A orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the altitude of the orbit (called a "synchronous orbit")?
Write the equation for the period of a satellite in terms of the distance from the Earth's center. It is a special form of Kepler's Third law. Then set the period equal to 24 hours and solve for the distance from the center. Subtract the Earth's radius to get the altitude.
This reference may help:
http://www.ctiinfo.com/SatControl/ComTrack/InclinedOrbitTutorial/satgeom1.htm
ok thanks.
To find the altitude of a synchronous orbit, we need to use Kepler's third law and set the period of the satellite equal to 24 hours. Let's break down the steps.
Step 1: Start with Kepler's Third Law
Kepler's third law states that the square of a satellite's orbital period is proportional to the cube of its average distance from the center of the Earth. Mathematically, it can be expressed as:
T^2 = k * r^3
where T is the period of the satellite, r is the distance from the center of the Earth to the satellite, and k is a constant.
Step 2: Set the satellite's period equal to 24 hours
Since we want the satellite to stay over the same spot on the equator, its period should match the Earth's rotational period, which is 24 hours.
T^2 = 24^2
Step 3: Solve for the distance from the center
Based on the equation from Step 1, we can solve for r:
24^2 = k * r^3
Step 4: Subtract the Earth's radius to get the altitude
The distance from the center of the Earth to the surface is the sum of the satellite's altitude and the Earth's radius. We need to subtract the Earth's radius to find the altitude:
Altitude = r - Earth's radius
You can input the equation from Step 3 into a calculator or use a computer algebra system to solve for r. Once you have the value of r, you can subtract the Earth's radius to find the altitude. Refer to the provided link for additional information and calculations.