If an object of mass `m` is taken from the surface of earth (radius R) to a height 5R, then the work done is

Options

A. `2mgR`
B. `mgR`
C. `(5)/(6)mgR`
D. `(3)/(2)mgR`

Explanation

To find the work done when an object of mass \( m \) is taken from the surface of the Earth (radius \( R \)) to a height of \( 5R \), we can use the concept of gravitational potential energy.

### Step-by-Step Solution:

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

2. **Calculate the initial potential energy (U_initial)**:
At the surface of the Earth, the distance from the center is \( R \):
\[ U_{\text{initial}} = -\frac{G M m}{R} \]

3. **Calculate the final potential energy (U_final)**:
When the object is at a height of \( 5R \), the distance from the center of the Earth is \( R + 5R = 6R \):
\[ U_{\text{final}} = -\frac{G M m}{6R} \]

4. **Calculate the change in potential energy (ΔU)**:
The work done (W) in moving the object is equal to the change in potential energy:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values:
\[ \Delta U = \left(-\frac{G M m}{6R}\right) - \left(-\frac{G M m}{R}\right) \]
\[ \Delta U = -\frac{G M m}{6R} + \frac{G M m}{R} \]
\[ \Delta U = \frac{G M m}{R} - \frac{G M m}{6R} \]
\[ \Delta U = \frac{G M m}{R} \left(1 - \frac{1}{6}\right) \]
\[ \Delta U = \frac{G M m}{R} \cdot \frac{5}{6} \]

5. **Final expression for work done (W)**:
Therefore, the work done in moving the object to a height of \( 5R \) is:
\[ W = \Delta U = \frac{5 G M m}{6R} \]

### Final Answer:
The work done in taking the object from the surface of the Earth to a height of \( 5R \) is:
\[ W = \frac{5 G M m}{6R} \]