How High Is a Geostationary Satellite Above Earth?

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SUMMARY

The height of a geostationary satellite from the Earth's center is calculated to be 2.16 x 109 meters using the formula r3/T2 = GM/4π2. The period (T) for a geostationary satellite is 86,400 seconds, which corresponds to one full rotation of the Earth. This ensures that the satellite remains stationary relative to a point on the Earth's surface. The discussion clarifies the importance of using the correct orbital period for accurate calculations.

PREREQUISITES
  • Understanding of gravitational constant (G = 6.67 x 10-11 N·m2/kg2)
  • Knowledge of Earth's mass (M = 5.98 x 1024 kg)
  • Familiarity with orbital mechanics and geostationary satellites
  • Ability to perform cube root calculations
NEXT STEPS
  • Research the concept of synchronous satellites and their applications
  • Learn about the Lagrangian points and their significance in satellite positioning
  • Explore the mathematical derivation of orbital mechanics equations
  • Read about the practical implications of geostationary satellites in telecommunications
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Students studying physics, aerospace engineers, satellite communication professionals, and anyone interested in orbital mechanics and satellite technology.

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Homework Statement



What is the height of a geostationary satellite measured from the Earth's centre?

Homework Equations



r^3/T^2 = GM/4pi^2

T of Earth = 3.16 * 10^7 s

The Attempt at a Solution



GM/4p2 = r3/T2
3root[T2 * GM/4p2 ] = r
3root[(3.16*107 s)2 * ((6.67*10^ -11N * m2 /kg2 )(5.98*1024 kg)/4p2 )] = r
2.16*109m = r
Therefore the height of the satellite from the Earth’s center is 2.16*109m.
 
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Why are you using a year as the period of the satellite?
 
gneill said:
Why are you using a year as the period of the satellite?

because in my textbook it gives 3.16*10^7 s as the period of revolution of orbit for earth, so since the satellite is geocentric its period is the same
 
A geostationary satellite has an orbit that keeps pace with the rotation of the Earth -- a daily rotation -- so that it remains over the same geographical location.
 
gneill said:
A geostationary satellite has an orbit that keeps pace with the rotation of the Earth -- a daily rotation -- so that it remains over the same geographical location.

ok well then i don't mean geocentric i just mean that it has the same orbit period as the Earth
 
Dude, your calculations are fine you are using the wrong period T.
Geostationary orbit means when you look up from Earth it appears the satellite isn't moving.
For this to be the case, the satellite must revolve around Earth at the same period as Earth rotates about it's axis.
This period is 24 hours x 60 minutes x 60 seconds = 86400 seconds.
T = 86400 seconds.
 
Jonnyb42 said:
Dude, your calculations are fine you are using the wrong period T.
Geostationary orbit means when you look up from Earth it appears the satellite isn't moving.
For this to be the case, the satellite must revolve around Earth at the same period as Earth rotates about it's axis.
This period is 24 hours x 60 minutes x 60 seconds = 86400 seconds.
T = 86400 seconds.

the question calls it a synchronous satellite...is this the same thing
 
I think you'd be better putting such a satellite at one of the http://en.wikipedia.org/wiki/Lagrangian_point" .
 
Last edited by a moderator:
Hi there,

For a full explanation of how to calculate the height of a geostationary satellite, check out a blog post I wrote on this very subject:

http://davechessgames.blogspot.com/2011/01/maths-problems-5-geostationary.html

This uses the formulae and the constants you've been given - it also calculates the distance from the centre of the Earth to the satellite, and also from the Earth's surface to the satellite.

Any problems, leave me a comment on the blog!

Cheers

Dave
 

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