Four particles, each of mass M and equidistant from each other,...

$\text{F} = {\text{F}}_{1} = {\text{F}}_{2}\text{=}\dfrac{{\text{GM}}^{2}}{2{\text{R}}^{2}}$

${\text{F}}_{3} = \dfrac{{\text{GM}}^{2}}{4 {\text{R}}^{2}}$

$∴$ F_Total towards the centre.

$= \sqrt{2}\text{F} + {\text{F}}_{3}$

$= \sqrt{2}\dfrac{{\text{GM}}^{2}}{{\text{2R}}^{2}} + \dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}$

$= \dfrac{{\text{GM}}^{2}}{4 {\text{R}}^{2}} \left[2 \sqrt{2} + 1\right]$

${\text{F}}_{\text{T}} {= \text{F}}_{\text{CP}}$

$\dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}\left[2\sqrt{2} + 1\right] = \dfrac{{\text{Mv}}^{2}}{\text{R}}$

$∴ \text{v} = \sqrt{\dfrac{\text{GM}}{4 \text{R}} \left[2 \sqrt{2} + 1\right]}$.