Question

# 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,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=2πrvT=2πa√3√aGmT=2π√a33Gm

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