Three bodies each of mass ‘m’ are placed at the three corners of a square of side ‘a’. The gravitational force on unit mass kept at the fourth corner is

${\text{m}}_{\text{1}}={\text{m}}_2={\text{m}}_3=\text{m}$

${\text{m}}_4=1$

${\text{F}}_{\text{1}}=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}{\text{ F}}_{\text{2}}=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}$

Resultant force of   ${\text{F}}_{\text{1}}{\text{ and F}}_{\text{2}}$

${\text{F}}_{\text{R}}=\sqrt{{\text{F}}_{\text{1}}{}^2+{\text{F}}_{\text{2}}{}^2}$

$=\sqrt{{\left(\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\right)}^2+{\left(\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\right)}^2}$

$=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\sqrt{2}$

Total net force   $\text{F}={\text{F}}_{\text{R}}+{\text{F}}_3$

$=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\sqrt{2}+\frac{\text{Gm}}{{\left(\sqrt{2}\text{a}\right)}^{\text{2}}}$

$=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\sqrt{2}+\frac{\text{Gm}}{{\text{2a}}^{\text{2}}}$

$=\frac{\text{Gm}}{{\text{a}}^{\text{2}}}\left[\sqrt{2}+\frac{1}{2}\right]$