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 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} ~ \times ~ t / {tmax}

distance =