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Q.

Three particles, each of mass M are moving in a circle under their mutual gravitational forces such that they always form an equilateral triangle of side a while rotating. Speed of each particle is:

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a

2GMa

b

3GMa

c

GMa

d

GM2a

answer is D.

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Detailed Solution

Question Image

2Fcos⁡60∘2=Mv2a/3
where F=GM2a2
So, v=GMa

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