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. `(GM)/(R )`
B. `sqrt(2sqrt(2)(GM)/(R ))`
C. `sqrt((GM)/(R)(2sqrt(2) + 1))`
D. `sqrt((GM)/(R)(2sqrt(2) + 1)/(4))`

Explanation

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

### Step 1: Identify the forces acting on a particle
Each particle experiences gravitational attraction from the other three particles. The gravitational force between any two particles of mass \( M \) separated by a distance \( r \) is given by Newton's law of gravitation:

\[ F = \frac{G M^2}{r^2} \]

### Step 2: Determine the distance between particles
Since the particles are positioned at the corners of a square (equidistant from each other), the distance \( r \) between any two adjacent particles is equal to the side of the square, which can be expressed in terms of the radius \( R \) of the circle in which they are moving. The diagonal distance between two opposite particles is \( \sqrt{2}R \).

### Step 3: Calculate the net gravitational force on one particle
For one particle, the net gravitational force due to the other three particles can be calculated. The forces from adjacent particles can be resolved into components. The net force can be found by considering the contributions from all three particles.

The force between two adjacent particles is:
\[ F_{adj} = \frac{G M^2}{R^2} \]
The force between two opposite particles is:
\[ F_{opp} = \frac{G M^2}{(R\sqrt{2})^2} = \frac{G M^2}{2R^2} \]

### Step 4: Calculate the resultant force
The net gravitational force acting on one particle can be calculated by vector addition of the forces from the other three particles. The resultant force \( F_{net} \) can be calculated as follows:

1. The components of the forces from the two adjacent particles will add up vectorially.
2. The force from the opposite particle will also contribute.

The total gravitational force acting on one particle can be expressed as:
\[ F_{net} = 2F_{adj} + F_{opp} = 2\left(\frac{G M^2}{R^2}\right) + \left(\frac{G M^2}{2R^2}\right) \]

### Step 5: Set the net force equal to centripetal force
The net gravitational force provides the necessary centripetal force for circular motion. The centripetal force required for a particle moving in a circle of radius \( R \) with speed \( v \) is given by:

\[ F_{centripetal} = \frac{M v^2}{R} \]

Setting the net gravitational force equal to the centripetal force gives:
\[ \frac{G M^2}{R^2} \left( 2 + \frac{1}{2} \right) = \frac{M v^2}{R} \]

### Step 6: Solve for speed \( v \)
Rearranging the equation to solve for \( v \):
\[ \frac{G M^2}{R^2} \cdot \frac{5}{2} = \frac{M v^2}{R} \]
\[ v^2 = \frac{5 G M}{2 R} \]
\[ v = \sqrt{\frac{5 G M}{2 R}} \]

### Final Answer
Thus, the speed of each particle is:
\[ v = \sqrt{\frac{5 G M}{2 R}} \]

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