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 km^{3}/s^{2}. The Earth rotates at 0.47 km/s at launch loop altitudes, so the surfacerelative 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 kilowatthour (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 farfromOregon 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 Tsiolkovskyexponentialcrippled 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 stationkeeping 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 1zone and 7zone 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.