The change in potential energy when a body of mass m is raised to a height n R from the earth surface is (R = Radius of earth )

Options

A. `mgR (n)/(n - 1)`
B. n m gR
C. `mgR (n^(2))/( n^(2) - 1)`
D. ` mgR (n)/(n + 1)`

Explanation

To find the change in potential energy when a body of mass \( m \) is raised to a height of \( nR \) from the Earth's surface, we can follow these steps:

### Step 1: Understand the Initial and Final Positions
- The initial position of the body is at the surface of the Earth, which is at a distance \( R \) (the radius of the Earth) from the center of the Earth.
- The final position of the body is at a height of \( nR \) above the Earth's surface. Therefore, the distance from the center of the Earth to the final position is \( R + nR = (n + 1)R \).

### Step 2: Write the Formula for Potential Energy
The gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the body.

### Step 3: Calculate Initial Potential Energy
The initial potential energy \( U_i \) at the surface of the Earth (distance \( R \)):
\[ U_i = -\frac{GMm}{R} \]

### Step 4: Calculate Final Potential Energy
The final potential energy \( U_f \) at a height of \( nR \) (distance \( (n + 1)R \)):
\[ U_f = -\frac{GMm}{(n + 1)R} \]

### Step 5: Find the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{GMm}{(n + 1)R}\right) - \left(-\frac{GMm}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{(n + 1)R} + \frac{GMm}{R} \]
Factoring out \( \frac{GMm}{R} \):
\[ \Delta U = \frac{GMm}{R} \left(1 - \frac{1}{n + 1}\right) \]

### Step 6: Simplify the Expression
Now, simplifying the term inside the parentheses:
\[ 1 - \frac{1}{n + 1} = \frac{(n + 1) - 1}{n + 1} = \frac{n}{n + 1} \]
Thus, we have:
\[ \Delta U = \frac{GMm}{R} \cdot \frac{n}{n + 1} \]

### Step 7: Express in Terms of \( g \)
We know that \( g = \frac{GM}{R^2} \). To express \( \Delta U \) in terms of \( g \), we can multiply and divide by \( R \):
\[ \Delta U = \frac{GMm}{R} \cdot \frac{n}{n + 1} = mgR \cdot \frac{n}{n + 1} \]

### Final Answer
Thus, the change in potential energy when a body of mass \( m \) is raised to a height of \( nR \) from the Earth's surface is:
\[ \Delta U = mgR \cdot \frac{n}{n + 1} \]