- #1
HarrisonColes
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How do you work out the distance of a geostationary satellite from the Earth's surface?
Cheers
Cheers
Geostationary Orbit: Calculating Distance from Earth
- Thread starter HarrisonColes
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Homework Help Overview
The discussion centers around calculating the distance of a geostationary satellite from the Earth's surface, specifically focusing on the principles of orbital mechanics and gravitational forces.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the relationship between centripetal acceleration and gravitational force for maintaining orbit. There are mentions of the necessary orbital speed and the concept of angular velocity in relation to the Earth's rotation.
Discussion Status
Some participants have provided insights into the equations governing geostationary orbits and the conditions required for a satellite to maintain such an orbit. There is an ongoing exploration of the mathematical relationships involved, but no consensus has been reached on a specific solution.
Contextual Notes
Participants reference the need for specific constants and values, such as the gravitational constant and the radius of the Earth, which may not have been provided in the original question. There is also a note about the distinction between a sidereal day and a solar day in the context of orbital calculations.
- #2
mgb_phys
Science Advisor
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Things stay in orbit when their centripedal accelearation ,pushing them out, matches the force of gravity pulling them in.
So for every orbital height there is a speed you have to travel out to maintain the orbit.
So you simply have to find the height at which the speed is exactly once around the Earth every 24hours (actaully slightly less than 24hours - see sidereal day )
The equations and constant you need are described here:
http://en.wikipedia.org/wiki/Geostationary_orbit
So for every orbital height there is a speed you have to travel out to maintain the orbit.
So you simply have to find the height at which the speed is exactly once around the Earth every 24hours (actaully slightly less than 24hours - see sidereal day )
The equations and constant you need are described here:
http://en.wikipedia.org/wiki/Geostationary_orbit
- #3
3029298
- 56
- 0
If a satellite is in a geostationary orbit, this means that the angular velocity of rotation of the satellite is the same as the angular velocity of rotation of the Earth. Because the satellite must be in a circular orbit to have a constant angular velocity, this means that the centripetal force [tex]m\omega^2r[/tex] on the satellite to remain in circular motion is in fact the gravity working on the satellite:
[tex] G\frac{Mm}{r^2}=m\omega^2r[/tex]
In this equation, G is the gravitational constant, M is the mass of the earth, m is the mass of the satellite, [tex]\omega[/tex] is the angular velocity of rotation of the satellite and r is the distance between the center of mass of the Earth and the satellite.
You want the angular velocity of rotation to be the same as for the Earth, so what you have to do is to plug in the known value for the angular velocity of rotation of the Earth, and then you can calculate r and thus you know the height of the satellite if you subtract the radius of the Earth from r.
[tex] G\frac{Mm}{r^2}=m\omega^2r[/tex]
In this equation, G is the gravitational constant, M is the mass of the earth, m is the mass of the satellite, [tex]\omega[/tex] is the angular velocity of rotation of the satellite and r is the distance between the center of mass of the Earth and the satellite.
You want the angular velocity of rotation to be the same as for the Earth, so what you have to do is to plug in the known value for the angular velocity of rotation of the Earth, and then you can calculate r and thus you know the height of the satellite if you subtract the radius of the Earth from r.
- #4
HarrisonColes
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Cheers guys, such good help!
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