... the number of chunks are highly dependent on the explosive change in velocity. So for the record, no significant piece of the station would actually reach the surface of Earth (assuming the presence of an atmosphere).
I was wondering how the size of the orbit might influence these numbers, however, but didn't have any purpose-built mathematical tools installed..
So, I quickly modded my KSP to mimic the real Sol system in terms of distances, sizes, and masses involved,
I put a small spaceship into a roughly 2000km orbit and used a manoeuvre node to find out the change in velocity required to reach the lower atmosphere (11km):
481m/s, assuming a retrograde trajectory, or approximately half of your mean change. The difference between a 1989km altitude change and a 2000km altitude change is largely irrelevant, I chose to put my target altitude at 11km because KSP doesn't display the periapsis if it's below the surface. By comparison, escape velocity requires an additional 2.8km/s (direct trajectory).
EDIT: Just to point it out, the only reason why the Moon appears to be on the edge of an escape orbit in the above image is the angle of the camera, and the relative inclinations of the orbit. The orbit of my spaceship is approximately equatorial, the Moon's very much is not. If I'd bothered getting them in the same plane it'd look something like this:
Source: Kichae's post below.
But would anyone really put a spacestation at a 2Mm orbit? Well, it's not a bad place to put it as such, the transition between LEO and MEO. Atmospheric resistance is all but gone (no, the atmosphere doesn't stop at 100km, it just grows thinner), and while you still have to deal with tidal effects these, too, are minimized with altitude.
But why not put it in high orbit? Geosynchronous or higher, where tidal forces actually help increase your orbital height rather than lower it. Well there's the obvious reason of "fuel requirements" but I took a look at it...
Getting into a Low Earth Orbit, according to wikipedia, takes a minimum of 9.6km/s+drag. I put my spaceship into a 200km orbit around kEarth and set up manoeuvre nodes to a 2Mm orbit:
Required delta-V: ~9.6km/s (to LEO) + 457m/s (first manoeuvre node, transition burn) + 433m/s (second node, in yellow text between navball and kEarth) = 10.5km/s. This is not an entirely accurate number, I probably skipped an important 100km or so, entirely ignored drag, and wasted a bunch of fuel circularizing in LEO, but for the purpose of comparison it'll do.
Compare this to a geosynchronous orbit of 35.7Mm or so:
~9.6km/s (to LEO) + 2455m/s + 1474m/s = 13.5km/s or about a 29% increase in delta-V. In the real world, this would be a much
bigger increase in cost due to the need to carry reaction mass with you for your engines, but in the Infinity universe this is not required. Oh yeah, and the spacestation, or what it's made of, might not even have visited the planet it orbits which would actually make a higher orbit cheaper. Oh, and keeping it in a geostationary orbit would greatly simplify flying to and from it from the surface of the planet. Oh, and...
In short, there's a lot of reasons why you'd want a station in GSO, and not a lot for MEO. So what's important about this? Minimum dV to get back to Earth from GSO:
Minimum dV for escape trajectory from GSO:
Of course, this balance would shift significantly if orbiting a larger, faster spinning planet (such as Jupiter). Now, for the real question: what would happen if a piece of the space station, say a crew capsule and large engine, was ejected at a vector that shifted it's periapsis into the planet's atmosphere?
The fall would accelerate it to over 10km/s.. it would enter the atmosphere..
Decelerating at a massive rate (exceeds 15g for most of the trip) it would burn ever brighter..
Until I realized I hadn't installed Deadly Reentry or anything similar and it calmly fell into the ocean...
(At a fatal velocity of 103.5m/s).