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

Posted April 04, 2010 5:01 PM

This month's Challenge Question:

When NASA sends flights to the moon, the path used is similar to the number eight. Why not travel in a straight line directly to the moon? Why not an ellipse around the earth and moon?

And the Answer is....

To save energy, the spacecraft should follow a path where it remains close to the line between the centers of the Earth and the Moon. Because of the continuous movement respect to each other, this path has a shape similar to the figure of the number eight. Following this path, the net force on the spacecraft is minimized so it consumes less energy.

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#1

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

04/04/2010 8:57 PM

Easy. First you aim ahead of your intended target, but most important is the entry orbit is designed to scrub off velocity by what is called gravitational assist.

Entry from beneath the orbit of the Moon trades off some of the energy in the spacecraft and transfers that energy into the Moon's orbit via conservation of momentum. This is much more efficient use of fuel.

Going the other way would increase the spacecraft's velocity, which makes attaining orbit more costly as far as fuel is concerned.

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#2

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

04/05/2010 10:44 AM

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.

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#3
In reply to #2

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

04/05/2010 12:36 PM

"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.

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#4
In reply to #3

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

04/05/2010 1:16 PM

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).

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#5
In reply to #4

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

04/05/2010 2:11 PM

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.

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#6
In reply to #5

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

04/05/2010 4:17 PM

"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...

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#7
In reply to #6

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

04/05/2010 5:26 PM

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?

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#8
In reply to #7

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

04/05/2010 5:40 PM

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...)

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#34
In reply to #2

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

04/14/2010 11:31 AM

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

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#9

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

04/05/2010 8:02 PM

Nasa was too cheap to buy a return ticket, so they needed a free ride back.

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#10

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

04/06/2010 12:07 AM

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.

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#12
In reply to #10

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

04/06/2010 6:44 AM

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..

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#13
In reply to #12

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

04/06/2010 8:01 AM

You made me check up on my expectations.
The orbital height above the moon would need to be less than about 60-miles for the direction of travel to reverse in an Earth-centred (pseudo-) inertial frame, and this would only happen when the mission is between the Earth and the moon - so we have no resemblance to a figure of eight. You really need to be using coordinates that rotate synchronously with the moon to see this.

BTW, and I think curiously: some of the orbits actually used for moon-missions do not seem to have been near the trajectories you might expect . The Luna-1 mission used a 34-hour transit that (prior to entering moon orbit) was very close to a straight line in the Earth-centered (pseudo-) inertial frame, though more recent missions have used times between 3 and 5 days.

Calculations are made relative to the Earth-centred non-rotating coordinates (not always...) - but illustrations are often relative to the Earth-moon axis, so you can see the position (relative to the moon) as the mission enters moon orbit.

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#16
In reply to #13

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

04/06/2010 9:25 AM

Nice find!

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#11

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

04/06/2010 12:18 AM

It is not figure of 8, it is figure of infinity! (See post 9)

This is to avoid use of infinite use of fuel

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#14

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

04/06/2010 8:25 AM

I am wondering who rates the answers? How can you give a good score for a stupid answer like # 9.

Is this a joke or do you look forward to take you webside to become a jokeside?

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#15
In reply to #14

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

04/06/2010 8:35 AM

That one at least included a diagram of sorts.

Why not register and try to improve matters... (You can even vote against, though some of us prefer to reserve this action for answers that appear plausible but are wrong).

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#17
In reply to #14

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

04/06/2010 9:25 AM

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!

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#18
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Re: Path To The Moon: Newsletter Challenge (04/06/10)

04/06/2010 9:31 AM

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.

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#21
In reply to #14

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

04/06/2010 7:25 PM

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.

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#22
In reply to #14

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

04/06/2010 11:22 PM

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.

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#19

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

04/06/2010 12:00 PM
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#20
In reply to #19

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

04/06/2010 6:26 PM

The path the Apollo mission followed was not a figure 8. It only appears to an Earthbound observer as a figure 8 because you are viewing it from a rotating reference frame - the line from the Earth to the moon. If you were to observe the path taken by Apollo from an inertial reference (or perhaps the Sun-Earth reference frame which rotates much slower) you would see that the path taken is actually an elliptical orbit with the Earth at one of the focii of the ellipse. The elliptical orbit is the classical Hohmann transfer orbit. It is a minimum-energy transfer orbit. The launch from earth orbit is timed so that when Apollo reaches it's apogee (farthest point from Earth) it has rendezvoued with the moon. Initially, Apollo's orbital speed (both angular rate and speed) is much faster than the moon's, but much slower once Apollo reaches its apogee point. The moon, in its nearly circular orbit has a speed of nearly 2100 mph. Apollo will be slowing down the whole way to its rendezvous point while the moon, at that point, will be moving at a greater orbital speed than Apollo. Roughly, if it took 4 days to reach the moon, which has a approx 28 day orbital period around Earth, 4/28 equates to approximately 51 degrees. So when Apollo "departs" Earth orbit for the moon, it does it's trans-lunar burn from the opposite side of the Earth from where it wants to rendezvous with the moon. And, the moon is 51 degrees away from the rendezvous point. So, initially, Apollo appears to the "right" of the Earth-moon line. As Apollo follows it's elliptical orbit it is initially moving faster than the moon (will pass the Earth-moon line and be "ahead" of the moon. But as it slows down, the moon starts to catch up and as Apollo's distance from the Earth becomes slightly greater than the moon's distance, the faster moving moon passes "below" Apollo and you have witnessed the first "half" of what looks like a figure 8. To enter lunar orbit, Apollo would do a burn to increase it's velocity to enter into lunar orbit. If Apollo did nothing at this point, the moon continue by and Apollo's elliptical orbit would continue with Apollo "falling" back towards the Earth. As it continues to move towards the Earth it's speed increases while its distance to Earth becomes less than the moon's distance. As it speeds up, Apollo appears to now move back towards the Earth-moon line and starts moving ahead of the moon as it returns to the Earth - thus completing what looks like a figure 8 trajectory from the perspective of the Earth-moon line. But it is an elliptical orbit in inertial space. Similar orbital dynamics are what cause geosynchronous (as opposed to geostationary) Earth satellites to appear to move in a figure 8 relative to a fixed observer on the ground.

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#36
In reply to #20

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

04/18/2010 6:32 PM

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.

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#37
In reply to #36

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

04/18/2010 9:15 PM

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.

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#23

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

04/07/2010 12:28 AM

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.

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#24
In reply to #23

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

04/07/2010 4:27 AM

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).

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#25
In reply to #24

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

04/08/2010 12:48 AM

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.

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#26
In reply to #25

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

04/08/2010 5:41 AM

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).

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#27
In reply to #26

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

04/08/2010 11:40 PM

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

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#28

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

04/13/2010 3:45 PM

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.

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#32
In reply to #28

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

04/14/2010 12:15 AM

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.

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#29

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

04/13/2010 7:24 PM

Here's a nice little lesson on the trajectory. You might not want to do all the math, but there's a good diagram that shows why we call it a figure-8 even though it isn't.

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#33
In reply to #29

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

04/14/2010 10:25 AM

Nicely done, and a very good link. Thanks TVP45.

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#35
In reply to #33

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

04/14/2010 7:36 PM

You're welcome.

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#30

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

04/13/2010 7:51 PM

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.

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#31
In reply to #30

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

04/13/2010 10:55 PM

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.

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#38

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

04/20/2010 3:02 PM

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.

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#39

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

04/21/2010 9:18 AM

There have been lots of good replies so far, but I don't think any have fully answered the question. I think my reply does. (IMHO.)

The path to the Moon involved many issues, of course, but here two issues are of major importance: A) safety and B) cost. To address item A) NASA wanted a 'free return' path that would automatically return the spacecraft home if a malfunction should occur and which did nor require any action by the astronauts to make the return possible. To address item B) NASA wanted a path that required the least amount of fuel, since fuel is expensive and since the greater the amount of fuel carried the bigger the booster has to be, etc.

The possible paths to the Moon were then considered and assessed via these two concerns. Since Gravity is a conservative, 1/r-squared force, all free-fall paths (orbits) in a gravity field follow some form of a conic section. It is important to note that the energy of the orbit is related to its semi-major axis – the longer the semi-major axis, the greater the amount of energy needed for the orbit.

The possible paths to the Moon fall into 3 main types: 1) a 'direct path' that lands the spacecraft on the Moon without orbiting the Moon; 2) an elliptical path where the apogee of the ellipse is on the far side of the Moon and the spacecraft orbits the Moon in a prograde direction; 3) an elliptical path where the apogee is on the near side of the Moon and the spacecraft orbits the Moon in a retrograde direction.

Paths like type 1 fail the A criteria. There is no free-return if a malfunction should occur. For a 'truly' straight-line path the B criteria is failed miserably. The semi-major axis of the conic section would be infinite; a straight line orbit would require infinite energy.

Paths like types 2 and 3 can both satisfy the A criteria for free-return. So the critical difference between a type 2 path and a type 3 path is that the type 3 path has its apogee point nearer to the Earth, meaning that its semi-major axis is smaller than a type 2 path. Thus a type 3 path (which yields the 'figure 8' pattern seen from the Earth's reference frame) requires less energy, and best satisfies the A and B criteria.

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#40
In reply to #39

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

04/21/2010 12:03 PM

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

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Anonymous Poster
#41
In reply to #40

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

04/21/2010 12:31 PM

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)

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#42
In reply to #40

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

04/21/2010 12:45 PM

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.

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#43
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Re: Path To The Moon: Newsletter Challenge (04/06/10)

04/21/2010 4:40 PM

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.

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#44

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

04/22/2010 4:27 PM

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.

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#45
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Re: Path To The Moon: Newsletter Challenge (04/06/10)

04/23/2010 8:23 PM

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.

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#46

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

04/25/2010 1:49 PM

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.

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