Three particles, each of the mass `m` are situated at the vertices of an equilateral triangle of side `a`. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation `a`. Find the initial velocity that should be given to each particle and also the time period of the circular motion. `(F=(Gm_(1)m_(2))/(r^(2)))`

Answer

(i)1(ii)

Explanation

In the figure three particles located at vertices A, B and C of equilateral triangle of side AB = BC = CA = a.

These particle move in a circle with O as the centre and radius
r=OA=OB=OC where
`r=(BD)/(cos30^@)=(a//2)/(sqrt3//2)=a/sqrt3`
The gravitational attraction force acting on a particle, say due to particle at B is `F_1=(Gmm)/a^2` and due to particle at C is `F_2` .
Here `F_1=F_2=F` (Say) =`(Gm^2)/a^2`
Therefore, the resultant force on the particles at A is
`=2Fcos30^@=(2Gm^2)/a^2 sqrt3/2 = sqrt3 (Gm^2)/a^2`
`F_r` is directed along AO..... Thus the net force on particle at A is radial. Similarly, the net force on particle at B and C at C is `F_r`, each directed towards centre O. This force provides the necessary centripetal force. If v is the required initial velocity of each particle , then `(mv^2)/r = sqrt3 (Gm^2)/a^2 ` or `v^2=sqrt3 (Gmr)/a^2`
Since `r=a/sqrt3`, we have `v^2=sqrt3(Gm)(sqrt3a)=(Gm)/a rArr v=sqrt((Gm)/a)`
Time period T = `(2pir)/v =(2pixxa//sqrt3)/sqrt((Gm)/a) =2pi(a^3/(3Gm))^(1//2)`