Question

Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is (A) $\sqrt{\frac{G M}{R}}$ (B) $\sqrt{2 \sqrt{2} \frac{G M}{R}}$ (C) $\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$ (D) $\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$

Name: Problem 80, Chapter 7: A Complete Resource Book in Physics for JEE Main
Uploaded: 2021-09-25T09:52:36-08:00
Duration: 1 min 43 s
Channel: Narayan Hari
Description: Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is (A) √((G M)/(R)) (B) √(2 √(2) (G M)/(R)) (C) √((G M)/(R)(1+2 √(2))) (D) (1)/(2) √((G M)/(R)(1+2 √(2)))

   Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is
(A) $\sqrt{\frac{G M}{R}}$
(B) $\sqrt{2 \sqrt{2} \frac{G M}{R}}$
(C) $\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$
(D) $\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$

A Complete Resource Book in Physics for JEE Main

Sanjeev Kumar 1st Edition

Chapter 7, Problem 80

Instant Answer

Solved by Expert Narayan Hari

09/25/2021

Step 1

We have four particles, each of mass $M$, moving along a circle of radius $R$. They are equidistant from each other, forming a square. The side of the square can be denoted as $x$. Since the particles form a square, we can use Pythagoras theorem to find $x$ in Show more…

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Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is (A) $\sqrt{\frac{G M}{R}}$ (B) $\sqrt{2 \sqrt{2} \frac{G M}{R}}$ (C) $\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$ (D) $\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}$

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Key Concepts

Newton's Law of Universal Gravitation

This law states that every two masses attract one another with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is crucial in calculating the gravitational forces present in a system of masses.

Centripetal Force in Circular Motion

To keep a body moving in a circle at a constant speed, its net inward force must equal the centripetal force needed to change the direction of its velocity. In gravitational systems, components of the gravitational attractions often sum to provide this necessary centripetal force.

Symmetry in Gravitational Systems

When masses are arranged symmetrically, such as being equally spaced on a circle, their individual gravitational forces can be combined using vector addition. This symmetry simplifies the analysis by reducing the complexity of the net force calculations needed to determine the dynamics of the system.

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