If a rocket is fired with a velocity, `V=2sqrt(2gR)` near the earth's surface and goes upwards, its speed in the inter-stellar space is

Options

A. `4sqrt(gR)`
B. `sqrt(2gR)`
C. `sqrt(gR)`
D. `sqrt(4gR)`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the rocket at the Earth's surface will be equal to the total mechanical energy of the rocket in interstellar space, where the gravitational potential energy is negligible.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The rocket is fired from the Earth's surface with an initial velocity \( V = 2\sqrt{2gR} \).
- At the Earth's surface, the potential energy \( U \) is given by:
\[ U = -\frac{GMm}{R} \]
- The kinetic energy \( K \) at the surface is:
\[ K = \frac{1}{2} m V^2 = \frac{1}{2} m (2\sqrt{2gR})^2 = \frac{1}{2} m (8gR) = 4mgR \]

2. **Calculate Total Energy at Earth's Surface**:
- The total mechanical energy \( E_{\text{surface}} \) at the surface is:
\[ E_{\text{surface}} = K + U = 4mgR - \frac{GMm}{R} \]
- Since \( g = \frac{GM}{R^2} \), we can substitute \( GM \) with \( gR^2 \):
\[ U = -\frac{gR^2 m}{R} = -gRm \]
- Thus, the total energy becomes:
\[ E_{\text{surface}} = 4mgR - gRm = (4m - m)gR = 3mgR \]

3. **Conditions in Interstellar Space**:
- In interstellar space, the gravitational potential energy \( U \) is approximately zero, and the total energy \( E_{\text{interstellar}} \) is purely kinetic:
\[ E_{\text{interstellar}} = K' = \frac{1}{2} m V^2 \]

4. **Set Total Energies Equal**:
- According to the conservation of mechanical energy:
\[ E_{\text{surface}} = E_{\text{interstellar}} \]
\[ 3mgR = \frac{1}{2} m V^2 \]

5. **Solve for Final Velocity \( V \)**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 3gR = \frac{1}{2} V^2 \]
- Rearranging gives:
\[ V^2 = 6gR \]
- Taking the square root:
\[ V = \sqrt{6gR} \]

### Final Answer:
The speed of the rocket in interstellar space is \( V = \sqrt{6gR} \).