A body of mass `m` taken form the earth's surface to the height is equal to twice the radius `(R)` of the earth. The change in potential energy of body will be

Options

A. `mg2R`
B. `(2)/(3)mgR`
C. `3mgR`
D. `(1)/(3)mgR`

Answer

To solve the problem of finding the change in potential energy of a body of mass \( m \) taken from the Earth's surface to a height equal to twice the radius of the Earth, we can follow these steps:

### Step 1: Understand the Initial and Final Positions
The body is initially at the Earth's surface, which is at a distance \( R \) (the radius of the Earth) from the center of the Earth. The final position is at a height of \( 2R \) above the Earth's surface. Therefore, the distance from the center of the Earth to the final position is:
\[ \text{Final distance} = R + 2R = 3R \]

### Step 2: Calculate Initial Potential Energy
The gravitational potential energy \( U \) 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, \( M \) is the mass of the Earth, and \( m \) is the mass of the body.

For the initial position (at the surface of the Earth, \( r = R \)):
\[ U_{\text{initial}} = -\frac{G M m}{R} \]

### Step 3: Calculate Final Potential Energy
For the final position (at a distance of \( 3R \)):
\[ U_{\text{final}} = -\frac{G M m}{3R} \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{R} \]
\[ \Delta U = \frac{G M m}{R} - \frac{G M m}{3R} \]
To combine these fractions, we find a common denominator:
\[ \Delta U = \frac{3G M m}{3R} - \frac{G M m}{3R} = \frac{(3G M m - G M m)}{3R} = \frac{2G M m}{3R} \]

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

### Final Result
Thus, the change in potential energy of the body is:
\[ \Delta U = \frac{2}{3} m g R \]