Four identical particles of equal masses 1\ {kg} made to move along the circumference of a circle of radius 1\ {m} under the action of their own mutual gravitational attraction. The speed of each particle will be 24 Feb 2021 Shift1

Correct Answer is : 12G(1+22)1/2 { G (1+2{2}) }21G(1+22)

Correct Answer is : 12G(1+22)1/2 { G (1+2{2}) }21G(1+22)

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Gravitation Part 1

Section: Physics

Question : 35 of 100

Marks: +1, -0

1\ \mathrm{kg}

1\ \mathrm{m}

Solution:

Given,

m=1\ \mathrm{kg},\ R=1\ \mathrm{m}

We know that,

F = \frac{G m_{1} m_{2}}{r^{2}}

\because\; F_1 = \frac{G m m}{(2R)^2} = \frac{G m^2}{4 R^2}

\text{ and } F_2 = \frac{G m m}{(\sqrt{2} R)^2} = \frac{G m^2}{2 R^2}

Net force on one particle,

F_{\text{net}} = F_1 + F_2 \cos 45^\circ + F_2 \cos 45^\circ

= F_1 + 2 F_2 \cos 45^\circ

= \frac{G m^2}{4 R^2} + 2 \left( \frac{G m^2}{2 R^2} \right) \cdot \frac{1}{\sqrt{2}}

= \frac{G m^2}{4 R^2} + \frac{G m^2}{\sqrt{2} R^2}

= \frac{G m^2}{R^2} \left[ \frac{1}{4} + \frac{1}{\sqrt{2}} \right]

As the gravitational force provides the necessary centripetal force, so

F_{\text{net}} = F_C = \frac{m v^2}{R}

Here,

F_C =

centripetal force.

\Rightarrow \frac{G m^2}{R^2} \left[ \frac{1}{4} + \frac{1}{\sqrt{2}} \right] = \frac{m v^2}{R}

\Rightarrow v = \frac{1}{2} \sqrt{ \frac{G m}{R} (1+2\sqrt{2}) }

\Rightarrow v = \frac{1}{2} \sqrt{ G (1+2\sqrt{2}) }

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