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
a |
vehicle acceleration |
|||
d |
vehicle distance along track |
|
d_e |
vehicle exit distance |
v |
vehicle velocity along track |
|
v_e |
vehicle exit velocity |
H |
vehicle heating (relative) |
|
H_e |
vehicle exit heating |
t |
time from start of launch run |
|
t_e |
exit time |
\rho |
atmospheric density |
|
\rho_e |
exit atmospheric density |
z |
vehicle/track altitude |
|
z_e |
exit altitude |
Assume constant 3 gee acceleration ( a = 29.4 m/s²) to earth-relative escape velocity at exit altitude z_e = 80 km ( v_x = 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 drag heating H = \rho v^3 proportional to density and velocity cubed.
An escape velocity run to v_e = 10800 m/s will last t_x = 10800/29.4 = 367 seconds. Most launches will be to high earth orbits, a slightly shorter launch run.
H ~=~ \rho(z) ~ v^3 ~=~ H_e ~ t / t_e