Differences between revisions 21 and 22
 ⇤ ← Revision 21 as of 2021-07-16 08:14:59 → Size: 1566 Editor: KeithLofstrom Comment: ← Revision 22 as of 2021-07-16 16:22:01 → ⇥ Size: 1909 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 5: Line 5: 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 Line 9: Line 10: || $\large R$ || 6378 || km || Equatorial radius of Earth || || $\large R$ || 6378 || km || Equatorial radius of Earth |||| $\large T$ || 6458 || km || Equatorial radius of launch track || Line 13: Line 15: || $\large v_R$ || 465.09 || m/s || Equatorial rotation velocity         || || $\large v_e$ || 465.09 || m/s || Equatorial surface rotation velocity |||| $\large v_t$ || 470.09 || m/s || 80 km track rotation velocity ||

# 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 5.972e+24 kg Mass of Earth \large \mu = G M 398600 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 (longer than sidereal day) sday 86141.1 s sidereal day (relative to fixed stars) \large\omega = 2\pi/sday 7.29216e-05 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

E<μ÷r (last edited 2021-07-17 07:19:46 by KeithLofstrom)