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

As the gravitational force between any two particles is F = G mm/ `a^(2)`, the resultant force on each particle due to the other two
`F_(R) = sqrt(F^(2) + F^(2) + 2F^(2) "cos" 60^(@))`

i.e., `" " F_(R) = sqrt3 F = sqrt3 Gm^(2)//a^(2) " " ...... (i)`
So , if the particles were at rest they will move under the action of `F_(R)` acting on each and will meet at the centre O . Now if each of the particle is given a tangential velocity v so that `F_(R)` acts as centripetal force , they will move in a circle of a radius r = (2/3)
a sin `60^(@) = (a//sqrt3)` maintaining the original mutual separation a.
i .e ., `" " F_(R) = ( mv^(2)//r)`
Which in the light of Eqn . (i) and above value of r yields
`sqrt3(Gm^(2))/(a^(2)) = (mv^(2))/(a) sqrt3 , ` i.e., `v = sqrt((Gm)/(a))`
So , that time of one revolution
`T = (2 pir)/(v) = (2pia)/(sqrt3) sqrt((a)/(Gm)) = 2pi sqrt((a^(3))/(3 Gm))`