Three masses, each equal to $M$ are placed at the three corners of a square of side $a$ . Calculate the force of attraction on unit mass placed at the fourth corner.

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$\frac{G M}{a^{2}} [\sqrt{2} + \frac{1}{2}]$

$\frac{G M}{a^{2}} [\sqrt{2} + \frac{1}{4}]$

$\frac{G M}{a^{2}} [\sqrt{2} + \frac{1}{5}]$

$\frac{G M}{a^{2}} [\sqrt{2} + \frac{1}{8}]$

answer is A.

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Detailed Solution

Force on $m = 1$ due to masses at corners $1$ and $3$ are $\vec{F_{1}}$ and $\vec{F_{3}}$ with $F_{1} = F_{3} = \frac{G M}{a^{2}}$ resultant of $\vec{F_{1}}$ and $\vec{F_{3}}$ is $F_{r} = \sqrt{2} \frac{G M}{a^{2}}$ and its direction is along the diagonal i.e. toward corner $2$

Force on $m$ due to mass $M$ at $2$ is $F_{2} = \frac{G M}{{(\sqrt{2} a)}^{2}} = \frac{G M}{2 a^{2}}; F_{r}$ and $F_{2}$ act in the same direction.
Resultant of these two is the net force:
$F_{n e t} = \frac{\sqrt{2} G M}{a^{2}} + \frac{G M}{2 a^{2}} = \frac{G M}{a^{2}} [\sqrt{2} + \frac{1}{2}];$ it is directed along the diagonal as shown in the figure.