Four identical particles of equal mass of 1 kg are 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:

Each particle experiences gravitational forces from the other three.

Net Gravitational Force ($$F_{net}$$)

A single particle is pulled by two adjacent particles at a distance $$d = \sqrt{2}R$$ and one opposite particle at a distance $$D = 2R$$. The resultant force acting towards the center is the sum of the components of these forces:

$$F_{net} = 2 \left( \frac{Gm^2}{(\sqrt{2}R)^2} \right) \cos 45^\circ + \frac{Gm^2}{(2R)^2}$$

$$F_{net} = \frac{Gm^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$$

This mutual attraction provides the centripetal force required for the circular motion of each particle:

$$\frac{mv^2}{R} = \frac{Gm^2}{R^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v^2 = G \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v = \frac{\sqrt{G(1 + 2\sqrt{2})}}{2}$$