NASA's RETURN TO THE MOON
Laurie Ann Holle, Russell Plummer, David C. Schlereth, Mark G. Strobel
(Advisor - Harald Schenk)
ABSTRACT
NASA is considering locating a base at the south pole of the Moon. The reason for this is that previous missions have detected possible ice deposits at both poles of our satellite. These volatiles would help to support a base by providing oxygen, water, and fuel. By not having to bring these items from Earth, a colony would be more economical to maintain.
Such a location may be hard to reach without performing costly energy inclination changes. Our objective was to present a possible alternative in reaching such a site.
Our alternative compares two possibilities for landing a vehicle on the lunar surface. Under the first, it is assumed that the inclination is changed directly while still in orbit. Under the second, the vehicle sets down anywhere that is convenient. It then launches to the base in a sub-orbital arc. The energy cost of each is compared.
Photo courtesy of NASA
CALCULATIONS
SCENARIO ONE: The vehicle reaches the Moon. It enters a lunar orbit that does not pass over the base. In order to do so after its flight from Earth, it would have had to be launched into a polar Earth orbit before being injected into a trans-lunar orbit. This has never been done before. Previous Apollo missions were launched from Florida, and were placed into an equatorial orbit around the Moon.
As a result of orbiting the Moon near the equator, a burn has to be performed that provides an INCLINATION change to the orbit. After completion of the burn, the vehicle will now pass over the base. It will be able to land at the destination. Here we assume that the inclination change will NOT be greater than 90 degrees.
This exercise involves the calculation of a simple PLANE CHANGE in km/s. The formula is
delta V(simple) = 2 V(initial) sin (Ø/2) where
delta V(simple) = delta V for simple plane change (km/s)
V(initial) = V(final) = velocities in initial and final orbits (km/s)
Ø = plane change angle (degree or radians)
If we're interested in changing only the orbit's inclination, then we must change the velocity at either the ASCENDING NODE, or at the DESCENDING NODE. When the delta V is done at one of these points, the orbit will pivot about a line connecting the two nodes. Thus, only the inclination will change.
The amount of velocity change we need to re-orient the orbit plane depends on two things. One is the ANGLE that we are turning through, and the other is the initial VELOCITY. As the turned angle increases, so does delta V(simple).
Since a satellite in an elliptical orbit slows down as it approaches apolune, we should perform the orbit change at that location. However, to be useful, the apolune must be located at one of the orbital nodes.
For simplicity sake, let's assume that the vehicle is in a CIRCULAR orbit around the Moon. We will put it at a distance of 170 km above the surface. (This was the furthest point that the Apollo missions traveled in their orbit around the Moon). We will calculate the MAXIMUM delta V that might be needed for the inclination change from this point. This would occur during a maximum 90 degree inclination change. The MINIMUM delta V would occur during a 0 degree inclination change.
PROBLEM STATEMENT: A lunar spacecraft is in a 170 km orbit. It needs to move from its current inclination of 0 degrees to an inclination of 90 degrees. What delta V is required?
CONCEPTUAL SOLUTION:
1) Solve for the orbit's energy and velocity.
€ = -µ/2a = -µ/2R (circular orbit)
€ = V²/2 - µ/R where V = { 2(µ/R + €) }½
2) Solve for the inclination change.
ð = | I (final) - I (initial) |
3) Find the change in velocity for a simple plane change.
Delta V (simple) = 2 V(initial) sin ð/2
ANALYTICAL SOLUTION
Given constants:
Radius of Moon = 1738 km
Gravitational parameter = µ(Moon) = 4.902794 x 10³ km³/s²
1) Solve for the energy and velocity of the orbit.
€ = -µ/2R = -4.902794 x 10³ km³/sec²/2(1738 + 170 km) = -1.285 km²/sec²
V (initial) = { 2(µ/R + €) }½ = { 2[(4.902794 x 10³ km³/sec²/1908 km) - 1.285 km²/sec²]}
= 2.569 km/sec
2) Solve for inclination change.
δ = | I (final) - I (initial) | = 90º - 0º
Find Delta V for the simple plane change
Delta V (simple) = 2 V (initial) sin δ/2 = 2(2.569 km/sec) sin 90º/2 = 4.372 km/sec
COMMENTS: From this we can see that an inclination change from this point can cost anywhere between 0 and 4.372 km/sec in Delta V. The exact amount will depend on the amount of inclination change required.
SCENARIO TWO: The vehicle reaches the Moon. It enters a lunar orbit that does not pass over the base. As before, it enters an orbit near the equator of the Moon. In this case, it simply lands anywhere on the surface. From the landing spot, a SUB-ORBITAL launch will bring it directly to the base. It is assumed that the sub-orbital path will cover LESS than 1/4th the circumference of the Moon. In this scenario, the most Delta V will equal twice the energy required for a single landing.
Apollo used the following orbits for the Moon:
Apollo 11 had an initial orbit of 61x169 nautical miles. After landing, the service module changed that to a 66x54 nautical mile orbit. The inclination to the lunar equator was 1.25 degrees.
Apollo 17 had an initial orbit of 52.6x170 nautical miles. We assume that the landings began at the lower altitude. Orbit inclination changes would have occured at the higher point.
ANALYSIS: Comparing the two scenarios, we determine that SCENARIO ONE involves one inclination change, plus a landing. SCENARIO TWO involves an initial landing. It also involves a sub-orbital launch from the lunar surface followed by a landing. The comparison in Delta V then becomes:
Inclination change + final lunar landing versus an initial landing + sub-orbital launch from the Moon with a landing.
Note that the sub-orbital launch from the Moon does not need to reach lunar orbit. The sub-orbital arc will cover no more than 1/4th the circumference of the Moon.
One lunar landing in Scenario One cancels out one lunar landing in Scenario Two. Both require the same delta V. We then have:
Inclination change versus sub-orbital launch B plus landing. This is the comparison between the two scenarios. Which one takes more energy?