Three particles, each of mass m, are situated at the 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 intended that each particle moves along a circle while maintaining their original separation 'a'. Determine the initial velocity that should be given to each particle and the time period of the circular motion. The resultant force on particle at A due to other two particles is

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")cos60^(@))=sqrt(3)("Gm"^(2))/(a^(2))` …. (i)
`[becuae 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))impliesv=sqrt(("Gm")/(a))`

Time period
`T=(2pi r)/(v)=(2 pi a)/(sqrt(3))sqrt((a)/("Gm"))=2 pi sqrt((a^(3))/(3"Gm"))`.