Escape velocity for a projectile at earth's surface is `V_(e)`. A body is projected form earth's surface with velocity `2 V_(e)`. The velocity of the when it is at infinite distance from the centre of the earth is :-

Options

A. `V_(e)`
B. `2V_(e)`
C. `sqrt(2) V_(e)`
D. `sqrt(3) V_(e)`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the body at the Earth's surface will be equal to the total mechanical energy of the body when it is at an infinite distance from the center of the Earth.

### Step-by-Step Solution:

1. **Identify Initial and Final States**:
- Initial state: The body is at the Earth's surface with an initial velocity of \(2V_e\).
- Final state: The body is at an infinite distance from the Earth's center.

2. **Write the Expression for Initial Energy**:
- The initial kinetic energy (KE) when the body is projected is given by:
\[ KE_i = \frac{1}{2} m (2V_e)^2 = \frac{1}{2} m (4V_e^2) = 2mV_e^2 \]
- The gravitational potential energy (PE) at the Earth's surface is:
\[ PE_i = -\frac{G M m}{R} \]
- Therefore, the total initial energy (E_i) is:
\[ E_i = KE_i + PE_i = 2mV_e^2 - \frac{G M m}{R} \]

3. **Write the Expression for Final Energy**:
- At an infinite distance, the potential energy (PE) is zero:
\[ PE_f = 0 \]
- Let \(v\) be the final velocity of the body at infinity. The kinetic energy (KE) at infinity is:
\[ KE_f = \frac{1}{2} mv^2 \]
- Therefore, the total final energy (E_f) is:
\[ E_f = KE_f + PE_f = \frac{1}{2} mv^2 + 0 = \frac{1}{2} mv^2 \]

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

5. **Cancel Mass (m)**:
- We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ 2V_e^2 - \frac{G M}{R} = \frac{1}{2} v^2 \]

6. **Substitute Escape Velocity**:
- The escape velocity \(V_e\) is given by:
\[ V_e = \sqrt{\frac{G M}{R}} \]
- Therefore, \(V_e^2 = \frac{G M}{R}\).

7. **Substitute \(V_e^2\) into the Equation**:
- Replace \(V_e^2\) in the energy equation:
\[ 2 \left(\frac{G M}{R}\right) - \frac{G M}{R} = \frac{1}{2} v^2 \]
\[ \frac{G M}{R} = \frac{1}{2} v^2 \]

8. **Solve for \(v^2\)**:
- Multiply both sides by 2:
\[ \frac{2 G M}{R} = v^2 \]

9. **Take the Square Root**:
- Therefore, the final velocity \(v\) is:
\[ v = \sqrt{\frac{2 G M}{R}} = \sqrt{2} V_e \]

### Final Answer:
The velocity of the body when it is at an infinite distance from the center of the Earth is \(v = \sqrt{2} V_e\).