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 Earth's surface to a height \( R \) (where \( R \) is the radius of the Earth), we can follow these steps:

### Step 1: Understand the Concept of Work Done
Work done against gravity when lifting an object is equal to the change in potential energy of the object. The potential energy \( U \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the object.

### Step 2: Calculate Initial Potential Energy
At the surface of the Earth (where \( r = R \)):
\[ U_i = -\frac{G M m}{R} \]

### Step 3: Calculate Final Potential Energy
When the object is lifted to a height \( R \) (so the distance from the center of the Earth becomes \( 2R \)):
\[ U_f = -\frac{G M m}{2R} \]

### Step 4: Calculate Work Done
The work done \( W \) in lifting the object is the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values we calculated:
\[ W = \left(-\frac{G M m}{2R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ W = -\frac{G M m}{2R} + \frac{G M m}{R} \]
\[ W = \frac{G M m}{R} - \frac{G M m}{2R} = \frac{G M m}{2R} \]

### Step 5: Relate \( G \) to \( g \)
We know that the acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
From this, we can express \( G M \) as:
\[ G M = g R^2 \]

### Step 6: Substitute \( G M \) in Work Done Formula
Now substituting \( G M \) in the work done equation:
\[ W = \frac{g R^2 m}{2R} = \frac{g R m}{2} \]

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