Three point masses are at the corners of an equilateral traingle of side `r`. Their separations do not change when the system rotates about the centre of the triangle. For this, the time period of rotation must be proportional to

Options

A. `r^(3//2)`
B. `r`
C. `m`
D. `m^(-1//2)`

Answer

A, D

Explanation

Refer to Fig. `O` is the centre of mass of the system of `3` particles present at `A, B` and `C`
repectively. Here `AD = AB sin 60^(@) = r sqrt(3)//2`
`OA = (2)/(3)AD = (2)/(3) xx (rsqrt(3))/(2) = (r )/(3) = OB = OC`
Gravitational force on the particle at `B` due to
particle at `C` is `F_(BC) = (Gm xx m)/(r^(2))` along `BC`
Gravitational force on the particle at `B` due to
particle at `A` is `F_(BA) = (GM xx m)/(r^(2))` along `BA`
The resultant force on particle at `B` will act along `BO` is
`F = sqrt((G^(2)m^(4))/(r^(4)) + (G^(2)m^(4))/(r^(r)) + (2Gm^(2))/(r^(2)) xx (Gm^(2))/(r^(2)) cos 60^(@))`
This force is providing the centripetal force, so
`(sqrt(3) Gm^(2))/(r^(2)) = (mr)/(sqrt(3)) (4pi^(2))/(T^(2))`
or `T = sqrt((4pi^(2)r^(3))/(3Gm))` or `T prop sqrt((r^(3))/(m))`