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

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

$$A) \space\sqrt{ (1+2\sqrt2)G\space \over2}$$

$$B) \sqrt{ G(1+2\sqrt2)}$$

$$C) \sqrt { {G\over2} (2\sqrt2-1)}$$

$$D) \sqrt {{G\over2} (1+2\sqrt2)}$$

1 Answer

verified

The correct option is A)$$\space\sqrt{ (1+2\sqrt2)G\space \over2}$$

explaination::

By resolving force F2, we get

F1 + F2 cos 45° + F2 cos 45° = Fc

F1 + 2F2 cos 45° = Fc

Fc = centripetal force = MV2 / R

$${ GM^2\over(2R)^2}={ {2GM^2 \over(2R)^2} \cos 45^o}={ MV^2\over R}$$

$${ GM^2\over 4R^2}+{ 2GM^2\over2\sqrt2R^2}={ MV^2\over R}$$

$${GM^2\over 4R^2}+{GM\over\sqrt {2R}}=V^2$$

$$V ={\sqrt{{ GM\over 4R}+{GM\over\sqrt{2R}}}}$$

$$V={1\over2}{ \sqrt{{ GM\over R} [ {1+2\sqrt2}]}}$$

now, mass=1kg and radius=1m

$$⇒ V={1\over2}{\sqrt{ G(1+2\sqrt2)}}$$