# How do you calculate the altitude of a satellite in geosynchronous orbit?

Using the equation for gravitational force and the equation for centripetal force?

So... m(v)^2/r = G(m1m2/d^2)

We started off in class with m(2*pi*r/86400 s)^2/r = G(m1m2/d^2)

What now? Thanks.

### 2 Answers

- kirchweyLv 710 years agoBest Answer
First, d = r, and let's use M = central mass, m = orbiting mass, so rewrite the equations and clean up the parentheses:

m(2*pi*r/86400)^2/r = GMm/r^2

Rearrange, grouping like terms, canceling where possible:

(2*pi/86400)^2 = GM/r^3

r^3 = GM/(2*pi/86400)^2

r = (GM/(2*pi/86400)^2)^(1/3)

Now set G = 6.67428E-11 N-m^2/kg^2, M(earth) = 5.9742E+24 kg, and (ta!)

r = (6.67428E-11*5.9742E+24 /(2*pi/86400)^2)^(1/3) = 42245843 m

Altitude h = r-re = 42245843-6378100 = 35867743 m

Now I hate to toss in a monkey wrench, but the actual rotation period of the earth is not exactly 24 h or 86400 s. The 24-h day is the average time between sightings of the sun in the same part of the sky, but each day our direction to the sun changes by about 1 degree because we are orbiting around it. A "sidereal day" is the time it takes to make one rotation relative to the distant stars, and it is this rotation rate that determines centripetal force. Google says the sidereal day is 86164.0905 s. If we plug that time into the equation, r = 42168907 m and h = 35790807 m, not a big difference but in better agreement with the more authoritative sources like wiki (ref.)..