How to determine the height of a geostationary satellite from the Earth’s surface using Kepler's 3rd Law
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How to determine the height of a geostationary satellite from the Earth’s surface using Kepler's 3rd Law

[From: ] [author: ] [Date: 11-08-29] [Hit: ]
This means that it has to orbit just as fast as Earth rotates, just to keep up.G = 6.You plug that in.........
Using...

Kepler's 3rd Law:

r = Radius from the *center* of the object being orbited (in meters)
T = Orbital period (in seconds)
G = Universal Gravitational Constant
M = Mass of the object being orbited (in kilograms)
π = pi

Solving for orbital radius

r = { [ (T^2) * G * M ] / [ 4 * (π^2) ] } ^ (3/2) (cube root)

Since you can look up the values for Earth

M = 5.9742E+24 kg
T = 86,400 s (seconds in one 24-hour day)
Re = 6,378,100 m

Note that we know time as 86,400 seconds, because geostationary means that it's orbiting above the same spot on Earth at all times. This means that it has to orbit just as fast as Earth rotates, just to keep up.

Look up:

G = 6.6726E-11 N-m^2/kg^2

You plug that in...

... all of the units cancel down to an *orbital* radius of

r = 42,242,298 m

Now to determine the height above the surface, just take out that Earth radius from above

H = r - Re

And you get your answer

Height of a geostationary satellite is 35,864,198 meters. Check that against reality of approximately 37,500 kilometers, or about 22,000 miles... and you see that that's the answer!
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