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 between two particles of mass $m$ separated by a distance $r$ is given by: <br/><br/> $F = \frac{Gm^2}{r^2}$ <br/><br/> where $G$ is the gravitational constant. In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational attraction. Therefore, the gravitational force between the two particles provides the necessary centripetal force to keep them in circular motion. <br/><br/> The centripetal force required for a particle of mass $m$ moving in a circle of radius $a$ with angular speed $\omega$ is given by: <br/><br/> $F_{\text{centripetal}} = m\omega^2a$ <br/><br/> Setting the gravitational force equal to the centripetal force, we get: <br/><br/> $\frac{Gm^2}{r^2} = m\omega^2a$ <br/><br/> Substituting $r = 2a$ (since the two particles are separated by a distance equal to twice the radius of the circle), we get: <br/><br/> $\frac{Gm^2}{(2a)^2} = m\omega^2a$ <br/><br/> Simplifying, we get: <br/><br/> $\omega^2 = \frac{Gm}{4a^3}$ <br/><br/> Taking the square root of both sides, we get: <br/><br/> $\omega = \sqrt{\frac{Gm}{4a^3}}$ <br/><br/> Therefore, the angular speed of each particle is $\sqrt{\frac{Gm}{4a^3}}$.