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)))`

Explanation

`F_(0)=(Gm^(2))/(a^(2))`
The resultant gravitational force acting on the particle
`F'=2F_(0)cos30^(@)=sqrt(3)F_(0)=(sqrt(3)Gm^(2))/(a^(2))`
To move in circle, necesary centripetal force is provided by gravitational force
`(mv^(2))/r=(sqrt(3)Gm^(2))/(a^(2))`
`v^(2)=(sqrt(3)Gm.(a//sqrt(3)))/(a^(2)) [since r=a/(sqrt(3))]`
`v=sqrt((Gm)/a)`
Time period of circular motion
`T=(2pir)/v=(2pi(a//sqrt(3)))/(sqrt(Gm//a))=(2pia^(3//2))/(sqrt(3Gm))`