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 :
2π√a33Gm
2π√3a4Gm
2π√a4Gm
2π√a23Gm
A
2π√a4Gm
B
2π√a23Gm
C
2π√a33Gm
D
2π√3a4Gm
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Solution
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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
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