Given: Mass of four identical particles placed at the corners of a square $= M$,

Edge length of square $= a$, the particles revolve in a circular path under the influence of each other's gravitational field.

To find: The speed of particles.

Let $F_{1},F_{2}$ and $F_{3}$ be the forces acting on particle at $A$ due to particles at $B,C$ and $D$ respectively. The net gravitational force acting on particle $A$,

$$F_{A} = F_{1}cos45^{\ {^\circ}} + F_{2}cos45^{\ {^\circ}} + F_{3}$$

[along $x$-direction]

The $y$-components of $F_{1}$ and $F_{2}$ cancels each other.

Net force on $A$ provides the centripetal force

$$\frac{Mv^{2}}{r} = 2F_{1}\cos 45^{\ {^\circ}} + F_{3}\ldots\ldots(i)$$

Let $r$ be the radius of the circle.

$$So,\ r^{2} + r^{2} = a^{2}$$

$$r = \frac{a}{\sqrt{2}}$$

Using it in equation (i),

$$\frac{Mv^{2}}{\frac{a}{\sqrt{2}}} = \frac{2GMM}{a^{2}}\frac{1}{\sqrt{2}} + \frac{GMM}{2a^{2}}$$

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

$$v = 1.16\sqrt{\frac{GM}{a}}$$