E < μ/r
Climbing out of the Earth's gravity well requires energy, but a launch loop on the rotating Earth can launch to infinity with less than the classical μ/r gravitational escape energy. The rest of the escape energy is taken from the rotational energy of the Earth itself. Not just the initial 0.11 MJ/kg from the Earth's rotation, but also because the vehicle "pushes against" the 80 km rotor/stator track.
\large G |
6.67408e-11 |
m³/kg/s² |
Gravitational constant |
\large M |
5.972e24 |
kg |
Mass of Earth |
\large \mu = G M |
398600.4418 |
km³/s² |
Standard gravitational parameter of Earth |
\large R |
6378 |
km |
Equatorial radius of Earth |
\large T |
6458 |
km |
Equatorial radius of launch track |
day |
86400 |
s |
solar day (relative to sun) |
sday |
86141.0905 |
s |
sidereal day (relative to fixed stars) |
\large\omega = 2\pi/sday |
7.292158e-5 |
radians/s |
Earth sidereal rotation rate |
\large v_e |
465.09 |
m/s |
Equatorial surface rotation velocity |
\large v_t |
470.09 |
m/s |
80 km track rotation velocity |
The launch loop track curves from slightly inclined (below orbital velocity) to mostly horizontal at higher speeds, to escape velocity and higher. Momentum is transmitted to the track and rotor, slightly displacing the track backwards and slowing the rotor; positions and velocities are soon restored by cable tension and surface motors, so the long term net energy change to the system is negligable.
Hence, we can approximate total launch energy:
Launch_Energy ~=~ Vehicle_Drive_Energy ~+~ Motor_Losses ~+~ Atmospheric_Drag_Energy
Motor_Losses include resistive losses in conductors, and hysteresis losses in the magnetics. $ MoreLater
and surface radius \large R . The standard gravitational parameter \large \mu for the planet is the product of the gravitational constant \large G and \large M : \large \mu ~=~ G M . The gravity at the surface of the planet is \large g(R) ~=~ \mu / R^2 , and the gravity at radius \large r above the surface is \large g(r) ~=~ \mu / r^2 .
For an