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(C) Let the three bodies be at the vertices of an equilateral triangle inscribed in a circle of radius $R$. The side length $L$ of this equilateral tr

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$\sqrt{\frac{GM}{\sqrt{3}R}}$

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(C) Let the three bodies be at the vertices of an equilateral triangle inscribed in a circle of radius $R$. The side length $L$ of this equilateral triangle is given by $L = 2R \sin(60^{\circ}) = R\sqrt{3}$.Consider one body of mass $M$. The gravitational force exerted on it by the other two bodies is directed towards the center of the circle.The force from one body is $F_1 = \frac{GM^2}{L^2}$.The component of this force towards the center is $F_c = F_1 \cos(30^{\circ}) = \frac{GM^2}{L^2} \cdot \frac{\sqrt{3}}{2}$.Since there are two such bodies,the total centripetal force is $F_{net} = 2 \cdot F_c = 2 \cdot \frac{GM^2}{L^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}GM^2}{L^2}$.Equating this to the required centripetal force $\frac{Mv^2}{R}$:$\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{(R\sqrt{3})^2} = \frac{\sqrt{3}GM^2}{3R^2} = \frac{GM^2}{\sqrt{3}R^2}$.Solving for $v$:$v^2 = \frac{GM}{\sqrt{3}R} \Rightarrow v = \sqrt{\frac{GM}{\sqrt{3}R}}$.

Three identical bodies of equal mass $M$ each are moving along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each body is

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