If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body ofmass m from the earth's surface to a height R equal to the radius of the earth is

Options

A. `1/2mgR`
B. `2mgR`
C. `mgR`
D. `1/4mgR`

Answer

To solve the problem of calculating the work done in slowly lifting a body of mass \( m \) from the surface of the Earth to a height \( R \) (where \( R \) is the radius of the Earth), we can follow these steps:

### Step 1: Understand the change in gravitational acceleration
At the surface of the Earth, the acceleration due to gravity is given as \( g \). When we lift the body to a height equal to the radius of the Earth \( R \), we need to find the new acceleration due to gravity \( g' \) at that height.

The formula for gravitational acceleration at a distance \( r \) from the center of the Earth is:
\[ g' = \frac{GM}{r^2} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Calculate \( g' \) at height \( R \)
When the body is lifted to a height \( R \), the distance from the center of the Earth becomes \( 2R \) (since the body is at the surface plus the height \( R \)):
\[ g' = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{g}{4} \]
This shows that the acceleration due to gravity at height \( R \) is one-fourth of that at the surface.

### Step 3: Calculate the initial and final potential energy
The potential energy \( U \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
- **Initial potential energy \( U_i \)** at the surface (distance \( R \)):
\[ U_i = -\frac{GMm}{R} \]
- **Final potential energy \( U_f \)** at height \( R \) (distance \( 2R \)):
\[ U_f = -\frac{GMm}{2R} \]

### Step 4: 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 we calculated:
\[ W = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) \]
\[ W = -\frac{GMm}{2R} + \frac{GMm}{R} \]
\[ W = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \]

### Step 5: Relate \( GM \) to \( g \)
Since \( g = \frac{GM}{R^2} \), we can express \( GM \) as \( gR^2 \):
\[ W = \frac{gR^2 m}{2R} = \frac{gRm}{2} \]

### Final Answer
Thus, the work done in slowly lifting the body to a height \( R \) is:
\[ W = \frac{gRm}{2} \]