If g be the acceleration due to gravity of the earth's surface, the gain is the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

Options

A. (1/2) m g R
B. 2 m g R
C. m g R
D. (1/4) m g R

Answer

To solve the problem of finding the gain in potential energy of an object of mass \( m \) raised from the surface of the Earth to a height equal to the radius \( R \) of the Earth, we can follow these steps:

### Step 1: Understand the Initial and Final Positions
- The object is initially at the surface of the Earth, which we can denote as height \( h_1 = 0 \).
- The object is raised to a height equal to the radius of the Earth, so the final height is \( h_2 = R \).

### Step 2: Calculate the Initial Potential Energy
The 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 object,
- \( r \) is the distance from the center of the Earth.

At the surface of the Earth, the distance \( r \) is equal to the radius \( R \):
\[ U_1 = -\frac{GMm}{R} \]

### Step 3: Calculate the Final Potential Energy
When the object is raised to a height equal to the radius of the Earth, the distance from the center of the Earth becomes \( r = 2R \):
\[ U_2 = -\frac{GMm}{2R} \]

### Step 4: Calculate the Gain in Potential Energy
The gain in potential energy \( \Delta U \) is the difference between the final and initial potential energies:
\[ \Delta U = U_2 - U_1 \]
Substituting the values from Steps 2 and 3:
\[ \Delta U = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} \]
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R} \]
\[ \Delta U = \frac{GMm}{2R} \]

### Step 5: Relate \( GM/R^2 \) 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:
\[ GM = gR^2 \]

### Step 6: Substitute \( GM \) in the Gain in Potential Energy
Now, substituting \( GM \) in the expression for \( \Delta U \):
\[ \Delta U = \frac{gR^2 m}{2R} = \frac{gRm}{2} \]

### Final Answer
The gain in potential energy of the object when raised to a height equal to the radius of the Earth is:
\[ \Delta U = \frac{gRm}{2} \]