A body of mass 2 m is placed one earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of h, where h=4R.

Options

A. `(2mgh)/R`
B. `2/3mgR`
C. `8/5 mgR`
D. `(mgR)/2`

Explanation

To solve the problem of calculating the change in gravitational potential energy when a body of mass \(2m\) is taken from the Earth's surface to a height of \(h = 4R\), we will follow these steps:

### Step 1: Understand the Initial and Final Positions
- The initial position of the body is at the Earth's surface, which is at a distance \(R\) from the center of the Earth.
- The final position of the body is at a height \(h = 4R\) above the Earth's surface. Therefore, the distance from the center of the Earth to the final position is:
\[ \text{Distance from center} = R + h = R + 4R = 5R \]

### Step 2: Write the Formula for 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:
\[ U = -\frac{GMm}{r} \]
where \(G\) is the gravitational constant and \(M\) is the mass of the Earth.

### Step 3: Calculate Initial Gravitational Potential Energy
The initial gravitational potential energy \(U_i\) when the body is on the Earth's surface (at distance \(R\)) is:
\[ U_i = -\frac{GM(2m)}{R} = -\frac{2GMm}{R} \]

### Step 4: Calculate Final Gravitational Potential Energy
The final gravitational potential energy \(U_f\) when the body is at height \(4R\) (at distance \(5R\) from the center) is:
\[ U_f = -\frac{GM(2m)}{5R} = -\frac{2GMm}{5R} \]

### Step 5: Calculate the Change in Gravitational Potential Energy
The change in gravitational potential energy \(\Delta U\) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{2GMm}{5R}\right) - \left(-\frac{2GMm}{R}\right) \]
\[ \Delta U = -\frac{2GMm}{5R} + \frac{2GMm}{R} \]
To combine these fractions, we need a common denominator, which is \(5R\):
\[ \Delta U = -\frac{2GMm}{5R} + \frac{10GMm}{5R} = \frac{8GMm}{5R} \]

### Step 6: Express the Result in Terms of \(g\)
We know that the acceleration due to gravity \(g\) is given by:
\[ g = \frac{GM}{R^2} \]
Thus, we can express \(GM\) in terms of \(g\):
\[ GM = gR^2 \]
Substituting this into our expression for \(\Delta U\):
\[ \Delta U = \frac{8(gR^2)m}{5R} = \frac{8g m R}{5} \]

### Final Answer
The change in gravitational potential energy when the body is taken from the Earth's surface to a height of \(4R\) is:
\[ \Delta U = \frac{8}{5} mgR \]