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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
$\sqrt {\frac{{GM}}{R}} $
$\sqrt {\frac{{GM}}{3R}} $
$\sqrt {\frac{{GM}}{{\sqrt 3 R}}} $
$\sqrt {\frac{{GM}}{{\sqrt 2 R}}} $
Solution
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.
$\mathrm{R}=\frac{2}{3} \mathrm{AD}=\frac{2}{3} \times \mathrm{L} \sin 60^{\circ}=\frac{2 \mathrm{L}}{3} \times \frac{\sqrt{3}}{2}$
$\frac{\mathrm{L}}{\sqrt{3}} \quad$ or $L=\sqrt 3$
Centripetal force on any one mass $\mathrm{M}$ is
$=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} \cos \left(\frac{60^{\circ}}{2}\right)$
$\frac{\mathrm{M} \mathrm{u}^{2}}{\mathrm{R}}=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} \times \frac{\sqrt{3}}{2}$
or $\frac{\mathrm{GM}^{2}}{(\sqrt{3} \mathrm{R})^{2}} \sqrt{3}=\frac{\mathrm{M} \mathrm{v}^{2}}{\mathrm{R}}$
$\therefore v=\sqrt{\frac{\mathrm{GM}}{\sqrt{3} \mathrm{R}}}$
