# how "off" (as a percentage) would an attempt at geostationary orbit need to be (in either distance or speed) for a satellite to fall

how "off" (as a percentage) would an attempt at geostationary orbit need to be (in either distance or speed) for a satellite to fall to earth in 100, 1000 or 10,000 years? (assumes initial setup only, and no subsequent adjustments).

• All satellites are falling continuously all the time. Nov 24 '17 at 17:50
• A satellite at the height of a geostationary orbit but with zero speed would fall straight down to Earth surface.
– Uwe
Nov 27 '17 at 20:54
• A satellite at the speed of geostationary orbit but with zero distance would already be at the surface of the Earth. Dec 12 '17 at 13:47
• Excluding drag from the atmosphere and (minor) interference from radiation/other solar bodies, your orbit will be an ellipse. A circle is just a special case of an elipse, and a geostationary orbit just just a special case of a circular orbit. If you miss your "ideal" orbit, you'll just be on a slightly different eliptical orbit. FWIW, grab a copy of "Kerbal Space Program" it's a really good intro to orbital mechanics that glosses over the maths Oct 12 '18 at 12:43

Being off the nominal geostationary orbital velocity won't cause the satellite to fall to Earth.

For a given altitude, there's one orbital velocity that yields a circular orbit. If your velocity is higher or lower, you instead get an elliptical orbit. As long as that elliptical orbit doesn't intersect the Earth or a substantial amount of the Earth's atmosphere, it will remain stable for quite a while (although, because of disturbances from solar wind, the unevenness of Earth's gravitational field, and the influence of other planets in the solar system, it may not stay up for tens of thousands of years).

Vanguard 1 has remained in orbit for over fifty years, and at the low end of its orbit its altitude is about ~660 km. Even at apogee it's much, much lower than geosynchronous. Vanguard is expected to stay in orbit for 240 years.

Circular orbit velocity for geosynchronous orbit altitude is 3.07 km/s. If you hit that speed, due East, at an altitude of 35,786km above the equator, you're in geostationary orbit. If you're going slower than that, you're in an elliptical orbit, with that point as your apogee.

Let's assume that Klassensat-1 reached geosynchronous altitude by doing a transfer from low Earth orbit but its circularization burn failed, leaving it in a 660km perigee, 35,786 km apogee orbit. This should last at least as long as Vanguard. The speed at apogee for this orbit works out to about 1644 m/s.

So if your orbital velocity is only 54% of geostationary velocity, you'll still stay up for well over 240 years.

If it stays above 900km, your satellite's lifetime will be on the order of 1000 years; this requires only a little more speed at apogee, 1668m/s.

• From what you are saying, am I correct to understand that a satellite that is slower than a geostationary velocity will eventually fall (through ever closer elliptical orbits)? I gather then, if it is going faster than a geostationary velocity, then it would slowly increase its altitude (?) I was thinking about the moon and its apparent retreat of 3cm/year (such an insignificant amount in relation to its altitude). I would think that if it has been orbiting for a long time without changing the orbit, and it is neither flying off into space nor crashing to the earth... Nov 27 '17 at 19:21
• No, that's not correct. The eventual fall comes about because these very eccentric orbits reach the uppermost, thinnest parts of Earth's atmosphere, which slows them a tiny, tiny bit more on each close approach. There's no corresponding mechanism to raise an orbit. At geosynchronous altitude, atmosphere is far less significant than the other perturbing factors (solar wind, other massive bodies in the solar system, etc). The moon's increasing altitude is due to an entirely different mechanism: en.wikipedia.org/wiki/Orbit_of_the_Moon#Tidal_evolution Nov 28 '17 at 21:29