A body of mass m is lifted up from the surface of earth to a height three times the radius of the earth . The change in potential energy of the body is (g - gravity field at the surface of the earth )

Options

A. `mgR`
B. `(3)/(4) mgR`
C. `(1)/(3) mgR`
D. `(2)/(3) mgR`

Explanation

To find the change in potential energy of a body of mass \( m \) lifted from the surface of the Earth to a height of three times the radius of the Earth, we can follow these steps:

### Step 1: Define the initial and final positions
- The initial position of the body is at the surface of the Earth, which is at a distance \( R \) from the center of the Earth.
- The final position of the body is at a height of \( 3R \) above the surface, making the total distance from the center of the Earth \( R + 3R = 4R \).

### Step 2: Write the formula for gravitational potential energy
The gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the body.

### Step 3: Calculate the initial potential energy \( U_i \)
At the initial position (at the surface of the Earth, \( r = R \)):
\[ U_i = -\frac{GMm}{R} \]

### Step 4: Calculate the final potential energy \( U_f \)
At the final position (at a height of \( 3R \), \( r = 4R \)):
\[ U_f = -\frac{GMm}{4R} \]

### Step 5: Calculate the change in potential energy \( \Delta U \)
The change in potential energy is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{GMm}{4R}\right) - \left(-\frac{GMm}{R}\right) \]
\[ \Delta U = -\frac{GMm}{4R} + \frac{GMm}{R} \]
To combine these fractions, we can express \(\frac{GMm}{R}\) with a common denominator:
\[ \Delta U = -\frac{GMm}{4R} + \frac{4GMm}{4R} = \frac{3GMm}{4R} \]

### Step 6: Express \( \Delta U \) in terms of \( g \)
We know that the gravitational acceleration at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
From this, we can express \( GM \) as:
\[ GM = gR^2 \]
Substituting this back into our expression for \( \Delta U \):
\[ \Delta U = \frac{3(gR^2)m}{4R} = \frac{3gRm}{4} \]

### Final Result
The change in potential energy of the body when lifted to a height of three times the radius of the Earth is:
\[ \Delta U = \frac{3}{4} mgR \]

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