The amount of work done in lifting a body of mass 'm' from the surface of the earth to a height equal to twice the radius of the earth is,

Options

A. `(2GMm)/(3R)`
B. `(3GMm)/(2R)`
C. `(5GMm)/(3R)`
D. `(3GMm)/(5R)`

Answer

To solve the problem of calculating the work done in lifting a body of mass 'm' from the surface of the Earth to a height equal to twice the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the mass of the body be \( m \).
- Let the radius of the Earth be \( R \).
- The height to which the body is lifted is \( 2R \).

2. **Calculate the Distance from the Center of the Earth**:
- When the body is lifted to a height of \( 2R \), the distance from the center of the Earth becomes:
\[ d = R + 2R = 3R \]

3. **Potential Energy Formula**:
- The gravitational potential energy \( U \) at a distance \( d \) from the center of the Earth is given by:
\[ U = -\frac{G M m}{d} \]
- Here, \( G \) is the universal gravitational constant, and \( M \) is the mass of the Earth.

4. **Calculate Initial Potential Energy**:
- The initial potential energy when the body is at the surface of the Earth (distance \( R \)):
\[ U_i = -\frac{G M m}{R} \]

5. **Calculate Final Potential Energy**:
- The final potential energy when the body is at a height of \( 2R \) (distance \( 3R \)):
\[ U_f = -\frac{G M m}{3R} \]

6. **Calculate the Work Done**:
- The work done \( W \) in lifting the body is equal to the change in potential energy:
\[ W = U_f - U_i \]
- Substituting the values:
\[ W = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
- Simplifying this gives:
\[ W = -\frac{G M m}{3R} + \frac{G M m}{R} \]
- Finding a common denominator:
\[ W = \frac{G M m}{R} - \frac{G M m}{3R} = \frac{G M m}{R} \left(1 - \frac{1}{3}\right) = \frac{G M m}{R} \cdot \frac{2}{3} \]
- Therefore, the work done is:
\[ W = \frac{2 G M m}{3R} \]

### Final Answer:
The amount of work done in lifting the body is:
\[ W = \frac{2 G M m}{3R} \]