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); 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:


 * $$X' = [K_1(K_{2+3} + 1) + K_{2+3}] \left\{ \left[ K_4 \frac{B + B_H} {1 - BK_4} \right](1 + a) + a\right\}$$ (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 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</SUB> = 384410 km
 * r<SUB>Earth</SUB> = 6378 km
 * r<SUB>Moon</SUB> = 1738 km
 * u<SUB>Earth</SUB> = 398600.3 km<SUP>3</SUP>/sec<SUP>2</SUP>
 * u<SUB>Moon</SUB> =4903 km<SUP>3</SUP>/sec<SUP>2</SUP>
 * 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<SUB>cir</SUB> = 7.7843 km/sec

The transfer orbit has


 * a<SUB>O</SUB> = [(r<SUB>e</SUB> + 200) + d<SUB>Moon</SUB>]/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 δV<SUB>1</SUB> = V<SUB>launch</SUB> - V<SUB>cir</SUB> = 3.2244 km/sec. This orbit has its apogee at the Moon's orbit and apogee velocity of V<SUB>apogee</SUB> = 0.18679 km/sec. If the Moon has a circular orbit, its orbital velocity is V<SUB>Moon</SUB> = 1.02453 km/sec, hence, spacecraft velocity relative to the Moon is V<SUB>Moon</SUB> - V<SUB>apogee</SUB> = V<SUB>infinity</SUB> = 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:


 * $$V = \sqrt{V^2_{infinity} + \frac{2\mu_{Moon}} {r_m + 50}} = 2.4872\ km/sec$$

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 δV<SUB>2</SUB> = V - V<SUB>apolune</SUB> = 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


 * $$\Delta V_4 = \sqrt{V^2_{infinity} + \frac{2\mu_{Moon}} {r_{Moon}}} = 2.5287\ km/sec$$