Why Is The Moon Not Spinning Then?

Discussion in 'Pseudoscience Archive' started by common_sense_seeker, Sep 6, 2008.

  1. wlminex Banned Banned

    Looks like this thread has gotten "off-topic' from the title . . . .

    Re OP . . . . Perhaps I missed as similar post (haven't read all the posts), but, IMO, the moon's center of mass is offset slightly toward the earth due to mutual gravitational attraction (during lunar igneous-phase formation) - much like 'tides' - a denser lunar interior (asthenosphere?) toward it's nearest neighbor - earth. A similar (but much less so) effect may contribute to the earth's 'wobble' (precession).
    Last edited: Jan 6, 2012
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  3. sadang Registered Member

    And still, the moon does not rotate around its own axis, but only around an external center of rotation, namely, the orbital revolution center of rotation. And Tesla, has argued very well this, in two articles published in 1919 in Electrical Experimenter magazine.
    - scribd. com/doc/90144454/1919-Electrical-Experimenter-Nikola-Tesla-Famous-Scientific-Illusions
    - scribd.com/doc/90097922/1919-Electrical-Experimenter-Nikola-Tesla-Moon-s-Rotation

    Since there are two main points of reference, one internal Earth-Moon system and the other outside of this system, and because the motion analysis of the moon will always lead to perfectly opposite interpretations of its movement, in the best case we can say that we have an apparent axial rotation. Just an apparent one, not a real one!
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  5. Billy T Valued Senior Member

    Thanks for the interesting links. In the second it is clear that Tesla is admitting the moon revolves. I.e. presents the opposite side to the sun (or the distant stars) approximately every 14 days; however he does not want to call this revolution, rotation. He reserves "rotation" for spin about an axis passing thur the center of mass or symmetry in case of a wagon wheel, etc.

    Thus the argument is semantics, not physics. Perhaps back in 1919 many made that rotation/revolve distinction.
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  7. sadang Registered Member

    Billy, I prefer to have reservations about the Moon's own axial rotation. Mainly, as I already specified, there are at least two main points of reference, one internal and the other external, from which, the Moon's movement has perfectly opposite results. Which one is the right one? Why science give credits only to the external point of reference, and argues in favor of its own axial rotation synchronous with its movement of revolution of the Moon?

    In the second article, Tesla demonstrated mathematically that the moon does not have its own axial rotation movement. And he wrote: "... In all the communications I have received, tho different in the manner of presentation, the successive changes of position in space are mistaken for axial rotation." and "... The view I have advanced is NOT BASED ON A THEORY but on facts demonstrable by experiment. It is not a matter of definition as some would have it. A MASS REVOLVING ON ITS AXIS MUST BE POSEST OF MOMENTUM. If it has none, there is no axial rotation, all appearances to the contrary notwithstanding." and "... From the foregoing it will be seen that in order to make one physical revolution on its axis the moon should have twice its present angular velocity" and again "... But from the character of motion of the satellite it may be concluded with certitude that it is devoid of momentum about its axis.".

    According to Tesla's claims, and the entire contents of the article, follows that he denies the existence of Moon's own axial rotation motion. Therefore, your interpretation, compared to those stated and demonstrated by Tesla, is inconsistent with the content and conclusions resulting from the article.
  8. Billy T Valued Senior Member

    What are your two “perfectly opposite results”? I don´t see two different facts about moon´s angular momentum, but of course angular momentum is ALWAYS wrt some axis, which is completely ARBITARY.

    For example, if we consider an axis, which is perpendicular to the plane of the ecliptic and passes thru the mass center of the sun, then the moon´s angular momentum (in strange units as I am lazy) is nearly constant and equals: 1m((pi)2/365.4) + S

    Angular momentum´s standard units are: (meters)Kg(meters/sec) or with moon´s mass called “m”, the current distance for some selected axis called “L’ (for "length") and moon´s speed (in reference frame where the axis is stationary) called “V” then, angular momentum is: LmV but the V is the component of the total velocity that is perpendicular to L.

    In my strange "lazy units", L = 1 astronomical unit, AU; m is the moon´s mass, say in Kg; and 2(pi) is the circumference in AU of the moon´s nearly circular orbit about the sun; and time is measured in days. Thus in my “lazy units” the moon´s V is approximately 6.28/365.4 astronomical units per day. The S is of course the tiny bit more angular momentum due to the fact moon is revolving (or rotating 360degrees if you prefer) about 13 times per year. (turning same side to a distant star ~13 times in 365 days)

    Another axis could be the black hole at center of the galaxy and perpendicular to the plain of the sun´s orbit. Then the moon´s angular momentum would be much larger.

    Just because the moon has translation motion as well as rotation (360degrees /27.5 days) does not prevent you from calculating its angular momentum wrt to any arbitrary axis you care to choose. For example, an axis perpendicular to the ecliptic and thru the mass center of Mars. That choice would make the moon´s angular momentum a strong function of time, but Martians might like that choice, just as many Earthlings* like an axis thru the mass center of Earth.

    As the sun is in a quasi-elliptic orbit about the central black hole, but with many other stars occasionally deflecting it each orbit about the black hole, that huge angular momentum about the black hole would also be strong function of time as the eons pass.

    Likewise as moon is in a slightly Earth disturbed elliptical orbit about the sun, so my stated: "1m((pi)2/365.4)" angular momentum is weak function of time (I.e. the "1" & "2" are slightly changing as Earth moon system orbit the sun - it is not a constant 1AU from the sun.)

    There is really only one axis for which the moon´s angular momentum is essentially the constant S. You know what that axis is, but may not want to admit the obvious best choice among all the infinite possibilities for stating the moon´s angular moment (as then, neglecting meteorite strikes, it is just the constant S) if we chose the axis passing thru the moon´s mass center.

    That is why any reasonable person who understands that there is only one axis for which the moon has constant angular momentum says: "The moon is spinning about its mass center axis with angular momentum S." (S is not hard to calculate in standard units, KgM^2/sec if you assume the moon is a sphere and uniform racially. I.e. the mass center is at the center of the sphere. Not a bad assumption, but the mass center is slightly denser than the average and is slightly closer to the earth but the spin axis passes thru it not the center of the “moon sphere.”

    Unfortunately there are a lot of unreasonable people who want to choose another axis and make the moon´s angular momentum a function of time.

    * And in their ignorance the usually do not include the angular momentum S. - Some even claim S does not exist! - Say silly ignorant things like "The moon is not spinning about it own axis."
    Last edited by a moderator: May 3, 2012
  9. sadang Registered Member

    I appreciate your explanations wrt to the moon's angular momentum, but all these explanations are wrt to an external frame of reference. I asked in my previous message which point of view related to a frame of reference, is the real one: the internal frame of reference point of view or the external frame of reference point of view? Because from an external frame of reference, the moon has an angular momentum (due to the revolution movement around the Earth, not to its own axial rotation), and from an internal frame of reference the moon has no angular momentum, as Tesla stated "... But from the character of motion of the satellite it may be concluded with certitude that it is devoid of momentum about its axis."

    Even if I understand your explanations about the angular momentum of the moon, I also have to observe that all these explanations are from different external frames of reference. If from any external frames of reference the moon has an angular moment, and form any internal frame of reference the moon has no angular momentum, I can't ask myself which one is the real and true one?
  10. Billy T Valued Senior Member

    Only if the "internal frame" is rotating wrt the fix stars (or the sun for practical considerations) at same rate moon is (spinning) and has it axis of rotation passing thru the moon´s mass center AND this frame has moon´s velocity thru space, would the angular momentum be zero.

    Angular momentum is ALWAYS wrt some axis. The amount angular momentum calculated will depend upon that ARBITARY choice of axis, but none is more correct or "true" or "real" than any other.

    How they differ and what makes for most purposes the non-rotating frame with the axis passing the mass center of the moon (and moving thru space with the moon) is for it the angular momentum of the moon is a non-zero constant, not a function of time (except for minor changes due to meteorite strikes).

    If you chose and axis passing thru the mass center of MARS, then approximately every half earth year the computed angular momentum of the moon wrt that axis will change from positive to negative (or conversely)* and is constantly a strong function of time. That is not a very useful choice (except perhaps for a Martian) but it is not a "false choice." There are no "false choices" only false calculations, if you don´t compute correctly.

    *Earth/moon has opposite velocity direction thru space every ~183 days.
    Last edited by a moderator: May 5, 2012
  11. sadang Registered Member

    An internal frame of reference means, according to Tesla or accordin to what I understood from Tesla's article, any point internal to the moon's circle of revolution around the Earth. The main condition for this point is to be outside the moon, but always located within the circle described by the motion of revolution of the moon. From this frame of reference (any point or axis internal to moon's revolution circle and the moon's mass center) the moon has no angular momentum.

    On the other hand, from an external frame of reference (any point or axis external to the moon's revolution circle) the moon will have always an aparent angular momentum. This angular momentum will be an aparent one, because the real angular momentum is wrt to any point internal to moon's revolution circle.

    In your last message you made again the entire analisys of the "internal frame of reference" from an external frame of reference, and you said "Only if the "internal frame" is rotating wrt the fix stars (or the sun for practical considerations) at same rate moon is (spinning) and has it axis of rotation passing thru the moon´s mass center AND this frame has moon´s velocity thru space, would the angular momentum be zero." I could say the same thing looking from an internal frame of reference, and your phrase will look like this "Only if the "external frame" is rotating wrt the fix planet (or the Earth for practical considerations) at same rate moon is (spinning) and has its axis of rotation passing thru the moon´s mass center AND this frame has moon´s velocity thru space, would the angular momentum be zero.". According to these assumptions, we can conclude that the Moon has angular momentum?

    I would like to see from you, an analisys of the moon's angular momentum from an internal frame of reference. Because, according to Tesla's articles, this angular momentum does not exist, but apparently, and only if the analysis is performed in a frame of reference, external to the motion of revolution of the Moon around the Earth. Do you agree?
  12. Janus58 Valued Senior Member

    You have got to be kidding me! Tesla's "argument" is bogus from the start. No disrespect intended, but this is an example of someone expounding on a subject on which he is clueless. The assertion that he makes to the extent that if the Earth were to disappear while the Moon is conjunction that the moon (after a couple of oscillations) would then keep one face towards the Sun is laughable, and displays a complete misunderstanding of angular momentum.

    In addition, he never accounts for, or even mentions the Moon's libration, and his whole treatment implies that the Moon always keeps exactly the same face towards the Earth, which leads me to believe that he was not aware of the effect.

    Libration is an effect that causes the Moon, if view from the Earth or even the center of the Moon's orbit, to rock slightly back and forth and nod slightly up and down. Both motions have a period equal to that of the Moon's orbit.

    The reasons for these librations are two-fold. The side to side motion is due to the elliptical nature of the Moon's orbit. When the Moon is closest to the Earth it has a higher orbital speed than when it is further away. This means that the angular speed of its orbit changes. However, the Moon's rotation around its own axis does not change and maintains a constant rate.

    As a result, at different points of the Moon's orbit, the Moon rotational angular speed can be slower than or faster than its orbital angular speed. This causes the Moon's rotation to fall behind or overshoot the orbital revolution, causing us to see a little more of its Eastern and Western limb over the course of a month.

    The up and down (or North and South) movement is cause by the fact that the Moon's internal rotational axis is not parallel to its orbital axis. In other words, just like the Earth has a axial tilt with respect to its orbit, which causes it to present first its North pole and then South pole towards the Sun over the course of the year, the Moon's axial tilt causes it to present a little more of its North and South pole over the course of the month.

    The only reasonable and logical conclusion is that the Moon rotates on its own axis and that this rotation is separate from and unconnected to its orbital motion.

    Tesla may have made some important contributions in the field of electricity, but he is not infallible. In this case, he has stepped beyond his field of expertise.
  13. Billy T Valued Senior Member

    Tesla knew less about angular momentum than any college student who had at least half of one course in physics. As Janus 58 told you, he was making foolish statement in part as he did not even know the actually dynamics of the moon, but mainly because he did not even know (based on his article you quoted) how to compute angular momentum.

    I told you in post 567 about the ONLY axis of a ROTATING coordinate frame in which the computed angular momentum is zero. In any non rotating frame, inside or outside of the moon, the moon has angular momentum.
    Perhaps you too, like Tesla, have no understanding of how one calculates the angular momentum about any arbitrary axis so keep making silly statements.

    Again, but for the last time as you do not seem inclined to learn, The angular momentum of a small volume with mass m and L from the chosen axis is mLs where speed "s" is that component of the total velocity which is perpendicular to the line (of length L) between the axis and the tiny mass.

    If the object is too large to be considered a "small volume" (a significant fraction of L) for example, with diameter (or greatest internal length) of 1% of L, then mentally subdivide it in a set of small volumes, compute the angular moment of each and then sum them.

    When the chosen axis is in the Earth or Sun or Mars, then L is so large that to the accuracy of the calculation the moon can be consider as one "small volume," but of course if the axis is insider the moon or only a few moon diameters away, then the angular momentum of small parts of the moon must be computed separately and summed.

    SUMMARY: At least one of m,L,s must be zero for the angular momentum to be zero, and non are for an extended object like the moon even if the axis passes thru the mass center of the moon (for all non-rotating coordinate systems). I stated in post 567 how proper choice of a particular ROTATING coordinate system and axis could make the moon´s angular momentum IN THAT COORDINATE system be zero.

    In it "s" is zero for all tiny parts of the moon. Just like the "s" of NYC or LA, etc. is zero for the rotating coordinate system we call latitude & longitude.

    I have little hope you will learn so will not try (post) again.
  14. sadang Registered Member

    No matter what you think about what Tesla said, but only for yourself. How well does not matter what I think about what Tesla said, but only for myself.

    If from Earth I can see only one side of the moon, means that the moon towards me has no angular momentum. Right? Of course! Is obvious. Further, I conclude that the moon perform only a natural movement of revolution around me. Right? Of course! Is obvious. In conclusion, towards me, the moon has no angular momentum, but only an external angular momentum (if I can say so), with the central point on the center of symmetry of the circle or ellipse of revolution. Right? Of course! Is obvious.

    You can contradict this point of view? No! You can expose any another argument to disprove all these exposed by me, making analysis of lunar motion, only in this frame of reference? No! If you have that kind of arguments, you're my guest!

    Your argument with the moon's libration, has no effect on the analysis of own axial rotation movement of the moon. Only an argument, really actual, but exposed only to complicate the discussion. Also the N-S oscillations of the moon, during a complete revolution's movement, has the same role, to complicate the discussion.

    The only point or frame of reference from which can be observed the apparent moon's own axial rotation, or the apparent angular momentum, is a point or frame of reference outside of its revolution around the Earth. And this motion of apparent axial rotation, is due precisely to the fact that the Moon is positioning successively in the circle or eclipse on its motion of revolution, keeping the same face to the center of revolution, with 0 angular momentum wrt the center of the circle or ellipse of revolution. And yes, from this point of view or frame of reference, the moon has a different angular momentum, but it appears ONLY due to successive positions taken by moon, during the revolution's movement, not due to its own axial rotation. This is the reason I said this is an apparent own axial rotation, or an apparent angular momentum! It is not a real one!

    I'm curious to see what is your point of view!

    @Billy T
    No matter what you think about what Tesla said, but only for yourself. How well does not matter what I think about what Tesla said, but only for myself.

    I think you're wrong when you say:
    because you develop your own reasoning, based on the presumption that the moon is spinning. This type of reasoning is wrong. This is why I gave that counterargument with the point of view from an internal frame of reference. And, from an internal frame of reference, the moon has no angular moment, not even an apparent one, as I already explained to Janus58.

    So my suggestion is to rethink your starting presumption in analysis of the moon's own axial rotation, trying to make this analysis from two different frames of reference and from the observable reality, not from the supposed one.

    For me, personally, does not matter the value of the angular momentum of the moon. For me count if it really exists or not! This is the real presumption from which we have to start our reasoning, and trying to develop further an correct reasoning. And I understand and appreciate your efforts to learn me how to calculate the angular momentum, but I think is useless to calculate it, if there are still doubts about the moon's spin.

    As yourself assert in SUMMARY, if "s" which represent the angular momentum is zero, means that the moon has no spin. And this without a rotating frame of reference with the same angular speed as the moon angular momentum, as your default presumption suggests!

    I'm curious to see what are your arguments!
  15. Billy T Valued Senior Member

    No I made no presumption. I did not apply my own ideas or assumptions as you do. I only applied the DEFINITION of Angular Momentum, AM, and gave you the formula by which is calculated, (mLs).

    AM is not my opinion or your opinion but defined and easy to calculate. It definitely is not your invented "towards me."

    I was aware that the moon´s OBSERVED libration proves the moon is spinning, but that is too complex an argument for one as ignorant as you to follow. I did not even use the vector cross product notation to tell you how to calculate AM so I could avoid the need to use any words when explaining what "s" is, etc. as I suspect you are ignorant of that very common vector notation too.

    Not for your benefit as you refuse to learn (preferring to remain ignorant it seems) but for other readers I will correct two of your other false statements:

    (1) The moon is spinning about its own axis as demonstrated by the fact that the length of the shadow the US flag post, planted in the moon by the astronauts grew longer during their brief stay on the moon - just as a flag pole shadow on Earth grows long (but about 27 times faster) due to fact Earth is spinning about its axis about 27 times faster than the moon is.

    It was necessary for the astronauts to leave the moon before the moon´s spin would take their landing site into the moon´s night as there they would soon freeze. If where and when they landed, the sun was at “high noon” (sun directly overhead) they had no more than 7 days to explore. By end of day 8 they would be frozen.

    (2) Then moon does not orbit the Earth. That is only the appearance the facts give to an observer on the Earth. The moon is in essentially the same orbit about the sun as the Earth is. As the moon travels around the sun, its path is ALWAYS turning towards the sun, the same as the Earth is. I.e. the trajectory of the moon is EVERY WHERE concave. (Towards the sun.) In fact, if you were to accurately plot a full year of BOTH the moon and Earth´s path around the sun the two ellipses would over lap, be indistinguishable one from the other, unless the paper you used is bigger than 8.5 by 11 inches and the pencil line used to make the graphs is exceptionally fine.

    I know you will not make that accurate plot and will not believe it is true that the moon goes around the sun, like the Earth does, always “falling” towards the sun under the influence of the sun´s strong gravity, so I will just tell you a couple of facts:

    (1) The earth is one AU from the sun (except with slight variation as orbit is not quite a perfect circle and it is the Earth moon barrie center that is on the elliptical orbit).

    (2) The distance from moon to earth is about 0.0025 AU with slight variation.

    That means that on a graph of the full ellipse trajectory for the moon at the point where the moon is in “full moon” (more distant than Earth from sun) position, the moon´s pencil line is less than 1/400 more distant from the sun at focus of the ellipse than earth´s pencil line is.

    E.g. if Earth´s location on the graph is then 4.000 inches from sun focus point, then the moon´s location on the graph is slightly less than 4.001 inches from the sun focus of the ellipse. To not have these two trajectories over lap, the width of your pencil line must be less than 0.0005 inches. I don´t think you can even buy such a fine pointed pencil.

    I.e. on the 8.5 by 11 paper an accurate drawing of the moon and Earth´s trajectory about the sun is ONE pencil line ellipse! Not distinguishable one line from the other! Both paths are ALWAYS turning towards the sun. If the earth were to magically cease to exist, the graph of the moon´s trajectory would not noticeable change. To conserve AM, it would continue to spin about its own axis, making about 13 complete 360 degree rotations each year, as it does now.
    Last edited by a moderator: May 6, 2012
  16. sadang Registered Member

    I see that you're upset. It's your choice. But here in this discussion is not about me, about you, or about anyone else. It's all about the reality of facts. They must derive from this discussion, no matter who is right or wrong, or in what proportion. I consider that trough the light of this vision, would be carried this discussion. Coming and throwing accusations or other epithets, when the interlocutor does not agree with those mentioned by you, denotes a certain temperament and way of thinking.

    Returning to our debate, and also for other readers, I have to see that the argument made by you ​​in support of moon's own their axial rotation, wrt to the elongated shadow left by the U.S. flag placed on the moon by astronauts, and the fact that it increased during their brief (7 days is not so brief, in this period of time can take place a moon's phase changing) stay on the moon, the real explanation of the elongation of the shadow, is not due to moon's own axial rotation, but to the fact that it moved on the elliptical path of revolution around the Earth, and changing the angle between the flag and the Sun. There are still seven days, my God!

    An the astronauts had to leave the moon, due to the same fact. That means the moon, in seven days, was moved by the gravitational force of the Earth, on its elliptical revolution's path around the Earth, changing its initial angle wrt the Sun, the landing place of astronauts going now in the unlit side of the Moon. The same effect like in the elongated flag shadow case.

    I think point 2 is a joke. How can you assert that the moon does not orbit around the Earth? What is the center of revolution of the moon? Sun or Earth? Don't tell about the barycenter. Its another story! Eventually you could say that the moon has a revolution movement around the Sun, but that's only because the Earth orbits the Sun. Please pay attention how you use words, does not induce readers into error!

    I appreciate again, the accurate calculations made ​​by you and the power of the example with pen thickness depending on the distance between the two path of revolutions, namely of the Moon and Earth around the Sun. But as I noted above, this is true because the Moon orbits the Earth, not around the Sun, and only the movement composed of two trajectories, ie the Earth and Moon around the Sun, make the appearance of the moon orbiting around the Sun. And that two revolution pats, are not concave, as you sustain, but for about 14 days the moon composed path around the Sun is concave, and for about other 14 days, the composed moon's path around the Sun is convex. And this is true, because the Moon really orbit just around the Earth. Around the Sun, it move on a consecutive concave-convex eliptical path, like o composed revolution path, due to the Earth really orbiting around the Sun.

    And yes, if the Sun will suddenly disappear, the Moon will continue to orbit around the Earth. Just orbit, not rotate. And just around the earth, not around the Sun.

    Anyway, if you have any photo, or other kind of arguments, that you can sustain your statements I am ready to take into account. And of course, I will express my opinion, because I am not a robot to just say "Sir, yes sir!" or to repeat what someone else say, or try to impose. The real truth, is formed from my truth, along with your truth, along with other truths. Does not exist someone who know everything. In every area we discuss. Everything is relative, even the truth, and it is real only in that finite and closed analyzed system.

    Hope to continue this discussion.
  17. Billy T Valued Senior Member

    OK, discuss the FACT that moon´s orbit about the sun differs from an ellipse by less than 0.0025 of 1 AU. That is slightly larger than the magnitude of "wobble" the Earth´s gravity wiggles the moon from a true ellipse.

    You are correct that the lengthening of flag pole shadow can be consider to be caused by the same side of the moon always facing Earth - i.e. the by the FACT that the moon´s rotation period is the same as it orbit period around the Earth, when view from the Earth.

    If moon did NOT rotate then the flag pole would always point in the same direction - perhaps as some particular distant star; or if planted vertically in moon at full moon, it would point in the direction of both the Earth and the sun just after being planted in moon soil. Without rotation about its axis, 7 days later it would still point at that fixed star or a few degrees from the center of the sun, but would be ~90 degrees turned from pointing at the Earth. Moon only can keep the same side facing the Earth because it is rotating about its own axis (and thus flag pole in it is not always pointing at the same fixed star.)

    Deal with this FACT too:
    The sun is so much more massive than the Earth that its force of gravity acting on the moon is > 200 times greater than the Earth´s gravity force acting on the moon is. Thus the sun controls the moon´s trajectory and the Earth´s gravity just makes relatively tiny wobbles from a perfect ellipse about the sun with the sun at the focus of that ellipse. The wobbles are not enough even to make any point of the moon´s trajectory about the sun cease from being concave towards the sun. If the Earth were not to exist, then the max change in the moon´s current trajectory or distance from the sun would be less than 1/4 of one percent.

    I´m sure you don´t believe this as it disagrees with your unsupported false opinion, but you said we should deal with the FACTS, not some one´s opinion, so lets calculate the gravity force of the sun, Fs, relative to that of the Earth, Fe, acting on the moon.

    It is a very simple calculation, working in AU units for distance. Recall that gravity goes as the inverse square of the separation distance and directly with the mass. I.e:

    Fs /Fe = [M / (1)^2] / [m / (0.0025)^2] = (M/m)(0.0025 /1)^2 where M is the mass of the sun and m is the mass of the Earth.

    Now (0.0025/1)^2 = 0.0025^2 = 0.000,625 so, Fs /Fe = 0.000,625(M/m). Ignorant people would assume that proves (Fs / Fe) is less than unity or that the Earth´s gravity dominates the trajectory of the moon but as often the case, they would be wrong.

    The mass of the sun is “around 330,000 times more massive than the Earth.” – Quote from any of several Google hits with search on: “how much more massive is the sun than the earth?” You don´t even need to open a few of the hits as Google gives 330,000 immediately when search is done.

    Thus the ratio Fs / Fe = 330 x 0.625 = 206.25
    I.e. the sun´s gravity acting on the moon is more than 200 times stronger than the earth´s gravity acting on the moon is. So the orbit of the moon is an ellipse about the sun with the sun at its focus, except for tiny (one half of one percent) force perturbations by the Earth.

    Again, if the Earth did not exist, the orbit of the moon would change only by not having any tiny perturbations from a perfect ellipse and it would still make ~13 full 360 degree rotations in each year.

    SUMMARY: The moon orbits the sun in a slightly Earth perturbed ellipse and rotates about its own axis 360 degrees every 27+ days, with (or without) an Earth existing!

    Deal with these FACTS. Don´t try to contradict them with only your opinion. Show an error in my calculation or make an analysis of your of own. – Don´t just kept repeating your false opinions with nothing but your verbal claims as support.
    Last edited by a moderator: May 7, 2012
  18. Billy T Valued Senior Member

    Completely a false assertion or opinion as shown by numerical calculation in my prior post.(574)

    The moon´s path is nearly a perfect ellipse ABOUT THE SUN, not the earth,* as the Earth´s gravity acting on the moon is more than 200 times weaker than that of the sun. All the Earth´s gravity does is make a tiny perturbation of the moon´s elliptical orbit about the sun (max amplitude deviation from perfect ellipse about the sun being LESS than 0.0025 AU or less than one fourth of one percent.)

    * if moon´s elliptical path about the sun is viewed from a moving and accelerating platform, then it can appear to be any smooth and continuous shape you like, but when viewed from any stationary (WRT the sun), inertial, platform moon has an elliptical path about the sun. The Earth is of course a "moving and accelerating platform" and from it the path of the moon does appear to orbit the earth.
    Last edited by a moderator: May 7, 2012
  19. sadang Registered Member

    Yes Billy, you are right wrt the flag shadow! But just on a foot, not on both of them, because the moon has no own axial rotation, not even one synchronous with the revolution movement! This is an example of what means to start working on a fact, having misconceptions, in your case, having well engraved in your mind the certainty that the moon has its own axial rotation synchronous with the Earth's revolution around it.

    And your description about the flag pole, pointing always in the same direction, represent a misunderstanding of the movement of an object on its path of revolution. A path of revolution is usually, a circle or an ellipse. Following this path, our object permanently change its angle wrt to an external point of reference, and when it close the loop of revolution, it made a 360 rotation wrt that external point of reference. But this 360 degrees rotation is not around its own center of mass, its just due to the consecutive placement in its movement, on its circular or elliptic path of revolution. So, in our discussion, the moon make exactly the same movement, and is changing the angle of the flag pole only wrt the Sun, because, in its consecutive placement on its revolution path, the moon has changed its angle wrt the Sun. And I from Earth, can see that flag shadow is stretching, even the moon maintain the same face towards the Earth. Hope you, and other readers, now understood how things run in the case of revolution movement of an fix (not rotating) object.

    Going forward to the next FACT suggested by you, the gravitational force acting on the moon, being 200 times greater than that of the Earth. Hmmm! I really can't contradict your calculations, but just a simple remark; if the moon is permanently between the Sun and The Earth, and not revolving around the Earth, how can we explain the moon phases?

    On the other hand, following your path of thinking, means that moon also does not revolve around the Earth, not even the Sun, but instead it rotate around that black hole (or maybe withe hole, depending on the point of reference) existing in the galactic center, because, certainly, that force of gravitational attraction is more greater than that of our petite Earth. Or, we can extrapolate this, and calculate the force of the gravitational attraction of the center of the galactic cluster, from which our galaxy is a part, and to see that actually the moon, in fact, rotate around that center. What you think about this! Can this be true? Of course, but only mathematically.

    This is an example of what mathematics can do when it is applied on imaginary facts, not to the real one. This happens when peoples are trying to model the real word, according to their calculations. Take a step back, detach the mathematical calculations, see the reality of the world, then look at what you calculated. Understand the error? It is not in your mathematically accurate calculations, but in the wrong understanding of the movement of celestial bodies, and the laws governing these movements. Here, I should talk about Newton's gravity, but that would require one or more topics. And for me it is enough if you make these calculations for the centers that I specified and if you can explain the moon phases as seen from Earth, when the moon revolve, for instance, around the galactic center.

    I know you will become even more upset, after will read this post, but I hope will motivate you more stronger, to understand what a crazy an maverick guy like me are trying to say! And forget to impose your FACTS, as you like to say them. These are just your mathematically or imaginary FACTS, not real FACTS. When you will learn to pay attention and take in consideration (even mathematically) what your interlocutor want to say, and to credit him with the same respect with which you yourself are treated, maybe will open your eyes and will understand reality through your own conscience and understanding, not trough one forced by current scientific dogma.

    I prefer to be a rebel or heretic guy wrt the scientific dogma, and to not accept what I don't understand, just because it was said from on a podium, or from someone with a scientific degree, or with white cuffs and bow tie. I am a human being, with will and mind, not a robot, to move or think as someone else wants, or order. PERIOD!

    And of course I am waiting for your new calculations, but based on what I suggested. Take it only as a simple exercise of kindness to a heretic.
  20. AlexG Like nailing Jello to a tree Valued Senior Member

    I really didn't see an answer there.

    Just a lot of hand waving and word salad.
  21. sadang Registered Member

    I tried to put some links with photos and documents to explain that the moon is orbiting around the Earth, and its composed movement around the Sun is as I already said: 14 days is concave and other 14 days is convex. But there is a limitation that says I can't post any link until my post's count will be greater than 20. That's it! Maybe some day I get to count over 20 posts.

    In the following I will make few statements, and I expect from you to confirm or rebut them.

    1. The Moon its revolving directly around the Earth in 27,3 terrestrial days, and indirectly around the Sun, around the Galactic Center, around... any other center of rotation, to which we can think, add and compose the movement of revolution of the moon around the Earth. And it not revolve directly around the Sun on an elliptical path "...being concave towards the sun", as you sustain. The simple argument for this is that we can see (with our eyes, not mathematically) the moon phases. If you know other explanations for moon phases please make them public.

    2. If we make the same calculations for the other moons of the existing planets in our solar system, probably (I say probably because I did not make these calculations), we find it all revolve on elliptical path around the Sun "...being concave towards the sun". Then how can we explain the fact that we can still see them revolving around their planets?


    Newtonian gravity equation describes the motion of material points simple, no-size, moving through empty space, through the universe. Under this gravitational laws, we are part from a solar system immutable, in a universe practically proved to be a changing and expanding one.

    On the other hand, today we know for sure that our universe is expanding, and aour solar system have to obey the same behavior, being part of this universe. But this is unacceptable, respecting Laplace-Lagrane-Poison theorem, and on an Earth having exactly the same size like in the first day of creation, or condemned to remain forever without real sizes, according to Newton's theory!

    And one more thing! Even we discuss about Newtonian or Einstenian gravity, both of them, don't take into account a permanently, real and universal movement of almost all celestial bodies, namely, the spin movement! Think deeply about this!

    I am sure this last paragraph will create even more confusion. Sorry!

    When I will can, I will put here some images, docs and other kind of information to sustain my claims.
  22. Robittybob1 Banned Banned

    @BillyT, _ In the thread "Life First Started On Planet Mercury?" http://www.physforum.com/index.php?showtopic=29842
    I show that Moon capture is possible by the early Earth. In that hypothesis the Moon was already orbitting the Sun on a similar orbital radius to the Earth, but over time the pertubations brought the Earth and Moon close enough that the Moon was drawn toward the Earth and captured in orbit around the Earth (but also as you tend to forget also orbiting the Sun as it had been doing prior). Thanks for that vital insight.
    If I find the area in the thread where this is being discussed I'll edit this post.

    http://www.physforum.com/index.php?act=Post&CODE=06&f=27&t=29842&p=504985 would be a start.

    There is not any one defining post but start around here http://www.physforum.com/index.php?act=Post&CODE=06&f=27&t=29842&p=500147
    Last edited: May 8, 2012
  23. Billy T Valued Senior Member

    Yes terminate your ignorance. Cease posting falsehoods, especially as the the truth and why it is the truth has been explained to you, in more than one way!
    The part I made bold is FALSE, based only in you ignorance and not supported by anything except your repeated false assertions of it. Here is the truth (& why it is so I have already twice explained to you, once with a mathematical discussion):

    “…From within the Earth-Moon system, the simplest way of picturing the situation is to have the Moon revolve about the Earth; but if you were to draw a picture of the orbits of the Earth and Moon about the Sun exactly to scale, you would see that
    the Moon's orbit is everywhere concave toward the Sun. It is always "falling toward" the Sun…

    From: http://en.wikipedia.org/wiki/Double_planet

    Billy T comment Exactly what I told and explained to you. I.e. the sun is ~400 times more distant than the separation between the Earth and moon. So as the Earth Moon “double planet” goes around the sun, both oscillate thru ~13 full cycles per year about their common elliptical path; slowly crossing (~26 times) from one side to the other side of their common elliptical path.

    This makes the moon closer to the sun for about 14 days (with the new moon in the middle of that half cycle of the oscillation, & etc. for the full moon in the middle of the other half of the crossing cycle when the moon is more distant from the sun than the Earth is).

    I.e. the amplitude of this oscillation is so tiny compared to the distance to the sun at the focus of the ellipse that, both moon and Earth are always turning towards (or falling towards) the sun.
    Graph from: http://en.wikipedia.org/wiki/File:Moon_trajectory1.svg

    If that enlarged image does not post well for you, there is the smaller image in the texts found at:
    http://en.wikipedia.org/wiki/Orbit_of_the_Moon#Path_of_Earth_and_Moon_around_Sun but then you must look very closely to even see any difference in the two trajectories about the sun. Recall, in an earlier post I told you that if the full ellipses of Earth and moon were accurately plotted, even with a very fine pointed pencil on 8.5 by 11 inch paper, as large as possible, the two paths completely over lap - your pencil graphs are indistinguishable!

    That text also states:
    “… The Moon's trajectory is always convex[13][14] (as seen when looking inward at the entire Moon/Earth/Sun system from a great distance off), and is nowhere concave (from the perspective just mentioned) or looped.

    If Wiki is not enough an authority for you, below is a quote from an article published in a respected astrophysics journal by Harvard University Astronomy Professor, which has also been republished by NASA (as they too are trying to correct the false, ignorant, POV you hold). Note that the first paragraph tells that the image of Fig 1 below (which, like your POV is not correct) was even found in some older physics books, when many were as ignorant as you still are about the facts. [​IMG]

    I don´t post these corrections of your false statements for your benefit as you have shown no inclination to learn the truth from my earlier demonstrations of the facts, but post for other readers as it is necessary to again correct your errors; However, you should at least stop posting your counter factual False nonsense. -I am growing tired of correcting it.
    Last edited by a moderator: May 8, 2012

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