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. `mgR2`
B. `(2)/(3)mgR`
C. `3mgR`
D. `(1)/(3) mgR`

Explanation

To find the change in potential energy of a body of mass \( m \) when it is 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 initial position of the body is at the surface of the Earth, which we can denote as \( r = R \) (where \( R \) is the radius of the Earth).
- The final position of the body is at a height of \( 2R \) above the Earth's surface. Therefore, the distance from the center of the Earth to the body at this height is \( R + 2R = 3R \).

### Step 2: Calculate the Initial Potential Energy
The gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where:
- \( G \) is the 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 surface of the Earth (initial position):
\[ U_i = -\frac{GMm}{R} \]

### Step 3: Calculate the Final Potential Energy
At the height of \( 2R \) (final position):
\[ U_f = -\frac{GMm}{3R} \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{GMm}{3R}\right) - \left(-\frac{GMm}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{3R} + \frac{GMm}{R} \]
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{3R} \]
To combine these fractions, we can find a common denominator:
\[ \Delta U = \frac{3GMm}{3R} - \frac{GMm}{3R} = \frac{2GMm}{3R} \]

### Step 5: Relate \( GM \) to \( g \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
Thus, we can express \( GM \) as:
\[ GM = gR^2 \]
Substituting this into our expression for change in potential energy:
\[ \Delta U = \frac{2gR^2 m}{3R} = \frac{2gRm}{3} \]

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

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