First slide

Universal law of gravitation

Question

Two particles each of mass m are placed at points A and C such that AB =BC =L as shown in fig.

The gravitational force on a third particle placed at D at a distance I metre on the perpendicular bisector of the line AC, is

Moderate

A

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

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

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

none of these

Solution

Here $F = \frac{{Gm}^{2}}{(\sqrt{2} L)^{2}} = \frac{{Gm}^{2}}{2 L^{2}}$
The resultant force on mass m placed at point D is F' = 2 F cos 45^o long DB
$= 2 (\frac{{Gm}^{2}}{2 L^{2}}) \times \frac{1}{\sqrt{2}} = \frac{{Gm}^{2}}{\sqrt{2} L^{2}} along DB$

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