Four identical particles of mass $M$ are located at the corners of a square of side ${}^{\prime }{a}^{\prime }$. 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?

$\text{Step1:Draw diagram and find radius of circular orbit}$

$(Hypotaneous{)}^2=a^2+a^2$

$Hypotaneous=\sqrt{2}$

$Diameter=\sqrt{2}$

$\text{Radius}=\frac{\text{Diameter}}{\sqrt{2}}$

$\text{Step2:Net force on particle towarda centre of circle}$

Both $F=\frac{G{M}^2}{a^2}$ are at $90^0$ so their resultant is $\frac{\sqrt{2}G{M}^2}{a^2}$ and act at angle bisector.

Net force on particle towards centre of circle is $F_C=\frac{G{M}^2}{2a^2}+\frac{G{M}^2}{a^2}\sqrt{2}$
$=\frac{G{M}^2}{a^2}\left(\frac{1}{2}+\sqrt{2}\right)$

$\text{Step3:Calculation of speed}$
This resultant  force will act as centripetal force. Distance of particle from centre of circle is $\frac{a}{\sqrt{2}}$.
$r=\frac{a}{\sqrt{2}},F_C=\frac{m{v}^2}{r}$
$\frac{m{v}^2}{\frac{a}{\sqrt{2}}}=\frac{G{M}^2}{a^2}\left(\frac{1}{2}+\sqrt{2}\right)$
$v^2=\frac{GM}{a}\left(\frac{1}{2\sqrt{2}}+1\right)$
$v^2=\frac{GM}{a}(1.35)$
$v=1.16\sqrt{\frac{GM}{a}}$.