If body of mass `m`has to be taken from the surface to the earth to a height `h=4R`, then the amount of energy required is (R = radius of the earth)

Options

A. `mgR`
B. `(mgR)/(5)`
C. `(4mgR)/(5)`
D. `(mgR)/(12)`

Explanation

To solve the problem of calculating the energy required to take a body of mass `m` from the surface of the Earth to a height `h = 4R` (where `R` is the radius of the Earth), we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
We will use the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) remains constant in a conservative force field like gravity.

### Step 2: Define the Initial and Final States
- **Initial State (Point 1)**: The body is at the surface of the Earth (height `h = 0`).
- **Final State (Point 2)**: The body is at a height `h = 4R` above the Earth's surface.

### Step 3: Write the Expression for Potential Energy
The gravitational potential energy (U) at a height `h` from the Earth's center is given by:
\[ U = -\frac{GMm}{R + h} \]
Where:
- \( G \) = universal gravitational constant
- \( M \) = mass of the Earth
- \( m \) = mass of the body
- \( R \) = radius of the Earth

### Step 4: Calculate Potential Energy at Initial and Final Points
- **At Point 1 (Surface)**:
\[ U_1 = -\frac{GMm}{R + 0} = -\frac{GMm}{R} \]

- **At Point 2 (Height \( h = 4R \))**:
\[ U_2 = -\frac{GMm}{R + 4R} = -\frac{GMm}{5R} \]

### Step 5: Apply Conservation of Energy
According to the conservation of energy:
\[ K_1 + U_1 = K_2 + U_2 \]
Assuming the body starts from rest, the initial kinetic energy \( K_1 = 0 \) and the final kinetic energy \( K_2 = 0 \) (as it comes to rest at height \( 4R \)):
\[ 0 + U_1 = 0 + U_2 \]
Thus:
\[ U_1 = U_2 \]

### Step 6: Substitute the Values of Potential Energy
Substituting the expressions for potential energy:
\[ -\frac{GMm}{R} = -\frac{GMm}{5R} \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ \frac{GMm}{R} - \frac{GMm}{5R} = 0 \]
This simplifies to:
\[ \frac{GMm}{R} \left(1 - \frac{1}{5}\right) = \frac{GMm}{R} \cdot \frac{4}{5} \]

### Step 8: Calculate the Energy Required
The energy required \( E \) to move the body from the surface to height \( 4R \) is:
\[ E = U_2 - U_1 = -\frac{GMm}{5R} + \frac{GMm}{R} \]
\[ E = \frac{GMm}{R} \left(1 - \frac{1}{5}\right) = \frac{4GMm}{5R} \]

### Step 9: Substitute Gravitational Acceleration
We know that the gravitational acceleration at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
Thus, we can express \( GM \) as:
\[ GM = gR^2 \]

### Step 10: Final Expression for Energy
Substituting \( GM \) back into the energy equation:
\[ E = \frac{4gRm}{5} \]

### Final Answer
The amount of energy required to take the body to a height \( h = 4R \) is:
\[ E = \frac{4}{5} gRm \]