Three particles each of mass m, are located at the vertices of an equilateral triangle of side a. At what speed must they move if they all revolve under the influence of their gravitational force of attraction in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ?
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Explanation

Let us consider the force acting on A. The force between A and `B =F_(1)` and that between A and C is `F_(2)`

`F_(1) = F_(2) = ( Gm^(2))/( a^(2)) = ` say x
The resultant of `F_(1)` and `F_(2)` is F which acts along AO.
`F = sqrt( F_(1)^(2) + F_(2)^(2) + 2F_(1) F_(2) cos theta ) = sqrt( x^(2) + x^(2) + 2x^(2) cos 60^(@))`
`F = sqrt( 3) x = sqrt( 3) ( Gm^(2))/( a^(2))`
For the system to rotate about O the force F provides the necessary centripetal force.
`sqrt( 3) ( Gm^(2))/( a^(2)) = ( mv^(2))/( r ) ` where `r = AO = ( a)/( sqrt( 3))`
`v^(2) = sqrt( 3) ( Gmr)/( a^(2)) = sqrt( 3) ( Gma)/( a^(2) sqrt(3))`
`v = sqrt((Gm)/( a))`
The time period of the circular motion is T
`T = ( 2pi r )/( v ) = ( 2pi ( a //sqrt(3)))/( sqrt( Gm//a)) = 2pi sqrt(( a^(3))/(2Gm))`