The radius of the earth`R` and acceleration due to gravity at its surface is `g`. If a body of mass `m` is sent to a height of `R//4` from the earth's surface, the potential energy increases by

Options

A. `(mgR)/3`
B. `(mgR)/4`
C. `(mgR)/5`
D. `(mgR)/16`

Explanation

To find the increase in potential energy when a body of mass \( m \) is raised to a height of \( \frac{R}{4} \) from the Earth's surface, we can follow these steps:

### Step 1: Understand the potential energy formula
The gravitational potential energy \( U \) of a mass \( m \) 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, and \( M \) is the mass of the Earth.

### Step 2: Calculate potential energy at the Earth's surface
At the surface of the Earth, the distance \( r \) is equal to the radius of the Earth \( R \). Thus, the potential energy \( U_{\text{surface}} \) is:
\[ U_{\text{surface}} = -\frac{G M m}{R} \]

### Step 3: Calculate potential energy at height \( \frac{R}{4} \)
When the body is raised to a height \( h = \frac{R}{4} \), the distance from the center of the Earth becomes:
\[ r = R + h = R + \frac{R}{4} = \frac{5R}{4} \]
Now, we can calculate the potential energy \( U_{\text{height}} \) at this new height:
\[ U_{\text{height}} = -\frac{G M m}{\frac{5R}{4}} = -\frac{4G M m}{5R} \]

### Step 4: Calculate the change in potential energy
The change in potential energy \( \Delta U \) when the body is raised to height \( \frac{R}{4} \) is given by:
\[ \Delta U = U_{\text{height}} - U_{\text{surface}} \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{4G M m}{5R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{4G M m}{5R} + \frac{5G M m}{5R} = \frac{G M m}{5R} \]

### Step 5: Express in terms of \( g \)
We know that at the surface of the Earth, the acceleration due to gravity \( g \) is given by:
\[ g = \frac{G M}{R^2} \]
Thus, we can express \( G M \) as:
\[ G M = g R^2 \]
Substituting this into our expression for \( \Delta U \):
\[ \Delta U = \frac{g R^2 m}{5R} = \frac{g R m}{5} \]

### Final Answer
The increase in potential energy when the body is raised to a height of \( \frac{R}{4} \) from the Earth's surface is:
\[ \Delta U = \frac{g R m}{5} \]