Two particles of equal mass $$m$$ move in a circle of radius $$r$$ under the action of their mutual gravitational attraction. The speed of each particle will be:

Two particles of equal mass $$m$$ move in a circle of radius $$r$$ under their mutual gravitational attraction. They are diametrically opposite on the circle, so the distance between them is $$2r$$.

$$F = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$$

This gravitational force provides the centripetal force for each particle moving in a circle of radius $$r$$:

$$\frac{Gm^2}{4r^2} = \frac{mv^2}{r}$$

$$\frac{Gm}{4r^2} = \frac{v^2}{r}$$

$$v^2 = \frac{Gm}{4r}$$

$$v = \sqrt{\frac{Gm}{4r}}$$

The correct answer is Option D: $$\sqrt{\dfrac{Gm}{4r}}$$.