A body of mass `m` is placed on the earth surface is taken to a height of `h=3R`, then, change in gravitational potential energy is

Options

A. `(mgh)/(R)`
B. `(2)/(3)mgR`
C. `(3)/(4)mgR`
D. `(mgR)/(2)`

Answer

To find the change in gravitational potential energy (ΔU) when a body of mass \( m \) is taken to a height \( h = 3R \) above the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We need to calculate the change in gravitational potential energy when a mass \( m \) is raised from the Earth's surface to a height \( h = 3R \), where \( R \) is the radius of the Earth.

2. **Identify the Relevant Formula**:
The change in gravitational potential energy when moving to a height \( h \) is generally given by:
\[ \Delta U = mg \Delta h \]
However, when \( h \) is significant compared to \( R \), we use the modified formula:
\[ \Delta U = mg \frac{h}{1 + \frac{h}{R}} \]

3. **Substitute the Values**:
Here, \( h = 3R \). We can substitute this into our formula:
\[ \Delta U = mg \frac{3R}{1 + \frac{3R}{R}} \]

4. **Simplify the Denominator**:
The denominator simplifies as follows:
\[ 1 + \frac{3R}{R} = 1 + 3 = 4 \]

5. **Final Calculation**:
Now substituting back into the equation:
\[ \Delta U = mg \frac{3R}{4} \]

6. **Conclusion**:
Therefore, the change in gravitational potential energy when the body is taken to a height of \( 3R \) is:
\[ \Delta U = \frac{3}{4} mgR \]

### Final Answer:
\[ \Delta U = \frac{3}{4} mgR \]