Difference between revisions of "Geosynchronous Orbit (Math)"

Revision as of 09:28, 15 June 2013

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 siderial time or siderial 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. As every orbit has to intersect the equator, a geosynchronous orbit only stays above one longitude but oscillate in latitude while a geostationary also stays above the same latitude (0°N/0°S).

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:

${\displaystyle F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}}$

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

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:

${\displaystyle F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}=M_{2}\cdot {\frac {v^{2}}{r}}=F_{C}}$

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

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

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:

${\displaystyle v={\frac {2\cdot \pi \cdot r}{t}}}$

But we are dealing with velocity squared, so we square our our equation and get:

${\displaystyle v^{2}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}}}}$

Plugging that back into the original equation, we get:

${\displaystyle {\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}\cdot r}}}$

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

${\displaystyle {\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r}{t^{2}}}}$

We now multiply by r²:

${\displaystyle G\cdot M_{1}={\frac {4\cdot \pi ^{2}\cdot r^{3}}{t^{2}}}}$

And multiply by t²/4π²:

${\displaystyle r^{3}={\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}$

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

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

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.

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

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.