Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be :

Options

A. `sqrt((Gm)/(r))`
B. `sqrt((Gm)/(4r))`
C. `sqrt((4Gm)/(r))`
D. `sqrt((Gm)/(2r))`

Answer

To find the speed of each particle moving in a circle of radius \( r \) under the action of their mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the System
We have two particles, each of mass \( m \), moving in a circular path due to their mutual gravitational attraction. The center of mass of the system lies at the midpoint between the two particles.

### Step 2: Determine the Distance Between the Particles
Since both particles are moving in a circle of radius \( r \) around the center of mass, the distance between the two particles is \( 2r \).

### Step 3: Write the Gravitational Force
The gravitational force \( F \) between the two particles can be expressed using Newton's law of gravitation:
\[ F = \frac{G m^2}{(2r)^2} = \frac{G m^2}{4r^2} \]
where \( G \) is the gravitational constant.

### Step 4: Relate Gravitational Force to Centripetal Force
The gravitational force acts as the centripetal force required to keep each particle moving in a circle. The centripetal force \( F_c \) for each particle is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the speed of each particle.

### Step 5: Set Gravitational Force Equal to Centripetal Force
Since the gravitational force provides the necessary centripetal force, we can set them equal:
\[ \frac{G m^2}{4r^2} = \frac{m v^2}{r} \]

### Step 6: Simplify the Equation
We can cancel one \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G m}{4r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{G m}{4r} = v^2 \]

### Step 7: Solve for \( v \)
Taking the square root of both sides gives us the speed of each particle:
\[ v = \sqrt{\frac{G m}{4r}} \]

### Final Answer
Thus, the speed of each particle is:
\[ v = \sqrt{\frac{G m}{4r}} \]