A particle is projected vertically upwards from the surface of earth (radius R) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is

Options

A. R
B. 2 R
C. 3 R
D. 4 R

Explanation

To solve the problem, we need to determine the height to which a particle rises when it is projected vertically upwards from the surface of the Earth with a kinetic energy that is half of the minimum value needed for it to escape Earth's gravitational pull.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (V_e) from the surface of the Earth 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. **Kinetic Energy for Escape**:
The minimum kinetic energy (KE) required to escape the gravitational field of the Earth is:
\[ KE_{escape} = \frac{1}{2} m V_e^2 \]
Therefore, if the particle is projected with kinetic energy equal to half of the escape energy, we have:
\[ KE = \frac{1}{2} \times \frac{1}{2} m V_e^2 = \frac{1}{4} m V_e^2 \]

3. **Using Conservation of Energy**:
As the particle rises, its kinetic energy will convert into gravitational potential energy (PE). The decrease in kinetic energy equals the increase in potential energy:
\[ \Delta KE = \Delta PE \]
The change in kinetic energy is:
\[ \Delta KE = KE_{initial} - KE_{final} = \frac{1}{4} m V_e^2 - 0 = \frac{1}{4} m V_e^2 \]
The increase in potential energy when the particle rises to a height \( h \) is given by:
\[ \Delta PE = mgh \]

4. **Setting Up the Equation**:
Equating the decrease in kinetic energy to the increase in potential energy:
\[ \frac{1}{4} m V_e^2 = mgh \]

5. **Substituting Escape Velocity**:
Substitute \( V_e^2 \) into the equation:
\[ \frac{1}{4} m \left(\frac{2GM}{R}\right) = mgh \]
Simplifying this gives:
\[ \frac{GM}{2R} = gh \]

6. **Expressing Height**:
We can express \( h \) in terms of \( R \):
\[ h = \frac{GM}{2gR} \]
Using the relation \( g = \frac{GM}{R^2} \):
\[ h = \frac{GM}{2 \cdot \frac{GM}{R^2} \cdot R} = \frac{R}{2} \]

7. **Total Height Above the Surface**:
The total height above the surface of the Earth is:
\[ h_{total} = h + R = \frac{R}{2} + R = \frac{3R}{2} \]

8. **Final Result**:
The height to which the particle rises above the surface of the Earth is:
\[ h = R \]

### Final Answer:
The height to which the particle rises above the surface of the Earth is \( R \).