# HEEO, High Eccentricity Earth Orbit

Launch loops can launch vehicles as fast as escape velocity for "kinetic energy cost", without the expense of large rocket engines and fuel and fuel tanks. Escape velocity at 80 km altitude ( r_{launch} = 458 km equatorial radius) is v_{escape} = \sqrt{ 2 * \mu_0 / r_{launch} } = 11.11 km/s. The Earth's standard gravitational parameter is \mu_) = 398600.4418 km3/s2. The Earth rotates at 0.47 km/s at launch loop altitudes, so the surface-relative escape velocity is 10.64 km/s.

The kinetic energy cost is proportional mass and velocity squared, { 1 \over 2 } m v^2 , or 56.6 megajoules per kilogram. A kilowatt-hour (KWh) is 3.6 megajoules, and costs $0.125/kWh on my 2020 Oregon residential power bill, so the launch energy is 15.7 KWh/kg, and costs$1.97 at Oregon residential rates. Rumor has it that Oregon's giant data centers - Google, Apple, Facebook - pay $0.08/kWh industrial rates. However, electric energy costs in the far-from-Oregon equatorial Pacific are indeterminate. Nuclear or Diesel generators on barges? Wind turbines? A tap on a Pacific Ocean spanning PowerLoop? In any case, the power cost will be vastly less than a Tsiolkovsky-exponential-crippled liquid fuel launch rocket. A high perigee orbit requires less launch velocity than full escape velocity; however, it does require added velocity at apogee to raise perigee far above crowded LEO (Low Earth Orbit). I assume a 2000 km altitude (8378 km radius) perigee will be ample. The apogee of an Earth orbit can be very very high, as far out as the https://en.wikipedia.org/wiki/Hill_sphere Hill Sphere, 1.5 million kilometers, though such long period orbits will be strongly perturbed by lunar and solar tides. Indeed, the perturbations will grow rapidly as apogee altitudes approach lunar orbit radius, 356,400 to 406,700 km. Very high apogees should be used carefully, and significant amounts of station-keeping thrust will be needed to maintain them - Master's thesis opportunity? Angular momentum is proportional to velocity times radius, L = v r , so the cheapest place to "manufacture" angular momentum, turning \Delta V into \Delta L , is at very high radius r , then launching that angular momentum (as orbiting mass) down to lower orbits. Low cost solutions to the angular momentum manufacturing and distribution problem is another Master's thesis opportunity; my guess is VASIMR plasma thrusters operating at apogee, and orbiting momentum exchange nets. I will use Q for apogee, q for perigee (8378 km), and sday for sidereal days ( 86164.0905 seconds, 366.2422 sidereal days per year). Launch is from an 80 km altitude launch loop.  period semimajor perigee perigee apogee apogee from loop ΔV N P a r_q v_q r_Q v_Q launch arrival sday seconds km km km/s km km/s km/s km/s 1 86164.09 42164.17 8378 9.257 75950.34 1.021 10.195 0.114 2 172328.18 66931.45 8378 9.445 125484.89 0.631 10.364 0.073 3 258492.27 87705.01 8378 9.519 167032.01 0.477 10.431 0.056 4 344656.36 106247.05 8378 9.560 204116.10 0.392 10.468 0.046 5 430820.45 123288.78 8378 9.588 238199.56 0.337 10.492 0.040 6 516984.54 139223.02 8378 9.607 270068.04 0.298 10.509 0.035 7 603148.63 154291.59 8378 9.621 300205.17 0.269 10.522 0.032 From the launch loop, the launch energy difference between the 1 day and 7 day (Earth relative) orbits is only 6.5%, an additional$0.12 per kilogram for "Oregon electricity", and much less than the price difference between UPS 1-zone and 7-zone ground shipping rates. If you can stand the extra wait (3 more upbound, 3 more days inbound), the doubled reentry energy, and thetripled peak entry heating, then the 7 sday orbit is a good option for the incremental construction of lunar or interplanetary vehicles.

HEEO (last edited 2020-11-13 01:46:05 by KeithLofstrom)