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Size: 1909
Comment:
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Size: 1909
Comment:
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| 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 |
6.67408e-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 |
\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 |
7.292158e-5 |
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