|
Size: 2132
Comment:
|
← Revision 8 as of 2018-07-10 19:27:29 ⇥
Size: 2011
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 3: | Line 3: |
| ---- Crude analysis, probably errors. ---- Start in circular equatorial orbit, decelerate for 160 seconds at 30 m/s² with 1 sec start and stop jerks. |
|
| Line 4: | Line 8: |
| ||<-3> '''Summary''' || ||<-3> Start in circular equatorial orbit, t=0 sec || || radius || 6600 || km || || altitude || 229 || km || || "horizontal" velocity || 7700 || m/s || || "vertical" velocity || 0 || m/s || || "vertical" acceleration || 0 || m/s || ||<-3> Decelerate for 160 seconds at 30 m/s², || ||<-)3> with 1 sec start and stop jerks || || 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 || 0 || m/s² || |
||<-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 || |
| Line 23: | Line 21: |
| 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 "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. |
| Line 25: | Line 23: |
| 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 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. |
| Line 29: | Line 27: |
| 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!). | 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!). |
| Line 31: | Line 29: |
| This is a crude approximation of a deceleration profile, not optimized. | This is a crude approximation of a deceleration profile, not optimized. |
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.