A body of mass m taken from the earth's surface to the height equal to the twice the earth radius R of the earth . The change in potential energy of the body will be

Options

A. `2mgR`
B. `(2)/(3) m g R`
C. `3 mg R`
D. `(mgR)/(3)`

Explanation

To find the change in potential energy of a body of mass \( m \) taken from the Earth's surface to a height equal to twice the Earth's radius \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Positions**:
- The initial position of the body is at the Earth's surface, which is at a distance \( R \) from the center of the Earth.
- The final position of the body is at a height equal to twice the Earth's radius, which means it is at a distance of \( R + 2R = 3R \) from the center of the Earth.

2. **Calculate Initial Potential Energy**:
- The formula for gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
- Therefore, the initial potential energy \( U_i \) at distance \( R \) is:
\[ U_i = -\frac{GMm}{R} \]

3. **Calculate Final Potential Energy**:
- The final potential energy \( U_f \) at distance \( 3R \) is:
\[ U_f = -\frac{GMm}{3R} \]

4. **Find Change in Potential Energy**:
- The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
- Substituting the values we calculated:
\[ \Delta U = \left(-\frac{GMm}{3R}\right) - \left(-\frac{GMm}{R}\right) \]
- Simplifying this expression:
\[ \Delta U = -\frac{GMm}{3R} + \frac{GMm}{R} \]
- Finding a common denominator (which is \( 3R \)):
\[ \Delta U = -\frac{GMm}{3R} + \frac{3GMm}{3R} = \frac{2GMm}{3R} \]

5. **Express in Terms of Acceleration Due to Gravity**:
- We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
- Rearranging gives \( GM = gR^2 \).
- Substituting this into our expression for \( \Delta U \):
\[ \Delta U = \frac{2(gR^2)m}{3R} = \frac{2g m R}{3} \]

### Final Result:
Thus, the change in potential energy of the body is:
\[ \Delta U = \frac{2}{3} mgR \]