choose the correct option:

A planet revolves around a star in a circular orbit of radius $$R$$ with period $$T$$, and the gravitational force is proportional to $$R^{-3/2}$$.

For a circular orbit, the gravitational force must provide the centripetal force, so we have $$F = \frac{mv^2}{R}$$.

Since the force varies as $$F \propto R^{-3/2}$$, it can be expressed as $$F = \frac{k}{R^{3/2}}$$ for some constant $$k$$.

Equating these expressions gives $$\frac{k}{R^{3/2}} = \frac{mv^2}{R}$$, which simplifies to $$v^2 = \frac{k}{m\sqrt{R}}$$ and hence $$v = \sqrt{\frac{k}{m}}\,R^{-1/4}$$.

The period $$T$$ is the orbital circumference divided by the speed, so $$T = \frac{2\pi R}{v} = 2\pi R \sqrt{\frac{m}{k}}\,R^{1/4} = 2\pi\sqrt{\frac{m}{k}}\,R^{5/4}$$.

It follows that $$T \propto R^{5/4}$$, and squaring this relationship yields $$T^2 \propto R^{5/2}$$.

The correct answer is $$T^2 \propto R^{5/2}$$ (Option A).