Two particles of equal mass go round a circle of radius R under action of their mutual gra

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7.Gravitation

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Two particles of equal mass go round a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

A

$v = \frac{1}{{2R}}\,\sqrt {\frac{1}{{Gm}}} $

B

$v = \sqrt {\frac{{Gm}}{{2R}}} $

C

$v = \frac{1}{2}\,\sqrt {\frac{{Gm}}{R}} $

D

$v = \sqrt {\frac{{4Gm}}{{R}}} $

Solution

$\frac{m v^{2}}{R}=\frac{G m \times m}{(2 R)^{2}}$

$\Rightarrow \quad \mathrm{v}=\frac{1}{2} \sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$

Standard 11

Physics

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