Four identical particles of equal masses 1kg 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 :

Options

A. `sqrt(5/2 (1+2sqrt2))`
B. `sqrt(G (1 + 2 sqrt2))`
C. `sqrt( G/2 (2 sqrt2 - 1))`
D. `sqrt(( (1+2 sqrt2)G)/(2))`

Answer

To solve the problem of finding the speed of each particle moving along the circumference of a circle under the action of their own mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the System
We have four identical particles, each with a mass \( m = 1 \, \text{kg} \), moving in a circle of radius \( r = 1 \, \text{m} \). The particles are influenced by their mutual gravitational attraction.

### Step 2: Calculate the Gravitational Force
The gravitational force \( F \) between any two particles can be calculated using Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{d^2} \]
where \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)), \( m_1 \) and \( m_2 \) are the masses of the particles, and \( d \) is the distance between the particles.

### Step 3: Determine Distances Between Particles
- The distance between any two adjacent particles (e.g., particle 1 and particle 2) is \( d = \sqrt{(1^2 + 1^2)} = \sqrt{2} \, \text{m} \).
- The distance between opposite particles (e.g., particle 1 and particle 3) is \( d = 2 \, \text{m} \).

### Step 4: Calculate the Forces Acting on One Particle
For one particle (let's say particle 1), the forces due to the other three particles are:
1. Force due to particle 2:
\[ F_{12} = \frac{G m^2}{(\sqrt{2})^2} = \frac{G m^2}{2} \]
2. Force due to particle 3:
\[ F_{13} = \frac{G m^2}{(2)^2} = \frac{G m^2}{4} \]
3. Force due to particle 4:
\[ F_{14} = \frac{G m^2}{(\sqrt{2})^2} = \frac{G m^2}{2} \]

### Step 5: Calculate the Net Gravitational Force
The net gravitational force acting towards the center of the circle can be calculated by vector addition of the forces. The forces \( F_{12} \) and \( F_{14} \) will have components that contribute to the net force towards the center.

### Step 6: Set Up the Centripetal Force Equation
The net gravitational force must equal the centripetal force required to keep the particle in circular motion:
\[ F_{\text{net}} = m \frac{v^2}{r} \]
where \( v \) is the speed of the particle.

### Step 7: Solve for Speed \( v \)
By equating the net gravitational force to the centripetal force, we can solve for \( v \):
\[ \frac{G m^2}{2} + \frac{G m^2}{4} + \frac{G m^2}{2} = m \frac{v^2}{1} \]
This simplifies to:
\[ \frac{5G m^2}{4} = m v^2 \]
Cancelling \( m \) from both sides gives:
\[ v^2 = \frac{5G m}{4} \]
Taking the square root gives:
\[ v = \sqrt{\frac{5G m}{4}} \]
Substituting \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) and \( m = 1 \, \text{kg} \):
\[ v = \sqrt{\frac{5 \times 6.674 \times 10^{-11} \times 1}{4}} = \sqrt{\frac{3.337 \times 10^{-10}}{4}} = \sqrt{8.3425 \times 10^{-11}} \approx 9.13 \times 10^{-6} \, \text{m/s} \]

### Final Answer
The speed of each particle will be approximately \( 9.13 \times 10^{-6} \, \text{m/s} \).

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