Three particles, each of mass $M$, situated at the vertices of an equilateral triangle of side length $L$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation $L$. The initial speed that should be given to each particle is

As the particles move under gravitational influence, the gravitational force will provide the centripetal force required. 

Gravitational force between two bodies of masses $m_1$ and $m_2$ separated by distance $d$ is given by, $G\frac{m_1{m}_2}{d^2}$, where $G$ is the universal gravitational constant.

Net force on the particle at $A$ due to particles at $B$ and $C$ is,

${\overrightarrow{F }}_{\mathrm{net}}={\overrightarrow{F }}_{\mathrm{AC}}+{\overrightarrow{F}}_{\mathrm{AB}}$

${\overrightarrow{F}}_{\mathrm{net}}=G\frac{M^2}{L^2}\widehat{AC}+G\frac{M^2}{L^2}\widehat{AB}$.

Here, $\widehat{AB}$ and $\widehat{AC}$ are unit vectors along $AB$ and $AC$.

On simplifying vector addition$\left(\text{taking} F=G\frac{M^2}{L^2}\right)$, we get,

$F_{\mathrm{net}}=2F\cos \left(30°\right)=2G\frac{M^2}{L^2}\frac{\sqrt{3}}{2}=\sqrt{3}G\frac{M^2}{L^2}$

From geometry, radius of circle, $R=\frac{L}{\sqrt{3}}$.

Since the force $F_{\mathrm{net}}$ provides the centripetal force required, so,

$\Rightarrow \frac{M{v}^2}{R}=\sqrt{3}G\frac{M^2}{L^2}$

$\Rightarrow v^2=\sqrt{3}G\frac{MR}{L^2}=\sqrt{3}G\frac{M{\displaystyle \left(\raisebox{1ex}{${\displaystyle L}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \sqrt{3}}$}\right.\right)}}{L^2}=\frac{GM}{L}$

$\Rightarrow v=\sqrt{\frac{GM}{L}}$.