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 where 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 formula for gravitational potential energy
The gravitational potential energy \( U \) of a body 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 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 potential energy at the Earth's surface
At the Earth's surface, the distance \( r \) is equal to \( R \) (the radius of the Earth). Therefore, the potential energy \( U_1 \) at the surface is:
\[ U_1 = -\frac{G M m}{R} \]

### Step 3: Calculate the potential energy at height \( h = 3R \)
When the body is taken to a height \( h = 3R \), the distance from the center of the Earth becomes:
\[ r = R + h = R + 3R = 4R \]
Thus, the potential energy \( U_2 \) at this height is:
\[ U_2 = -\frac{G M m}{4R} \]

### Step 4: Calculate the change in gravitational potential energy
The change in gravitational potential energy \( \Delta U \) is given by:
\[ \Delta U = U_2 - U_1 \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{G M m}{4R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{4R} + \frac{G M m}{R} \]
To combine these fractions, we can express \( \frac{G M m}{R} \) with a common denominator of \( 4R \):
\[ \Delta U = -\frac{G M m}{4R} + \frac{4G M m}{4R} = \frac{3G M m}{4R} \]

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

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