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. $\frac{24m^{2}G}{a^2}$
• B. $\frac{6m^{2}G}{a^2}$
• C. $\frac{4\sqrt{2}G{m}^2}{a^2}$
• 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 $F_{PD}=\frac{G4mm}{x^2}=4F$

Hence the net force on $P$

${\vec{F}}_{net}={\vec{F}}_{PA}+{\vec{F}}_{PB}+{\vec{F}}_{PC}+{\vec{F}}_{PD}$

$=2\sqrt{2}F$

$\therefore {\vec{F}}_{net}=2\sqrt{2}\frac{Gmm}{x^2}$

$=2\sqrt{2}\frac{G{m}^2}{(a/\sqrt{2}{)}^2}$

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

$=\frac{4\sqrt{2}G{m}^2}{a^2}$