Consider the movement of three bodies along a circle centered at as shown in the diagram alongside.
Gravitational force between A and B FBA=
Gm2
(AB)2
Gravitational force between B and C FBC=
Gm2
(BC)2
From geometry of the figure, we can write BO‌cos‌30°+CO‌cos‌30°=BC ⇒AB=BC=CA=2R‌cos‌30°=√3R Net force on B, FB=FBA‌cos‌30°+FBC‌cos‌30° =
2Gm2
(AB)2
×
√3
2
=
2Gm2
(√3R)2
×
√3
2
=
2Gm2
3R2
×
√3
2
This force will provide centripetal force for circular motion of B FB=