Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is :

Question

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

$\sqrt {2\sqrt 2 {{GM} \over R}} $

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

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

Solution

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