# 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" (specifically the sidereal time or sidereal rotation period), so you remain in the same spot over the planet consistently. Also the gravitational force and the centripetal force needs to be equal, which is the case for 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.

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:

$F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}$ where G is the gravitational constant ($6.67384\cdot 10^{-11}{\frac {m^{3}}{kg\cdot s^{2}}}$ ), 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.

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

$F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}=M_{2}\cdot {\frac {v^{2}}{r}}=F_{C}$ The masses of the satellite cancel out, so we are left with:

${\frac {G\cdot M_{1}}{r^{2}}}={\frac {v^{2}}{r}}$ 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:

$v={\frac {2\cdot \pi \cdot r}{t}}$ But we are dealing with velocity squared, so we square our equation and get:

$v^{2}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}}}$ Plugging that back into the original equation, we get:

${\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}\cdot r}}$ The r2 cancels out the r in the denominator and becomes a plain old r. Thus, we have:

${\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r}{t^{2}}}$ We now multiply by r2:

$G\cdot M_{1}={\frac {4\cdot \pi ^{2}\cdot r^{3}}{t^{2}}}$ And multiply by t2/4π2:

$r^{3}={\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}$ And by taking the cube root of that, we arrive at our answer (well, sort of):

$r={\sqrt[{3}]{\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}$ 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:

$a={\sqrt[{3}]{\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}-R_{p}$ Where $a$ is the altitude from the sea level of the planet, and $R_{p}$ is the radius of the planet. This formula calculates the altitude above sea level for any given orbital period $t$ . To get the height for a stationary orbit, the orbital period must be as long as the sidereal rotation period, the time of a full revolution of the planet relative to the sky.