A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of earth. What is the potential due to earth's gravity at the site of this satellite? ( Take the potential energy at infinity to be zero). Mass of the earth `6.0 xx 10^(24) kg`, radius = 6400km.

Explanation

To find the gravitational potential at the site of a geostationary satellite, we can use the formula for gravitational potential \( V \) at a distance \( r \) from the center of the Earth:

\[ V = -\frac{G M}{r} \]

where:
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the Earth, given as \( 6.0 \times 10^{24} \, \text{kg} \)
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 1: 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.

Given:
- Radius of the Earth \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- Height of the satellite \( h = 36000 \, \text{km} = 36000 \times 10^3 \, \text{m} \)

Now, calculate \( r \):

\[ r = R + h = (6400 \times 10^3) + (36000 \times 10^3) = 6400 \times 10^3 + 36000 \times 10^3 = 42600 \times 10^3 \, \text{m} \]

### Step 2: Substitute values into the potential formula

Now we substitute \( G \), \( M \), and \( r \) into the potential formula:

\[ V = -\frac{G M}{r} \]

Substituting the values:

\[ V = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24})}{42600 \times 10^3} \]

### Step 3: Calculate the potential \( V \)

Now we perform the calculation:

\[ V = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24})}{42600 \times 10^3} \]

Calculating the numerator:

\[ 6.67 \times 10^{-11} \times 6.0 \times 10^{24} = 4.002 \times 10^{14} \]

Now calculate the denominator:

\[ 42600 \times 10^3 = 4.26 \times 10^7 \]

Now divide the two results:

\[ V = -\frac{4.002 \times 10^{14}}{4.26 \times 10^7} \approx -9.39 \times 10^6 \, \text{J/kg} \]

### Final Answer

The potential due to Earth's gravity at the site of the geostationary satellite is approximately:

\[ V \approx -9.39 \times 10^6 \, \text{J/kg} \]