Three point masses each of mass `m` rotate in a circle of radius `r` with constant angular velocity `omega` due to their mutual gravitational attraction. If at any instant, the masses are on the vertices of an equilateral triangle of side `a`, then the value of `omega` is

Options

A. `sqrt(((Gm)/(a^(3))))`
B. `sqrt(((3GM)/(a^(3))))`
C. `sqrt(((Gm)/(3a^(3))))`
D. None

Answer

B

Explanation

Here, `F_(0) = (Gm^(2))/(a^(2))`
`:. F = 2F_(0) cos 30^(@) = (2Gm^(2))/(a^(2)) (sqrt(3))/(2)`

`=(sqrt(3)Gm^(2))/(a^(2))` or `m r omega^(2) = (sqrt(3)Gm^(2))/(a^(2))`
or `m(a)/(sqrt(3)) omega^(2) = (sqrt(3)Gm^(2))/(a^(2)) :. omega = sqrt(((3Gm)/(a^(3))))`