Question

three identical point of masses are fixed symmetrical on the periphery of a circle obatin the resultant gravitational force on any point masses M at the centre of the circle extend this idea to mora than three identical masses symmetrically located on the periphery how far can you extend tbis concepts​

Answer 1

Explanation:

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}

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