Wikipedia claims (although with no citation) that in order to make the space shuttle land, an initial powered delta-v of 322 km/h was applied in orbit, retrograde to the shuttle's orbit. 322 km/h is equal to 89.4 m/s. This caused the orbit to be lowered into the atmosphere, ultimately causing the shuttle to come to a full stop intact on the ground (or that was the general idea; as we know, it didn't work out perfectly every time).
The deorbit burn usually decreases the vehicle's orbital velocity anywhere from 200 to 550 fps, depending on orbital altitude.
where 200 fps is about 61 m/s, and 550 fps is about 168 m/s. Given this data and the space shuttle's operational range, 90 m/s seems a reasonable figure to use as a median-mission deorbit-burn delta-v.
What I fail to understand is how this relatively small (about 1% velocity change: in the case of Wikipedia's figure, 90 m/s out of the on the order of 7 km/s orbital velocity in a low Earth orbit) could be enough to sufficiently lower the orbit to commit the orbiter to landing, rather than being just a small change in the shuttle's orbit.
Why was such a small delta-v applied under power in orbit sufficient to commit the orbiter to landing?
I expect that good answers will draw on orbital mechanics and atmospheric gas density (aerobraking) to show why the small change was sufficient.