First slide

Universal law of gravitation

Question

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

Difficult

A

$\sqrt{[\frac{GM}{\sqrt{3} R}]}$
B

$\sqrt{(\frac{GM}{3 R})}$
C

$\sqrt{(\frac{GM}{\sqrt{3} R})}$
D

$\sqrt{(\frac{GM}{\sqrt{2} R})}$

Solution

Due to mutual attraction, let every body is moving with a velocity v.
From figure, $AD = \frac{3}{2} R$
or $Lsin 60^{\circ} = \frac{3}{2} R or L = \frac{3}{2} \frac{R \times 2}{\sqrt{3}} = \sqrt{3} R$
Centripetal force on an one mass M is
$2 (\frac{{GM}^{2}}{L^{2}}) \cos (60^{\circ} / 2) = \frac{{Mv}^{2}}{R}$
$2 G \frac{M^{2}}{L^{2}} \times \frac{\sqrt{3}}{2} = \frac{{Mv}^{2}}{R} or \frac{GM}{(\sqrt{3} R)^{2}} \times \sqrt{3} = \frac{v^{2}}{R}$
$∴ V = \sqrt{[\frac{GM}{\sqrt{3} R}]}$

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