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 track.

 \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 Earth standard gravitational parameter \large R 6378 km Equatorial surface radius of Earth \large \mu/R 62.50 MJ/kg Equatorial gravity well energy \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_R 465.09 m/s Equatorial surface rotation velocity \large E_{lift} = \mu ( 1/R-1/T ) 0.774 MJ/kg Lift energy from surface to 80 km \large v_T = \omega T 470.09 m/s 80 km track rotation velocity \large E_\inf = \mu/r -½{ v_R }^2 61.72 ??? MJ/kg Surface launch energy \large v_\inf = \sqrt{ 2 E_\inf } 11110.53 m/s 80 km earth centered escape velocity \large v_X = v_\inf - v_T 10640.44 m/s Track relative earth escape velocity \large E_{launch} = ½{v_X}^2 + E_{lift} 57.38 MJ/kg Track lift + launch energy \large E_\inf / (\mu/R ) 98.8 % ??? percent surface launch energy advantage \large E_{launch} / (\mu/R ) 91.8 % percent launch loop energy advantage

The launch loop track curves from slightly inclined (below orbital velocity) to mostly horizontal at higher speeds, to escape velocity and higher. Momentum is transmitted to the track and rotor, slightly displacing the track backwards and slowing the rotor; positions and velocities are soon restored by cable tension and surface motors, so the long term net energy change to the system is negligable.

A vehicle launching to escape from loop exit altitude must leave the Earth with a velocity (relative to the center of the Earth) of 11110.53 m/s, but the 80 km exit point on the track is already moving 470.09 m/s. Hence the track/motor system only needs to release the vehicle at 10640.44 m/s ... and accelerate the vehicle (plus launch sled) to that lower velocity.

We can approximate total launch energy:

• Launch.Energy ~=~ Vehicle.Acceleration.Energy ~+~ Motor.Losses ~+~ Atmospheric.Drag.Energy

Motor.Losses include resistive losses in conductors, and hysteresis losses in the magnetics. \$

Surface.Driver.Losses ~=~ Rotor.Restoration.Losses ~+~ Power.Transmission.Losses

Standing.Power ~=~ Turnaround.Power ~+~ Residual.Gas.Drag ~+~ Stabilization.Actuator.Power ~+~ Control.System.Power

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)