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 :

Two identical particles of mass $$m$$ go round a circle of radius $$a$$. They are diametrically opposite, so the distance between them is $$2a$$.

The gravitational force between them provides the centripetal force:

$$\frac{Gm^2}{(2a)^2} = m\omega^2 a$$

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

$$\omega^2 = \frac{Gm}{4a^3}$$

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

This matches option 3.

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