If a body of mass m is raised to height 2 R from the earth`s surface, then the change in potential energy of the body is (R is the radius of earth)

Options

A. mgR
B. 12 mgR
C. `2/3 mgR`
D. 9 mgR

Answer

To find the change in potential energy of a body of mass \( m \) raised to a height of \( 2R \) from the Earth's surface, 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 we can denote as height \( h_1 = 0 \).
- The final position of the body is at a height \( h_2 = 2R \) from the Earth's surface.

### Step 2: Calculate the gravitational potential energy at the initial position
The gravitational potential energy (U) 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.

At the Earth's surface, the distance \( r_1 \) is equal to the radius of the Earth \( R \):
\[ U_1 = -\frac{G M m}{R} \]

### Step 3: Calculate the gravitational potential energy at the final position
At the height of \( 2R \), the distance \( r_2 \) from the center of the Earth is:
\[ r_2 = R + 2R = 3R \]
Thus, the gravitational potential energy at this height is:
\[ U_2 = -\frac{G M m}{3R} \]

### Step 4: Calculate the change in potential energy
The change in potential energy (\( \Delta U \)) is given by:
\[ \Delta U = U_2 - U_1 \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{R} \]
\[ \Delta U = G M m \left(-\frac{1}{3R} + \frac{1}{R}\right) \]
\[ \Delta U = G M m \left(\frac{-1 + 3}{3R}\right) \]
\[ \Delta U = G M m \left(\frac{2}{3R}\right) \]

### Step 5: Express in terms of \( g \)
We know that \( g = \frac{G M}{R^2} \). Thus, we can express \( G M \) as \( g R^2 \):
\[ \Delta U = \frac{2}{3R} \cdot g R^2 \cdot m \]
\[ \Delta U = \frac{2}{3} g m R \]

### Final Answer
The change in potential energy of the body when raised to a height of \( 2R \) from the Earth's surface is:
\[ \Delta U = \frac{2}{3} g m R \]