A body is projected from the surface of the earth with thrice the escape velocity `(V_(e))` from the surface of the earth. What will be its velocity, when it will escape the gravitational pull?

Options

A. `2sqrt(2)V_(e)`
B. `2V_(e)`
C. `(V_(e))/(2)`
D. `4V_(e)`

Explanation

To solve the problem, we need to determine the velocity of a body projected from the Earth's surface with a velocity three times the escape velocity when it escapes the gravitational pull of the Earth.

### Step-by-Step Solution:

1. **Define Escape Velocity**:
The escape velocity \( V_e \) from the Earth's surface is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Velocity of the Body**:
The body is projected with a velocity \( V_p = 3V_e \).

3. **Conservation of Energy**:
The total mechanical energy of the body at the surface of the Earth must equal the total mechanical energy when it escapes the gravitational pull. At the surface, the potential energy \( U \) is given by:
\[ U = -\frac{GMm}{R} \]
and the kinetic energy \( K \) at the surface is:
\[ K = \frac{1}{2} m V_p^2 \]

4. **Total Energy at the Surface**:
The total energy \( E \) at the surface is:
\[ E = K + U = \frac{1}{2} m V_p^2 - \frac{GMm}{R} \]

5. **Total Energy at Escape**:
When the body escapes the gravitational pull, its potential energy becomes zero (as it is at an infinite distance), and its kinetic energy is \( K' = \frac{1}{2} m V^2 \) where \( V \) is the velocity at escape. Thus, the total energy at escape is:
\[ E' = K' + U' = \frac{1}{2} m V^2 + 0 = \frac{1}{2} m V^2 \]

6. **Setting Total Energies Equal**:
Since energy is conserved, we set the total energy at the surface equal to the total energy at escape:
\[ \frac{1}{2} m V_p^2 - \frac{GMm}{R} = \frac{1}{2} m V^2 \]

7. **Substituting \( V_p \)**:
Substitute \( V_p = 3V_e \):
\[ \frac{1}{2} m (3V_e)^2 - \frac{GMm}{R} = \frac{1}{2} m V^2 \]
Simplifying gives:
\[ \frac{1}{2} m (9V_e^2) - \frac{GMm}{R} = \frac{1}{2} m V^2 \]

8. **Canceling Mass**:
Cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{9}{2} V_e^2 - \frac{GM}{R} = \frac{1}{2} V^2 \]

9. **Expressing \( GM/R \)**:
Recall that \( \frac{GM}{R} = \frac{1}{2} V_e^2 \):
\[ \frac{9}{2} V_e^2 - \frac{1}{2} V_e^2 = \frac{1}{2} V^2 \]
This simplifies to:
\[ \frac{8}{2} V_e^2 = \frac{1}{2} V^2 \]
or:
\[ 4V_e^2 = \frac{1}{2} V^2 \]

10. **Solving for \( V \)**:
Multiply both sides by 2:
\[ 8V_e^2 = V^2 \]
Taking the square root gives:
\[ V = \sqrt{8} V_e = 2\sqrt{2} V_e \]

### Final Answer:
The velocity of the body when it escapes the gravitational pull of the Earth is:
\[ V = 2\sqrt{2} V_e \]