Question

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. $\sqrt{\frac{ G }{2}(1+2 \sqrt{2})}$
• B. $\sqrt{ G (1+2 \sqrt{2})}$
• C. $\sqrt{\frac{ G }{2}(2 \sqrt{2}-1)}$
• D. $\sqrt{\frac{(1+2 \sqrt{2}) G }{2}}$

Answer 1

The Correct Option is D

To solve the problem of determining the speed of each particle in the system of four identical particles, we must analyze the gravitational forces acting on them and how this affects their motion around the circle as given:

1. Begin by considering the gravitational force between any two particles. Since all particles are identical and arranged symmetrically along a circle, the force on any particle will be due to the other three particles.
2. The gravitational force between two particles separated by a distance \( r \) is given by the formula: \(F = \frac{G m^2}{r^2}\) where \( G \) is the gravitational constant, \( m \) is the mass of each particle, and \( r \) is the distance between them.
3. For particles placed in a circle of radius \( R = 1 \, \text{m} \), the particles at each vertex of a square inscribed in the circle will be primarily influenced by their nearest neighbors.
4. The side length \( L \) of the square (i.e., the distance between nearest neighbors) will be: \(L = \sqrt{2} \, R = \sqrt{2} \times 1 = \sqrt{2} \)\)
5. The distance from a particle to its non-adjacent particle across the square is \( 2R = 2 \, \text{m} \).
6. The total force acting on a particle due to the others can be calculated by summing up the forces from these distances: 

\[F_{\text{total}} = 2 \times \frac{Gm^2}{L^2} + \frac{Gm^2}{(2R)^2} = 2 \times \frac{G \cdot 1^2}{(\sqrt{2})^2} + \frac{G \cdot 1^2}{2^2}\]

1. Simplifying these forces: 

\[F_{\text{total}} = 2 \cdot \frac{G}{2} + \frac{G}{4} = G + \frac{G}{4} = \frac{5G}{4}\]

1. To maintain circular motion, the centrifugal force due to particle motion must balance the inward gravitational pull: 

\[\frac{m v^2}{R} = F_{\text{total}}\]

1. where \( v \) is the speed of the particle.
2. Substitute \( m = 1 \, \text{kg} \), \( R = 1 \, \text{m} \), and compute: 

\[v^2 = \frac{5G}{4} \]

1. Thus, the speed \( v \) of each particle is: 

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

The correct answer is \(\sqrt{\frac{(1+2 \sqrt{2}) G }{2}}\), matching the given options.