Two particles of equal mass `'m'` go around a circle of radius `R` under the action of their mutual gravitaitonal attraction. The speed of each particle with respect to their centre of a mass is -

Options

A. `sqrt((GM)/(4R))`
B. `sqrt((GM)/(3R))`
C. `sqrt((GM)/(2R))`
D. `sqrt((GM)/(R))`

Answer

A

Explanation

Here, centripetal force will be given by the gravitational force between the two particles.
`(Gm^(2))/((2R)^(2)) = m omega^(2)`R
`rArr (Gm)/(4R^(3)) = omega^(2) rArr omega = sqrt((Gm)/(4R^(3)))`
If the velocity of the two particles with respect to the centre of gravity is v then
v = `omega`R
v = `sqrt((Gm)/(4R^(3))) xx R = sqrt((Gm)/(4R))`
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