Why is geosynchronous orbit an altitude, rather than a velocity?

begin{align} T&=24times60^2&&=86400,s\ omega&=2pi f&&={2piover T}\ F&={m v^2over r}&&=momega^2r\ therefore F&=mleft({2piover T}right)^2r&&= {4pi^2mrover T^2}\ text{And }F&={GMmover r^2}\ & text{For height to be maintained: }sum f=0\ {4pi^2mrover T^2}&={Gmover r^2}\ therefore r^3&={T^2GMover4pi^2}\ therefore r&=root 3of{T^2GMover4pi^2}\ T&=86400, G=6.67times10^{-11},M=5.97times10^{24}\ therefore r&=root 3of{86400^2times6.67times10^{-11}times5.97times10^{24}over4pi^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.

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.

(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.

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?

Splat

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?

WHOOOSH

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.

A satellite in a geosynchronous geostationary orbit is both at specific altitude (26199 miles high), specific direction (equatorial orbit going from west to east), and specific velocity (1.91 miles per second). The altitude implies the velocity because if the velocity were incorrect, the satellite would not stay in orbit.

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, $2pi$ 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 $F_c=mv^2/r = momega ^2r$ 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 $momega^2 r = GMm/r^2$ or $r = sqrt[3]{GM/omega^2}$. When you plug in the numbers, you get $r simeq 4.23times 10^7$ 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.

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.