# Advanced Automation for Space Missions/Appendix 4A

## 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:

MLAN + MPL + MPR4 = (MPL + MLAN)eδV4/c = (MPL + MLS+ aMPL + BMPR4)eδV4/c

Therefore,

${\displaystyle M_{PR_{4}}={\frac {K_{4}[(1+a)M_{PL}+M_{LS}]}{1-BK_{4}}}}$

where c is exhaust velocity. Therefore,

Kn = eδVn/c-1

Since the OTV and LANDER are fueled at LEO, the only hydrogen carried to the Moon is that required in MPR4. If MH is defined as the mass of hydrogen carried to the lunar surface, then

${\displaystyle M_{H}=B_{H}M_{PR_{4}}=B_{H}K_{4}{\frac {(1+a)M_{PL}+M_{LS}}{1-BK_{4}}}}$

where BH is the hydrogen fraction in the propellant.

The mass landed on the Moon must be:

${\displaystyle M_{LAN}+M_{H}=M_{LS}+aM_{PL}+BM_{PR_{4}}+B_{H}M_{PR_{4}}={\frac {(B+B_{H})K_{4}[(1+a)M_{PL}+M_{LS}]}{1-BK_{4}}}+M_{LS}+aM_{PL}}$

The payload for the OTV is therefore

(MLAN + MH)eδV2+3/c

where:

δV2+3 = δV2 + δV3
MPR2+3 = (MLAN + MH)(eδV2+3/c - 1) = K2+3(MLAN + MH)

and

MOTV = MOS + BMPR1

for MOS defined as OTV structure mass.

The mass leaving LEO is therefore:

MOTV + MPR1 + (MLAN + MH)eδV2 + 3/c

where:

MPR1 = [MOTV + (MLAN + MH)eδV2+3/c](eδV1/c - 1)
= K1[MOS + BMPR1 + (MLAN + MH)eδV2+2/c]
${\displaystyle ={\frac {K_{1}[M_{OS}+(M_{LAN}+M_{H})e^{\delta V_{2+3}/c}]}{1-BK_{1}}}}$ (Equation 1)

The amount of material lifted off the Earth is:

${\displaystyle M_{H_{lift}}=M_{H}+B_{H}(M_{PR_{1}}+M_{PR_{2+3}})}$
${\displaystyle =B_{H}(M_{PR_{1}}+M_{PR_{2+3}}+M_{PR_{4}})}$
${\displaystyle =B_{H}\left\{{\frac {K_{1}}{1-BK_{1}}}\left[M_{OS}+\left(M_{LAN}+M_{H}\right)\left(K_{2+3}+1\right)\right]+K_{2+3}\left(M_{LAN}+M_{H}\right)+M_{PR_{4}}\right\}}$
${\displaystyle =B_{H}\left\{{\frac {K_{1}}{1-BK_{1}}}M_{OS}+\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\times \left(M_{LAN}+M_{H}\right)+M_{PR_{4}}\right\}}$
${\displaystyle =B_{H}\left\{{\frac {K_{1}}{1-BK_{1}}}M_{OS}+\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\times \left[{\frac {\left(B+B_{H}\right)K_{4}\left[\left(1+a\right)M_{PL}+M_{LS}\right]}{1-BK_{4}}}+M_{LS}+aM_{PL}\right]+{\frac {K_{4}}{1-BK_{4}}}\left[\left(1+a\right)M_{PL}+M_{LS}\right]\right\}}$

If we define A, b, and C as follows:

${\displaystyle A\equiv B_{H}\left\{\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\left[{\frac {\left(B+B_{H}\right)K_{4}\left(1+a\right)}{1-BK_{4}}}+a\right]+{\frac {K_{4}\left(1+a\right)}{1-BK_{4}}}\right\}}$
${\displaystyle b\equiv {\frac {B_{H}K_{1}}{1-BK_{1}}}}$
${\displaystyle C\equiv B_{H}\left\{\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\left[{\frac {\left(B+B_{H}\right)K_{4}}{1-BK_{4}}}+1\right]+{\frac {K_{4}}{1-BK_{4}}}\right\}}$
${\displaystyle M_{H_{lift}}=AM_{PL}+bM_{OS}+CM_{LS}}$

From MPL is taken material sufficient to replace the nonhydrogen part of the fuel supply. The amount of payload left over is P, hence:

${\displaystyle M_{PL}=P+(1-B_{H})(M_{PR_{1}}+M_{PR_{2+3}})}$
${\displaystyle =P+(1+B_{H})\left[{\frac {M_{H_{lift}}}{B_{H}}}-M_{PR_{4}}\right]}$
${\displaystyle =P+{\frac {1+B_{H}}{B_{H}}}\left\{M_{H_{lift}}-{\frac {B_{H}K_{4}}{1-BK_{4}}}[(1+a)M_{PL}+M_{LS}]\right\}}$
${\displaystyle M_{PL}={\frac {P+{\frac {1-B_{H}}{B_{H}}}\left[M_{H_{lift}}-{\frac {B_{H}K_{4}}{1-BK_{4}}}M_{LS}\right]}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}}$

and

${\displaystyle M_{H_{lift}}=A{\frac {P+{\frac {1-B_{H}}{B_{H}}}\left[M_{H_{lift}}-{\frac {B_{H}K_{4}M_{LS}}{1-BK_{4}}}\right]}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}+bM_{OS}+CM_{LS}}$
${\displaystyle {\frac {dM_{H_{lift}}}{dP}}=A{\frac {1+\left({\frac {1-B_{H}}{B_{H}}}\right)\left({\frac {dM_{H_{lift}}}{dP}}\right)}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}}$
${\displaystyle ={\frac {\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}{1-\left[{\frac {1-B_{H}}{B_{H}}}\right]\left[{\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}\right]}}}$
${\displaystyle ={\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)-{\frac {1-B_{H}}{BH}}A}}}$

But if

${\displaystyle X={\frac {A}{B_{H}}}-{\frac {K_{4}(1+a)}{1-BK_{4}}}}$
${\displaystyle =\left[{\frac {K_{1}(K_{2+3}+1)}{1-BK_{1}}}+K_{2+3}\right]\left[{\frac {K_{4}(B+B_{H})}{1-BK_{4}}}(1+a)+a\right]}$ (Equation 2)

then

${\displaystyle {\frac {dM_{H_{lift}}}{dP}}={\frac {B_{H}\left[X+{\frac {K_{4}(1+a)}{1-BK_{4}}}\right]}{1-(1-B_{H})X}}}$ (Equation 3)

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:

${\displaystyle X'=[K_{1}(K_{2+3}+1)+K_{2+3}]\left\{\left[K_{4}{\frac {B+B_{H}}{1-BK_{4}}}\right](1+a)+a\right\}}$ (Equation 4)

and

${\displaystyle {\frac {dM_{H_{lift}}}{dP}}={\frac {B_{H}\left[X'+{\frac {K_{4}(1+a)}{1-BK_{4}}}\right]}{1-(1-B_{H})X'}}}$ (Equation 5)

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.

### 4A.1 Numerical Equations

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:

${\displaystyle V_{launch}={\sqrt {{\frac {2\mu _{e}}{r_{e}+200}}-{\frac {\mu _{e}}{a_{O}}}}}=11.0087\ km/sec}$

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:

${\displaystyle V={\sqrt {V_{infinity}^{2}+{\frac {2\mu _{Moon}}{r_{m}+50}}}}=2.4872\ km/sec}$

The semimajor axis of the orbit about the Moon is 1763 km and so the velocity of the LANDER at apolune is

${\displaystyle V_{apolune}={\sqrt {{\frac {2\mu _{Moon}}{r_{Moon}+50}}-{\frac {\mu _{Moon}}{a_{O}}}}}}$

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:

${\displaystyle \Delta V_{3}=V_{perilune}={\sqrt {{\frac {2\mu _{Moon}}{r_{Moon}}}-{\frac {\mu _{Moon}}{a_{O}}}}}}$

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

${\displaystyle \Delta V_{4}={\sqrt {V_{infinity}^{2}+{\frac {2\mu _{Moon}}{r_{Moon}}}}}=2.5287\ km/sec}$

### 4A.2 Propellants

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.

(a) Lunar oxygen, terrestrial hydrogen propellant option

The relevant chemical propellant combustion reaction is:

2H2 + O2 → 2H2O

The molecular weight of H2 is 2 and of O2 is 32, so:

BH = MH2/(MH2 + MO2) = 1/9

The achievable specific impulse of LOX - LH2 is about 450 sec, using heat of formation data from Weast (1978) and assuming 75% thermal efficiency. This yields an exhaust velocity of 4.41 km/sec.

(b) Lunar oxygen, Earth/lunar silane propellant option

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.

### 4A.3 References

Lunar and Planetary Institute: Extraterrestrial Materials Processing and Construction, Final Report NSR 09-051-001 Mod #24, Houston, Texas, 1980.

Weast, Robert C., ed.: Handbook of Chemistry and Physics, CRC Press, West Palm Beach, Florida, 1978. Fifty-ninth Edition.