If a body is thrown at speed of triple as that of escape speed `(V_e)` from earth`s surface, then at what speed it will move in interstellar space?

Options

A. `2sqrt2V_e`
B. `sqrt2V_e`
C. `V_e`
D. `sqrt3V_e`

Answer

To solve the problem, we need to determine the speed of a body thrown at three times the escape speed from the Earth's surface as it moves into interstellar space.

### Step-by-Step Solution:

1. **Understand Escape Speed**:
The escape speed \( V_e \) is the minimum speed required for an object to break free from the gravitational pull of the Earth without any further propulsion. The formula for escape speed is given by:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Speed**:
According to the problem, the body is thrown at a speed of three times the escape speed:
\[ V = 3V_e \]

3. **Energy Conservation**:
We will use the principle of conservation of mechanical energy. The total mechanical energy at the Earth's surface is the sum of kinetic energy (KE) and gravitational potential energy (PE):
\[ \text{Total Energy} = KE + PE \]
At the surface of the Earth:
\[ KE = \frac{1}{2} m V^2 = \frac{1}{2} m (3V_e)^2 = \frac{9}{2} m V_e^2 \]
\[ PE = -\frac{GMm}{R} \]

4. **Total Energy at the Surface**:
Combining the kinetic and potential energy:
\[ E_{\text{initial}} = \frac{9}{2} m V_e^2 - \frac{GMm}{R} \]

5. **Total Energy in Interstellar Space**:
In interstellar space, the potential energy is zero (as the gravitational influence of Earth is negligible), and the total energy will be equal to the kinetic energy at that point:
\[ E_{\text{final}} = KE_{\text{final}} = \frac{1}{2} m v^2 \]

6. **Setting Initial Energy Equal to Final Energy**:
Since total energy is conserved:
\[ \frac{9}{2} m V_e^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 \]

7. **Substituting for \( V_e \)**:
We know that \( V_e^2 = \frac{2GM}{R} \). Substituting this into the equation gives:
\[ \frac{9}{2} m \left(\frac{2GM}{R}\right) - \frac{GMm}{R} = \frac{1}{2} m v^2 \]
Simplifying this:
\[ \frac{9GMm}{R} - \frac{GMm}{R} = \frac{1}{2} m v^2 \]
\[ \frac{8GMm}{R} = \frac{1}{2} m v^2 \]

8. **Solving for \( v^2 \)**:
Dividing both sides by \( m \) and multiplying by 2 gives:
\[ \frac{16GM}{R} = v^2 \]

9. **Finding \( v \)**:
Taking the square root:
\[ v = \sqrt{\frac{16GM}{R}} = 4V_e \]

### Final Answer:
The speed at which the body will move in interstellar space is:
\[ v = 4V_e \]