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Offline Centres

Two particles each of mass ‘m’ are placed at A and C are such AB=BC=L. The gravitational force on the third particle placed at D at a distance L on the perpendicular bisector of the line AC is

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$\frac{{Gm}^{2}}{L^{2}} along BD$

$\frac{{Gm}^{2}}{\sqrt{2} L^{2}} along DB$

$\frac{{Gm}^{2}}{L^{2}} along AC$

$\frac{{Gm}^{2}}{L^{2}} along BD$

answer is B.

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Detailed Solution

$F^{1} = \sqrt{2} F, F = \frac{{Gm}^{2}}{2 L^{2}}$

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