Four identical particles of mass M are located at the corners of a...

Gravitational force on each mass 'M'
$= \sqrt{2}\dfrac{G{M}^{2}}{{a}^{2}} + \dfrac{G{M}^{2}}{{\left(\sqrt{2}a\right)}^{2}} = \left(\sqrt{2} + \dfrac{1}{2}\right)\dfrac{G{M}^{2}}{{a}^{2}}$
This force will provide required centripetal force
\left(\sqrt{2} + \dfrac{1}{2}\right)\dfrac{G{M}^{2}}{{a}^{2}} = \dfrac{M{v}^{2}}{( a / \sqrt{2} )}$<br />$v = \sqrt{\left(1 + \dfrac{1}{2\sqrt{2}}\right)\dfrac{GM}{a}} = 1.16\sqrt{\dfrac{GM}{a}}