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= Eμr = | #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 $ || 6.67408e-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) || || $\large\omega = 2\pi/day$ || 7.292158e-5 || radians/s || Earth sidereal rotation rate || 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 |
6.67408e-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) |
\large\omega = 2\pi/day |
7.292158e-5 |
radians/s |
Earth sidereal rotation rate |
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