Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be

If two particles of mass m are placed x distance apart then force of attraction $\frac{\mathrm{Gmm}}{{\mathrm{x}}^2} = \mathrm{F}\left(\mathrm{Let}\right)$

Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces

${\mathrm{F}}_{\mathrm{PA}} = \mathrm{force} \mathrm{at} \mathrm{point} \mathrm{P} \mathrm{due} \mathrm{to} \mathrm{particle} \mathrm{A} = \frac{\mathrm{Gmm}}{{\mathrm{x}}^2} = \mathrm{F}$

Similarly, ${\mathrm{F}}_{\mathrm{PB}} = \frac{\mathrm{G}2\mathrm{mm}}{{\mathrm{x}}^2} = 2\mathrm{F}, {\mathrm{F}}_{\mathrm{PC}} = \frac{\mathrm{G}3\mathrm{mm}}{{\mathrm{x}}^2} = 3 \mathrm{F}$ and ${\mathrm{F}}_{\mathrm{PD}} = \frac{\mathrm{G}4\mathrm{mm}}{{\mathrm{x}}^2} = 4\mathrm{F}$

Hence the net force on P

${\overrightarrow{\mathrm{F}}}_{\mathrm{net}} = {\overrightarrow{\mathrm{F}}}_{\mathrm{PA}}+{\overrightarrow{\mathrm{F}}}_{\mathrm{PB}}+{\overrightarrow{\mathrm{F}}}_{\mathrm{PC}}+{\overrightarrow{\mathrm{F}}}_{PD} = 2\sqrt{2} \mathrm{F}$

$\therefore {\overrightarrow{\mathrm{F}}}_{\mathrm{net}} = 2\sqrt{2}\frac{\mathrm{Gmm}}{{\mathrm{x}}^2} = 2\sqrt{2}\frac{{\mathrm{Gm}}^2}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}$

     [x = $\frac{\mathrm{a}}{\sqrt{2}} = \mathrm{half} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{of} \mathrm{the} \mathrm{square}$]

   = $\frac{4\sqrt{2}{\mathrm{Gm}}^2}{{\mathrm{a}}^2}$