Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :

Question

${1 \over 2}\sqrt {{{GM} \over {R(2\sqrt 2 + 1)}}} $

${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 + 1)} $

${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 - 1)} $

$\sqrt {{{GM} \over R}} $

Solution

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