Two identical particles each of mass $m$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be : 

The gravitational force $\left(F_G\right)$ between the particles is given by

$\begin{array}{rcl}{F}_G& =& \frac{G{m}^2}{{\left(2a\right)}^2}\\ & =& \frac{G{m}^2}{4a^2} ...\left(1\right)\end{array}$

The centripetal force $\left(F_C\right)$ of each particle can be written as

$F_C=m{\omega }^{2}a ...\left(2\right)$

Under balanced condition, equate equation (1) and equation (2) and simplify to obtain the angular speed of each particle.

$\begin{array}{rcl}m{\omega }^{2}a& =& \frac{G{m}^2}{4a^2}\\ & \Rightarrow & {\omega }^2=\frac{Gm}{4a^3}\\ & \Rightarrow & \omega =\sqrt{\frac{Gm}{4a^3}}\end{array}$