Four identical particles of mass \(M\) are located at the corners of a square of side \(‘a’\). What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?

\({F_c} = \frac{{G{M^2}}}{{2{a^2}}} + \frac{{G{M^2}}}{{{a^2}}}\sqrt 2 \)

\( = \frac{{G{M^2}}}{{{a^2}}}\left( {\frac{1}{2} + \sqrt 2 } \right)\)

This force will act as centripetal force.

Diatance of particle from center of circle is \(\frac{a}{{\sqrt 2 }}.\)

\(r = \frac{a}{{\sqrt 2 }},{F_c} = \frac{{m{v^2}}}{r}\)

\(\frac{{m{v^2}}}{{\frac{a}{{\sqrt 2 }}}} = \frac{{G{M^2}}}{{{a^2}}}\left( {\frac{1}{2} + \sqrt 2 } \right)\)

\({v^2} = \frac{{GM}}{a}\left( {\frac{1}{{2\sqrt 2 }} + 1} \right)\)

\({v^2} = \frac{{GM}}{a}\left( {1.35} \right)\,\,;\,\,v = 1.16\sqrt {\frac{{GM}}{a}} \)