Question

Four particles of masses $m,2m,3m$ and $4m$ are placed at the corners of a square of side length $a$. The gravitational force on a particle of mass $m$ placed at the centre of the square is

• A. $\frac{4\sqrt{2}G{m}^2}{a^2}$
• B. $\frac{3\sqrt{2}G{m}^2}{a^2}$
• C. $\frac{2\sqrt{2}G{m}^2}{a^2}$
• D. $\frac{\sqrt{2}G{m}^2}{a^2}$

Answer 1

The correct option is A $\frac{4\sqrt{2}G{m}^2}{a^2}$

Force is given by,
$F=\frac{G{m}_1{m}_2}{r^2}$
Force of attraction between $m$ and $m$ is given by,
$F_1=\frac{G\left(m\right)\left(m\right)}{(\frac{a}{\sqrt{2}}{)}^2}=\frac{2G{m}^2}{a^2}$
Force of attraction between $m$ and $2m$ is given by,
$F_2=\frac{G\left(m\right)\left(2m\right)}{(\frac{a}{\sqrt{2}}{)}^2}=\frac{4G{m}^2}{a^2}$
Force of attraction between $m$ and $3m$ is given by,
$F_3=\frac{G\left(m\right)\left(3m\right)}{(\frac{a}{\sqrt{2}}{)}^2}=\frac{6G{m}^2}{a^2}$
Force of attraction between $m$ and $4m$ is given by,
$F_3=\frac{G\left(m\right)\left(4m\right)}{(\frac{a}{\sqrt{2}}{)}^2}=\frac{8G{m}^2}{a^2}$
The resultant of $F_4$ and $F_2$ is given by,
$F_5=F_4-F_2=\frac{4G{m}^2}{a^2}$
The resultant of $F_3$ and $F_1$ is given by,
$F_6=F_3-F_1=\frac{4G{m}^2}{a^2}$
Net resultant force is given by,
$F=\sqrt{F_5^2+F_6^2}$
$F=\sqrt{\frac{4G{m}^2}{a^2}+\frac{4G{m}^2}{a^2}}=4\sqrt{2}\frac{G{m}^2}{a^2}$