Calculating Altitude in Geostationary Orbit

In summary, the satellite is orbiting the Earth at a height of 10,000 km above the surface. To find the height of the satellite above the Earth, you need to use Kepler's third law equation and solve for r.
  • #1
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Homework Statement



A 10,000 kg satellite is rbiting the Earth in a geostationary orbit. The height of the satellite above the surface of the Earth is ?


Homework Equations



[tex] V = \omega r [/tex]

Newtons gravitational force equation

Keplers third law equation

The Attempt at a Solution



I really don't understand how to set this problem up. Here is what I am thinking. We know the radius of the Earth ( 6.37 x 10 ^6 m) so all we need to find is how much above is the satellite outside of earth. That plus the radius of the Earth will give me to total radius. But what equations should I use. I don't think I know [tex] \omega [/tex] nor do I know the velocity. If I use Kepler's third law equation all I get is the distance to the geosynchronous orbit. Don't know what to do.
 
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  • #2
In orbit the force pulling the satelite down due to gravity is equal to the outward (centripetal force)

So for any orbital height there is a speed you need to go at to provide enough outward force. The only special thing about GSO is that the speed is exactly that needed to go around the Earth on 24hours

This is basically what Kepler's third law says - although you need the value of the constant in this case.
 
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  • #3
Yes, I kind of figured that since there was no other quantity given for time, but I still don't see or understand how we would find the radius - the height of the satellite above the earth.
 
  • #4
Kepler's law

[tex] T^2 = [ (4pi^2) / (GM_E) ] r^3 [/tex]

I would solve for r then?

but I am not sure if that gives me the distance TO the satellite.
 
  • #5
Correct - or you can just use:
Force due to gravity depends on radius F = GMm/r^2
Centripetal force depends on radius F = m w^2/r
(M is mass of Earth, m is mass of satelite)


Set the forces equal to each other and find the raidus
Note that r is radius of the orbit - ie measured from the centre of the Earth, to get orbital height (above sea level) you need the raious of the Earth as well
 
  • #6
WOW

I was making such a silly mistake.
I was doing what I said above - using Kepler's third law - but when I solved for r I just used r as my answer. Which was not correct at all. I needed to subtract The r I got from Kepler's law from the radius of the Earth to get my distance.

Such a silly mistake

Thanks for helping me out, much appreciated.
 

What is a geostationary orbit?

A geostationary orbit is a type of orbit around the Earth in which a satellite appears to stay in the same position above the Earth's surface. This is achieved by having the satellite orbit at the same speed and direction as the Earth's rotation, allowing it to maintain a fixed position relative to the ground.

Why is a geostationary orbit important?

A geostationary orbit is important because it allows satellites to provide continuous coverage of a specific region on Earth. This is particularly useful for communication and weather satellites, as well as for various types of Earth observation and remote sensing activities.

How high is a geostationary orbit?

A geostationary orbit is approximately 35,786 kilometers (22,236 miles) above the Earth's equator. This distance allows satellites to maintain a fixed position relative to the Earth's surface, as the gravitational pull from the Earth's mass is balanced by the centrifugal force of the satellite's orbital speed.

How long does it take for a satellite to complete one orbit in a geostationary orbit?

A satellite in a geostationary orbit takes approximately 24 hours to complete one orbit around the Earth. This is the same amount of time it takes for the Earth to complete one rotation on its axis, which is why the satellite appears to stay in the same position above the Earth's surface.

Are there any disadvantages to using a geostationary orbit?

One potential disadvantage of a geostationary orbit is that it is limited to specific latitudes near the Earth's equator. Satellites in this orbit cannot provide continuous coverage of the Earth's poles, making it less useful for certain types of Earth observation and remote sensing activities. Additionally, the high altitude of this orbit can lead to a longer signal delay, which can be problematic for certain types of communication applications.

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