How do they get it in the exact spot to stay over the same spot on the surface of the Earth? At a high level, it can be described easily. To maintain an orbit, (not falling back or speeding away) there is a basic formula which gives the relationship between the distance from the Earth and how fast the object must orbit. The lower the orbit, the faster the orbit. So you just have to provide enough life to get it to the point where the speed of the orbit matches the ground rotational speed of the Earth's surface.
Actually the faster the orbit the higher the orbit. The way to achieve a higher orbit is to apply thrust on a prograde vector. To get lower you do a retrograde burn. Orbital mechanics are all about velocity vectors.
A relative question perhaps. I think what I said was correct, relative to the surface of the Earth. (low orbit recon sats going around multiple times a day, geosync sats going around once per day). But I think you are correct in terms of the actual radial velocity of the speed around a larger and higher orbit.
Hello guys, did you forget the third law of Kepler? It says : cube of the main axis proportional to the square of the period. Then if for sake of simplicity you take a circular orbit, with a diameter 2R and and period T = 2*pi/Omega, it translates into : R^3*Omega^2 = constant. There is at least 2 ways to read this formula :
A) Omega = constant*R^(-3/2), telling you the higher the orbit, the lower the angular velocity of the sat,
or
B) R*Omega = V = constant*R^(-1/2), telling you the higher the orbit, the lower the
radial tangential velocity.
So whatever is your point of view, angular velocity or tangential velocity, the higher the orbit, the slower the satellite!...