# Geostationary Orbit Question

1. Oct 20, 2009

### Want to learn

1. The problem statement, all variables and given/known data

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 ???

2. Relevant equations

$$V = \omega r$$

Newtons gravitational force equation

Keplers third law equation

3. 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 $$\omega$$ 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.

2. Oct 20, 2009

### mgb_phys

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.

Last edited: Oct 20, 2009
3. Oct 20, 2009

### Want to learn

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. Oct 20, 2009

### Want to learn

Kepler's law

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

I would solve for r then?

but I am not sure if that gives me the distance TO the satellite.

5. Oct 20, 2009

### mgb_phys

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. Oct 20, 2009

### Want to learn

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.