With what minmum speed should `m` be projected from point `C` in presence of two fixed masses `M` each at `A` and `B` as shows in the figure such that mass `m` should escape the gravitational of `A` and `B` ?

Options

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

Explanation

To solve the problem of determining the minimum speed \( v \) with which mass \( m \) should be projected from point \( C \) to escape the gravitational influence of two fixed masses \( M \) located at points \( A \) and \( B \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the System**:
- We have two fixed masses \( M \) at points \( A \) and \( B \).
- The distance between the two masses is \( 2R \).
- The point \( C \) is located at the midpoint between \( A \) and \( B \), at a distance \( R \) from each mass.

2. **Initial Energy Calculation**:
- The initial kinetic energy (KE) of mass \( m \) when projected with speed \( v \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The initial potential energy (PE) due to the gravitational attraction from both masses \( M \) is:
\[ PE = -\frac{GMm}{R} - \frac{GMm}{R} = -\frac{2GMm}{R} \]
- Therefore, the total initial energy \( E_i \) is:
\[ E_i = KE + PE = \frac{1}{2} mv^2 - \frac{2GMm}{R} \]

3. **Final Energy Calculation**:
- For mass \( m \) to escape the gravitational influence of \( M \), it must reach a point where the potential energy is zero and the kinetic energy is also zero (at infinity):
\[ E_f = 0 \]

4. **Applying Conservation of Energy**:
- According to the conservation of energy, the initial energy must equal the final energy:
\[ E_i = E_f \]
\[ \frac{1}{2} mv^2 - \frac{2GMm}{R} = 0 \]

5. **Solving for Minimum Speed**:
- Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{2GMm}{R} \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{2GM}{R} \]
- Multiplying both sides by 2:
\[ v^2 = \frac{4GM}{R} \]
- Taking the square root:
\[ v = \sqrt{\frac{4GM}{R}} = 2\sqrt{\frac{GM}{R}} \]

### Final Answer
The minimum speed \( v \) with which mass \( m \) should be projected from point \( C \) is:
\[ v = 2\sqrt{\frac{GM}{R}} \]