Flight Question Newbie question: Left or Right of the Moon?

goretteux

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Hi all,

Please forgive this simple question if the answer is obvious to anyone around here who isn't a newbie like myself...

I've been looking into information about lunar insertion for a few days now for a personal project, and it seems all the illustrations I find on the net about your typical Apollo flight path show an "S-shaped" curve:

- slingshotting out of Earth along the direction of the Earth's rotation (emerging from the "right-hand side of Earth" when looking at the Moon)
- being captured on what I guess I would call the leading hemisphere of the Moon ("to the left of the Moon" as seen from Earth)
- wrapping around the Moon in the direction of the body's rotation...

As shown here:

Fig1a.jpg


Now, I've come across this 1 illustration (following next) showing a "C-shaped" trajectory, where the takeoff and Earth departure are the same as what I previously described, but the capture by the Moon occurs "to the right of the Moon" as seen from Earth... I'm probably wrong here, but it strikes me as strange that one would manage to "go against" the rotation of the Moon. Unless this is a better way to lose energy for landing?

diagrammes.jpg


Could anyone confirm that this is not a fluke, and that such insertions actually do exist?

If so, what is their purpose in comparison to the typical "S-shaped" Apollo-era insertion?

Benefits and drawbacks?

Again, please forgive my layman's terminology, and many thanks in advance for any help that can be provided.

Michel
 

the.punk

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The S-shaped trajectory is called a free return trajectory. When your TLI burn is complete you would fly around the backside of the moon and return to earth without any TEI burn if you don't land on moon. This would then look like an 8-shaped trajectory when not inserting to moon orbit and landing. In apollo 13 they had to continue the flight without landing and then flying around the backside to sling around the moon.
 

Evil_Onyx

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The only reason that i can think of to go to the moon and approach from behind (ie. the "C-shaped" trajectory) is to use the moon to slingshot you out of Earths SOI, as the moons movement adds more energy to your trajectory. Coming in front requires less DV to attain a stable orbit.

I think the second image needs to be take in contex of its origanal document, or the artist had no idea that he did it wrong till it was too late to change it.
 
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goretteux

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Yup I do remember Apollo 13 and the whole going-behind-the-Moon part... thanks for the information, at least I know now what a free-return trajectory is :)

I have to say that still leaves me puzzled as to what that "C-shaped" trajectory is though :/

I've been wondering whether this C/S-shape nuance is what some documents I've come across call Prograde/Retrograde approach... but I probably have it mixed with something else.

---------- Post added at 08:24 PM ---------- Previous post was at 08:16 PM ----------

The only reason that i can think of to go to the moon and approach from behind (ie. the "C-shaped" trajectory) is to use the moon to slingshot you out of Earths SOI, as the moons movement adds more energy to your trajectory.

But then in that particular case, that's going *against* the direction of the Moon's rotation - wouldn't approaching it on the trailing edge ("right-hand side seen from Earth") like it is pictured mean that the Moon would be drawing from your own energy, not the other way around as you suggest?

In any case, I imagine the amount of energy either gained or lost to the Moon by its radial velocity would be negligeable compared to the gravity bump that you would get from flying close to any celestial body...

I think the second image needs to be take in contex of its origanal document, or the artist had no idea that he did it wrong till it was too late to change it.

I've thought about that too, but honestly I find it hard to believe that whomever created that illustration (be it a scientific/technical journalist from the Apollo era, or some NASA illustrator for the Apollo project) would make such a gross mistake delivering such an atypical trajectory without checking it first...
 
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Hielor

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The second picture also shows the capsules as going to the "left" of the moon, as looking from Earth...
 

goretteux

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The second picture also shows the capsules as going to the "left" of the moon, as looking from Earth...

Well... not as far as I can see it myself:

  • The first picture, if seen from above with the Moon at the top of the picture, depicts an S-shaped trajectory, so the capsule goes behind the Moon from left (leading hemisphere) to right (trailing hemisphere)
  • The 2nd picture shows a "mirrored C-shaped" trajectory, so actually more like a ")-shaped" trajctory, with the capsule going behind the Moon from right (trailing hemisphere) to left (leading hemisphere)
So there's definitely a difference between the 2.

Let me put it some other way maybe - has any of you users of Orbiter managed to approach the Moon "from behind", as Evil_Onyx described it, and is it even possible to do so, or does the software only let you shoot "ahead" of the Moon?

It seems to me to be a fairly straightforward question, but there doesn't seem to be any answer to it anywhere on the internet, and I've been looking for several days now :p

My guess is that, again as Evil_Onyx mentioned, since the S-shaped solution is the safest one (free-return), even if the C-shaped solution is possible, it is pretty much dismissed out of hand and not bothered with in discussions...

Still, I wouldn't mind knowing if it is even remotely possible... just for the sake of knowing.

Does the direction of the rotation of a planet have an impact on an incoming capsule? Or does it not matter, and the effect imparted by the gravity well is all that registers initially, with an additional curving inwards of the existing capsule trajectory, regardless of its direction in relation to the planet's direction of rotation?

From what I've found out, the Moon's radial velocity at the equator is 100 times less than that of Earth... so if the latter of the 2 previous hypotheses is valid (planet's radial velocity direction and intensity don't matter), and even if the former hypothesis is the correct one, given the weakness of the Moon's rotation then I'd be inclined to believe it would be possible to "come from behind", even if at greater cost in terms of expanded fuel...

And of course there's the 3rd (not unlikely) hypothesis, that something eluded my comprehension totally and so I'm talking out of my own bum :(

Some expert opinion please?
 
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Urwumpe

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You also overread one important detail:

Delta-V, the impulse needed for each maneuver.

For the S-shaped trajectory, the Delta-V is lower, less fuel is needed, because the spacecraft enters the gravity well of the moon with less relative velocity, it is simpler to enter lunar orbit. The S-Shape would look completely different when seen from the moon, which is the important part for LOI. Also, if the LOI maneuver fails, you would be less far away from getting back to Earth with alternative methods, while the C-shaped trajectory would require a direct crash into the moon, since you enter it head-on, instead of getting overtaken by the moon.
 
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goretteux

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Awesome - that's the most useful answer so far (no offense to the other two), thanks Urumpwe.

So you're saying the C-shaped trajectory *is* possible, just way way more hazardous than the S-shaped one given the chances of adding another crater on the Moon?

One additional question: assuming you had the means to slow down massively just prior to impact, any chance such a C-shaped trajectory would get cargo (as opposed to people) to a point on the trailing hemisphere of the Moon *faster* than with the S-shaped trajectory, where you have to go around the Moon once?
 
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Andy44

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Also, it's no big deal to have to land on the moon from a retro-grade orbit, since the moon's rotation is so slow. The surface of the moon at the equator moves about the moon at about the speed of a man on a fast bicycle, which isn't much to overcome for a vehicle that has to brake out of orbit.
 

goretteux

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Also, it's no big deal to have to land on the moon from a retro-grade orbit, since the moon's rotation is so slow. The surface of the moon at the equator moves about the moon at about the speed of a man on a fast bicycle, which isn't much to overcome for a vehicle that has to brake out of orbit.

So that's what a retro-grade orbit means... thanks Andy44, I think I might finally be getting somewhere.

Would you lose more energy in a retro-grade approach, and get an added energy boost in a pro-grade one, or does direction and intensity of rotation not matter and it's only the intensity of the gravity well that does?

Any PDF primer to recommend where this type of layman questions like mine might be answered?

Thanks in advance
 

Urwumpe

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Any PDF primer to recommend where this type of layman questions like mine might be answered?

Not pdf, but the basic concepts are here:

http://www2.jpl.nasa.gov/basics/

Essentially all things in the mission planning for Apollo are just implications of these basic things.

Since the lunar trajectory is a three-body problem, there is no simple solution to it, but the book "Fundamentals of Astrodynamics" has a whole chapter dealing with this topic on a pretty harmless level. Well. Harmless for orbital mechanics.
 

Andy44

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Would you lose more energy in a retro-grade approach, and get an added energy boost in a pro-grade one, or does direction and intensity of rotation not matter and it's only the intensity of the gravity well that does?

Yes, in a retro-grade approach (orbting against a planet's rotation) you need to not only slow down to zero inertial (correction, not ground) speed for a landing, but also accelerate in the opposite direction a bit. Likewise, launching against a planet's rotation requires more delta-V.

This is why Cape Canaveral is the preferred launch site in the US, and most launches are pro-grade (to the east with Earth's rotation). It also means that a heat shield has an easier time braking a vehicle out of orbit when it's flying in a easterly direction. Most polar orbits out of Vandenberg are slighly retrograde, having an inclination greater than 90 deg.

Like I said, the moon's rotation is so mild that retrograde landings and takeoffs exact only a trivial delta-V penalty, but on Earth the equator gives you over a 400 m/s kick for a direct-east launch.

Also, the definition of "prograde orbit" is an orbit with an inclination less than 90 deg, while "retrograde orbit" means inclination greater than 90 deg.

Orbiter also uses the words "prograde" and "retrograde" to describe orientation with or against the velocity vector, respectively, so the context the word is used in is important to avoid confusion.
 

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What i meant by behind the moon was approaching the moon, from the side of the moon that is retro-grade to the moons VV from earth.

I think i am lacking a the proper words that i require to explain what im trying to say.

Urwumpe, explained it better than me it think.

Need to dig out one of my reference books and acctually make some of it stay in my head this time :)

I've thought about that too, but honestly I find it hard to believe that whomever created that illustration (be it a scientific/technical journalist from the Apollo era, or some NASA illustrator for the Apollo project) would make such a gross mistake delivering such an atypical trajectory without checking it first...

I've seen examples of this sort of thing get published before been picked up, by that point they could have a 1000 copys of a 400 page book on the shelves...
 

goretteux

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Yes, in a retro-grade approach (orbting against a planet's rotation) you need to not only slow down to zero ground speed for a landing, but also accelerate in the opposite direction a bit. Likewise, launching against a planet's rotation requires more delta-V.

Thanks Andy44

Hmmm you answer brings more questions to my mind I'm afraid...

First, I'm not sure I understand why you'd need to accelerate past zero ground speed in the other direction if you were to prepare for a landing?

Second, the impact of a planet's rotation on launch makes perfect sense, but I have to admit I'm still somewhat confused about the effect on a potential retrograde approach to a Moon landing...

In a retrograde scenario I understand you would still need a delta-v correction in order to match speed with the surface, but isn't that delta-v *lessened*, not increased, by the fact the approach is retrograde? I mean in the sense that the vector for the Moon's speed at the equator gets substracted from your own insertion vector, as opposed to added if you were to approach prograde?

To add a bit to the confusion, I've just jumped to a Physics forum where they discuss gravity drag on prograde and retrograde satellites:

http://www.advancedphysics.org/forum/showthread.php?t=1715

I can't really see everyone agreeing on whether this has an impact or not, however one of the posters seems adamant that once you're in a stable orbit it doesn't matter whether you're going pro- or retrograde, your orbit will decay just as fast in either case, assuming it's not geostationary...

The way I understand the discussion, combined with some elements gathered here, once you have reached and gone past escape velocity it doesn't matter whether you are going (and so presumably also coming in) prograde or retrograde, you've entered the realm of freefall and only your energy (mass times squared speed, regardless of direction) matters.

It's below escape velocity that the Earth's own velocity at equator seems to matter (at least to the extent of my understanding) in the form of that additional vector, whether adding or substracting to a vehicle's own...

So would it be the same for approaching the Moon retrograde from a distance? No impact from the direction of the rotation of the Moon on an incoming capsule, unless you fall under Moon escape velocity, in which case vectors start to add up and that's where your delta-v budget begins to suffer?

I think I might still have some things mixed up :p
 
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Urwumpe

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Retrograde orbits are more stable in general, since the three body problem gets solved better, essentially your speeds relative to the central planet are such, that the effects of the planet stabilize your orbit, instead of pulling you slowly out of the orbit around the moon.
 

goretteux

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I think i am lacking a the proper words that i require to explain what im trying to say.

No that's fine I actually got your point the first time :thumbup:

I've seen examples of this sort of thing get published before been picked up, by that point they could have a 1000 copys of a 400 page book on the shelves...

That's the sort of thing I'm trying to avoid myself by posting here, albeit with much smaller circulation involved, but probably the same amount of self-esteem at risk ;)

---------- Post added at 12:44 AM ---------- Previous post was at 12:37 AM ----------

Retrograde orbits are more stable in general, since the three body problem gets solved better, essentially your speeds relative to the central planet are such, that the effects of the planet stabilize your orbit, instead of pulling you slowly out of the orbit around the moon.

That actually makes sense to me. I suppose the "retrograde-ness" reduces the length of involved vectors and so contributes to dampen the effects of orbital disturbances, when they would be magnified in a prograde orbit where vectors length would be increased.

But what of a retrograde insertion on the Moon then? Is it like I suspect, negligeable effect when the capsule is still above Lunar escape velocity?
 
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Andy44

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Thanks Andy44

Hmmm you answer brings more questions to my mind I'm afraid...

First, I'm not sure I understand why you'd need to accelerate past zero ground speed in the other direction if you were to prepare for a landing?

I was mistaken; I should've said you need to accelerate past zero inertial speed in the other direction. The object of landing, of course, is to get to zero ground speed.
 

goretteux

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I was mistaken; I should've said you need to accelerate past zero inertial speed in the other direction. The object of landing, of course, is to get to zero ground speed.

This is going to sound really stupid, but I still gotta ask... if you maintain just enough of a vector opposed to the Moon's equatorial rotation one, so you have constant zero inertial speed... do you just start falling vertically towards the surface?
 

Andy44

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Yes, that is correct. "Inertial" means a frame of reference which is fixed in 3-space, as opposed to a body-fixed frame which is fixed to the body's surface.

Think of the body-centered inertial frame as being fixed to the starfield. (In real life the stars do move slowly, but for a first-order approximation this is good.)

The body-fixed frame, on the other hand, rotates with the body in question.

The difference between the two frames must be considered for launches, landings, and for plotting ground tracks.
 

goretteux

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Thanks to everyone.

I didn't quite get answers to all of my questions :p but at least I've got a few solid facts + documents to start figuring out from.
 
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