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(C) Consider three bodies of mass $M$ at the vertices of an equilateral triangle inscribed in a circle of radius $R$. The side length $L$ of the trian

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

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(C) Consider three bodies of mass $M$ at the vertices of an equilateral triangle inscribed in a circle of radius $R$. The side length $L$ of the triangle is related to the radius $R$ by $L = \sqrt{3}R$.The gravitational force exerted on one body by the other two 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 force towards the center $O$ is $F_{net} = 2F \cos(30^{\circ}) = 2 \left( \frac{GM^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3}GM^2}{L^2}$.This net force provides the necessary centripetal force for circular motion: $\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{L^2}$.Substituting $L^2 = 3R^2$,we get $\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{3R^2} = \frac{GM^2}{\sqrt{3}R^2}$.Solving for $v$,we get $v^2 = \frac{GM}{\sqrt{3}R}$,so $v = \sqrt{\frac{GM}{\sqrt{3}R}}$.

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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