Let escape velocity od a body a kept surface of a planet is u , If it is projected at a speed of 200 % more than the escape speed , then its speed in interstellar space will be

Options

A. u
B. `sqrt(3)` u
C. `2u`
D. `2sqrt(2)u`

Answer

To solve the problem, we need to find the speed of a body in interstellar space after it has been projected at a speed that is 200% more than the escape velocity of the planet.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (u) is the minimum speed required for an object to break free from the gravitational attraction of a planet without any additional propulsion.

2. **Calculating the Projected Speed**:
The body is projected at a speed that is 200% more than the escape velocity.
\[ \text{Projected speed} = u + 200\% \text{ of } u = u + 2u = 3u \]
So, the initial speed (v) of the body when projected is \(3u\).

3. **Understanding the Energy Conservation**:
When the body is projected, it has both kinetic and potential energy. As it moves away from the planet, the potential energy approaches zero when it reaches interstellar space.

4. **Setting Up the Energy Equation**:
The total mechanical energy at the surface of the planet is the sum of kinetic energy (KE) and gravitational potential energy (PE):
\[ \text{Total Energy at surface} = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{R} \]
where \(v = 3u\), \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.

5. **Calculating Initial Total Energy**:
Substitute \(v = 3u\):
\[ \text{Total Energy} = \frac{1}{2}m(3u)^2 - \frac{GMm}{R} = \frac{9}{2}mu^2 - \frac{GMm}{R} \]

6. **Final Energy in Interstellar Space**:
In interstellar space, the potential energy is zero, and the kinetic energy is given by:
\[ \text{Total Energy in interstellar space} = \frac{1}{2}mV_1^2 \]
where \(V_1\) is the speed in interstellar space.

7. **Setting the Energies Equal**:
Since energy is conserved, we can set the total energy at the surface equal to the total energy in interstellar space:
\[ \frac{9}{2}mu^2 - \frac{GMm}{R} = \frac{1}{2}mV_1^2 \]

8. **Substituting for \(GM/R\)**:
From the escape velocity formula, we know:
\[ u^2 = \frac{2GM}{R} \implies \frac{GM}{R} = \frac{u^2}{2} \]
Substitute this into the energy equation:
\[ \frac{9}{2}mu^2 - \frac{m}{2}u^2 = \frac{1}{2}mV_1^2 \]
Simplifying gives:
\[ \frac{8}{2}mu^2 = \frac{1}{2}mV_1^2 \implies 4mu^2 = \frac{1}{2}mV_1^2 \]

9. **Solving for \(V_1\)**:
Cancel \(m\) from both sides:
\[ 4u^2 = \frac{1}{2}V_1^2 \implies V_1^2 = 8u^2 \implies V_1 = \sqrt{8}u = 2\sqrt{2}u \]

10. **Final Result**:
The speed of the body in interstellar space is:
\[ V_1 = 2\sqrt{2}u \]