Question

Four particles having masses, m, 2m, 3m and 4m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre.

Answer 1

To calculate the gravitational force on 'm' due to other force,

Distance of the corner from the mid point of the square is half of the diagonal

So, $r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$
Now,Force due to the mass m placed at the corner A

${\overrightarrow{F}}_{OA}=\frac{\text{G.m.m}}{r^2}$

Force due to the mass 2m placed at the corner B

${\overrightarrow{F}}_{OB}=\frac{\text{G.m.2m}}{r^2}$

Force due to the mass 3m placed at the corner C

${\overrightarrow{F}}_{OC}=\frac{\text{G.m.3m}}{r^2}$

Force due to the mass 4m placed at the corner D
and ${\overrightarrow{F}}_{OD}=\frac{G.m.4m}{r^2}$

Now,
Resultant force $={\overrightarrow{F}}_{OA}+{\overrightarrow{F}}_{OB}+{\overrightarrow{F}}_{OC}+{\overrightarrow{F}}_{OD}$

Or,

We have already calculated the value of distance $r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$

$=\frac{2Gmm}{a^2}[-\frac{\overrightarrow{i}}{\sqrt{2}}+\frac{\overrightarrow{j}}{\sqrt{2}}]+\frac{4Gmm}{a^2}[\frac{\overrightarrow{i}}{\sqrt{2}}+\frac{\overrightarrow{j}}{\sqrt{2}}]+\frac{6Gmm}{a^2}[\frac{\overrightarrow{i}}{\sqrt{2}}-\frac{\overrightarrow{j}}{\sqrt{2}}]+\frac{8Gmm}{a^2}[\frac{-\overrightarrow{i}}{\sqrt{2}}-\frac{-\overrightarrow{j}}{\sqrt{2}}]$

After solving
$F=\frac{4\sqrt{2}G{m}^2}{a^2}$.

Hence, the force at the mass placed in the middle of the square is $F=\frac{4\sqrt{2}G{m}^2}{a^2}$.