When people talk about geosynchronous orbit--an orbit in which the satellite continuously remains "directly overhead" for the same ground position on Earth--they talk about it being at a specific altitude, approximately 22,000 miles.

Intuitively, this doesn't seem to make any sense. You would think that a geosynchronous orbit would be attainable at any altitude, by flying exactly fast enough that the satellite keeps pace with the rotation of the earth beneath it, and therefore the required speed would be greater the higher up you go. What's so special about the magic number 22,000 that makes it possible to do a geosynchronous orbit at that altitude but not at any arbitrary altitude?

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    $\begingroup$ an orbit in which the satellite continuously remains "directly overhead" for the same ground position on Earth This is a description of a geostationary orbit, which is a special case of a geosynchronous orbit. $\endgroup$ Aug 31, 2015 at 21:34
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    $\begingroup$ desmos.com/calculator/pxdeyiunxz $\endgroup$
    – TildalWave
    Aug 31, 2015 at 22:27
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    $\begingroup$ Satellites don't fly, they continuously fall. If they are in true orbit, the speed at which they fall is dependant on their height above the Earth. $\endgroup$
    – CJ Dennis
    Sep 1, 2015 at 4:11
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    $\begingroup$ Picture what would happen with an orbit 1m above the ground, not moving sideways relative to the ground. $\endgroup$ Sep 1, 2015 at 7:40
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    $\begingroup$ Is geosynchronous orbit an altitude or a velocity? . . . Yes. $\endgroup$
    – geometrian
    Sep 1, 2015 at 22:21

7 Answers 7


I quite agree that it is not intuitive. However, orbital mechanics are frequently not intuitive, probably because we don't get to experience an orbital environment on a regular basis (if ever).

Let's just assume we're talking about circular orbits for the remainder of my post, since you are a beginner in orbital mechanics.

There is only one speed that a given circular orbit of a certain altitude can go. Keep in mind that stable orbits do not require any force from an engine to keep going as they have been. Basically, in a circular orbit, the falling-toward-the-planet motion is matched exactly by the moving-forward motion.

Sir Issac Newton figured this out, and exemplified it with a thought experiment called Newton's Cannonball.

Note that if the orbital speed is too slow for that altitude, the cannonball crashed into the planet.

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And if the orbital speed is too high for the altitude, the orbit will be an ellipse, rather than circular, or the cannonball may even escape Earth altogether!


Finally, if the cannonball is launched at the 'correct' orbital speed to be in a circular orbit at that altitude, it will neither crash, nor fly away, but will remain stable, traveling around earth at that particular velocity.


At different altitudes, this Goldilocks velocity is different. If the orbit is closer to the planet, the effect of gravity is higher, so the orbiting object must be moving faster to counteract the falling. When the orbiting object is further away, there is less falling force due to gravity (because gravitational force is based on distance), and so the object does not need to be moving as fast to counteract the falling force.

From Wikipedia's Geocentric Orbit article, we know that Low Earth Orbit could be, for example, an altitude of 160km. At this altitude, the Goldilocks velocity to keep a circular orbit is about 8000 m/s, and takes about 90 minutes.

Now what happens if we look at a slightly higher altitude? Well the velocity is lower, and the path the orbiting object travels gets bigger (the circle is bigger), so both of those factors make the orbit take longer. A slightly higher orbit might take 100 minutes instead of 90.

For a geosynchronous orbit, the orbit has to take 24 hours instead of 90 minutes, because the earth takes 24 hours to spin. This happens when the circle is expanded to an altitude of about 35000 km. The Goldilocks velocity at this altitude is about 3000 m/s.

This is all somewhat simplified, but the broad strokes are all there. As Organic Marble pointed out, you could try to force a craft to orbit at a different altitude in a 24 hour period, but it would not be a stable orbit, you would need engines to keep it going.

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    $\begingroup$ Please note-- Goldilocks velocities do not guarantee that your ship will remain too hot, too cold nor just right. (Sorry, I've never heard the term Goldilocks velocity and needed to make a pun). $\endgroup$ Jun 27, 2018 at 19:42

Simply put, for a circular orbit and a given central body, the orbital period is solely a function of the radius. A geosynchronous orbit is just the orbital radius at which the corresponding period is equal to the rotational period of the Earth.

You could fly around the Earth in 24 hours at any altitude, but not without propulsion.

See this question for the math.


Think of it this way. A circular orbit is characterized by the fact that the fictitious centrifugal force is exactly canceled out by the (centripetal) force of gravity. If that wasn't the case, if gravity was stronger, the satellite would begin to sink; if gravity was weaker, it would begin to rise. In either case, it would no longer be in a circular orbit.

A geostationary orbit is characterized by its angular velocity (specifically, $2\pi$ radians per day). The centrifugal force for circular motion at constant angular velocity is proportional to the radius. The gravitational force is proportional to the inverse square of the radius. So you have an equation in the (generic) form, $Ar = B/r^2$ where $A$ and $B$ are some numbers. This equation is not valid for arbitrary $r$; rather, you can calculate the value of $r$ by solving the equation for it.

When you plug in the numbers, this is exactly what happens. The centrifugal force for a mass $m$ is given by $F_c=mv^2/r = m\omega ^2r$ where $\omega$ is the angular velocity. The gravitational force for a mass $m$ is $F_g = GMm/r^2$ where $G$ is Newton's constant of gravity and $M$ is the Earth's mass. When these two are equal, you have $m\omega^2 r = GMm/r^2$ or $r = \sqrt[3]{GM/\omega^2}$. When you plug in the numbers, you get $r \simeq 4.23\times 10^7$ meters, or after subtracting the radius of the Earth, an altitude of approximately 36,000 km. This is the only value for which the two forces cancel at an angular velocity of one full revolution per day, so this is the geostationary altitude.


\begin{align} T&=24\times60^2&&=86400\,s\\ \omega&=2\pi f&&={2\pi\over T}\\ F&={m v^2\over r}&&=m\omega^2r\\ \therefore F&=m\left({2\pi\over T}\right)^2r&&= {4\pi^2mr\over T^2}\\ \text{And }F&={GMm\over r^2}\\ & \text{For height to be maintained: }\sum f=0\\ {4\pi^2mr\over T^2}&={Gm\over r^2}\\ \therefore r^3&={T^2GM\over4\pi^2}\\ \therefore r&=\root 3\of{T^2GM\over4\pi^2}\\ T&=86400, G=6.67\times10^{-11},M=5.97\times10^{24}\\ \therefore r&=\root 3\of{86400^2\times6.67\times10^{-11}\times5.97\times10^{24}\over4\pi^2}\\ r&=42,226km\;\text{from centre of Earth}\\ h&=r-R\\ \therefore h&=42,226km-6370km=35856km \end{align} $M$ is the mass of Earth. $R$ is the radius of Earth.

This is my attempt at getting the value. It is off by a little bit but this may be due to accuracy of numbers used and considering the orbit perfectly circular.

Basically, in order for it to orbit correctly it must have the same angular velocity as earth (rotate at the same speed), which means having the same frequency or time period of rotation as the earth.

The weight of the object orbiting must then be equal to the centripetal force it has acting on it due to the circular motion. As others have said if these two forces are not equal then it will either crash into earth or fly off.

From this point onward it is just maths to calculate the actual value, remembering that this value of r gives the radius of orbit which is distance from the centre of the earth, so you must subtract R to get the height above earth.

From this you could calculate a velocity that the satellite is traveling at but in this area generally angular velocity is used more. Most people would not know what to do with this velocity either as it doesn't mean much and isn't useful.

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    $\begingroup$ Thank you! The maths are appreciated, and understated in other answers. $\endgroup$ Jun 27, 2018 at 19:44

What's so special about the magic number 22,000 that makes it possible to do a geosynchronous orbit at that altitude but not at any arbitrary altitude?

Lift an object to an orbital altitude of 1 meter. Let it go. What happens?


The centrifugal force of a geosynchronous orbit of 1 meter cannot support an object against gravity.

Then assume that Pluto is in a geosynchronous orbit... that is to say the dwarf planet needs to revolve around Earth in 24 hours. The speed it would need for that is approximately light speed. What happens?


Pluto will disappear out into the big black yonder, because Earth's gravity cannot possibly contain an object in a geosynchronous orbit of 7.5 billion kilometers.

Somewhere in between these two extremes is the altitude where gravity and the centrifugal force of a 24 hour orbit are equal and balance each other out.

That - special - altitude is 22 000 miles.

Move higher up and the centrifugal force of a 24 hours orbit is too strong... it will overcome gravity and result in an elliptic orbit, or make the object break away from Earth all-together. Move lower, and the centrifugal force is too weak to balance out gravity and the object will begin to lose altitude, again resulting in an eccentric orbit, or possibly even crash into the atmosphere.

  • $\begingroup$ "Then assume that Pluto is in a geosynchronous orbit... that is to say the dwarf planet needs to revolve around Earth in 24 hours. The speed it would need for that is approximately light speed." What do you mean? In its current orbit, Pluto obviously isn't orbiting Earth, so the question is moot. For an object in geostationary or geosynchronous orbit around Earth, the size of the object is irrelevant: a speck of dust or a huge rock, doesn't matter, the orbit is the same. $\endgroup$
    – user
    Sep 4, 2015 at 13:34
  • $\begingroup$ I meant exactly what I wrote - "Assume that..." - in the sense "Make the thought experiment that Pluto is in a geosynchronous orbit around Earth". No of course that is not what is happening in real life, but for the sake of examining the original poster's assumption that any orbit can be geosynchronous we can toy with the idea - that Pluto is in a geosynchronous orbit - for a moment and see what the consequences of it are. They are a) at that distance the gravity of Earth have a next to negligible effect on Pluto and b) Pluto would need to move at light speed. I.e: OP's assumption is wrong. $\endgroup$
    – MichaelK
    Sep 4, 2015 at 14:05
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    $\begingroup$ To be clear, there is an important but unspoken assumption here with the Pluto thought experiment that Pluto's orbital distance from Earth initially set at some number. Since both Earth and Pluto orbit the sun (and at very different orbital periods, plus Pluto's orbit being elliptical), the distance between the Earth and Pluto varies significantly. I assume that @MichaelKarnerfors just chose an average Earth-Pluto distance, or something, for computing the speed Pluto would need for a 24-hour Earth-centric orbit. $\endgroup$
    – CBHacking
    Oct 15, 2017 at 20:50

(No-math answer)

You're falling around the earth at any altitude at any speed. Even if you throw a ball, it is falling around the earth. It just doesn't have enough velocity to keep from hitting it. So the sweet spot is for an orbit that you travel far enough that the curvature of the earth is equal to how far you fell. The closer you are the more gravity, the less distance you have to fall before you hit, the faster you have to go for the earth to curve away from / out of your fall. The higher you are the slower you can go as the earth curves out of your way — less gravity. This way you don't have to add any energy — you just keep falling. At a certain altitude, your speed exactly matches the rotation of the earth. This is great because we can point our satellite dish at it. If you want to be geosync at any other altitude, you can be — but you will need fuel/energy and a lot of it to do it and you won't be weightless. You are only weightless because you are falling. If there was a tower built up that high, you would stand on it with gravity just as you would down here. A little less gravity — but still gravity. Hence the falling. You are weightless when you fall down here too. You're just too worried about sticking the landing to notice.


There is no magical number 22,000.

If, as you say, you could achieve geostationary orbit at any altitude, then you could go to any location on Earth's equator, hold an object at arm's length, release it, and it expect it to remain in place, essentially hovering in the air. After all, you and the object are travelling some 1,000 miles per hour around Earth's axis. We all know the object would simply fall to ground.

We also know that objects in low-Earth orbit must travel at some 17,000 miles per hour to remain in orbit, taking some 90 minutes to complete one orbit. We also know that the Moon is in orbit around the Earth (strictly speaking, the Earth-Moon barycenter), is about 240,000 miles away, and completes one orbit in about 27 days, travelling something like 2,500 miles per hour. We also know that gravity follows the inverse-square law, diminishing in proportion to the square of the distance.

What does this tell us about orbits in general? For one thing, the closer an object to the body it is orbiting, the more it must oppose gravity, which it can only do by traveling faster, which requires greater acceleration to remain on the closed, curved path we call an orbit. Given the two examples of low Earth orbit and the Moon, there must be an infinite range of orbital distances, each of which has an associated velocity and period. There must therefore be an orbit where the period coincides with the rotation of the Earth, and it will have its own specific distance.

Given the above, knowing Earth's gravitational acceleration (~9.8 m/s/s at the surface), the radius of the Earth (the point at which gravity has that value), the inverse-square law, and the formula for circular motion relating radius and period to acceleration, we can calculate the distance at which an orbit will have a desired period. It turns out that the orbital distance at which the period coincides with the Earth's rotation occurs some 22,000 mile up.


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