A body of mass is taken from earth surface to the height `h` equal to twice the radius of earth `(R_(e))`,the increase in potential energy will be : `(g=` acceleration due to gravity on the surface of Earth)

Options

A. `(1)/(2)mgR_(e)`
B. `3mgR_(e)`
C. `(2)/(3)mgR_(e)`
D. `(1)/(3)mgR_(e)`

Answer

To solve the problem of calculating the increase in potential energy when a body of mass \( m \) is taken from the Earth's surface to a height \( h \) equal to twice the radius of the Earth \( (R_e) \), we can follow these steps:

### Step 1: Understand the Initial and Final 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:
\[ U = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

### Step 2: Calculate the Initial Potential Energy
At the Earth's surface, the distance from the center of the Earth is \( R_e \). Therefore, the initial potential energy \( U_i \) is:
\[ U_i = -\frac{G M m}{R_e} \]

### Step 3: Calculate the Final Potential Energy
When the body is raised to a height \( h = 2R_e \), the distance from the center of the Earth becomes:
\[ r = R_e + h = R_e + 2R_e = 3R_e \]
Thus, the final potential energy \( U_f \) is:
\[ U_f = -\frac{G M m}{3R_e} \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = \left(-\frac{G M m}{3R_e}\right) - \left(-\frac{G M m}{R_e}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{3R_e} + \frac{G M m}{R_e} \]
\[ \Delta U = \frac{G M m}{R_e} - \frac{G M m}{3R_e} \]
\[ \Delta U = \left(\frac{3G M m}{3R_e} - \frac{G M m}{3R_e}\right) \]
\[ \Delta U = \frac{2G M m}{3R_e} \]

### Step 5: Express in Terms of \( g \)
We know that the acceleration due to gravity at the Earth's surface is given by:
\[ g = \frac{G M}{R_e^2} \]
Thus, we can express \( G M \) as:
\[ G M = g R_e^2 \]
Substituting this into our expression for \( \Delta U \):
\[ \Delta U = \frac{2(g R_e^2) m}{3R_e} = \frac{2g m R_e}{3} \]

### Final Answer
The increase in potential energy when the body is raised to a height \( h = 2R_e \) is:
\[ \Delta U = \frac{2g m R_e}{3} \]