Path To The Moon: Newsletter Challenge (04/06/10)

We don't actually make a figure of eight - it only looks like this relative to the Earth-moon axis.

In other respects, Anonymous Hero is absolutely correct (I gave a G.A.), but I would doubt whether even given a factor of two more energy there is a realistic way of landing from an orbit that circles the moon in a constant direction.
So here is a different way of arriving at this:

First, we may assume that we use the ground-sling to contribute to our orbiting velocity, so we will be orbiting the Earth in the same direction as the moon.
Now we observe that we can afford to make only relatively slow changes to our orbit (once we have jettisoned the booster stages), so locally our path is always very similar to a satellite ellipse.

Ignoring the effect of the moon itself (we would have to if we are to treat our orbit as an ellipse), if we try to make an (approximate) ellipse beyond the moon, our angular velocity at the apogee of our orbit will inevitably be less than the angular velocity of the moon. This means that if we want to meet the moon we have to start from somewhat ahead of it. However, at low altitude our angular velocity is much greater than that of the moon (greater than once per day...), so we inevitably first overtake the moon (this being between Earth and moon) and then let it catch up again.

"but I would doubt whether even given a factor of two more energy there is a realistic way of landing from an orbit that circles the moon in a constant direction."

Thank you for the GA.

I am a little confused about the above quote. Could you clarify?

I am thinking that for intents and purposes, we are simply trying to insert the space craft into a lunar orbit about the Moon. Landing is another matter. A trans lunar injection is the mechanism to get to the point where we can actually achieve lunar orbit.

The mechanism to do that looks like this illustration from Wikipedia:

TLI

I was interpreting the challenge question as to why the injection takes you to a path that is beneath the Moon and then follows a retrograde motion backwards. If the spacecraft velocity is neither too high nor too low it can be "slung" around the far side of the Moon and return toward Earth in that figure eight motion. The gravity assist trades some of the spacecraft's momentum to the Moon's orbital momentum (with the inertial frame of reference being the Earth).

Approaching the Moon from the other direction and over top of the Moon takes orbital momentum from the moon and transfers it toward the spacecraft, which would be counterproductive if you are either trying to inject into an orbit or a direct return mission to Earth.

I see what you mean.

So I'll rephrase that: "but I would doubt whether even given a factor of two more energy there is a realistic way of getting into a lunar orbit from an Earth orbit that circles the moon in a constant direction."

On the other hand, I had stated that you would still use the Earth's rotation to help with your velocity, so you would still be orbiting the Earth in the same direction as the moon does; the problem is that your orbit would need to be capable of taking you way beyond the moon if you were to be overtaking it as you reached its orbit (which would be necessary if you were to avoid the apparent figure of eight).

Yes, I assumed that like all manned launches you launch eastward, so you would be constrained to either an approach beneath the Moon or from behind and above the Moon's orbital path about Earth. The latter would provide a positive gravity assist, which is not what you would want.

If I understand the lunar orbit approach you speak of you are talking about raising the Earth orbit of the spacecraft to match the Moon's orbital speed? That would be expensive. :-)

Lunar missions do a trans lunar injection that just reaches Earth's escape velocity. So if you miss you could be a "Lost in Space" reality show.

Update:

Just did a historical check and the Apollo did a total of an 8-minute burn in two separate burns starting from an orbit of 115 miles at just over 15,000 mph to get to a final delta-V of the Earth's escape velocity using a single J-2 rocket engine.

"Lunar missions do a trans lunar injection that just reaches Earth's escape velocity. So if you miss you could be a "Lost in Space" reality show."

I think you will find that the nominal escape velocity is reached when the craft is orbiting the moon and is between the moon and the Earth. So it is subject to both gravitational fields and its potential energy is significantly lower than you would normally expect at that distance from the Earth - which means that it would need to be moving at an appreciably higher velocity in order to have enough energy to escape. Now, if the moon and its gravitational field were to vanish in a whiff of magic...
(This is of course so far from my areas of work and knowledge that I'm resting entirely on theory, and miss steaks are maw than possible...)

P.S. Yes, it would be expensive - I guessed at a minimum of 1.2x the rocket-applied momentum (and cost rises disproportionately). On the other hand, going in a westward direction would be very much worse...

While orbiting the Moon implies a much lower delta-V because the escape velocity of the Moon is 2.37 km/sec, much lower than Earth's 11.2 km/s.

So, where is the point of maximum?

The maximum velocity would be when directly between the moon and the Earth, as this is the point where the gravitational potential energy is lowest. To my mind, the simplest way to calculate the potential energy is relative to the orbiter being free of both the Earth's and the moon's gravity. So it is simply the sum of the (negative) potential energy due to its finite distance from the Earth and the (also negative) potential energy due to its finite distance to the moon (each of these being inverse linear laws...)

What approach would best have the effect of retarding the moons drift away from earth? Milton Babb 720gunby(at)gmail(dot)com

Well, that's not a problem really, if you want to travel to the moon in a straight line you can....provided that your speed of travel is pretty close to the speed of light.

The reason we travel in a weird (not really a figure 8) remember that the earth and the moon are not static, the moon rotates around the earth in 28 days and the earth rotates around the sun in 365 days.

Since we don't yet have the capability to travel at the speed of light nor even close to that, we need to use balistics to calculate the trajectory required to hit the target, that means that we have to know where the moon is going to be in the future so that we can reach it with the velocities that we can achieve today.

The actual figure is closer to a spiral than a elipse again due to the maximum velocity we can achieve today.

You are correct about the movement of the Moon. However, the inertial frame of reference in this instance is the Earth, so the Earth's movement is not a factor in the calculations. All lunar trajectories are relative to the Earth rather than the Sun or Milky Way, etc..

Nice find!

Chill out dude, humour is a part of normal life.

Life would be very dull without it. It has also thrown up some valid points and alternative ideas in the past.

Remember your blood pressure!

I can't see the diagram he posted clearly, but there is a trajectory that NASA used on Apollo 13 to return the spacecraft to Earth without a trans-Earth injection burn. If memory serves, that brought the spacecraft as close as 60 miles to the lunar surface. They also used that same trajectory on some earlier missions and it saves a lot of fuel when they do that because there is no additional burn required to get back to Earth orbit.

I think that was what poster #9 was trying to show that, but if you have evidence that is not the case, please elaborate.

I don't care two hoots about the GA, but the answer really wasn't stupid. That configuration is called a free-return trajectory. Sheesh.

1. Post 9 is not stupid answer. Thats the fact. Some such trajectories are designed have very small fuel consumption on return path

2. If you want to say it is Stupid, be courageous and do not hide behind name Guest. Even with your name, nobody knows you personally. This is coward action. My protest.

http://en.wikipedia.org/wiki/Trans_Lunar_Injection

Your answer is good, but I wonder if it is complete.

Once the spacecraft rendezvoused with the Moon, the Moon's gravity pulled the spacecraft around to the far side of the Moon. To enter lunar orbit the Apollo spacecraft actually had to reduce its velocity to go into orbit around the Moon. The engine faced the direction of motion when fired, to allow it to 'fall' into lunar orbit. As you pointed out, from the reference frame of the Earth this gives the figure 8 appearance. I'm not sure where the 'true' apogee point was of the original orbit, but one could imagine the apogee point was just this side of the Moon's orbit, tangent to the lunar orbit that the craft had as it orbited the Moon; thus giving a 'Figure 8'.

Why was the apogee point not on the far side of the Moon, with Apollo orbiting in a prograde orbit (with the loop of the lunar orbit tangent to the inside of the original orbital path)? If such orbital mechanics were possible, I don't think the path would have had the appearance of a figure 8.

My answer was incomplete. My primary objective was to explain that all objects that are not under continuous propulsive force move in orbital trajectories that are principally conic sections – i.e., they are trajectories that are circular, elliptical, parabolic, or hyperbolic. They emphatically are not straight lines from any inertial reference frame. The simplest problem is analyzed as a two-body problem – the spacecraft is one body and the second body is the larger one that you are orbiting around. More complex is a three-body problem which in this case would be the Earth-Moon-spacecraft situation. If one wanted to include the perturbing influences of other bodies (such as the Sun, Jupiter, etc), the complexity increases. Where I was incomplete and perhaps a little sloppy in my first explanation is that this is a three-body problem. As Apollo neared the moon it entered the Moon's gravitational sphere of influence. The first several Apollo missions followed a "free-return" trajectory so that if there were any problem the astronauts would loop behind the moon without entering lunar orbit and would, without needing a velocity correcting rocket burn, be on a return trajectory back to Earth. As Apollo looped around the far-side of the moon in what would appear as a clockwise orbit around the moon (looking down from the moon's north pole), it would have too much energy (kinetic as in velocity) to enter into orbit around the moon. So, yes, Apollo would have to do a burn to reduce it's velocity to enter into a lunar orbit. And, that is also why when Apollo was ready to return to Earth they did a burn on the far side of the Moon to increase their velocity to "escape" the moon. If you are old enough to recall, we all waited for Apollo to reappear from the far side of the moon before communication could be reestablished to learn whether their burn was successful. Later missions reflected NASA's increased confidence and were not "free return" trajectories - but they were still basically the same trajectory and would look like a figure 8 from the rotating reference frame defined by the line from the Earth to the Moon. Apollo 13 did not start out on a free-return trajectory. It experienced it's crippling malfunction shortly after departing Earth orbit for the moon and had to do a corrective burn during its trans-lunar coast to place it into a free-return trajectory to, at a minimum, ensure that the crew would be returned to Earth's immediate neighborhood. To your last point, keep in mind that the Apollo trajectory cannot be viewed as a pure elliptical trajectory around the Earth because Apollo has to enter into the Moon's sphere of influence (approximately 5/6-ths of the Earth-Moon distance). Different strategies for rendezvousing with moon would have been simulated/modeled with all of the needed rocket burns, the timing and orientation of the burns, and the total energy required determined. NASA selected a minimal-energy strategy that also minimized risk with the "free return" feature. My apologies for being long-winded, but I cannot do this explanation in a more concise manner. Orbits are conic sections. Energy and momentum are conserved. Non-circular orbits trade kinetic for potential energy, and vice-versa.

I'm sure: NASA took the shortest way to the moon with the Apollo-Flights!

And this is allways still a stright line - if it looks like an eight this is because the moon moves around the earth and the velocity of the command modul at the trip is not constant.

Regrettably, in this case your certainty is unwarranted.

There are two main constraints - time and energy. The paths will be designed to compromise between these.

The earliest moon missions (unmanned and Russian) were close to straight-line, because this had the simplest control requirements. But they were not capable of entering moon orbit, in spite of the relatively high initial rocket-to-payload ratio.

In order to enter lunar orbit from a "straight-line" path you would need rather high acceleration to match your velocity to initial orbit. This in turn would require large rocket motors, which would increase your mass, and so the total fuel needed for the mission.

This is one of the reasons that missions no longer fly linearly from atmospheric exit to the impact/orbit region - and there may well be others. FYI, the direct ascent took about 1.5 days, NASA missions have taken between 3 days and 5.

If you can afford to take months for the flight, you can use gravitational perturbations to gradually place you in a suitable position for lunar injection - and substantially reduce the energy requirement. The details of this are beyond me - but it has actually been done (the rescue of the otherwise failed Japanese Hiten mission).

SURE: a straight line in a dynamic system is not realistic - every change in the system basics will change the line to a curve. It's not the point. But i's a simple way to understand the potential of the moon-earth-system.

Clearly this is at best a communications problem.

Are you saying that you meant that the "figure of eight" would appear even if the actual path was along a straight line? Because that too would be false; the straight line would start ahead of the moon, be coincident as you come level with the moon, and get progressively further behind thereafter - so there could be no crossover in that curve (unless the time exceeded an inertial month, of course).

have a look at mars for the motion - there is a backward move while moving around the sun seen from the earth.

It must have to do with using the gravitational pull of the earth and moon to sort of sling shot the vehicles in a trajectory ahead to where the next object will be. So I guess you swing around the planet or moon one way and release on the trailing side of the orbit to be flung toward the leading edge of the other object, which would allow you to pick up the most speed before being released into space.

shooting from earth in direction to the moon in front of the moon path creates an gravitational force from the moon to the projectile that the moon accelerates the projectile on the path to the moon and increases the speed-difference between projectile and moon, shooting behind to the moon decreases the relative speed between moon and projectile.

my quess is that the moon rotates in the opposite direction and you need to change directions in order for the spacecraft to orbit at a relatively slower speed.

No.

When seen from North (from say pole star, earth rotates anticlockwise, moon rotates anticlockwise and moon revolves around earth anticlockwise. Even the earth revolves around sun in anticlockwise direction.

The reason is to establish a Hohmann Transfer Orbit. Also it allows for lunar orbit injection without having to use another main engine burn to put the craft in lunar orbit.

Second, it has to do with the narrow re-entry corridor for the command module to enter into the Earth's atmosphere. This corridor is only about 2 degrees if I remember right. Screaming in at 25,000mph requires a very precise approach to keep from either skipping off the atmosphere or burning up instantly in the higher layers of the atmosphere.

Also, the figure 8 trajectory allows for mid course corrections due to thruster action or other trajectory errors caused during the TLI (trans lunar injection) burn.

GA = Great Answer. Very well constructed and highly informative.

but even in the sixties, computers helped with this right? (vanguard)

while looking at this question, I came across this resource.

Chris

Thanks.

I'd guess back then they would have used their slide rules to first work out the rough details, then used the computers to get everything worked out precisely -- at least, in the early years. I'd suppose by the late 60s they probably had the computer programs pretty well set to run the calculations.

Interesting links, by the way.

(Guest 9:21 AM)

I think it was well answered too.

However, the relevance of lunar orbit insertion as far as energy is concerned is in error, I believe.

The direction of passing over or under the Moon matters in terms of what you get from the gravity assist maneuver.

Passing over the top of the Moon would add spacecraft velocity versus passing under the Moon scrubs space craft velocity off.

I think that gravity assist will turn out to be the key for this puzzle. We will just have to wait and see if that turns out to be true or not.

The other half of the puzzle may have to do with what is determined to be the inertial frame of reference for the trajectory. As pointed out, depending on the frame of reference, the spacecraft path can take various shapes.

For an interesting twist on gravity assist, look up Epimetheus and Janus, which are two small moons of Saturn.

The interesting thing about these two moons is that they share the same orbit! They continuously trade orbits as they approach each other and exchange orbital momentum. The exchange takes about 4 years and appears to be stable.

By passing under or over the Moon, I assume you mean passing by the leading edge of the Moon or the trailing edge of the Moon -- or equivalently, the nearside (hence giving a retrograde orbit around the Moon) or the farside (hence giving a prograde orbit). I suppose it's semantics concerning the phrase 'passing by'; it depends on what the reference point is.

You may have a point, I'm just not sure. I'm trying to decide -- once the spacecraft is within the Moon's gravitational influence it has to shed energy to fall lower into the Moon's gravitational well. I'm thinking that this energy has to be regained when it needs to leave the Moon's gravity well, so that's maybe a zero sum game. [I.e. perhaps the total delta-V is the same for either scenario.] One has to be careful in shifting from an Earth-centered reference frame to a Moon-centered reference frame when looking at the total energy in the system, so I know I need to be cautious in my thought process.

Guest

NASA likes orbital paths of minimum energy (fuel) required. They use earth spin (launch east) to assist climbing out of earth gravity well; Lunar approach offers choice of flying in front or behind the moon; if behind the moon was chosen spacraft is sling-shot to a higher velocity and would then escape lunar gravity well (without large retro-fire). The 'figure eight' permits the spacecraft to pass in front of the moon so as to slow the spacecraft (transfer momentum to the moon) which results in minimum thruster burn to achieve lunar orbit. Less fuel required.

The trajectory of the spacecraft after a pass behind the Moon would depend on the path during its approach to the Moon's gravitational region of influence. Here are a few examples of possible paths that might occur following a pass 'behind' the Moon. Black lines indicate the path if the Moon weren't there, the blue lines show the Moon's effect. It is often possible for a change in velocity to have no net effect on speed, but the change in direction accounts for the increase (or decrease) in energy.

The believe the path chosen was to use the moon's gravity to return the spacecraft to the earth if the rocket burn in lunar orbit failed. This, in fact, happened in the case of Apollo 13.