Three masses, each equal to M, are placed at the three corners of a square of side a. The force of attraction on unit mass at the fourth corner will be

${\mathrm{F}}_{1 }={\mathrm{F}}_2= \frac{\mathrm{GM}}{{\mathrm{a}}^2}$

Resultant of ${\mathrm{F}}_1 \mathrm{and} {\mathrm{F}}_2 \mathrm{is} \sqrt{2}\frac{\mathrm{GM}}{{\mathrm{a}}^2}$

Now, ${\mathrm{F}}_3 = \frac{\mathrm{GM}}{{\left(\sqrt{2}\mathrm{a}\right)}^2} = \frac{\mathrm{GM}}{2{\mathrm{a}}^2}$

Now, $\frac{\sqrt{2} \mathrm{GM}}{{\mathrm{a}}^2} \mathrm{and} \frac{\mathrm{GM}}{2{\mathrm{a}}^2}$ act in the same direction.

Their resultant is $=\frac{\sqrt{2} \mathrm{GM}}{{\mathrm{a}}^2}+\frac{\mathrm{GM}}{2{\mathrm{a}}^2} = \frac{\mathrm{GM}}{{\mathrm{a}}^2}[\sqrt{2}+\frac{1}{2}]$