# Calculating the height of a geostationary satellite of Earth

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1. Feb 24, 2016

### kokodile

1. The problem statement, all variables and given/known data
Calculate the height of a geo-stationary satellite of earth.

Gravitational force of earth=6.667 x 10^-11 nm^2/kgm^2
Mass of earth=6x10^24 kgm
V=86400
2. Relevant equations
GM/r=v^2
r=R+h

3. The attempt at a solution
I plugged everything into the equation and got 53,583.6 for r. Then since I need only height of the satellite to earth, I subtracted 6400 km from 53,583.6 and I got 47,183 for h. However, my professor said the answer for r should be around 36,000 km. Here is what I did.

(6.667x10^-11)(6x10^24)/r=86400^2

From multiplying, I got 4x10^14/r=7464960000

From this, I got r=53,583.6
Then h=47,183.

Are my calculations incorrect?

2. Feb 24, 2016

### haruspex

Looks to me like you calculated the number of seconds in a day and set that as a velocity in m/s. Have I misunderstood your working?

3. Feb 24, 2016

### kokodile

I just realized that as well. I can't believe I've been staring at this problem for half an hour without seeing this. Thanks

4. Feb 24, 2016

### SteamKing

Staff Emeritus
That's why it's important to carry the units through your calculations. If you don't know what the numbers represent, how can anyone else know?

5. Feb 24, 2016

### kokodile

Thansk everyone

Last edited: Feb 24, 2016
6. Feb 24, 2016

### kokodile

I still can't figure this problem out. If I need to find the velocity to calculate r, how can I do that when I need r to calculate velocity?

7. Feb 24, 2016

### haruspex

How are the radius and velocity related for a geostationary satellite?

8. Feb 24, 2016

### kokodile

Well the larger the radius, the smaller the velocity of the satellite will be.

9. Feb 24, 2016

### haruspex

And yet remain geostationary?

10. Feb 24, 2016

### kokodile

Yes. So it has the same velocity as Earth. But I'm still confused as to how to find the velocity of earth. Even if I use the equation 2pir/T, I get .465 m/s. And when I plug that into the equation GM/r=v^2, I get a very large number that must be incorect.

11. Feb 24, 2016

### haruspex

No, not the same velocity. What must be the same?

12. Feb 24, 2016

### kokodile

The time it takes for one orbit?

13. Feb 24, 2016

### haruspex

Well, not quite right. The time for one orbit of the satellite must equal what?
From that, write an equation relating radius of orbit to velocity.