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^(2)))`
D. zero

Answer

B

Explanation

`F=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos theta)`
`(sqrt(3)GMm)/(a^(2))=mr omega^(2)`, here `r=a/(sqrt(3))`