- #1
HarrisonColes
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How do you work out the distance of a geostationary satellite from the Earth's surface?
Cheers
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Geostationary Orbit: Calculating Distance from Earth
- Thread starter HarrisonColes
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To calculate the distance of a geostationary satellite from Earth's surface, one must ensure that the centripetal acceleration matches gravitational force. This involves determining the orbital speed required for a satellite to complete one orbit every 24 hours, factoring in the sidereal day. The relationship between gravitational force and centripetal force can be expressed using the equation G(Mm/r^2) = m(ω^2r), where G is the gravitational constant, M is Earth's mass, and ω is the satellite's angular velocity. By substituting the known angular velocity of Earth into this equation, one can solve for r, the distance from Earth's center, and subtract Earth's radius to find the satellite's height above the surface. Understanding these calculations is essential for positioning satellites in geostationary orbits.
- #2
mgb_phys
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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
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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 m\omega^2r on the satellite to remain in circular motion is in fact the gravity working on the satellite:
<br /> G\frac{Mm}{r^2}=m\omega^2r<br />
In this equation, G is the gravitational constant, M is the mass of the earth, m is the mass of the satellite, \omega 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.
<br /> G\frac{Mm}{r^2}=m\omega^2r<br />
In this equation, G is the gravitational constant, M is the mass of the earth, m is the mass of the satellite, \omega 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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