Geosynchronous Orbit (Math)

A geosynchronous or, more specifically, geostationary orbit is an orbit where your orbital period is equal to that of the gravitational body's "day", so you remain in the same spot over the planet consistently (This means that the gravitational force and the centripetal force are equal, which is true in any circular orbit.). The difference between geosynchronous and geostationary orbits is that geosynchronous orbits have an orbital period of 1 day, but geostationary orbits are equatorial as well.

So, in order to calculate the geostationary orbit around any given body, we must first create an equation with the force of gravity and the centripetal force. The force of gravity is equal to:

(GM₁M₂)/r²

where G is the gravitational constant (6.67384 • 10-11 m3 kg-1 s-2), M₁ is the mass of the body, M₂ is the mass of the satellite, and r is the distance between the center of mass of the planet and that of the satellite.

Now, because in a geostationary orbit the the gravitational force is the same as the centripetal force, we can put them in opposite sides of the equation:

GM₁M₂/r² = M₂ • v²/r

The masses of the satellite cancel out, so we are left with:

GM₁/r² = v²/r

So, no matter how large your satellite is, the geostationary altitude will be the same.

We know that velocity is equal to distance divided by time. In this case, the distance is the circumference of your orbit, which is 2πr, and time in this instance is your orbital period, or one day for the planetary body. So, we now have:

v = 2πr/t But we are dealing with velocity squared, so we square our our equation and get:

v² = 4π²r²/t²

Plugging that back into the original equation, we get:

GM₁/r² = 4π²r²/t²r

The r² cancels out the r in the denominator and becomes a plain old r. Thus, we have:

GM₁/r² = 4π²r/t²

We now multiply by r³:

GM₁ = 4π²t² r³

And multiply by t²/4π²:

r^{3 = GM₁t2/4π2}

And by taking the cube root of that, we arrive at our answer (well, sort of):

r = ³√(GM₁t²/4π²)

But remember how r is the distance from the center of mass of the planet to that of the satellite? Well, it turns out that, unlike satellites, the center of mass of a planet is rather far away from its surface.

Always remember to subtract the radius of the planet from your answer once you find r.

Note: You probably noticed that the gravitational constant is in three units at once, two of them to a negative power. Don't panic, it will all work out when you plug in real numbers.