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Three particles, each of mass M, are situated at the vertices of an equilateral triangle of side 'a. Theonly forces acting on the particles are their mutualgravitational forces. It is desired that each particlemoves in a circle while maintaining their original separation 'a'. The initial velocity that should be given to each particle and time period of circular motion are respectively.

A
2GMa,2Πa33GM
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B
GMa,12Π3GMa3
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C
GMa,2Πa33GM
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D
GM3a,2Πa33GM
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Solution

The correct option is A GMa,2Πa33GM
As per the figure on the left hand side.
Resultant Gravitational force acting on each particle is:
F=(Gm2a2)2+(Gm2a2)2+2(Gm2a2)(Gm2a2)cos60=3Gm2a2

As per figure on right side, as the particles are expected to move in a circle of radius r (say) while maintaining original separation a,
a/2r=cos30
r=a3

Because of circular motion, the centripetal force is balanced by resultant gravitational force.
If v is the velocity:
mv2r=3Gm2a2

3mv2a=3Gm2a2
v=Gma
Time period T=2πrv

T=2πa3aGm

T=2πa33Gm

624266_282161_ans.png

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