Two particles of equal mass m go around a circle of radius R under the action the of their mutual gravitational attraction . The speed v of each particle is

Options

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

Answer

To find the speed \( v \) of each particle in circular motion under the influence of their mutual gravitational attraction, we can follow these steps:

### Step 1: Understand the System
We have two particles, each of mass \( m \), rotating around a common center in a circular path of radius \( R \). The distance between the two particles is \( 2R \).

### Step 2: Identify Forces Acting on the Particles
Each particle experiences two forces:
1. **Gravitational Force (\( F_G \))**: This force acts between the two masses and is given by Newton's law of gravitation:
\[ F_G = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2} \]
where \( G \) is the gravitational constant.

2. **Centripetal Force (\( F_C \))**: This is the force required to keep the particle moving in a circular path and is provided by the gravitational force. The centripetal force can be expressed as:
\[ F_C = \frac{m v^2}{R} \]

### Step 3: Set the Forces Equal
In circular motion, the gravitational force provides the necessary centripetal force. Therefore, we can set these two forces equal:
\[ F_G = F_C \]
Substituting the expressions for \( F_G \) and \( F_C \):
\[ \frac{G m^2}{4R^2} = \frac{m v^2}{R} \]

### Step 4: Simplify the Equation
To simplify the equation, we can multiply both sides by \( 4R^2 \) to eliminate the fraction:
\[ G m^2 = 4 m v^2 R \]

### Step 5: Cancel Out Common Terms
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ G m = 4 v^2 R \]

### Step 6: Solve for \( v^2 \)
Now, we can solve for \( v^2 \):
\[ v^2 = \frac{G m}{4R} \]

### Step 7: Take the Square Root
Finally, we take the square root to find \( v \):
\[ v = \sqrt{\frac{G m}{4R}} = \frac{1}{2} \sqrt{\frac{G m}{R}} \]

Thus, the speed \( v \) of each particle is:
\[ v = \frac{1}{2} \sqrt{\frac{G m}{R}} \]