1909
Comment:

1909

Deletions are marked like this.  Additions are marked like this. 
Line 12:  Line 12: 
 $ day $  86400  s  solar day (longer than sidereal day)    $ day $  86400  s  solar day (relative to sun)  
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
\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 
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