Question

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)))`

Answer 1

In Fig. gravitational force between any two particle is
`F = (Gm m)/(a^(2))`
Resultant force on each particle du eto other two particle is
`R = sqrt(F^(2) + F^(2) + 2F F cos 60^(@)) = sqrt(F^(2) + F^(2) + F^(2)) = F sqrt(3)`
`R = (sqrt(3)Gm^(2))/(a^(2))`
If particles were at rest, each particle would move under the action of resultant force `R` (on each) and meet at the centroid `O` of the triamgle.
Let each particle be given a tangential velocity `v` so that `R` acts as the centripetal force, they would move in
a circle of radius `= OA = OB = OC = r = (2)/(3)a sin 60^(@) = (2)/(3)a(sqrt(3))/(2) = a//sqrt(3)`
The original mutual separation will be maintained.
As `R = (mv^(2))/(r ) :. v = sqrt((R r)/(m)) = sqrt((sqrt(3)Gm^(2))/(a^(2)).(a)/(sqrt(3))m) or v = sqrt(Gm//a)`
Time period of circular motion `T = (2pi r)/(v) = 2pi (a)/(sqrt(3)) sqrt((a)/(Gm)) = 2pi sqrt((a^(3))/(3Gm))`