DecelDescent1
Crude analysis, probably errors.
Start in circular equatorial orbit, decelerate for 160 seconds at 30 m/s² with 1 sec start and stop jerks.
Summary (approximate!!!) 


start 
end 

time 
0 
160 
seconds 
radius 
6600 
6591 
km 
altitude 
229 
220 
km 
"horizontal" velocity 
7700 
3000 
m/s 
"horizontal" distance 
0 
856 
km 
"vertical" acceleration 
0 
7.8 
m/s² 
"vertical" velocity 
0 
710 
m/s 
"vertical" distance down 
0 
9 
km 
Start in a circular orbit at 6600 km altitude, 7770 m/s orbital velocity. Slow to 3000 m/s (approximately mach 10) at 30 m/s² horizontal deceleration in 160 seconds, including 1 second beginning and end jerks. Shave off the jerks, make the "end time" 159 seconds. Ignore Earth rotation (pessimistically) .
The "horizontal" velocity is 7770  30 t m/s. The horizontal distance travelled is the integral of that, 7770 t  15² t meters, or 856 kilometers. The static vertical gravity acceleration a_g is 398600.44/6600² km/s² = 9.15 m/s².
The vertical ( ≈radial ) downward acceleration is a_g  v^2/r = ( 9.15  ( 7770  30 t )² / 6600000 ) m/s² = ( 777 t 1.5 t² ) / 11000 m/s², or 7.8 m/s² after 16.
The vertical ( ≈radial ) downward velocity is the integral of the downward acceleration, ( 777 t²  t³ )/ 22000 m/s , or 710 m/s after 160 ( ≈159 ) seconds.
The vertical ( ≈radial ) distance dropped is the integral of the downward velocity, ( 1036  t ) t³ / 264000 meters, or 8900 meters after 160 ( ≈159 ) seconds, an terminal altitude of 220 km (it doesn't drop much!).
This is a crude approximation of a deceleration profile, not optimized.