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)))`

Options

A. `sqrt((3GM)/(a)), 2pi sqrt((a^(2))/(3GM))`
B. `sqrt((GM)/(a)),(1)/(2pi)sqrt((3GM)/(a^(3)))`
C. `sqrt((GM)/(a)),2pi sqrt((a^(3))/(3GM))`
D. `sqrt((GM)/(3a)), 2pi sqrt((a^(3))/(3GM))`

Answer

C

Explanation

The resultant force on each particle due to other two particles is,
`F_(R)=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos theta)=sqrt3F_(1)`
`=sqrt3(GM^(2))/(a^(2))" …(i)"`
If each particle is given a tangential velocity v, so that F acts as the centripetal force, they will move in a circle of radius,
`r=((2)/(3))a sin 60^(@)=(a)/(sqrt3)`

Now, `F_(R)=(Mv^(2))/(r)=sqrt3(Mv^(2))/(a)" ...(ii)"`
From eqn. (i) and (ii),
`sqrt3(Mv^(2))/(a)=(GM^(2)sqrt3)/(a^(2))" or "v=sqrt((GM)/(a))`
Time period,
`T=(2pir)/(v)=(2pia)/(sqrt3)sqrt((a)/(GM))=2pi sqrt((a^(3))/(3GM))`