1145
Comment:
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1140
|
Deletions are marked like this. | Additions are marked like this. |
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|| $ day $ || 86400 || sec || solar day (longer than sidereal day || | || $ day $ || 86400 || sec || solar day (longer than sidereal day || |
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For an | 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 |
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