Page:Advanced Automation for Space Missions.djvu/164

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

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