A body of mass m is placed on the earth's surface . It is taken from the earth's surface to a height h = 3 R when R is the radius of the earth. The change in gravitational potential energy of the body is

Options

A. `3/2mgR`
B. `3/4mgR`
C. `1/2mgR`
D. `1/4mgR`

Answer

To solve the problem of finding the change in gravitational potential energy when a body of mass \( m \) is taken from the Earth's surface to a height \( h = 3R \) (where \( R \) is the radius of the Earth), we can follow these steps:

### Step 1: Understand the Initial and Final Positions
- The initial position of the body is at the Earth's surface, which is at a distance \( R \) from the center of the Earth.
- The final position of the body is at a height of \( 3R \) above the Earth's surface, which means the total distance from the center of the Earth is \( R + 3R = 4R \).

### Step 2: Calculate the Initial 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, and \( m \) is the mass of the body.

At the Earth's surface (initial position):
\[ U_{\text{initial}} = -\frac{GMm}{R} \]

### Step 3: Calculate the Final Gravitational Potential Energy
At the height of \( 3R \) (final position):
\[ U_{\text{final}} = -\frac{GMm}{4R} \]

### Step 4: Calculate the Change in Gravitational Potential Energy
The change in gravitational potential energy \( \Delta U \) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{GMm}{4R}\right) - \left(-\frac{GMm}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{4R} + \frac{GMm}{R} \]
To combine these fractions, we can find a common denominator (which is \( 4R \)):
\[ \Delta U = -\frac{GMm}{4R} + \frac{4GMm}{4R} = \frac{3GMm}{4R} \]

### Step 5: Convert to Terms of \( g \)
We know that the acceleration due to gravity \( g \) at the Earth's surface is given by:
\[ g = \frac{GM}{R^2} \]
From this, we can express \( GM \) as:
\[ GM = gR^2 \]
Substituting this back into our expression for \( \Delta U \):
\[ \Delta U = \frac{3(gR^2)m}{4R} = \frac{3mgR}{4} \]

### Final Answer
Thus, the change in gravitational potential energy when the body is taken to a height of \( 3R \) is:
\[ \Delta U = \frac{3}{4} mgR \]