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Three equal masses m are placed at the three vertices of an equilateral triangle of side a. The gravitational force exerted by this system on another particle of mass m placed at the mid-point of a side

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\frac{{4Gm}^{2}}{3 a^{2}}

\frac{{Gm}^{2}}{3 a^{2}}

\frac{{Gm}^{2}}{a^{2}}

answer is B.

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Detailed Solution

$F_{1} {and F}_{2}$ act equally in opposite direction. So they cancel each other.

Remaining force $F_{3} = G \frac{m \times m}{d^{2}}$

$F_{3} = \frac{{Gm}^{2}}{\frac{3 a^{2}}{4}}$ $d = \frac{\sqrt{3}}{2} a$

$F_{3} = \frac{4 {Gm}^{2}}{3 a^{2}}$

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