#format jsmath
= DecelDescent1 =
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Crude analysis, probably errors.
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Start in circular equatorial orbit, decelerate for 160 seconds at 30 m/s² with 1 sec start and stop jerks.

||<-4> '''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.