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

Linear Heating Profile

• 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 = { 1 over 2 } \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.

v ~~=~ a ~ t

v^2 ~=~ 2 ~ a ~ x ~~~~~~~~~~~~ {v_e}^2 ~=~ 2 ~ a ~ x_e

Chosen heating rate over the launch run, ramping up proportional to time and velocity:

H ~~=~ H_e ~ t / t_e ~=~ H_e ~ v / v_e

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

H = { 1 \over 2 } \rho_e ~ {v_e}^3 ~ t / t_e ~=~ \rho(z) ~ a^2 ~ x ~ t

H_e ~=~ { 1 \over 2 } \rho_e ~ {v_e}^3 \rho(z) ~ a^2 ~ x_e ~ t_e

rearranging:

$ \rho(z)