How High Is a Geostationary Satellite Above Earth?

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Homework Help Overview

The discussion revolves around determining the height of a geostationary satellite from the Earth's center, utilizing gravitational equations and the period of Earth's rotation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to use the gravitational relationship involving the period of the Earth to calculate the satellite's height. Some participants question the choice of period used in the calculations, specifically whether it should be based on the Earth's rotation period.

Discussion Status

Participants are actively discussing the correct period to use for a geostationary satellite, with some clarifying that it should match the Earth's rotational period of 86400 seconds. There is an exploration of terminology, such as the distinction between geostationary and synchronous satellites.

Contextual Notes

There is mention of a textbook reference for the period of revolution, which may lead to confusion regarding the correct value to use in calculations. Additionally, a suggestion is made regarding Lagrangian points, although its relevance to the current problem is not fully explored.

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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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