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← Revision 29 as of 20210717 07:19:46 ⇥
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=E<μ/r=  #format jsmath = 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  Earth standard gravitational parameter   $\large R $  6378  km  Equatorial surface radius of Earth   $\large \mu/R $  62.50  MJ/kg  Equatorial gravity well energy   $\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_R $  465.09  m/s  Equatorial surface rotation velocity   $\large E_{lift} = \mu ( 1/R1/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 + $E_{lift}  57.38  MJ/kg  Track lift + launch energy   $\large E_\inf / (\mu/R ) $  98.8 % ???  percent  surface launch energy advantage   $\large E_{launch} / (\mu/R ) $  91.8 %  percent  launch loop energy advantage  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 $ 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 
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 
Earth standard gravitational parameter 
\large R 
6378 
km 
Equatorial surface radius of Earth 
\large \mu/R 
62.50 
MJ/kg 
Equatorial gravity well energy 
\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_R 
465.09 
m/s 
Equatorial surface rotation velocity 
\large E_{lift} = \mu ( 1/R1/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 + E_{lift} 
57.38 
MJ/kg 
Track lift + launch energy 
\large E_\inf / (\mu/R ) 
98.8 % ??? 
percent 
surface launch energy advantage 
\large E_{launch} / (\mu/R ) 
91.8 % 
percent 
launch loop energy advantage 
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