The decorative Death Pit feature is on the top of a huge tower, falling to the bottom of the Toronto Tower would take about 10.8 seconds or less so it would take something in the same ball park (which matches what we see on the screen) to hit the bottom of the tower.
The air friction on a small object like the Death Star (with a surface gravity like Earth's) would be relatively greater so Palpatine would experience re-entry like friction effects after falling only a few hundred feet. In short he would explode (which is what we see).
The Death Star has an artificial Earth like gravity through most levels (as far as we know) and is full of air but it's about the size of a small moon, falling to the centre would possibly never happen.
On Earth gravity pulls things to centre of it's mass, the nearer you get to the centre the higher the pull.
The artificial gravity on the Death Star would bugger that up in the long term.
On the Death Star he would fall so far, slowed by the air and then reach a lower gravity zone and the air currents would cause him to slowly go up and down and up and down, essentially he would stop in mid air (assuming what we have seen of the station is carried through all layers).
Once again friction would be a factor so the poor guy wouldn't be bouncing for very long.
In any case it's complicated and almost certainly not what we see happening in the film.