Appendix 4A: Lunar Supply Of A Low Earth Orbit Station: Derivation Of Formulas
The mass brought to LEO from the Moon is MPL + MLAN where MPL is the mass of the payload of lunar soil and MLAN is the mass of the LANDER system that carries it. The LANDER must have sufficient tankage to carry payload plus the propellant to lift off from the Moon (MPR4), or to carry the hydrogen required on the Moon plus the propellant to carry the system to the Moon from the OTV (MPR2+3 = MPR2 + MPR3 the propellant requirements for burns two and three), whichever is greater. The fact that δV4 ~ δV2 + δV3 and that MPL >> MH where MH is the mass of hydrogen carried to the Moon, makes it clear that the former tankage requirement is the more stringent. It has therefore been assumed that:
MLAN = MLS + aMPL + BMPR4
where MLS is the mass of the LANDER structure and a and B are the tankage fractions for the payload and propellant, respectively. For all burns and for both the OTV and the LANDER B is assumed to be the same.
On the lunar surface prior to takeoff, the mass of the LANDER system is:
From MPL is taken material sufficient to replace the nonhydrogen part of the fuel supply. The amount of payload left over is P, hence:
This is the mass of hydrogen that must be uplifted from Earth to gain 1 kg of extra lunar payload to LEO. If no OTV is to be used, return to equation (1); MOS is now zero. If it is assumed that the payload tankage is more than enough to hold MPR1, then the term BMPR1 also disappears. Following through with these changes, X becomes:
The text shows that this reduces the marginal propellant cost by a small amount. If extra tankage is required to hold MPR1 the advantage is probably wiped out.
For simplicity, assume that the OTV starts in a circular 200 km orbit in the Earth-Moon plane and just reaches the Moon. Various relevant parameters used in the calculations are listed below.
dMoon = 384410 km
rEarth = 6378 km
rMoon = 1738 km
uEarth = 398600.3 km3/sec2
uMoon =4903 km3/sec2
LEO at 200 km altitude in place of lunar orbit
Perilune of transfer orbit at 50 km altitude
The circular orbital velocity at 200 km altitude is:
Vcir = 7.7843 km/sec
The transfer orbit has
aO = [(re + 200) + dMoon]/2 = 195,494 km
Therefore the spacecraft velocity upon leaving LEO is:
so δV1 = Vlaunch - Vcir = 3.2244 km/sec. This orbit has its apogee at the Moon's orbit and apogee velocity of Vapogee = 0.18679 km/sec. If the Moon has a circular orbit, its orbital velocity is VMoon = 1.02453 km/sec, hence, spacecraft velocity relative to the Moon is VMoon - Vapogee = Vinfinity = 0.8377 km/sec.
While passing 50 km above the lunar surface the OTV releases LANDER, which at once performs a burn to place it into a 1738 X 1788 km orbit around the Moon. The OTV's velocity relative to the Moon prior to separation is:
The semimajor axis of the orbit about the Moon is 1763 km and so the velocity of the LANDER at apolune is
The magnitude of the required orbital injection burn is therefore δV2 = V - Vapolune = 0.84303 km/sec. The LANDER then performs a half-orbit of the Moon and lands:
The lunar processor refuels the LANDER and loads it payload tanks with lunar soil. Takeoff from the Moon on trajectory that returns to LEO by way of aerobraking requires
There are two most promising propellant options for lunar-LEO transport systems. The first is an oxy-hydrogen combination using lunar-derived oxygen and hydrogen imported from Earth. The second option again requires native lunar oxygen as the oxidant but combines terrestrial-imported hydrogen with silicon purified on the Moon to produce a more powerful silane rocket fuel.
The silane produced on the Moon is assumed here for simplicity to be entirely SiH4. The propellant chemical reaction is:
SiH4 + 2O2 → SiO2 + 2H2O
The molecular weight of SiH4 is 32, so BH = 1/24. The achievable vacuum specific impulse is within the range 328-378 sec (Lunar and Planetary Institute, 1980). Assuming the middle of the range, Isp = 353 and C = 3.46 km/sec.