A particle is projected from the mid-point of the line joining two fixed particles each of mass `m`. If the separation between the fixed particles is `l`, the minimum velocity of projection of the particle so as to escape is equal to

Options

A. `sqrt((G)/(l))`
B. `sqrt((G)/(2l))`
C. `sqrt((2Gm)/(l))`
D. `2sqrt((2Gm)/(l))`

Answer

To solve the problem of finding the minimum velocity of projection of a particle so as to escape from the gravitational influence of two fixed particles, we can follow these steps:

### Step 1: Understand the setup
We have two fixed particles, each of mass \( m \), separated by a distance \( l \). A third particle is projected from the midpoint of the line joining these two masses.

### Step 2: Determine the potential energy
The potential energy \( U \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula:
\[ U = -\frac{G m_1 m_2}{r} \]
In our case, the potential energy of the particle at the midpoint (which is at a distance \( \frac{l}{2} \) from each mass) can be calculated as follows:
- The potential energy due to the first mass \( m \) is:
\[ U_1 = -\frac{G m m}{\frac{l}{2}} = -\frac{2G m^2}{l} \]
- The potential energy due to the second mass \( m \) is:
\[ U_2 = -\frac{G m m}{\frac{l}{2}} = -\frac{2G m^2}{l} \]
Thus, the total potential energy \( U \) at the midpoint is:
\[ U = U_1 + U_2 = -\frac{2G m^2}{l} - \frac{2G m^2}{l} = -\frac{4G m^2}{l} \]

### Step 3: Apply conservation of energy
The total mechanical energy (kinetic energy + potential energy) must be conserved. At the point of projection, the kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 \]
At infinity, the potential energy is zero, and the kinetic energy is also zero. Therefore, we can set up the equation:
\[ K + U = 0 \]
Substituting the expressions for kinetic and potential energy:
\[ \frac{1}{2} mv^2 - \frac{4G m^2}{l} = 0 \]

### Step 4: Solve for the velocity \( v \)
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{4G m^2}{l} \]
Multiplying both sides by 2:
\[ mv^2 = \frac{8G m^2}{l} \]
Dividing both sides by \( m \):
\[ v^2 = \frac{8G m}{l} \]
Taking the square root:
\[ v = \sqrt{\frac{8G m}{l}} \]

### Step 5: Relate \( G \) to \( g \)
We know that \( G = g \cdot R^2 \) where \( R \) is the radius of the Earth or the distance from the center of mass. However, since we are looking for the escape velocity in terms of \( g \), we can express it as:
\[ v = \sqrt{\frac{8g m}{l}} \]

### Conclusion
The minimum velocity of projection \( v \) for the particle to escape the gravitational influence of the two masses is:
\[ v = 2\sqrt{\frac{2g m}{l}} \]

### Final Answer
Thus, the minimum velocity of projection of the particle so as to escape is:
\[ v = 2\sqrt{\frac{2g m}{l}} \]