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

Options

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

Explanation

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

### Step 1: Understand the formula for gravitational potential energy
The gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the body,
- \( r \) is the distance from the center of the Earth.

### Step 2: Calculate the initial potential energy at the surface of the Earth
At the surface of the Earth, the distance \( r \) is equal to the radius of the Earth \( R \). Therefore, the initial potential energy \( U_1 \) is:
\[ U_1 = -\frac{GMm}{R} \]

### Step 3: Calculate the potential energy at the height \( h \)
When the body rises to a height \( h = \frac{R}{5} \), the new distance from the center of the Earth becomes:
\[ r + h = R + \frac{R}{5} = \frac{6R}{5} \]
Now, we can calculate the potential energy \( U_2 \) at this height:
\[ U_2 = -\frac{GMm}{\frac{6R}{5}} = -\frac{5GMm}{6R} \]

### Step 4: Calculate the increase in potential energy
The increase in potential energy \( \Delta U \) as the body moves from the surface to the height \( h \) is given by:
\[ \Delta U = U_2 - U_1 \]
Substituting the values from Steps 2 and 3:
\[ \Delta U = -\frac{5GMm}{6R} - \left(-\frac{GMm}{R}\right) \]
\[ \Delta U = -\frac{5GMm}{6R} + \frac{6GMm}{6R} \]
\[ \Delta U = \frac{(6 - 5)GMm}{6R} = \frac{GMm}{6R} \]

### Step 5: Express the result in terms of \( g \)
At the surface of the Earth, the acceleration due to gravity \( g \) is given by:
\[ g = \frac{GM}{R} \]
Thus, we can rewrite the increase in potential energy:
\[ \Delta U = \frac{g m}{6} \]

### Final Answer
The increase in potential energy when the body rises to a height \( h = \frac{R}{5} \) is:
\[ \Delta U = \frac{g m}{6} \]