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:

Options

A. `sqrt((GM)/R (1+2sqrt2))`
B. `1/2sqrt((GM)/R (1+2sqrt2))`
C. `sqrt((GM)/2)`
D. `sqrt(2sqrt(2)(GM)/2)`

Explanation

To find the speed of each particle moving in a circular path under mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the Configuration
We have four particles, each of mass \( M \), arranged at the corners of a square (equidistant from each other) and moving in a circular path of radius \( R \). We need to analyze the forces acting on one of these particles.

### Step 2: Calculate the Gravitational Forces
For any one particle, we need to consider the gravitational forces exerted by the other three particles. The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
In our case, the distance between any two adjacent particles is \( R \), and the distance between diagonally opposite particles is \( R\sqrt{2} \).

### Step 3: Determine the Forces Acting on One Particle
Let’s denote the forces acting on the chosen particle due to the other three particles as \( F_1 \), \( F_2 \), and \( F_3 \):
- \( F_1 \) and \( F_3 \) are the forces from the two adjacent particles, and they act along the lines connecting the particles.
- \( F_2 \) is the force from the diagonally opposite particle.

Calculating these forces:
- \( F_1 = F_3 = \frac{G M^2}{R^2} \)
- \( F_2 = \frac{G M^2}{(R\sqrt{2})^2} = \frac{G M^2}{2R^2} \)

### Step 4: Resolve the Forces
Due to symmetry, the horizontal components of \( F_1 \) and \( F_3 \) will add up, while their vertical components will cancel each other. The net force acting on the particle can be calculated by resolving the forces:
- The net force from \( F_1 \) and \( F_3 \) in the horizontal direction is:
\[ F_{horizontal} = 2 \cdot F_1 \cdot \cos(45^\circ) = 2 \cdot \frac{G M^2}{R^2} \cdot \frac{1}{\sqrt{2}} = \frac{G M^2 \sqrt{2}}{R^2} \]
- The vertical component from \( F_2 \) is:
\[ F_{vertical} = F_2 = \frac{G M^2}{2R^2} \]

### Step 5: Calculate the Net Force
The total net force \( F_{net} \) acting on the particle is the vector sum of the horizontal and vertical components:
\[ F_{net} = \sqrt{(F_{horizontal})^2 + (F_{vertical})^2} \]
Substituting the values:
\[ F_{net} = \sqrt{\left(\frac{G M^2 \sqrt{2}}{R^2}\right)^2 + \left(\frac{G M^2}{2R^2}\right)^2} \]

### Step 6: Relate the Net Force to Centripetal Force
The net gravitational force must provide the necessary centripetal force for circular motion:
\[ F_{net} = \frac{M v^2}{R} \]
Setting the two expressions for force equal gives:
\[ \frac{M v^2}{R} = F_{net} \]

### Step 7: Solve for Speed \( v \)
From the above equation, we can solve for \( v \):
\[ v^2 = \frac{R F_{net}}{M} \]
Substituting \( F_{net} \) into this equation allows us to find \( v \).

### Final Result
After simplifying the expression, we find:
\[ v = \sqrt{\frac{G M}{R} \cdot \left(\frac{4 + \sqrt{2}}{4}\right)} \]