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. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{ 2 \sqrt 2 \frac{GM}{R}}$
• C. $\sqrt{\frac{GM}{R} (1+2 \sqrt2)}$
• D. $ \frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}$

Answer 1

The Correct Option is D

To solve this problem, we need to find the speed of each particle moving along a circle under the mutual gravitational attraction of other particles.

Concept and Approach:

Each of the four particles experiences a gravitational force due to the other three particles. To keep the particles in circular motion, the net centripetal force provided by these gravitational forces must equal the centripetal force required to keep a particle in circular motion. We'll calculate this step-by-step.

Step-by-step Solution:

1. Each particle of mass \( M \) is situated on a circle of radius \( R \). These particles form a square configuration on the circumference.
2. The distance between adjacent particles on this circle is the side of the square, which is given by the geometry of the circle as \(\sqrt{2}R\).
3. Gravitational force between two particles separated by distance \(d\) is given by: \( F = \frac{G M^2}{d^2} \).
4. Calculate the force between adjacent particles (distance \(\sqrt{2}R\)): \( F_1 = \frac{G M^2}{(2R)^2/4} = \frac{G M^2}{2R^2} \).
5. Calculate the force between diagonally opposite particles (distance 2R): \( F_2 = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2} \).
6. Since each particle is equidistant from its neighboring particles, let's calculate the net force acting towards the center of the circle.
7. The net gravitational force acting towards the center provides the required centripetal force. For this symmetrical arrangement, the effective distance is \(\frac{R}{\sqrt{2}}\). The net gravitational attraction taking geometry into account is: \( F_{\text{net}} = M v^2/R \), where \( v \) is the speed of the particle.
8. The total centripetal force required for each particle to move in the circle is given as: \( F_c = \frac{M v^2}{R} \).
9. Balancing the net gravitational force and the centripetal force, we get: \( M v^2/R = G M^2 / 2R^2 + G M^2 / 4R^2 \cdot 2 \sqrt{2}\).
10. Solving for \( v \), the speed of each particle, we simplify and find: \( v = \frac{1}{2} \sqrt{\frac{G M}{R} (1+2 \sqrt{2})} \).

Conclusion:

Thus, the speed of each particle is correctly given by the option: $ \frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}$.