Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side a. At what speed must they move if they all revolve under the influence of one another's gravitation in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ?

Three identical bodies, each of mass m, are located at the vertices of

1.	Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side a. At what speed must they move if they all revolve under the influence of one another's gravitation in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ?
Answer» Solution :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^(@))` The resultant force acting along CD, Here `OC = (a)/(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 radius of the circular orbit is `OC = a//sqrt(3)`, If V is the speed of the body in circular orbit, then Centripetal force = resultant GRAVITATIONAL force `(mV^(2))/(a//sqrt(3)) = (sqrt(3)Gm^(2))/(a^(2))` or `V = sqrt((Gm)/(a))`