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 finding 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 use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two fixed masses \( M \) at points \( A \) and \( B \).
- The distance between \( A \) and \( B \) is \( 2r \).
- The distance from \( C \) to both \( A \) and \( B \) is \( r\sqrt{2} \) (using the Pythagorean theorem).

2. **Conservation of Energy**:
- The total mechanical energy at point \( C \) (initial) must equal the total mechanical energy when mass \( m \) is at infinity (final).
- At infinity, the potential energy is zero, and we want the kinetic energy to also be zero for escape.

3. **Initial Energy Calculation**:
- The initial kinetic energy (KE) when mass \( m \) is projected from point \( C \) is given by:
\[ KE_i = \frac{1}{2} mv^2 \]
- The initial potential energy (PE) due to the gravitational attraction from both masses \( M \) is:
\[ PE_i = -\frac{GMm}{r\sqrt{2}} - \frac{GMm}{r\sqrt{2}} = -\frac{2GMm}{r\sqrt{2}} \]

4. **Final Energy Calculation**:
- At infinity, the final kinetic energy \( KE_f = 0 \) and the potential energy \( PE_f = 0 \).

5. **Setting Up the Energy Equation**:
- According to the conservation of energy:
\[ KE_i + PE_i = KE_f + PE_f \]
- Substituting the values:
\[ \frac{1}{2} mv^2 - \frac{2GMm}{r\sqrt{2}} = 0 \]

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

### Final Answer:
The minimum speed \( v \) required for mass \( m \) to escape the gravitational influence of the two masses \( M \) is:
\[ v = \frac{2\sqrt{GM}}{\sqrt[4]{2} \cdot \sqrt{r}} \]