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 find the speed of a body in interstellar space when it is thrown with a speed three times the escape speed from the Earth's surface.

### Step-by-Step Solution:

1. **Understanding 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 Speed**:
The body is thrown at a speed of three times the escape speed:
\[ V_i = 3V_e \]

3. **Conservation of Mechanical Energy**:
In the absence of external forces, the mechanical energy is conserved. The total mechanical energy at the Earth's surface (initial) is equal to the total mechanical energy in interstellar space (final):
\[ \text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy} \]

4. **Calculating Initial Energies**:
- The initial potential energy \( U_i \) at the Earth's surface is:
\[ U_i = -\frac{GMm}{R} \]
- The initial kinetic energy \( K_i \) when the body is thrown is:
\[ K_i = \frac{1}{2} m (3V_e)^2 = \frac{1}{2} m \cdot 9V_e^2 = \frac{9}{2} m V_e^2 \]

5. **Final Energies**:
- The final potential energy \( U_f \) in interstellar space is:
\[ U_f = 0 \]
- The final kinetic energy \( K_f \) is:
\[ K_f = \frac{1}{2} mv^2 \]
where \( v \) is the speed we want to find.

6. **Setting Up the Equation**:
Using conservation of energy:
\[ U_i + K_i = U_f + K_f \]
Substituting the values:
\[ -\frac{GMm}{R} + \frac{9}{2} m V_e^2 = 0 + \frac{1}{2} mv^2 \]

7. **Simplifying the Equation**:
We can cancel \( m \) (assuming \( m \neq 0 \)):
\[ -\frac{GM}{R} + \frac{9}{2} V_e^2 = \frac{1}{2} v^2 \]

8. **Substituting \( GM/R \)**:
From the escape velocity formula, we know:
\[ \frac{GM}{R} = \frac{1}{2} V_e^2 \]
Therefore:
\[ -\frac{1}{2} V_e^2 + \frac{9}{2} V_e^2 = \frac{1}{2} v^2 \]
Simplifying gives:
\[ \frac{8}{2} V_e^2 = \frac{1}{2} v^2 \]
or:
\[ 4 V_e^2 = \frac{1}{2} v^2 \]

9. **Finding \( v^2 \)**:
Multiplying both sides by 2:
\[ 8 V_e^2 = v^2 \]

10. **Finding \( v \)**:
Taking the square root of both sides:
\[ v = \sqrt{8} V_e = 2\sqrt{2} V_e \]

### Final Answer:
The speed at which the body will move in interstellar space is:
\[ v = 2\sqrt{2} V_e \]