Three identical particles each of mass "m" are arranged at the corners of an equiliteral triangle of side "L". If they are to be in equilibrium, the speed with which they must revolve under the influence of one another's gravity in a circular orbit circumscribing the triangle is

Explanation

Refer to Fig. force of attraction on body at `C` due to body at `A` is,
`F_(1) = (Gmm)/(r^(2)) = (Gm^(2))/(r^(2))` along `CA`

Force of attraction on body at `C` due to body at
`B` is, `F_(2) (G mm)/(r^(2)) = (G m^(2))/(r^(2))` along `CB`
These force `vec(F_(1))` and `vec(F_(2))` are inclined at an
angle `60^(@)`, the resultant force on the body at `C` is
`F = sqrt(F_(1)^(2) + F_(2)^(2) + 2F_(1) F_(2) cos 60^(@))`
`= sqrt(F_(1)^(2) + F_(2)^(2) + 2 F_(1) F_(2) (1//2))`
`= sqrt(3) F_(1) = sqrt(3) (Gm^(2))/(r^(2))` acting along `CD`
`[ :.F_(1) = F_(2) = (Gm^(2))/(r^(2))]`
Here, `CO = (2)/(3) CD = (2)/(3) AC sin 60^(@)`
`= (2)/(3) r xx sqrt(3)/(2) = (r)/(sqrt(3))`
When each body is describing a circular orbit with centre of orbit at `O`, the force `F` provides the required centripetal force. The redius of the circular orbit is `OC (= r//sqrt(3))`. If `upsilon` is the speed of the body in the circular orbit, then
`(m upsilon^(2))/((r//sqrt(3))) = (sqrt(3) Gm^(2))/(r^(2))` or `upsilon = sqrt((Gm)/(r))`