Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be:
Answer: (D) $\dfrac{\sqrt{(1+2\sqrt2)G}}{2}$
The particles sit at the corners of a square inscribed in the circle: the sides are $\sqrt2 R$ and the diagonal is $2R$.
Net force on one particle towards the centre:
$$F = 2\cdot\frac{Gm^2}{2R^2}\cos45^\circ + \frac{Gm^2}{4R^2} = \frac{Gm^2}{R^2}\left(\frac{1}{\sqrt2} + \frac14\right) = \frac{Gm^2}{R^2}\cdot\frac{2\sqrt2+1}{4}$$
This provides the centripetal force $mv^2/R$:
$$v^2 = \frac{Gm}{R}\cdot\frac{1+2\sqrt2}{4}$$
With $m = 1$ kg and $R = 1$ m, $v = \dfrac{\sqrt{(1+2\sqrt2)G}}{2}$.
Solution by Sreeraj P, M.Sc Physics