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 |
Earth standard gravitational parameter |
\large R |
6378 |
km |
Equatorial surface 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_R |
465.09 |
m/s |
Equatorial surface rotation velocity |
\large E_{lift} = \mu ( 1/R-1/T ) |
0.774 |
MJ/kg |
Lift energy from surface to 80 km |
\large v_T = \omega T |
470.09 |
m/s |
80 km track rotation velocity |
|| \large E_\inf = \mu/r -½{ v_R }^2 || 61.72 || MJ/kg || Surface launch energy ||
\large v_\inf = \sqrt{ 2 E_\inf } |
11110.53 |
m/s |
80 km earth centered escape velocity |
\large v_X = v_\inf - v_t |
10640.44 |
m/s |
Track relative earth escape velocity |
\large E_{launch} = ½ { v_X }^2 |
56.61 |
MJ/kg |
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.
A vehicle launching to escape from loop exit altitude must leave the Earth with a velocity (relative to the center of the Earth) of 11110.53 m/s, but the 80 km exit point on the track is already moving 470.09 m/s. Hence the track/motor system only needs to release the vehicle at 10640.44 m/s ... and accelerate the vehicle (plus launch sled) to that lower velocity.
We can approximate total launch energy:
Launch.Energy ~=~ Vehicle.Acceleration.Energy ~+~ Motor.Losses ~+~ Atmospheric.Drag.Energy
Motor.Losses include resistive losses in conductors, and hysteresis losses in the magnetics. $
Surface.Driver.Losses ~=~ Rotor.Restoration.Losses ~+~ Power.Transmission.Losses
Standing.Power ~=~ Turnaround.Power ~+~ Residual.Gas.Drag ~+~ Stabilization.Actuator.Power ~+~ Control.System.Power
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