When a body id lifted from surface of earth height equal to radius of earth, then the change in its potential energy is

Options

A. `mg R`
B. `2mg R`
C. `(1)/(2) mg R`
D. `4mg R`

Explanation

To find the change in potential energy when a body is lifted from the surface of the Earth to a height equal to the radius of the Earth, we can follow these steps:

### Step 1: Understand the formula for gravitational potential energy
The gravitational potential energy (U) of an object at a distance (r) from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( r \) is the distance from the center of the Earth.

### Step 2: Calculate the initial potential energy
When the body is on the surface of the Earth, the distance from the center of the Earth is equal to the radius of the Earth (R). Therefore, the initial potential energy (U_initial) is:
\[ U_{\text{initial}} = -\frac{GMm}{R} \]

### Step 3: Calculate the final potential energy
When the body is lifted to a height equal to the radius of the Earth (R), the new distance from the center of the Earth becomes \( R + R = 2R \). Thus, the final potential energy (U_final) is:
\[ U_{\text{final}} = -\frac{GMm}{2R} \]

### Step 4: Calculate the change in potential energy
The change in potential energy (ΔU) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) \]
\[ \Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} \]
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R} \]
\[ \Delta U = \frac{GMm}{2R} \]

### Conclusion
The change in potential energy when a body is lifted from the surface of the Earth to a height equal to the radius of the Earth is:
\[ \Delta U = \frac{GMm}{2R} \]