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

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

A) \( -\frac{mgR}{3} \)
B) \( \frac{2mgR}{3} \)
C) \( \frac{mgR}{3} \)
D) \( \frac{2mgR}{3} \)

Answer: B) \( \frac{2mgR}{3} \)

Step 1: Change in gravitational potential energy

The gravitational potential energy of a body at a distance \( r \) from the center of the earth is given by:

\[ U = -\frac{GMm}{r} \]

Here, \( M \) is the mass of the earth, \( G \) is the gravitational constant, \( m \) is the mass of the body, and \( r \) is the distance from the earth's center.

Step 2: Potential energy at the earth's surface

At the earth's surface, the distance from the center is \( r = R \). Therefore:

\[ U_1 = -\frac{GMm}{R} \]

Step 3: Potential energy at height \( h = 2R \)

At a height equal to twice the earth's radius, the total distance from the center is:

\[ r = R + h = R + 2R = 3R \]

The potential energy at this height is:

\[ U_2 = -\frac{GMm}{3R} \]

Step 4: Change in potential energy

The change in potential energy is given by:

\[ \Delta U = U_2 - U_1 \]

Substitute \( U_2 = -\frac{GMm}{3R} \) and \( U_1 = -\frac{GMm}{R} \):

\[ \Delta U = -\frac{GMm}{3R} - \left( -\frac{GMm}{R} \right) \]

Simplify the terms:

\[ \Delta U = -\frac{GMm}{3R} + \frac{GMm}{R} \]

Take a common factor of \( \frac{GMm}{R} \):

\[ \Delta U = \frac{GMm}{R} \left( 1 - \frac{1}{3} \right) \]

Simplify the bracket:

\[ \Delta U = \frac{GMm}{R} \left( \frac{2}{3} \right) \]

Final Answer: B) \( \frac{2mgR}{3} \)