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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 difference is taken from the rotational energy of the Earth itself.  $\large G $  6.67408e11  m³/kg/s²  Gravitational constant   $\large M $  5.972e24  kg  Mass of Earth   $\large \mu$  398600.4418  km³/s²  Standard gravitational parameter of Earth   $\large R $  6378  km  Equatorial radius of Earth   $ day $  86400  sec  solar day (longer than sidereal day  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 difference is taken from the rotational energy of the Earth itself.
\large G 
6.67408e11 
m³/kg/s² 
Gravitational constant 
\large M 
5.972e24 
kg 
Mass of Earth 
\large \mu 
398600.4418 
km³/s² 
Standard gravitational parameter of Earth 
\large R 
6378 
km 
Equatorial radius of Earth 
day 
86400 
sec 
solar day (longer than sidereal day 
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