Work done in taking a body of mass m to a height nR above surface of earth will be : (R = radius of earth)

Options

A. mgn R
B. mg R(n/n+1)
C. `mgR((n+1))/(n)`
D. `(mgR)/(n(n+1))`

Answer

To find the work done in taking a body of mass \( m \) to a height \( nR \) above the surface of the Earth (where \( R \) is the radius of the Earth), we can follow these steps:

### Step 1: Understand the Potential Energy Change
The work done in lifting the mass \( m \) to a height \( nR \) is equal to the change in gravitational potential energy. The gravitational potential energy \( U \) at a distance \( d \) from the center of the Earth is given by:
\[ U = -\frac{G M m}{d} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Calculate Initial Potential Energy \( U_1 \)
At the surface of the Earth (distance \( R \) from the center), the potential energy \( U_1 \) is:
\[ U_1 = -\frac{G M m}{R} \]

### Step 3: Calculate Final Potential Energy \( U_2 \)
When the body is lifted to a height \( nR \), the distance from the center of the Earth becomes \( R + nR = (n + 1)R \). Thus, the potential energy \( U_2 \) at this height is:
\[ U_2 = -\frac{G M m}{(n + 1)R} \]

### Step 4: Calculate the Work Done \( W \)
The work done \( W \) in lifting the mass is the change in potential energy:
\[ W = U_2 - U_1 \]
Substituting the values of \( U_1 \) and \( U_2 \):
\[ W = \left(-\frac{G M m}{(n + 1)R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ W = -\frac{G M m}{(n + 1)R} + \frac{G M m}{R} \]
\[ W = G M m \left(\frac{1}{R} - \frac{1}{(n + 1)R}\right) \]
\[ W = G M m \left(\frac{(n + 1) - 1}{(n + 1)R}\right) \]
\[ W = G M m \left(\frac{n}{(n + 1)R}\right) \]

### Step 5: Relate \( G \) and \( g \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
Thus, we can express \( G M \) as:
\[ G M = g R^2 \]
Substituting this back into our equation for work done:
\[ W = \frac{n}{(n + 1)R} \cdot g R^2 m \]
\[ W = \frac{n g R m}{n + 1} \]

### Final Result
The work done in taking a body of mass \( m \) to a height \( nR \) above the surface of the Earth is:
\[ W = \frac{n g R m}{n + 1} \]