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

Options

A. m g h
B. (4/5) m g h
C. (5/6) m g h
D. (6/7) m g h

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 Initial and Final Potential Energy
The potential energy \( P \) of a mass \( m \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ P = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

At the surface of the Earth (where \( r = R \)):
\[ P_{\text{initial}} = -\frac{G M m}{R} \]

### Step 2: Calculate the Final Potential Energy
When the body rises to a height \( h = \frac{R}{5} \), the distance from the center of the Earth becomes:
\[ r + h = R + \frac{R}{5} = \frac{6R}{5} \]

Thus, the final potential energy is:
\[ P_{\text{final}} = -\frac{G M m}{\frac{6R}{5}} = -\frac{5 G M m}{6R} \]

### Step 3: Calculate the Change in Potential Energy
The change in potential energy \( \Delta P \) is given by:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} \]
Substituting the values we found:
\[ \Delta P = \left(-\frac{5 G M m}{6R}\right) - \left(-\frac{G M m}{R}\right) \]
\[ \Delta P = -\frac{5 G M m}{6R} + \frac{6 G M m}{6R} \]
\[ \Delta P = \frac{G M m}{6R} \]

### Step 4: Relate to Acceleration due to Gravity
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
We can express \( G M \) in terms of \( g \) and \( R \):
\[ G M = g R^2 \]

### Step 5: Substitute \( G M \) into the Change in Potential Energy
Substituting \( G M \) into our expression for \( \Delta P \):
\[ \Delta P = \frac{g R^2 m}{6R} = \frac{g m R}{6} \]

### Step 6: Substitute \( R \) in terms of \( h \)
Since \( h = \frac{R}{5} \), we can express \( R \) as:
\[ R = 5h \]
Substituting this into the expression for \( \Delta P \):
\[ \Delta P = \frac{g m (5h)}{6} = \frac{5 g m h}{6} \]

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