Sigma Percentile
JEE Main 2021, 24 Feb Shift-I
LEVELJEE Advanced

Animated Solution for Physics - Gravitation: Four identical particles of equal masses made to move along the circumference of a circle of radius under the action of their own mutual gravitational attraction. The speed of each particle will be

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Visualized Solution

  • Four identical particles of mass are moving in a circle of radius .

  • The mutual gravitational attraction provides the necessary centripetal force.
  • F_{net} = F_C = \frac{mv^2}{R}

  • Let's analyze the forces acting on one particle.
  • It experiences force from the diametrically opposite particle and from the two adjacent particles.

  • Distance to the opposite particle is .
  • F_1 = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}

  • Distance to the adjacent particles is .
  • F_2 = \frac{Gmm}{(\sqrt{2}R)^2} = \frac{Gm^2}{2R^2}

  • The components of perpendicular to the center cancel out.
  • Net force towards the center:
  • F_{net} = F_1 + 2F_2 \cos 45^\circ

  • F_{net} = \frac{Gm^2}{4R^2} + 2 \left(\frac{Gm^2}{2R^2}\right) \frac{1}{\sqrt{2}}
  • F_{net} = \frac{Gm^2}{R^2} \left[ \frac{1}{4} + \frac{1}{\sqrt{2}} \right]

  • \frac{mv^2}{R} = \frac{Gm^2}{R^2} \left[ \frac{1}{4} + \frac{1}{\sqrt{2}} \right]
  • v^2 = \frac{Gm}{R} \left[ \frac{1+2\sqrt{2}}{4} \right]

  • v = \frac{1}{2} \sqrt{\frac{Gm}{R} (1+2\sqrt{2})}

  • Given and :
  • v = \frac{\sqrt{G(1+2\sqrt{2})}}{2}
  • This matches option (a).

The Sigma Insight: Kepler's Law and Universal Law of Gravitation

Solution Diagram

Visualizing the Cosmic Dance

Imagine a beautifully symmetric cosmic ballet: four identical particles, each with a mass , are placed at equal distances along the circumference of a circle of radius . They aren't attached to any physical string or track. Instead, they are bound together by the invisible, relentless pull of their own mutual gravitational attraction.
Because the setup is perfectly symmetric, every particle experiences the exact same environment. As they pull on each other, they begin to move. To maintain this perfect square formation while moving, they must travel along the circular path. This means the net gravitational force acting on any single particle must act exactly towards the center of the circle, providing the necessary centripetal force for circular motion.

The Gravitational Tug-of-War

Let's zoom in and focus on just one of these particles. It is being pulled by three other particles.
First, there is the particle sitting diametrically opposite to it. The distance between them is simply the diameter of the circle, . According to Newton's Law of Universal Gravitation, the force exerted by this opposite particle is:
Next, we have the two adjacent particles. If we draw lines from the center of the circle to our chosen particle and one of its adjacent neighbors, we form a right-angled triangle with two sides of length . The distance between these adjacent particles is the hypotenuse, which is . The force exerted by each of these adjacent particles is:

The Centripetal Connection

Now, we need to find the net force directed towards the center of the circle. The force is already pointing straight at the center. However, the two forces are pulling at an angle. Because the particles form a square, the angle between the force and the line pointing to the center is exactly .
The components of the two forces that are perpendicular to the center line are equal and opposite, so they perfectly cancel each other out. The components pointing towards the center, however, add up. Therefore, the total net force towards the center is:
Substituting our expressions for and , and knowing that , we get:
This net gravitational force is the sole reason the particle is moving in a circle. Therefore, it must equal the required centripetal force, :

The Final Reveal

We are now in the home stretch! Let's solve for the orbital speed . We can cancel one mass and one radius from both sides of the equation:
Taking the square root of both sides gives us the general formula for the speed of any particle in this configuration:
The problem gives us specific values: the mass and the radius . Substituting these into our master equation, we find the final speed:
This elegant result perfectly matches option (a). It's a beautiful demonstration of how symmetric gravitational forces can perfectly orchestrate uniform circular motion!