Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is:

As shown in figure, each mass experiences three forces namely F, F and F, Hence, F= force between two masses at R separation.
As the all particles move in the circular path of radius R due their mutual gravitational attraction.
Then net force on mass m = mass x centripetal acceleration
$\Rightarrow \frac{\mathrm{F}}{\sqrt{2}}+\frac{\mathrm{F}}{\sqrt{2}}+{\mathrm{F}}^{\mathrm{\prime }}=\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}}$

$\Rightarrow \mathrm{F}\sqrt{2}+{\mathrm{F}}^{\mathrm{\prime }}=\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}}$

$\Rightarrow \frac{\sqrt{2}{\mathrm{Gm}}^2}{(\mathrm{R}\sqrt{2}{)}^2}+\frac{{\mathrm{Gm}}^2}{4{\mathrm{R}}^2}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

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

$\Rightarrow \mathrm{v}=\frac{1}{2}\sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}(1+2\sqrt{2})}$