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| Density at 80 km is |
|| 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 || |
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* 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} $ |
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}