Q.
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?
Solution:
Net force on particle towards centre of circle is $F_{C} = \frac{GM^{2}}{2a^{2}} + \frac{GM^{2}}{a^{2}} \sqrt{2}$
$ = \frac{GM^{2}}{a^{2}} \left( \frac{1}{2} + \sqrt{2}\right)$
This force will act as centripetal force. Distance of particle from centre of circle is $ \frac{a}{\sqrt{2}}$
$ r = \frac{a}{\sqrt{2}} , F_{C}= \frac{mv^{2}}{r} $
$ \frac{mv^{2}}{\frac{a}{\sqrt{2}}} = \frac{GM^{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}} $
