Three particles, each of mass M, situated at the vertices of an equilateral triangle of side length ' l '. 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 separation ' l '. The initial speed that should be given to each particle is
The situation given in the question is depicted below
Let the speed of particle is v for circular motion. Clearly the gravitational force acting all the 3 masses will be same due to symmetry and let it be F . Now; considering the mass M placed at point B . Since, it is performing circular motion in radius say R , hence the net gravitational pull on it due to the other two masses are providing the required centripetal force. Now, net force on M at B, F′=√F2+F2+2F2‌cos‌60∘ ⇒‌‌F′=√3F‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) and ‌‌F=‌
GM2
l2
(gravitational force of Newton) Also, by symmetry of forces the F′ will be acting along the bisector of angle at point B i.e. at 30∘ . from each force F. Now, in △OBP , cos‌30∘‌=‌
‌
l
2
R
‌‌(∵BP=‌
BC
2
) ‌
√3
2
‌=‌
l
2R
⇒‌‌R=‌
l
√3
‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) Now, since F′=F‌centripetal ‌ i.e. ‌‌F′=Fc ⇒‌‌√3F=‌