JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Gravitation

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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) $\frac{( 1 + 2 2 ) G}{2}$ ✓ Correct
(B) $\frac{G}{2} (1 + 22)$
(C) $\frac{G}{2} (22 - 1)$
(D) $G (1 + 22)$

Answer: (A)

Solution

For a particle at one vertex, the net centripetal force comes from three other particles. Two adjacent particles are at distance

2 r

and one opposite particle is at distance

2 r

.

The centripetal force equation:

\frac{G m ^{2}}{2 r ^{2}} \cdot \frac{1}{2} + \frac{G m ^{2}}{2 r ^{2}} \cdot \frac{1}{2} + \frac{G m ^{2}}{4 r ^{2}} = \frac{m v ^{2}}{r}

Simplifying:

v^{2} = \frac{G m}{r} (\frac{1}{2 2} + \frac{1}{2 2} + \frac{1}{4}) = \frac{G m}{4 r} (22 + 1)

With

m = 1

kg and

r = 1

v = \frac{1}{2} G (1 + 22)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Gravitation

Solution

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