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Comment:

2575

Deletions are marked like this.  Additions are marked like this. 
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$ Launch_Energy ~=~ Vehicle_Drive_Energy ~+~ Motor_Losses ~+~ Atmospheric_Drag_Energy $  $ Launch.Energy ~=~ Vehicle.Drive.Energy ~+~ Motor.Losses ~+~ Atmospheric.Drag.Energy $ 
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$ Motor_Losses $ include resistive losses in conductors, and hysteresis losses in the magnetics. $ MoreLater  $ Motor.Losses $ include resistive losses in conductors, and hysteresis losses in the magnetics. $ MoreLater 
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 
6.67408e11 
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 
7.292158e5 
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