Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

   
Formula: Gravitational force b/w two particles:
                                                                        ​$F_G = \frac{G{m}_1{m}_2}{r^2}$  

(1). Gravitational Force b/w particle A & B :
                                                                        $F_{AB} = \frac{G\times M\times M}{(\sqrt{2R}{)}^2} = \frac{G{M}^2}{2R^2}$         [Distance AB = $\sqrt{2R}$]
(2).  Gravitational Force b/w particle A & C :
                                                                        $F_{AC} = \frac{G \times M \times M}{(2R{)}^2} = \frac{G{M}^2}{4R^2}$      ​[Distance AC = 2R]
(3).  Gravitational Force b/w particle A & D :
                                                                        $F_{AD} = \frac{G\times M\times M}{(\sqrt{2R}{)}^2} = \frac{G{M}^2}{2R^2}$          ​[Distance AD = $\sqrt{2R}$]
As per given that the four particles are moving in a circular path, so each particle will experience gravitational force due to the other three particles. And as we know that all particles are of similar masses, so they will be arranged as shown in figure.
    Now suppose that particle A is moving with 'v' speed, therefor a centrifugal force($F_C$) will be acting on the particle, which is balanced by the gravitational force applied by the other three particle.(Shown in figure)
     * Formula : Centrifugal Force ($F_C$) :
                                                              ​$F_C = \frac{M{V}^2}{R}$
So According to the free body diagram shown in figure:
(1). net force acting along horizontal direction is Zero so:
                                                              ​$$\begin{gathered} F_C = F_{AB} Cos\left({45}^0\right) + F_{Ac} Cos\left(0^0\right) + F_{AD} Cos\left({45}^0\right) \\ \frac{M{V}^2}{R} = \frac{G{M}^2}{2R^2}\times \frac{1}{\sqrt{2}} + \frac{G{M}^2}{4R^2} \times 1 + \frac{G{M}^2}{2R^2}\times \frac{1}{\sqrt{2}} \\ \frac{M{V}^2}{R}= \frac{G{M}^2}{2R^2}(\frac{1}{\sqrt{2}}+ \frac{1}{2} + \frac{1}{\sqrt{2}}) \\ {V}^2 = \frac{GM}{2R}(\frac{1}{2} + \sqrt{2}) \\ V= \sqrt{\frac{GM}{2R}(\frac{1}{2} + \sqrt{2})} \end{gathered}$$
According to the figure All four particles will have same speed.