# Calculating the height of a geostationary satellite of Earth

• kokodile

## Homework Statement

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

GM/r=v^2
r=R+h

## 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?

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?

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?
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

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
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?

Thansk everyone

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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?

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?

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

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

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

Well the larger the radius, the smaller the velocity of the satellite will be.
And yet remain geostationary?

And yet remain geostationary?
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.

So it has the same velocity as Earth.
No, not the same velocity. What must be the same?

No, not the same velocity. What must be the same?
The time it takes for one orbit?

The time it takes for one orbit?
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.