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 positions
- The initial position (A) of the body is at the Earth's surface, which is at a distance \( R \) from the center of the Earth.
- The final position (B) of the body is at a height of \( 2R \) above the Earth's surface. Therefore, the distance from the center of the Earth to this position is \( R + 2R = 3R \).

### Step 2: Write the formula for gravitational potential energy
The gravitational potential energy (U) of a mass \( m \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 3: Calculate the potential energy at position A
At position A (Earth's surface, distance \( R \)):
\[ U_A = -\frac{G M m}{R} \]

### Step 4: Calculate the potential energy at position B
At position B (height \( 2R \), distance \( 3R \)):
\[ U_B = -\frac{G M m}{3R} \]

### Step 5: Find the change in potential energy
The change in potential energy (\( \Delta U \)) is given by:
\[ \Delta U = U_B - U_A \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{R} \]
\[ \Delta U = \frac{G M m}{R} - \frac{G M m}{3R} \]
Finding a common denominator:
\[ \Delta U = \frac{3G M m}{3R} - \frac{G M m}{3R} = \frac{2G M m}{3R} \]

### Step 6: Express in terms of \( mg \)
We know that \( g = \frac{G M}{R^2} \). Therefore, \( G M = g R^2 \).
Substituting this into the equation:
\[ \Delta U = \frac{2(g R^2) m}{3R} = \frac{2g m R}{3} \]

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