Question

Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Answer 1

$\begin{array}{c}f=\frac{Gm.m}{{\left(2R\right)}^2}=\frac{G{m}^2}{\left(4R\right)}\\ f=\frac{m{v}^2}{R}\\ \frac{G{m}^2}{\left(4R\right)}=\frac{m{v}^2}{R}\\ v=\frac{1}{2}\sqrt{\frac{Gm}{R}}\end{array}$