Question

Four particles of masses $m, 2m, 3m$ and $4m$ are kept in sequence at the comers 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

• A. $\frac{24m^2G}{a^2}$
• B. $\frac{6m^2G}{a^2}$
• C. $\frac{4\sqrt{2}Gm^2}{a^2}$
• D. Zero

Answer 1

Answer: (C)

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

Now according to problem particle of mass $m$ is placed at the centre $(P)$ of square. Then it will experience four forces
$F_{PA} =$ force at point $P$ due to particle $A = \frac{Gmm}{x^2} = F$
Similarly, $F_{PB} = \frac{G2mm}{x^2} = 2F,$
$F_{PC} = \frac{G3mm}{x^2} = 3F$
and $F_{PD} = \frac{G4mm}{x^2} = 4\,F$
Hence the net force on $P$
$\vec{F}_{net} =\vec{F}_{PA} + \vec{F}_{PB} + \vec{F}_{PC} + \vec{F}_{PD}$
$ = 2\sqrt{2}F$
$\therefore \vec{F}_{net} = 2\sqrt{2} \frac{Gmm}{x^2} $
$ = 2\sqrt{2} \frac{Gm^2}{(a/\sqrt{2})^2}$
$[ x = \frac{a}{\sqrt{2}} =$ half of the diagonal of the square]
$ = \frac{4\sqrt{2}Gm^2}{a^2}$