Three particle 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 ?

Explanation

The force of attraction on body C due to bodies at A and B are
`F_(1)=(Gm^(2))/a^2 ` along `CA and F_(2) = (Gm^2)/a^2 ` along CB
The resultant force on the body at C is
`F=sqrt(F_1^2+F_2^2+2F_(1)F_(2)cos 60^@)=sqrt(3)F_(1)=sqrt3(Gm^2)/a^2`
`( :. F_(1)=F_(2)=(Gm^2)/a^2)`
The resultant force acting along CD, Here OC = `a/sqrt3`
When each body is describing a circular orbit with centre of orbit at O, the force F provides the required centripetal force. The radius of the circular orbit is OC = a/ `sqrt3` , If V is the speed of the body in circular orbit, then Centripetal force = resultant gravitational force
`(mV^2)/(a//sqrt3)=(sqrt3Gm^2)/a^2" or " V = sqrt((Gm)/a)`