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Size: 1580
Comment:
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Size: 1582
Comment:
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| Deletions are marked like this. | Additions are marked like this. |
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| $ Drag = \rho(alt) v^3 = Drag(max) * t / tmax $ | $ Drag = \rho({alt}) v^3 = Drag{max} * t / tmax $ |
Track Slope
Assume a maximum exit velocity a little larger than Earth escape (11.2 km/s) at 6458 km radius, 80 km altitude, near the equator. At that radius, the Earth's rotation velocity is 0.47 km/s ( 2π × 6458 km / 86414s ), so the maximum atmosphere-relative velocity (ignoring wind and adding drag loss) is 10.8 km/s. We will compute track slope backwards from that.
Loop inclination - the angle to the equator - will probably be between 10 and 30 degrees, To Be Determined.
Atmosphere Density Model
Altitude |
Density |
Scale Height |
Average |
km |
kg/m3 |
km |
km to 80 |
80 km |
1.85e-5 |
6.33 |
6.33 |
70 km |
8.28e-5 |
7.14 |
6.7 |
60 km |
3.10e-4 |
8.02 |
7.1 |
50 km |
1.03e-3 |
8.14 |
7.5 |
Linear Heating Profile
- Assume constant 3 gee acceleration (a = 29.4 m/s²) to earth-relative escape velocity at 80 km ( 10.8 km/s ), and a heating rate proportional to time on the track.
- Assume that the track altitude is lower near the beginning of the acceleration run, and is designed to increase with distance (and thus velocity) to produce that heating profile.
- Assume classical form drag proportional to density and velocity cubed $
An escape velocity run to vmax = 10800 m/s will last tmax = 10800/29.4 = 367 seconds. Many launches will be to high earth orbits, a slightly shorter launch run.
v = a t
Drag = \rho({alt}) v^3 = Drag{max} * t / tmax