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(C) Consider three bodies of mass $M$ placed at the vertices of an equilateral triangle inscribed in a circle of radius $R$. Let $L$ be the side lengt

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

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(C) Consider three bodies of mass $M$ placed at the vertices of an equilateral triangle inscribed in a circle of radius $R$. Let $L$ be the side length of the equilateral triangle.The distance from the center $O$ to any vertex is $R$. In an equilateral triangle,the distance from the center to a vertex is related to the side length $L$ by $R = \frac{L}{\sqrt{3}}$,which implies $L = \sqrt{3}R$.The gravitational force exerted on one body by the other two bodies is the vector sum of two forces,each of magnitude $F = \frac{GM^2}{L^2}$. The angle between these two forces is $60^{\circ}$.The resultant gravitational force $F_{net}$ is given by:$F_{net} = 2F \cos(30^{\circ}) = 2 \left( \frac{GM^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3}GM^2}{L^2}$.This resultant force provides the necessary centripetal force for circular motion:$\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{L^2}$.Substituting $L^2 = 3R^2$:$\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{3R^2} = \frac{GM^2}{\sqrt{3}R^2}$.Solving for $v$:$v^2 = \frac{GM}{\sqrt{3}R} \implies v = \sqrt{\frac{GM}{\sqrt{3}R}}$.

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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