Its not possible or is it ?

[kevb]

As the moon orbits the earth every 27 days at a speed of 1.01kms
Not taking earths gravity into account
What would be the speed of Apollo when it gets back to earths orbit
When taking into account the pirouette effect
Regard kev

[brotherofthemoon]

There's no reasonable answer to that question, as the CSM never actually "returned to Earth orbit."

Unless you consider slamming into the Earth's atmosphere at 35,000 feet per second "returning to orbit."

[Bob B.]

It is not very clear to me what you are asking. It sounds like you want to know what Apollo’s speed would be after swinging by the Moon (as if on a free return trajectory) and traveling out to a distance equal to the Earth-Moon distance but considering only the Moon’s gravity, i.e. neglecting the presence of Earth. Is this correct? If not, can you better explain your question?

[Jason Thompson]

How exactly do you propose to calculate the speed of an object without taking into account Earth's gravity, and why do you presume the answer, if such a calculation could be done, would be relevant to Apollo?

Since Earth's gravity slows the capsule down on the outward journey and accelerates it again on the inbound journey, discounting it is ridiculous.

[brotherofthemoon]

So is the question "what if we sent a man to the Moon and the Earth didn't want to have anything to do with it?"

Or "how fast would the spacecraft be traveling during a free-return trajectory when it got back to Earth and the Earth wasn't there?"

The few references I found on Google for "pirouette effect" related to:

-"anomalies associated with rotating magnetic systems"
-the "squeezatron" -"A mechanical device for showing the pirouette effect."

[pzkpfw]

I would guess that he thinks that as the Apollo craft, at or near the moon, were going at some speed, and "circling" Earth; that when they got back to Earth their speed would have increased "too much".

I think he is trying to draw a parallel with what happens when you sit on a chair and spin with your arms out; how conservation of angular momentum (?) means that you speed up when you draw your arms in.

That is, the "pirouette effect" is what you see with ice-skaters... and he thinks that would happen to spacecraft too?

(I'm only trying to interpret the question - it's part of the fun I have with CT's; I have zero HB)

[nomuse]

Hrm. My best guess is he means "slingshot" effect -- but couldn't remember the word. I am crossing my fingers that Moon Brother's find of "anomalies associated with rotating magnetic systems" has nothing to do with the OP's query; I don't know if I could deal with another Electric Universe thread!

Read the post one more time. Perhaps he is talking about the moon's orbital velocity, and wondering if it should be added in?

[Bob B.]

nomuse said:

Hrm. My best guess is he means "slingshot" effect -- but couldn't remember the word.

That was my guess as well.

[kevb]

OK I let every one think about this before I replied
PZKPFW was rite he knows the pirouette effect is the same as when
You spin your self on a swiveling chair then put your legs out then rotation slows
Bring your legs in and rotation speeds up.
The same affect happens when ice skaters spin themselves with there arms out
And when they put there arms close to their body they can spin at tremendous speed
The ice skating name for this is the pirouette hence the pirouette effect
Now any craft coming from the moon starts off the same speed as the moon orbit around
The earth 1.01 kms
Even if the craft is in moon orbit it is still traveling more or less 1.01 kms around the earth.
Now when the craft powered its way out of the moons orbit towards earth the pirouette effect Will Kick in and as soon as the burn stops the centrifugal force will be greater than
Earths gravity and the craft will simply return back to a similar orbit as the moon Around the earth the speed will be back to 1.01kms
So am I rite in thinking it is impossible to power you way back to earth from the moon
Regards kev

[nomuse]

I may be reading too much into one statement here.... but are you claiming that it is impossible to break orbit? If what you describe were to hold true for attempting to leave an orbit around Earth at Lunar distance, it would also hold true for attempting to leave an orbit around Earth at, say, the distance of the ISS or your average shuttle flight.

You are at least partially correct in your description, however. A single burn does not create a different circular orbit. Any short burn can only increase or decrease the eccentricity of an orbit; lengthen or shorten the ellipse, basically. Within a range of velocities, your spacecraft will still return to the same starting point. It is the other end of the ellipse that changes.

And now I'll bow out and let people who can actually calculate (as well as better describe) orbital mechanics....

[Jason Thompson]

Why would it be impossible to break out of lunar orbit and return to Earth when it is patently not impossible to break out of Earth orbit and go to Mercury, Venus, Mars, Jupiter, Saturn.....?

[brotherofthemoon]

Well, KevB I think you better get on the phone with NASA. They've been planning a manned mission to Mars for years now and apparently don't realize that it's orbiting the sun at 24 kilometers a second. If you can't get back to Earth from the Moon there's no way you'd get back from Mars.

[JayUtah]

Well, Kev, I think your notion of celestial mechanics with more than two bodies is, to put it politely, way the heck out in outer space.

Closed-form (i.e., easy) formulations for the velocity and position of orbiting bodies exist only for two bodies. When you have three or more bodies, the problem is much more complex and can't be solved in closed form.

Navigating in cislunar space is generally formulated as a restricted three-body problem. You have three bodies: the Earth, the moon, and the spacecraft. If you were scrupulous, you'd consider these all as free bodies acted upon only by momentum and gravitation. But fortunately for us we can neglect the effect of the spacecraft's gravity upon either Earth or moon and consider that Earth and moon would behave for us essentially as they would if there were no spacecraft.

So we can consider that the two bodies -- Earth and moon -- pursue their orbits oblivious of our presence, so then it becomes a matter of plotting our course through two moving and gravitating bodies. The path of a third body in such a case is computable, but only by iteration. Again, no closed forms.

An object appearing to orbit the moon is indeed tracing out an orbit around it, but with effects from the Earth also computable and observable. The same is true with objects we put into orbit about the Earth. Although they appear to follow orbits only around the Earth, they are pulled this way and that inexorably by the much fainter gravity of the moon (and also of the Sun, if you want to be more general). There is no orbit around the Earth that fully disregards effects from the other bodies. Nor is there such an orbit around the moon.

For orbits very close to the primary, we can in some cases disregard the gravitational effects of other bodies in our computations, accepting that the computations will be simplified and thus somewhat inaccurate. They may simply be accurate enough for our purposes. In a close orbit around the moon, it is habitual to consider the problem not in the general (or even restricted) three-body situation, but in the closed-form two-body case with "perturbations" deriving from the other bodies. The moon is considered fixed in space. The paths of the other bodies are described in moon-centric fashion and their gravity forces accounted for by adjustments in the closed form.

But this doesn't hold for orbits sufficiently distant. Where an impulse raises an orbit about the moon to an altitude where the perturbations increase from second- or third-order effects fully to first-order (remember gravity is an inverse square law), then the notion of "orbiting around the moon" no longer applies. The spacecraft is then embroiled in the full computational fury of the restricted three-body problem. It no longer orbits the moon. It pursues an orbit in the multivariate regions of cislunar space.

[kevb]

For those who say if you can get from the earth to the moon then it is the same to get
From the moon to earth then you are forgetting one thing
The moon orbits the earth the earth dose not orbit the moon
The pirouette effect only works one way and that is from the moon to earth
So my question to you is how did they return from the moon
I like to talk to you in a simplest way possible to try and get my message across
Unlike JAYUTAH posted previously he talks in riddles I am not impressed by that
So come on in simple terms how was it done
Regards kevin

[gonzo]

kevb said:

For those who say if you can get from the earth to the moon then it is the same to get
From the moon to earth then you are forgetting one thing
The moon orbits the earth the earth dose not orbit the moon
The pirouette effect only works one way and that is from the moon to earth
So my question to you is how did they return from the moon
I like to talk to you in a simplest way possible to try and get my message across
Unlike JAYUTAH posted previously he talks in riddles I am not impressed by that
So come on in simple terms how was it done
Regards kevin

You were provided in the last post with sufficient information to answer your own question.
However, here's a thought. The Earth orbits the sun. Your analogy holds true then for the Earth-Sun system does it?
Do celestial mechanics prevent us from launching probes into a 'higher' orbit to Mars, Jupiter etc, or to lower orbits such as Venus, Mercury? If not, why not?