Understand the configuration

Two identical particles of mass $m$ each revolve in a circle of radius $a$ due to their mutual gravitational attraction.

Since the masses are identical, they revolve about their common center of mass, which lies midway between them.

Hence:

• Radius of each circular path = $a$
• Distance between the two particles = $2a$

Gravitational force between the particles

The magnitude of gravitational force is $F = \frac{Gm^2}{(2a)^2} = \frac{Gm^2}{4a^2}$

Required centripetal force

For each particle moving in a circle of radius $a$ with angular speed $\omega$, $F_c = m\omega^2 a$

This centripetal force is provided entirely by the mutual gravitational force.

Therefore, $m\omega^2 a = \frac{Gm^2}{4a^2}$

Solve for $\omega$

Divide both sides by $ma$: $\omega^2 = \frac{Gm}{4a^3}$

Thus, $\omega = \sqrt{\frac{Gm}{4a^3}}$

Match with the options

This corresponds to: $\boxed{\sqrt{\frac{Gm}{4a^3}}}$

So the correct option is D.