First slide

Universal law of gravitation

Question

Four particles of equal mass M move along a circle of radius .R under the action of their mutual gravitational attraction [Fig.]. The speed of each particle is

Difficult

A

$\frac{GM}{R}$
B

$\sqrt{(2 \sqrt{2} GM / R)}$
C

$\sqrt{[(2 \sqrt{2} + 1) GM / R)]}$
D

$\sqrt{[(2 \sqrt{2} + 1) GM / 4 R]}$

Solution

The gravitational force on each particle due to other three particles provides the necessary centripetal force.
$\begin{array}{l} \frac{{GM}^{2}}{(\sqrt{2} R)^{2}} \cos 45^{\circ} + \frac{{GM}^{2}}{(\sqrt{2} R)^{2}} \cos 45^{\circ} + \frac{{GM}^{2}}{(2 R)^{2}} = \frac{{Mv}^{2}}{R} \\ 2 [\frac{{GM}^{2}}{2 R^{2}} \times \frac{1}{\sqrt{2}}] + \frac{{GM}^{2}}{4 R^{2}} = \frac{{Mv}^{2}}{R} \\ \frac{\sqrt{2} GM}{2 R} + \frac{GM}{4 R} = v^{2} ∴ v = \sqrt{\frac{GM}{R} [\frac{2 \sqrt{2} + 1}{4}]} \end{array}$

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