Question

# Three particles, each of mass m are situated at the vertices of an equilateral triangle of side a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle should move in a circle while maintaining the original mutual separation a. Then their time period of revolution is :

A
2πa23Gm
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B
2πa33Gm
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C
2π3a4Gm
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D
2πa4Gm
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Solution

## The correct option is A 2π√a33GmAs 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=√3Gm2a2As 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=cos30r=a√3Because of circular motion, the centripetal force is balanced by resultant gravitational force. If v is the velocity:mv2r=√3Gm2a2√3mv2a=√3Gm2a2v=√GmaTime period, T=2πrv=2πa√3√aGm=2π√a33Gm

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