Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. On what power of r will the square of time period will depend if the gravitational force of attraction between the planet and the star is proportional to `r^(-5//2)`.

Answer

C

Explanation

As gravitation provides centripetal force `(mv^2)/r=K/r^(5//2)`
i.e., `v^2 K/(mr^(3//2))`
So that `T=(2pir)/v=2pirsqrt((m r^(3//2))/K)` or `T^2=(4pi^2m)/K r^(7//2)` , so `T^2 prop r^(7//2)`