A body of mass `m` rises to a height `h=R//5` from the earth's surface where `R` is earth's radius. If `g` is acceleration due to gravity at the earth's surface, the increase in potential energy is

Options

A. `mgh`
B. `4/5mgh`
C. `5/6mgh`
D. `6/7mgh`

Explanation

To solve the problem of finding the increase in potential energy when a body of mass \( m \) rises to a height \( h = \frac{R}{5} \) from the Earth's surface, we can follow these steps:

### Step 1: Understand the change in height
The height \( h \) from the Earth's surface is given as:
\[ h = \frac{R}{5} \]
where \( R \) is the radius of the Earth.

### Step 2: Determine the distance from the center of the Earth
When the body is raised to height \( h \), the distance from the center of the Earth becomes:
\[ d = R + h = R + \frac{R}{5} = R \left(1 + \frac{1}{5}\right) = R \left(\frac{6}{5}\right) = \frac{6R}{5} \]

### Step 3: Write the formula for 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 4: Calculate initial and final potential energy
1. **Initial potential energy \( U_i \)** at the surface of the Earth (where \( d = R \)):
\[ U_i = -\frac{G M m}{R} \]

2. **Final potential energy \( U_f \)** at height \( h \) (where \( d = \frac{6R}{5} \)):
\[ U_f = -\frac{G M m}{\frac{6R}{5}} = -\frac{5 G M m}{6R} \]

### Step 5: Calculate the increase in potential energy
The increase in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{5 G M m}{6R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{5 G M m}{6R} + \frac{6 G M m}{6R} = \frac{G M m}{6R} \]

### Step 6: Relate \( G M \) to \( g \)
We know that at the surface of the Earth:
\[ g = \frac{G M}{R^2} \]
Thus,
\[ G M = g R^2 \]
Substituting this into our expression for \( \Delta U \):
\[ \Delta U = \frac{g R^2 m}{6R} = \frac{g m R}{6} \]

### Step 7: Substitute \( R \) in terms of \( h \)
Since \( h = \frac{R}{5} \), we can express \( R \) as:
\[ R = 5h \]
Substituting this into the potential energy equation gives:
\[ \Delta U = \frac{g m (5h)}{6} = \frac{5 g m h}{6} \]

### Final Answer
Thus, the increase in potential energy is:
\[ \Delta U = \frac{5 g m h}{6} \]