Geosynchronous and Geostationary Orbits

Objects can orbit the Earth in different ways. Most orbits look the same from above, a sine curve (or if you like different phasing, a cosine curve) with the Earth’s equator as the x-axis. The difference is how fast the satellite or spacecraft or space station takes to orbit the Earth.

However, there is a special orbit which does not orbit the entire Earth, but stays above a particular longitude. These orbits are called geosynchronous orbits. These orbits have a period that is just equal to the Earth’s sidereal day (23 h, 56 min, and 4 seconds). Because of this orbit, they tend to remain around the same longitude on Earth and if you were to look down on this orbit, the satellite would trace out something called an analemma, which is just a fancy term for the figure 8. Depending on the inclination of the orbit, these satellites are not visible from all parts of the Earth. These orbits are used mostly for communications and weather satellites. This is why you do not have to move your satellite dish if you have satellite television, as a non-geosynchronous orbit would be a pain if you are watching your favorite TV shows.

There is a special geosynchronous orbit called a geostationary orbit. Not only does this have a period of one sidereal day, but a satellite in this orbit does not move at all. It is always above the same place on Earth, and by definition, the location in the sky must be above the equator. If we were to build a space elevator (more on this concept later), the receiving station for the elevator must in a geostationary orbit. All geostationary orbits are geosynchronous, but not all geosynchronous orbits are geostationary.

How far up is an object in a geosynchronous/geostationary orbit? Just using some basic concepts from Newtonian mechanics, the calculation is relatively simple.

First, to be in a stationary orbit, the force of gravity on the satellite must be counteracted by the centripetal force, i.e.:

Where:

• G is the gravitational constant, 6.67x10^-11 m³/kg·s²
• M_E is the mass of the Earth, 5.972x10²⁴ kg
• m is the mass of the satellite
• v is the orbital velocity, m/s
• R is the radius of the orbit (assuming circular orbit), in m

Equating these two and we get:

We know the period of the orbit (P) has to be one sidereal day, 23h56m4s, which in seconds is 86,164 seconds (60 seconds in a min, 60 min in an hour) and the orbital velocity is just the length of the orbit (the circumference of the orbit, 2πR) divided by the period, P.

Plug v=2πR/P into the above equation and simplifying, we get:

and plugging in all the constants, we find that the orbital radius is 42,164 km. (If you like, you can solve this yourself and see if I’m right.) Note, that this is the radius of the orbit from the center of the Earth. If we take into account the Earth’s radius, the orbital altitude is 35,786 km (R_E = 6378 km at the equator).