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 this problem, we need to analyze the motion of the four identical particles moving along the circumference of a circle due to their mutual gravitational attraction.

**Step 1: Understanding the System**

Four identical particles, each with a mass of \(1 \, \text{kg}\), are located at the vertices of a square inscribed in a circle of radius \(1 \, \text{m}\). The gravitational force between the particles provides the centripetal force required for circular motion.

**Step 2: Calculate Effective Gravitational Force**

The arrangement of particles forms a square with a diagonal equal to the diameter of the circle i.e., \(2 \, \text{m}\).

The side of the square \(s\) can be found using Pythagoras' theorem in the diagonal:

\(s = \sqrt{2^2 - (2 \cdot 1^2)} = \sqrt{2} \, \text{m}\)

**Step 3: Compute Centripetal Force**

Each particle is subjected to the gravitational forces along the edges and the diagonal of the square.

The gravitational force between two masses \(m\) separated by distance \(r\) is:

\(F_g = \frac{G m^2}{r^2}\)

Where \(G\) is the gravitational constant.

**Step 4: Calculate Effective Forces**

• Force between two adjacent particles:
• Force between two diagonally opposite particles:

**Step 5: Net Centripetal Force and Speed Calculation**

Considering symmetry, the net force providing centripetal acceleration for each particle comes from all other three particles:

\(F_{\text{net}} = 2F_{\text{adj}} \cos(45^\circ) + F_{\text{diag}} = \frac{G}{\sqrt{2}}(1+2\sqrt{2})\)

Setting \(F_{\text{net}}\) as the centripetal force:

\(m v^2 = F_{\text{net}}\)

Solving for velocity \(v\):

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

This matches option D: \(\sqrt{\frac{(1+2 \sqrt{2}) G }{2}}\), which is the correct answer.