The only atoms for which the Schrodinger equation has an analytic solution are the one electron atoms i.e. H, He$^+$, Li$^{2+}$ and so on. That's because with more than one electron the forces between electrons make the equation too hard to solve analytically. However, over the 90 or so years since Schrodinger proposed his equation a vast array of numerical methods for solving it have been developed, and of course modern computers are so powerful they can calculate the (electronic) structure of any atom with ease. This applies even to heavy atoms where relativistic effects need to be taken into account.
The Rydberg equation is an approximation because it does not take the electronic fine structure into account. However it's a pretty good approximation. It works because for a one electron atom the energy of the orbitals (ignoring fine structure) is proportional to 1/$n^2$, where $n = 1$ is the lowest energy orbital, $n = 2$ is the second lowest and so on.
First, how do you actually determine the position of the electron without "kicking" it out of the atom?
When talking of quantum mechanical entities, as the atom and the electron are, one has to keep clearly in mind that our well validated models that allow us to probe their behavior are probabilistic, the probability given by the square of the wave function.
The wavefunction is a function of (x,y,z,t) . The way it has been validated is by making probability distributions and checking them against the data. The only way of measuring positions for an electron in the atom is by the electron interacting. This might be by its being kicked off and measured, giving one point eventually in the probability distribution under measrurement, or in fitting weak scattering data, for example, like light through a crystal, or x-rays , the interferences of light giving information of position . Again a statistical distribution. In this case the end result is a probing of the atom's position as a whole, as the electron orbitals define the size of the atoms.
Second, if you were able to determine its position very precisely, wouldn't its momentum be so high that it would exceed the speed of light? (or does it just become more massive? Either way, it doesn't seem like it could remain bound to the nucleus.
If you only determine the position, it could be as precise as your measurement capabilities. The Heisenberg uncertainty constrains one only if both momentum and position are required together.
Third, if you were able to determine its position, how does your knowledge of its position degrade with time? It would appear that to get back to its original probability distribution (over all space) it would need a great deal of time, again so as not to violate the speed of light (unless it can pop in and out of existence far, far away).
Again, please note that experiments are one off for individual interactions. One photon goes through the crystal and interacts with the field of the electron and is registered as one point in a probability distribution. Or one electron is kicked off and its track is measured and projected back to its position , as in this recent expreriment, giving the distribution of the electron in the hydrogen orbitals.
hydrogen orbitals
In the case of photons probing non destructively the atom there is no way that one can know what an individual electron is doing in its orbital after that slight interaction. So there is no degradation detectable with time as the electron is still in its orbital.
In the case of scattering electrons off the hydrogen atom, the process is completely destructive of the atom, the electron flies off and is detected in an appropriate detector system, and the hydrogen becomes an ion, a proton seeking for an electron from the environment to return to neutrality.
Best Answer
The problem is that you're thinking of the electron as a particle. Questions like "what orbit does it follow" only make sense if the electron is a particle that we can follow.
But the electron isn't a particle, and it isn't a wave either. Our current best description is that it's an excitation in a quantum field (philosophers may argue about what this really means; the rest of us have to get on with life). An electron can interact with its environment in ways that make it look like a particle (e.g., a spot on a photographic plate) or in ways that make it look like a wave (e.g., the double slits experiment) but it's the interaction that is particle-like or wave-like, not the electron.
If we stick to the Schrödinger equation, which gives a good description of the hydrogen atom, then this gives us a wavefunction that describes the electron. The ground state has momentum zero, so the electron doesn't move at all in any classical sense. Excited states have a non-zero angular momentum, but you shouldn't think of this as a point like object spinning around the atom. The angular momentum is a property of the wavefunction as a whole and isn't concentrated at any particular spot.