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Size: 2132
Comment:
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Size: 2132
Comment:
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| Deletions are marked like this. | Additions are marked like this. |
| Line 19: | Line 19: |
| || "vertical" acceleration || 0 || m/s² || | || "vertical" acceleration || 7.8 || m/s² || |
DecelDescent1
Summary |
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Start in circular equatorial orbit, t=0 sec |
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radius |
6600 |
km |
altitude |
229 |
km |
"horizontal" velocity |
7700 |
m/s |
"vertical" velocity |
0 |
m/s |
"vertical" acceleration |
0 |
m/s |
Decelerate for 160 seconds at 30 m/s², |
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with 1 sec start and stop jerks |
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radius |
6591 |
km |
altitude |
220 |
km |
"horizontal" distance |
856 |
km |
"horizontal" velocity |
3000 |
m/s |
"vertical" distance |
9 |
km |
"vertical" velocity |
710 |
m/s |
"vertical" acceleration |
7.8 |
m/s² |
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
The vertical ( ≈radial ) downward acceleration is
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.