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Three bodies of each mass m are placed at the three corners of the square of side ‘a’. The gravitational force on unit mass kept at the fourth corner is

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$\frac{G m}{3 a^{2}}$

$\frac{G m}{\sqrt{3} a^{2}}$

$\frac{3 G m}{a^{2}}$

$\frac{G m}{a^{2}} (\sqrt{2} + \frac{1}{2})$

answer is D.

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Detailed Solution

$F_{R} = F^{'} + F_{3} F_{1} = \frac{Gm}{a^{2}}$

$\begin{array}{l} F^{'} = \sqrt{F_{1}^{2} + F_{2}^{2}} F_{2} = \frac{Gm}{a^{2}} \\ = \frac{\sqrt{2} Gm}{a^{2}} F_{R} = \frac{\sqrt{2} Gm}{a^{2}} + \frac{Gm}{2 a^{2}} \Rightarrow \frac{Gm}{a^{2}} (\sqrt{2} + \frac{1}{2}) \end{array}$

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