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

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.

v ~~=~ a ~ t

v^2 ~=~ 2 ~ a ~ d ~~~~~~~~~~~~ {v_e}^2 ~=~ 2 ~ a ~ d_e

H ~~=~ H_e ~ t / t_e

H ~~=~ { 1 \over 2 } \rho(z) ~ v^3 ~=~ { 1 \over 2 } \rho(z) ~ ( 2 ~ a ~ d ) ~ ( a ~ t ) ~=~ \rho(z) ~ a^2 d ~ t

H_e ~=~ { 1 \over 2 } \rho_e ~ {v_e}^3

H_e ~ t / t_e ~=~ \rho(z) ~ ( 2 a d ) ~ ( a t )

rearranging:

$ \rho(z)