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(C) As the particles move under gravitational influence,the gravitational force provides the necessary centripetal force.The gravitational force betwe

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$\sqrt{\frac{G M}{L}}$

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(C) As the particles move under gravitational influence,the gravitational force provides the necessary centripetal force.The gravitational force between two bodies of mass $M$ separated by distance $L$ is $F = \frac{G M^2}{L^2}$.The net force on a particle at vertex $A$ due to particles at $B$ and $C$ is the vector sum of the forces $\vec{F}_{AB}$ and $\vec{F}_{AC}$.Since the angle between these forces is $60^\circ$,the magnitude of the resultant force is $F_{net} = 2 F \cos(30^\circ) = 2 \left( \frac{G M^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G M^2}{L^2}$.The particles move in a circle of radius $R$. For an equilateral triangle,the distance from the centroid to a vertex is $R = \frac{L}{\sqrt{3}}$.Equating the net force to the centripetal force,$\frac{M v^2}{R} = F_{net}$.Substituting the values,$\frac{M v^2}{L/\sqrt{3}} = \frac{\sqrt{3} G M^2}{L^2}$.Solving for $v$,$v^2 = \frac{\sqrt{3} G M^2}{L^2} \cdot \frac{L}{\sqrt{3} M} = \frac{G M}{L}$.Therefore,$v = \sqrt{\frac{G M}{L}}$.

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