What is the acceleration due to gravity in the orbit of a communication satellite at a height of 36,000 km?

Steps to Solve

1. Determine the radius of the Earth

The average radius of the Earth is approximately $R = 6,371$ km.

2. Calculate the distance from the center of the Earth

To find the total distance from the center of the Earth to the geostationary satellite, we need to add the radius of the Earth to the height of the satellite.

This gives us:

$$ d = R + \text{height} $$ $$ d = 6,371 \text{ km} + 36,000 \text{ km} $$ $$ d = 42,371 \text{ km} $$

3. Use the formula for gravitational acceleration

The formula for gravitational acceleration $g$ at a distance $d$ from the center of the Earth is given by:

$$ g = \frac{GM}{d^2} $$

Where:

• $G$ is the gravitational constant, approximately $6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}$
• $M$ is the mass of the Earth, approximately $5.972 \times 10^{24} \text{ kg}$

4. Convert distance to meters

Since the gravitational constant is in SI units, we must convert $d$ to meters:

$$ d = 42,371 \text{ km} = 42,371,000 \text{ m} $$

5. Calculate gravitational acceleration

Now we can calculate $g$ using the previously mentioned values:

$$ g = \frac{(6.674 \times 10^{-11}) (5.972 \times 10^{24})}{(42,371,000)^2} $$

6. Perform the final calculation

After performing the calculations, we find the value for $g$.