# 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.

DecelDescent1 (last edited 2018-07-10 19:27:29 by KeithLofstrom)