If two particles each of mass $^{'} m^{'}$ are placed at the two vertices of an equilateral triangle of side $^{'} a^{'}$ , then the resultant gravitational force on mass $m$ placed at the third vertex is

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Solution

$F_{1} = F_{2} =$
$∴ F_{n e t} = \sqrt{F^{2} + F^{2} + 2 F^{2} cos 60}$
$⟹ \sqrt{3} F$
$⟹ \sqrt{3} \frac{G m^{2}}{a^{2}}$ , vertically downward

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If two particles each of mass ′m′ are placed at the two vertices of an equilateral triangle of side ′a′ , then the resultant gravitational force on mass m placed at the third vertex is

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$⟹ \sqrt{3} \frac{G m^{2}}{a^{2}}$ , vertically downward