A satellite orbits the earth at a height of 400 km, above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence ? Mass of the satellite=200 kg, mass of the earth=`6.0xx10^(24)`kg, radius of the earth=`6.4xx10(6)`m, G=`6.67xx10^(-11)Nm^(2)Kg^(-2)`.

Answer

To find the energy required to rocket the satellite out of the Earth's gravitational influence, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the energy required to escape the Earth's gravitational influence for a satellite that is currently in orbit at a height of 400 km above the Earth's surface.

### Step 2: Convert Height into Meters
The height of the satellite above the Earth's surface is given as 400 km. We need to convert this into meters:
\[ h = 400 \, \text{km} = 400 \times 10^3 \, \text{m} = 4.0 \times 10^5 \, \text{m} \]

### Step 3: Calculate the Distance from the Center of the Earth
The radius of the Earth is given as \( R = 6.4 \times 10^6 \, \text{m} \). The total distance from the center of the Earth to the satellite is:
\[ r = R + h = 6.4 \times 10^6 \, \text{m} + 4.0 \times 10^5 \, \text{m} = 6.8 \times 10^6 \, \text{m} \]

### Step 4: Calculate the Gravitational Potential Energy
The gravitational potential energy (U) of the satellite in orbit can be calculated using the formula:
\[ U = -\frac{G M m}{r} \]
Where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant)
- \( M = 6.0 \times 10^{24} \, \text{kg} \) (mass of the Earth)
- \( m = 200 \, \text{kg} \) (mass of the satellite)
- \( r = 6.8 \times 10^6 \, \text{m} \)

Substituting the values:
\[ U = -\frac{(6.67 \times 10^{-11}) (6.0 \times 10^{24}) (200)}{6.8 \times 10^6} \]

Calculating this gives:
\[ U \approx -1.76 \times 10^{10} \, \text{J} \]

### Step 5: Calculate the Kinetic Energy
The kinetic energy (K) of the satellite in orbit can be calculated using the formula:
\[ K = \frac{1}{2} m v^2 \]
Where \( v \) is the orbital velocity given by:
\[ v = \sqrt{\frac{G M}{r}} \]

Calculating \( v \):
\[ v = \sqrt{\frac{(6.67 \times 10^{-11}) (6.0 \times 10^{24})}{6.8 \times 10^6}} \]
Calculating this gives:
\[ v \approx 7.9 \times 10^3 \, \text{m/s} \]

Now calculating the kinetic energy:
\[ K = \frac{1}{2} (200) (7.9 \times 10^3)^2 \]
Calculating this gives:
\[ K \approx 6.24 \times 10^6 \, \text{J} \]

### Step 6: Total Energy
The total mechanical energy (E) of the satellite is given by:
\[ E = K + U \]
Substituting the values:
\[ E = 6.24 \times 10^6 + (-1.76 \times 10^{10}) \]
Calculating this gives:
\[ E \approx -1.75 \times 10^{10} \, \text{J} \]

### Step 7: Energy Required to Escape
The energy required to escape the Earth's gravitational influence is equal to the negative of the total energy:
\[ E_{\text{escape}} = -E = 1.75 \times 10^{10} \, \text{J} \]

### Final Answer
The energy that must be expended to rocket the satellite out of the Earth's gravitational influence is approximately:
\[ E_{\text{escape}} \approx 1.75 \times 10^{10} \, \text{J} \]

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