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

The resultant force on particle at a due to other two particles is
`F_(A)=sqrt(F_(AB)^(2)+F_(AC)^(2)+2F_(AB)F_(AC) cos 60^(@))=sqrt(3) (Gm^(2))/a^(2) ...(i)" "[ :' F_(AB)=F_(AC)=(Gm^(2))/a^(2)]`
Radius of the circle `r=a/sqrt(3)`
If each particle is given a tangential velocity v, so that the resultant force acts as the centripetal force, then `(mv^(2))/r=sqrt(3) (mv^(2))/a` ...(ii)
From (i) and (ii), `sqrt(3) (mv^(2))/a= (Gm^(2) sqrt(3))/a^(2) implies v=sqrt((Gm)/a)`
Time period `T=(2pir)/v=(2pia)/sqrt(3) sqrt(a/(Gm))=2pi sqrt(a^(3)/(3Gm))`