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=‌ (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
â–³OBP ,
cos‌30∘‌=‌‌‌(∵BP=‌) ‌‌=‌ ⇒‌‌R=‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)Now, since
F′=F‌centripetal ‌i.e.
‌‌F′=Fc ⇒‌‌√3F=‌‌‌ [from Eq. (i)
][∵Fc=‌] ⇒√3‌=‌Now, from Eq. (ii),
√3‌=‌√3 ⇒‌‌v2=‌ ⇒‌‌v=√‌