Two identical particles of mass 1kg each go round a circle of...

Two identical particles of mass

1kg

each go round a circle of radius

R

, under the action of their mutual gravitational attraction. The angular speed of each particle is: Two identical particles of mass 1kg each go round a circle of radius R , under the action of their mutual gravitational

Two identical particles of mass 1kg each go round a circle of radius R , under the action of their mutual gravitational

Options

A $\sqrt{\dfrac{G}{2{R}^{3}}}$
B $\dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$
C $\dfrac{1}{2R}\sqrt{\dfrac{1}{G}}$
D $\sqrt{\dfrac{2G}{{R}^{3}}}$

Correct answer

B. $\dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$

Step-by-step solution

Two identical particles of mass 1kg each go round a circle of radius R , under the action of their mutual gravitational

$F = \dfrac{G{m}^{2}}{( 2R{)}^{2}} = mR{\omega}^{2}$

$\omega = \dfrac{1}{2}\sqrt{\dfrac{G}{{R}^{3}}}$

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