Question

Assuming that the law of gravitation is of the form $F=\frac{GMm}{r^3}$ and attractive. A body of mass m revolves in a circular path of radius r around a fixed body of mass M. Find on what power of r will the square of time period depend.

Answer 1

The centrepetal acceleration is due to the gravitational force

$\therefore \frac{m{v}^2}{r}=\frac{GMm}{r^3}$

$\therefore v^2=\frac{GM}{r^2}\Rightarrow v=\frac{\sqrt{GM}}{r}$

$TimeperiodT=\frac{2\Pi r}{v}=\frac{2\Pi r^2}{\sqrt{GM}}$

$\therefore T\propto r^2$