Three identical bodies of equal mass M each a | vedclass

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Question

Consider the circle with centre at $\mathrm{O},$ and having radius $\mathrm{R}$. In it consider three bodies of masses $\mathrm{M}$ each moving with v

Vedclass · Accepted Answer

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

Vedclass · Answer

Consider the circle with centre at $\mathrm{O},$ and having radius $\mathrm{R}$. In it consider three bodies of masses $\mathrm{M}$ each moving with velocity $v$ under the action of their gravitational attraction.

$\mathrm{R}=\frac{2}{3} \mathrm{AD}=\frac{2}{3} 	imes \mathrm{L} \sin 60^{\circ}=\frac{2 \mathrm{L}}{3} 	imes \frac{\sqrt{3}}{2}$

$\frac{\mathrm{L}}{\sqrt{3}} \quad$ or     $L=\sqrt 3$

Centripetal force on any one mass $\mathrm{M}$ is

$=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} \cos \left(\frac{60^{\circ}}{2}ight)$

$\frac{\mathrm{M} \mathrm{u}^{2}}{\mathrm{R}}=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} 	imes \frac{\sqrt{3}}{2}$

or  $\frac{\mathrm{GM}^{2}}{(\sqrt{3} \mathrm{R})^{2}} \sqrt{3}=\frac{\mathrm{M} \mathrm{v}^{2}}{\mathrm{R}}$

$	herefore v=\sqrt{\frac{\mathrm{GM}}{\sqrt{3} \mathrm{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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