Potential energy of a satellite having mass `m` and rotating at a height of `6.4xx10^(6)m` from the earth surface is

Options

A. `-0.5mgR_(e)`
B. `-mgR_(e)`
C. `-2mgR_(e)`
D. `4mgR_(e)`

Explanation

To find the potential energy of a satellite of mass \( m \) rotating at a height of \( 6.4 \times 10^6 \) m from the Earth's surface, we can follow these steps:

### Step 1: Identify the relevant formula for gravitational potential energy
The gravitational potential energy \( U \) of a satellite in orbit is given by the formula:
\[ U = -\frac{G M m}{r} \]
where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)),
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Calculate the distance \( r \)
The distance \( r \) is the sum of the Earth's radius and the height of the satellite above the Earth's surface. The average radius of the Earth \( R_e \) is approximately \( 6.4 \times 10^6 \, \text{m} \). Therefore:
\[ r = R_e + h \]
Given that \( h = 6.4 \times 10^6 \, \text{m} \):
\[ r = 6.4 \times 10^6 + 6.4 \times 10^6 = 12.8 \times 10^6 \, \text{m} \]

### Step 3: Substitute the values into the potential energy formula
Now, substituting \( r \) into the potential energy formula:
\[ U = -\frac{G M m}{12.8 \times 10^6} \]

### Step 4: Express the potential energy in terms of \( g \)
To express the potential energy in terms of the acceleration due to gravity \( g \), we know that:
\[ g = \frac{G M}{R_e^2} \]
Thus, we can rewrite \( U \):
\[ U = -\frac{G M m}{r} = -\frac{G M m}{2 R_e} \quad \text{(since } r = 2R_e \text{)} \]
This can be expressed as:
\[ U = -\frac{1}{2} g m R_e \]

### Final Answer
The potential energy of the satellite is:
\[ U = -0.5 \, g \, m \, R_e \]