# 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 |
6.67408e-11 |
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 |

day |
86400 |
s |
solar day (longer than sidereal day) |

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 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