four particles each of mass m placed at the four corner of square of edge a,move with velocity v,along a circle,which circumscribes the square,under the influence of their mutul gravitational force.find the speed v

The gravtiatation force on the a single particle is Gm²/a² + Gm²/a² + Gm²/2a² = 5Gm²/2a²

= 5Gm²/2a²= root(2)mv²/a (Centripetal force)

= (5Gm/2a(root(2)))^1/2

Sorry for my mistake,

The gravtiatation force on the a single particle is Gm²root(2)/a² + Gm²/2a² = Gm²(2root(2)+1)/2a²

= Gm²(2root(2)+1)/2a²= root(2)mv²/a (Centripetal force)

= (Gm(1+1/2(root(2))/a)^1/2

Daer Aashish,

The gravtiatation force on the a single particle is Gm²root(2)/a² + Gm²/2a² = Gm²(2root(2)+1)/2a²

= Gm²(2root(2)+1)/2a²= root(2)mv²/a (Centripetal force)

= (Gm(1+1/2(root(2))/a)^1/2

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