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 M_PL + M_LAN where M_PL is the mass of the payload of lunar soil and M_LAN 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 (M_PR4), or to carry the hydrogen required on the Moon plus the propellant to carry the system to the Moon from the OTV (M_PR2+3 = M_PR2 + M_PR3 the propellant requirements for burns two and three), whichever is greater. The fact that δV₄ ~ δV₂ + δV₃ and that M_PL >> M_H where M_H 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:

M_LAN = M_LS + aM_PL + BM_PR4

where M_LS 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:

M_LAN + M_PL + M_PR4 = (M_PL + M_LAN)e^δV₄/c = (M_PL + M_LS+ aM_PL + BM_PR4)e^δV₄/c

Therefore,

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

where c is exhaust velocity. Therefore,

K_n = e^δV_n/c-1

Since the OTV and LANDER are fueled at LEO, the only hydrogen carried to the Moon is that required in M_PR4. If M_H is defined as the mass of hydrogen carried to the lunar surface, then

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

where B_H is the hydrogen fraction in the propellant.

The mass landed on the Moon must be:

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

The payload for the OTV is therefore

(M_LAN + M_H)e^{δV₂₊₃/c}

where:

δV₂₊₃ = δV₂ + δV₃

M_PR2+3 = (M_LAN + M_H)(e^{δV₂₊₃/c} - 1) = K₂₊₃(M_LAN + M_H)

and

M_OTV = M_OS + BM_PR1

for M_OS defined as OTV structure mass.

The mass leaving LEO is therefore:

M_OTV + M_PR1 + (M_LAN + M_H)e^{δV_{2 + 3}/c}

where:

M_PR1 = [M_OTV + (M_LAN + M_H)e^{δV₂₊₃/c}](e^δV₁/c - 1)

= K₁[M_OS + BM_PR1 + (M_LAN + M_H)e^{δV₂₊₂/c}]

$={\frac {K_{1}[M_{OS}+(M_{LAN}+M_{H})e^{\delta V_{2+3}/c}]}{1-BK_{1}}}$ (1)

The amount of material lifted off the Earth is:

$M_{H_{lift}}=M_{H}+B_{H}(M_{PR_{1}}+M_{PR_{2+3}})$

$=B_{H}(M_{PR_{1}}+M_{PR_{2+3}}+M_{PR_{4}})$

$=B_{H}\left\{{\frac {K_{1}}{1-BK_{1}}}\left[M_{OS}+\left(M_{LAN}+M_{H}\right)\left(K_{2+3}+1\right)\right]+K_{2+3}\left(M_{LAN}+M_{H}\right)+M_{PR_{4}}\right\}$

$=B_{H}\left\{{\frac {K_{1}}{1-BK_{1}}}M_{OS}+\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\times \left(M_{LAN}+M_{H}\right)+M_{PR_{4}}\right\}$

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

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

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

$b\equiv {\frac {B_{H}K_{1}}{1-BK_{1}}}$

$C\equiv B_{H}\left\{\left[{\frac {K_{1}\left(K_{2+3}+1\right)}{1-BK_{1}}}+K_{2+3}\right]\left[{\frac {\left(B+B_{H}\right)K_{4}}{1-BK_{4}}}+1\right]+{\frac {K_{4}}{1-BK_{4}}}\right\}$

$M_{H_{lift}}=AM_{PL}+bM_{OS}+CM_{LS}$

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

$M_{PL}=P+(1-B_{H})(M_{PR_{1}}+M_{PR_{2+3}})$

$=P+(1+B_{H})\left[{\frac {M_{H_{lift}}}{B_{H}}}-M_{PR_{4}}\right]$

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

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

and

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

${\frac {dM_{H_{lift}}}{dP}}=A{\frac {1+\left({\frac {1-B_{H}}{B_{H}}}\right)\left({\frac {dM_{H_{lift}}}{dP}}\right)}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}$

$={\frac {\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}{1-\left[{\frac {1-B_{H}}{B_{H}}}\right]\left[{\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)}}\right]}}$

$={\frac {A}{1+{\frac {1-B_{H}}{1-BK_{4}}}K_{4}(1+a)-{\frac {1-B_{H}}{BH}}A}}$

But if

$X={\frac {A}{B_{H}}}-{\frac {K_{4}(1+a)}{1-BK_{4}}}$

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

then

${\frac {dM_{H_{lift}}}{dP}}={\frac {B_{H}\left[X+{\frac {K_{4}(1+a)}{1-BK_{4}}}\right]}{1-(1-B_{H})X}}$ (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); M_OS is now zero. If it is assumed that the payload tankage is more than enough to hold M_PR1, then the term BM_PR1 also disappears. Following through with these changes, X becomes:

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

and

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

The text shows that this reduces the marginal propellant cost by a small amount. If extra tankage is required to hold M_PR1 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.

• d_Moon = 384410 km
• r_Earth = 6378 km
• r_Moon = 1738 km
• u_Earth = 398600.3 km³/sec²
• u_Moon =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:

V_cir = 7.7843 km/sec

The transfer orbit has

a_O = [(r_e + 200) + d_Moon]/2 = 195,494 km

Therefore the spacecraft velocity upon leaving LEO is:

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

so δV₁ = V_launch - V_cir = 3.2244 km/sec. This orbit has its apogee at the Moon's orbit and apogee velocity of V_apogee = 0.18679 km/sec. If the Moon has a circular orbit, its orbital velocity is V_Moon = 1.02453 km/sec, hence, spacecraft velocity relative to the Moon is V_Moon - V_apogee = V_infinity = 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:

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

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

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

The magnitude of the required orbital injection burn is therefore δV₂ = V - V_apolune = 0.84303 km/sec. The LANDER then performs a half-orbit of the Moon and lands:

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

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

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

B_H = M_H2/(M_H2 + M_O2) = 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 B_H = 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, I_sp = 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.