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. `2/3mgR`
C. `3mgR`
D. `1/3mgR`

Explanation

To find 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 \( (2R) \), we can follow these steps:

### Step 1: Understand the Potential Energy Formula
The gravitational potential energy \( U \) of a mass \( m \) 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,
- \( r \) is the distance from the center of the Earth.

### Step 2: Calculate Initial Potential Energy at the Earth's Surface
At the surface of the Earth, the distance \( r \) is equal to the radius of the Earth \( (R) \). Therefore, the initial potential energy \( U_i \) is:
\[ U_i = -\frac{G M m}{R} \]

### Step 3: Calculate Potential Energy at Height \( 2R \)
When the body is taken to a height equal to twice the radius of the Earth, the total distance from the center of the Earth becomes \( R + 2R = 3R \). Thus, the potential energy \( U_f \) at this height is:
\[ U_f = -\frac{G M m}{3R} \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by the final potential energy minus the initial potential energy:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{R} \]
To combine these fractions, we can write:
\[ \Delta U = \frac{G M m}{R} - \frac{G M m}{3R} = \frac{G M m}{R} \left(1 - \frac{1}{3}\right) = \frac{G M m}{R} \cdot \frac{2}{3} \]
Thus, we have:
\[ \Delta U = \frac{2 G M m}{3R} \]

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

### Final Answer
The change in potential energy of the body when taken to a height equal to twice the radius of the Earth is:
\[ \Delta U = \frac{2 g m R}{3} \]