Question

Three identical bodies of equal mass m each are moving along a circle of radius r under the action of their mutual gravitational attraction. the speed of each body is :-

Answer 1

First understand what is mutual gravitational force ?
When two massive bodies have equal mass then they attract with equal gravitational force , this gravitational force is called as mutual gravitational force.
Because here all masses are equal e.g., M so, gravitational forces act on each are equal in magnitude.

Better understanding, I attached a rough daigram
Here you see side length of triangle = 2rcos30° = 2r × √3/2 = √3r
Now, force due C on B , F₁ = GMM/(√3r)²
F₁ = GM²/3r²
Similarly force due A on B , F₂ = GMM/(√3r)²
F₂ = GM²/3r²
Here it is clear that , F₁ = F₂ = GM²/3r² = F{assume}
angle between F₁ and F₂ = 60°
∴ Fnet = √{F₁² + F₂² + 2F₁F₂cos60°}
= √{F² + F² + 2F² × 1/2} = √{F² + F² + F²}
= √3F
= √3GM²/3r² = GM²/√3r²
∴ Fnet = GM²/√3r²
But you see net force is acting along centre of circle
so, Fnet = centripetal force
GM²/√3r² = Mv²/r
v² = GM/√3r
v = √{GM/√3r}

Hence, speed of each body = √{GM/√3r}

Answer 2

Explanation:

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