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 find the amount of 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 Gravitational Potential Energy
The gravitational potential energy \( U \) of a mass \( m \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{G M_e m}{r} \]
where \( G \) is the gravitational constant and \( M_e \) is the mass of the Earth.

### Step 2: Calculate Initial Potential Energy
At the surface of the Earth (where \( r = R \)):
\[ U_{\text{initial}} = -\frac{G M_e m}{R} \]

### Step 3: Calculate Final Potential Energy
At a height \( h = 4R \), the distance from the center of the Earth is \( r = R + h = R + 4R = 5R \):
\[ U_{\text{final}} = -\frac{G M_e m}{5R} \]

### Step 4: Calculate Change in Potential Energy
The change in gravitational potential energy \( \Delta U \) when moving from the initial position to the final position is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M_e m}{5R}\right) - \left(-\frac{G M_e m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M_e m}{5R} + \frac{G M_e m}{R} \]
\[ \Delta U = \frac{G M_e m}{R} - \frac{G M_e m}{5R} \]
Finding a common denominator (which is \( 5R \)):
\[ \Delta U = \frac{5G M_e m}{5R} - \frac{G M_e m}{5R} = \frac{4G M_e m}{5R} \]

### Step 5: Relate Gravitational Constant to Acceleration Due to Gravity
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M_e}{R^2} \]
Thus, we can express \( G M_e \) as:
\[ G M_e = g R^2 \]

### Step 6: Substitute \( G M_e \) into the Energy Equation
Substituting \( G M_e \) into our change in potential energy:
\[ \Delta U = \frac{4(g R^2) m}{5R} = \frac{4g m R}{5} \]

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