Why Is The Moon Not Spinning Then?

We all know how you feel about math, but my response was to CheskiChips, and they used math, so I responded with math.

 

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The laser reflector data analysis should be able to show whether the Moon's velocity is increasing or not. I'd rather trust that than your maths.

 

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that tells the distance to the moon, to a few mm, I think, not its speed, which has an extremely low change as a percent /year. Also has the complexity of the changes due to moon not being in circular orbit. - Makes much greater speed changes than could be observed with lasers. To get the average speed from distance measurments you would need all the math I and Janus58 explained to you and more. You and some others active in this thread simply do not have the slightest knowledge of what you post about.*

The best way to get the average speed over full orbit period I bet has nothing to do with lasers and a lot to due with photographs against the back ground stars. (And correcting for Earth's motion between exposures, with more math, of course.)

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*Too many posting here seem to think orbital mechanic is something you guess at, or has probable answers. It was the queen of the precise sciences and certainly held that position uniquely even 2000 years ago, despite not know how it worked as we do now. Now we can send space craft to Pluto or to skim thru the solar corona! (both of which my former employer, APL/JHU has or is doing.) It also was the reason for/ motivation of / most of the advance in math were made pre 1850. (things like calculus, logrythmic computations, etc.)

 

That's absolutely NOT true!!!

Since I don't have full posting privileges yet and can't post your images side by side for easier comparison, why don't you do that for us all!

I dispute that your 1st and 4th graphics are the same with their only differences the observer's viewpoints (i.e., center-point versus sidereal.)

Uranus' polar axis is essentially on its side, but its polar axis is NOT "stationary with respect to the fixed stars!"

Can you provide a citation?

The force causing torque can also be parallel to the spin axis, as is the case with a toy top's procession caused by the Earth's gravity.

There is both torque-free precession as well as procession caused by torque.

Not so - since our Earth rotates on its polar axis and has constant torque being applied by the sun and moon at right angles:

---snip---
The Earth goes through one complete precession cycle in a period of approximately 25,800 years, during which the positions of stars as measured in the equatorial coordinate system will slowly change; the change is actually due to the change of the coordinates. Over this cycle the Earth's north axial pole moves from where it is now, within 1° of Polaris, in a circle around the ecliptic pole, with an angular radius of about 23.5 degrees (or approximately 23 degrees 27 arcminutes [2]). The shift is 1 degree in 180 years, where the angle is taken from the observer, not from the center of the circle.

Discovery of the precession of the equinoxes is generally attributed to the ancient Greek astronomer Hipparchus (ca. 150 B.C.), though the difference between the sidereal and tropical years was known to Aristarchus of Samos much earlier (ca. 280 B.C.). It was later explained by Newtonian physics. The Earth has a nonspherical shape, being oblate spheroid, bulging outward at the equator. The gravitational tidal forces of the Moon and Sun apply torque as they attempt to pull the equatorial bulge into the plane of the ecliptic. The portion of the precession due to the combined action of the Sun and the Moon is called lunisolar precession.

Revolution of a planet in its orbit around the Sun is also a form of rotary motion. (In this case, the combined system of Earth and Sun is rotating.) So the axis of a planet's orbital plane will also precess over time.

The major axis of each planet's elliptical orbit also precesses within its orbital plane, partly in response to perturbations in the form of the changing gravitational forces exerted by other planets. This is called perihelion precession or apsidal precession (see apsis). Discrepancies between the observed perihelion precession rate of the planet Mercury and that predicted by classical mechanics were prominent among the forms of experimental evidence leading to the acceptance of Einstein's Theory of Relativity, which predicted the anomalies accurately.[3][4]

These periodic changes of Earth's orbital parameters, combined with the precession of the equinoxes and of the inclination of the Earth's axis on its orbit, are an important part of the astronomical theory of ice ages. For precession of the lunar orbit see lunar precession.

(Add the http only): en.wikipedia.org/wiki/Precession#Of_the_Earth.27s_axis

Stays fixed in relation to what?

That simply is not true! Venus nearly does keep one face pointed towards the stars as it rotates clockwise (retrograde) on its polar axis nearly one time per orbit.

(NOTE: Venus completes one counter-clockwise orbit every 224.65 days and Venus rotates once clockwise every 243 days.)

In any event, I've already covered that topic well, and an astronomical body that *appears* from the sidereal perspective to not be rotating, actually does have one polar axial spin (in a clockwise direction) per each complete orbit, as can plainly be seen in the top graphic here (don't add www):

community-2.webtv.net/kdine5/Lunacy/index.html

That's an informal explanation of precession often demonstrated in high school classes:

"In a classic beginning physics demonstration, the instructor stands on a swiveling platform and holds a spinning bicycle wheel at arm's length. The wheel is vertical and the instructor is standing still. The instructor then tilts the wheel toward horizontal. This causes the instructor to start spinning slowly on the platform. Bringing the wheel back to vertical and tilting it the other way makes the instructor spin the other way.

"Since the forces on opposite sides are in opposite directions, the result is torque. Each pair of opposite particles on the wheel contributes to the torque that causes the instructor to turn on the platform. Tilting the wheel the other direction produces torque in the opposite direction, slowing the instructor's spin and eventually reversing it."

As for a sidereal perspective being a "preferred" perspective (it obviously has its uses in astronomy), that doesn't turn its perspective into an absolute reality.

Absolutely not true!!! I have pointed out that a large astronomical body rotating around its polar axis assumes the shape of an oblate sphere (as the Earth is shaped.) WHEREAS, our non-polar-rotating moon has solidified into the shape of a football as it cooled over 3 billion years ago freezing its two tidal bulges in place.

When the moon still had axial rotation its tidal bulges wouldn't have been fixed in one location on its surface. When the moon lost its polar rotation over 3 billion years ago, it wound down to ZERO polar axial rotations per orbit, NOT one (1) polar axial rotation per orbit!

Today the moon ONLY revolves around an exterior axis, the Earth-moon barycenter!

Models prove that an orbiting astronomical body CANNOT spin on two axes at the same time and still keep one of its faces constantly pointed towards the center point!

That's all quite a bit more than mere tautology!!!

A classic example of tautology would be claiming that, in order to keep one face pointing towards Earth, the moon rotates once per each orbit.

Rotates around WHAT?? If the axis for that claimed rotation is the Earth-moon barycenter, which is the same axis of the moon's orbit, then that's classic tautology!!!

That's your apparent argument, not mine!

I say that one complete moon orbit is one (1) 360 degree revolution around the Earth. The moon's ONLY axis for that orbit is the Earth-moon barycenter located within the Earth.

Libration is interesting, but it doesn't account for any 360 degree rotations of the moon. What is at issue here is the location of the moon's axis as it orbits 360 degrees around the Earth.

If you claim the moon has two separate 360 degree spin axes and it still keeps one face pointed towards us, then I dispute that!

Thank you for correcting my sloppy verbiage – you are 100% right about the transferred energy being stored as gravitational potential as the moon slowly recedes from the Earth.

Ken

 

To Ken Dine:

You have a strong bias that drives you to claim the moon rotates about the Earth moon barycenter without spin. Others have one that prefers to consider the axis of rotation passes thru the center of mass of the rotating moon and when we do this we call the rotation spin, or even polar axis spin also.

In truth the moon has two types of momentum, angular and linear.

In the case of angular momentum of any body with mass M there is always an axis of rotation implied when stating the angular momentum, but where it is, is completely arbitrary in the sense that if you take each tiny differential mass element m and multiply it by its speed orthogonal to the line between m and the line A, which is the selected axis of rotation, and then sum up all such products for all m in the mass M and set it equal with some specified angular momentum, you will find (if memory serves me correctly - too lazy to do this all again, showing the proof, after so many years.) that the location of A can be anywhere. For example, I would have no trouble in using an axis of rotation for the moon, which passes thru Mars.

Now no one does an analysis of the moon's angular momentum about Mars* about an axis of rotation passing thru Mars because it does not seem that way to humans but the math is OK with that axis. People speak of what they know and what seems reasonable to them.

We all agree that the moon is turning the side never seen by Earthlings to point to all the distant stars in a circular arc. So we all agree the moon is turning. You unfortunately know about the barycenter and are fixated that must be the "true" axis of this rotation. Most people have never heard of the barycenter so they do not do that. Most people, myself included, think that any object steadily swinging around to sequentially present one side to all the distant stars along a great circle in the heavens is rotating or spinning and chose the spin axis as passing the center (of mass, if they have a little knowledge of physic) of the turning object.

Your stubborn dogmatic POV that only the barycenter is correct as the axis of rotation (another error not listed below) has caused you to make many statements which are clearly false and to ignore many simple examples which show them to be false. Instead of admitting or discussing your errors you are now engaged in discussion about what some of Janus58 drawings do or do not show.

Your more serious errors include:

(1) Treating as an absolute property of rotation the term "clockwise" and not realizing that "clockwise" switches to "counter -clockwise" as the observer changes from one side to the plane of rotation to the other. (Associated with your second "shooting of self in the foot" with a false "proof" that you were correct.)

(2) Not realizing the Foucault Pendulum operating above either pole of the moon does make a 360 circle in about 28 days - you offered the fact that it would not as if it did that would indicate moon was spinning about the polar axis. (Associated with your first "shooting of self in the foot" with a false "proof" that you were correct.)

(3) Not realizing that the sun's gravity is stronger force at the moon than the earth's gravity is. (I worked it out for you to show sun is 8.45 times stronger, but received no thanks or even an acknowledgement that you were wrong. - So I assume you still hold that false POV.)

(4) Not realizing that the tides are made by the gradient of gravity, not by the force of gravity. Or that your argument based on moon making stronger tides than the sun does not show the moon has stronger gravity at the moon.

(5) Not realizing the moon only appears to orbit the Earth for a small set of reference frames, but all others in the universe, for example for a frame based on any planet or asteroid, or the sun etc. see it is in orbit about the sun with about a 1/4 of one percent wobble. I illustrated this 1 part in 400 radial variation with 8 inch diameter circular drawn by a line only 0.01 inches wide. Something approximately this width| forming a circle filling your laptop computer screen and told you the wobble stayed inside the line's width. –I.e. was not even noticeable in that scale drawing!)

Hey, you do know what the barycenter is and most do not; but as they say: "A little knowledge is a dangerous thing", especially for one lacking the grace to admit their errors.

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*Except Martians, of course.

 

They are already posted one under the other, side by side would make no difference. I'm afraid that you are seeing what you want to see.

There is nothing to dispute, it is fact. These animations were done with a ray-tracer. Ray-tracing works like this: You place objects into a three dimensional area by defining their x,y and z coordinates. You also place a light source and "camera" in the area the same way. The computer then renders the image as it would appear from the "camera's" position. You can also translate and rotate objects in the scene by using a "clock" variable, causing the program to render successive frames of an animation. The only difference in scene language between the 1st and 4th graphic is the position and aiming of the Camera. The translation and rotation parameters of the satellite and its axis remain the same.

So, are you calling me a liar?

http://www.apl.ucl.ac.uk/iopw/uworkshop_060905.pdf

http://www.news.wisc.edu/releases/12826.html

Why would you think that Uranus would be any different from any other planet in that respect?

if the force were applied parallel to the axis it will not precess. A toy top will only precess if its axis is tilted from vertical, in which case the force of gravity is not parallel to the axis, and there is no torque. No torque, no precession.

Wow, you can quote Wikipedia.
The Earth's precession is caused by the uneven gravitational pull of the Moon and Sun on its equatorial bulges. And as according to wiki, this precession has a period of over 25,000 years. The satellite in the graphic would need to precess with a period equal to its orbital period, quite a different matter.( and also an amazing coincidence.) Also, since the satellite in the graphic has exactly a 90° axial tilt, any uneven pull on it's equatorial bulges would cause a torque within the plane of the orbit, not at right angles, as would be needed to cause the apparent precession seen in the graphic.

The fixed stars. And yes, Uranus probably undergoes a small precession, but since its moons all orbit in or nearly in the plane of its equator, and any torque on its equatorial bulges caused by the Sun would be extremely small (Such a torque is caused by tidal forces, and tidal forces fall of by the cube of the distance), it would have an extremely long period.

which has absolutely nothing to do with the principles of the discussion at hand. You do seem to have a habit of flooding a post with irrelevant data.
Besides which, the 243 rotational period is Venus sidereal period. IOW, the time it takes for Venus to rotate once relative to the fixed stars.
Why would you think that the very astronomers that you claim are wrong about the rotation of the Moon would list a rotational period for Venus based on your perception of planetary rotation?

Once again, you are simply imposing your personal perception of "rotation of an astronomical body. What you "plainly see" in that graphic is not what everyone else plainly sees. We plainly see a moon that does not rotate in the top graphic, and one that rotates counter-clockwise once per orbit in the bottom graphic.

Once again, some pasted material that does not add anything to the discussion. The ability to paste does not infer understanding.

That is a silly argument. If enough time were to pass that the Earth becomes tidally locked to the Moon such that its tidal bulges remained fixed on its surface, would you then claim that the Earth had stopped rotating?

The fact that the moon is oblong is simple due to the fact that because of its slow rate of rotation, the tidal forces of the Earth had a greater effect on its shape. It is also related to the fact that the moon likely had a much more eccentric orbit when it cooled into its present shape.

Again, it seems like you are using your own peculiar notion of rotation here. It can rotate around its own axis and revolve around a orbital axis while constantly facing the axis of revolution. In fact, it must rotate once per orbit to do so.

No it isn't because you are assuming that your notion of "rotation on an axis" is true to "prove" that it is correct.

It rotates around it own polar axis and revolves around the Earth-moon barycenter Oh, and by the way, the Moon's polar axis is tilted 6.5° to that of its orbit. Because of this, as seen from the Earth, the Moon's axis traces out a cone as with a period of once per orbit. This is what causes the libration of latitudes

Learn to read. I never said it was your argument, I said it was the reverse of your argument, And as just as strong as your argument on their own merits.

And in doing so you are making yourself look foolish.

 

I'm gone for a week because some silly hurricane passed right overhead and filled my yard with trash, and this silly thread has itself become full of garbage! The normal, everyday garbage from those who don't think the Moon is "spinning" is one thing. The garbage from our more educated members is another.

The Moon orbits the Earth. The relative strength of the Sun's gravitational force is not the metric used to determine whether a body is a satellite of some other body. The Sun's gravitational force is also stronger than is Jupiter's for the six outermost moons of Jupiter -- yet they IAU still classifies these as satellites of Jupiter, just as it classifies the Moon as a satellite of Earth. You are using the wrong metric, Billy. The Moon's velocity relative to the Earth is well below Earth escape velocity, and the Moon's orbit about the Earth lies well within the Earth's gravitational sphere of influence with respect to the Sun.

Even those who dig themselves in deep embarrassing holes are right sometimes. This is one of those times. Any non-spherical body will undergo torque-free precession if its angular velocity vector is not aligned with one of the body's principal axes of rotation. Wikipedia discusses this briefly here. The Earth undergoes a torque-free nutation called the Chandler wobble. It is very tiny compared to the torque-induced lunisolar precession.

 

This should be good.

 

I'm glad I started the thread though. From what I've seen above it might be that the maths isn't quite as cut and dried as you may like to think.

 

Or it might be that Billy is right and that you "simply do not have the slightest knowledge of what you post about." The math is very cut and dried, and has been since Newton's time. Read post #137, for instance, which contains the equation for the velocity of a small body in a circular orbit.

More generally, the vis-viva equation describes the magnitude of the velocity vector for the orbit of one point mass about another (or two bodies with spherical mass distribution):

\(v^2 = G\,(m_1+m_2)\,\left(\frac 2 r - \frac 1 a\right)\)

Back to the circular orbital velocity equation presented by Janus,

\(v=\sqrt{\frac{GM}r}\)

The circumference of a circle is \(c = 2\pi r\). The time taken to complete one revolution at a constant speed \(v\) is

\(P = \frac c v = \frac{2\pi r}{v} = 2\pi \sqrt{\frac {r^3}{GM}}\)

or

\(\frac{P^2}{r^3} = \frac{4 \pi^2}{GM}\)

which is Kepler's third law. In other words, this is not just cut-and-dried mathematics, it is 400 year-old cut-and-dried mathematics.

 

Well, the mention of barycenter and lunar axis left me looking up a few words in the dictionary.

The overall perspective on terms could use some clearing up, relavant to the effect of various points in such interations between bodies. It also seems to be part the case in this disscussion, a little confusion about the significence of various points.

For example Barycenter in my memory meant the gravitional interaction that exist between to bodies at a point between to bodies. I have since found other defintions.

there are more than 5 points of interest in the relations of the moon and the earth.

The point of gravitional interactance that exist between the two bodies, and being in between the two physical bodies exist approximatly 17,280 miles from the moons surface. (being relative to the apogee and perigee). A point of reaction appeantly missed by a great deal of people.


DwayneD.L.Rabon

 

Granted. I wasn't considering this precession as I was treating the satellite in my example as spherical. But even if it weren't, such a precession would not account for the preceived motion shown in the graphic as Ken is implying it could. The precession axis would have to be at a 90 degree angle to the spin axis, which give an angular velocity for the precesssion of zero.

 

Lol... yes it is.

 

I think you misunderstood why I mentioned that the sun's gravity was ~8.45 times stronger at the moon than the Earth's is. There were two reasons:

(1) Ken was erroneously using the true fact that the moon makes greater tides in Earth's oceans to falsely assert that therefore the sun's gravity had to be less. (He apparently did not understand then at least, that the tides are caused by the gradient of gravity, not the strength of gravity, but I think he does now.) By calculating the ratio of sun to Earth gravity (my 8.45) I demonstrated his logic was false, as based on ignorance of what cases the tides.

(2) I want to show all, who might not already know, that the sun controls the motion of the moon much more than the earth does. I.e. the solar acceleration of the moon is ~8.45 times larger. This relates to your objection.

I did not use the 8.45 as my "metric" for statement that moon is in orbit about the sun. My metric was the fact that the moon is always curving towards the sun. The moon only curves towards the Earth when the Earth is closer to the sun. Then the constant curving towards the sun means it is also curving towards the Earth. When the Earth is near the line between sun & moon, the radius of curvature towards the sun is less than 1AU. Two weeks later the moon's radius of curvature is more than 1AU and then moon is curving AWAY from the Earth. Something orbiting the Earth should never be curving AWAY from the Earth, at least that is not what I call "orbiting." (As the Earth is also curving towards the sun, the earth/ moon separting may not be increasing despite the moon curving away from where the Earth is.)

My "metric" for stating the moon is NOT orbiting the earth is not only the fact that the moon is always a curving towards the sun but also that for more than half the time the moon's trajectory thru space is actually curving AWAY for the Earth! Here is another attempt to describe my "metric.":

If the moon's trajectory about the sun were a perfect ellipse, would you not agree it is in orbit about the sun? I will assume you agree and continue. So the question becomes to what extent can it deviate from a perfect ellipse and still be consider to be orbiting the Sun along with the Earth? Clearly a low Earth orbit satellite should be considered to orbit the Earth even though it is also circumnavigating the sun. I think even a geostationary satellites orbit the Earth. I make the division based on whether the sun or the Earth exerts the greater gravitational force.

At some altitude Earth satellite has portions of their trajectory where Earth's is the stronger (trajectory curving towards Earth) and others portions where the sun's is so it curves towards the sun. Here one could say part of orbit is about the sun and part is about the Earth. The moon however is ALWAYS TURNING TOWARDS THE SUN, thus I consider it to have such small deviations from the perfect ellipse that it must be considered to be orbiting the sun. For example, a scale drawing on an 8.5 by 11 inch paper with even a fine (0.01 inchwide) pencil lead as a PERFECT ellipse with 8 inch major axis wound not need anymodification to accurately represent the moon's trajectory - only a statement that the precise trajectory remains within the width of the pencil line and has approximately 13 complete oscillations within that pencil line's width! On my laptop screen, 0.01 inch is about the width of: | To me that is dam little diviation on an 8 inch ellipse from a perfectly elliptical orbit about the sun.

You, in contrast, seem to think that even though moon is always turning towards the sun and only accidently towards the Earth when the Earth happens to be on the sun-side of the moon, the moon is orbiting the Earth as it is bound to the Earth. This despite the fact it's trajectory deviates by only ~1 part in 400 radially from the perfect ellipse. It really comes down to the definition of "orbiting." For me an body that always curves to another is orbiting that other body, even if it is also bound to some third body. earth is also bound to this galaxy, but neither of us would say it is orbiting the galaxy, I think. Orbiting has to do with shape of the trajectory and completing the trajectory many times, for me at least.

Janus58, can if he chooses, state my metric better than me, but I think you understand it.

I of course agree the moon has less than the Earth escape velocity -i.e. is bound to the Earth. I am not sure but think it will be for long time all time. (Assuming no large third body comes near.) If we ignore the sun's gravity, then as I DERIVE from F=ma in post 140, r (v^2) is a constant. Thus, as the moon recedes from the Earth its kinetic energy, KE, goes down linearly as r goes up. Or KE ~ 1/r but this is just the way potential energy, PE, scales with r in any inverse square force field like gravity. So the ratio KE/PE is a constant independent of r for circular orbits. (That constant is -0.5, but I will not derive that now. The negative sign is, as I am sure you know, due to the fact the PE is negative - Moon is trapped in a negative energy well.) However, there is a solar gravity field and I think when the moon goes farther way from Earth there will be a time when it can escape for Earth's grasp.

When it does so it will still retain the spin about its polar axis but that spin rate will be less than the current one 360 turn in ~28 days as for quite some time moon will remain tide-locked to Earth so the spin rate will drop to the slower orbit rate and then reamain that forever (until sun goes red giant at least).

BTW do you or Janus know what approximately the spin rate of the moon will be when it is no longer bound to Earth?

It would be interesting to hear Ken Dines try to explain how it suddenly began to spin about it polar axis when it ceases to rotate about the barycenter. I will not hold my breath until he tries as I have already asked (and been ignored) essentially this in post 124, partly reproduced below:

I should also note that we see more than half of the moon because it is spinning about an axis that is not exactly perpendicular to the orbit plane. - The same reason the "sunmen" would see nearly all of the earth even if its polar spin (on a 23.5degrees tilted axis) were one 360 turn in 356.25 days.

 

NO, I am instead saying that the moon ONLY revolves around an exterior axis once each 27.3 day orbit, and that axis the moon is revolving around is the common mass of the Earth & moon, which is called a barycenter, which is a foci located within the Earth.

AGAIN, draw a happy-face on an orange, then hold the orange in your outstretched hand and spin on your heels - to any observer watching you the orange will likewise TURN 360 since they'll see all sides of the orange once per each spin of your body, but you (the person in the center) will NOT see the orange rotate in your outstretched hand.

The orange is NOT *rotating* around its internal polar axis any more than the moon is rotating around its internal polar axis. The only necessary spin axis is the barycenter exterior to the moon!

In orbiting around that barycenter the moon merely turns around its polar axis, which is NOT the same thing as a true polar rotation. The moon lost all of its true polar rotations over 3 billion years ago!

E.g. the Earth has 365.25 true polar rotations per year, but from a sidereal perspective, the Earth *appears* to rotate 366.25 times per year. If the Earth has 366.25 sidereal rotations per year, then why do we only have 365 days per year?

Clearly, a sidereal rotation is NOT the same thing as true polar rotation, otherwise we'd have 366 days per year.

Please ducate yourself about the Right-hand rule:

(Add the http & www): studio4learning.tv/sub_subject.php?pl=72&id=210

(Add only the HTTP): en.wikipedia.org/wiki/Right_hand_rule#Direction_associated_with_a_rotation

Foucault Pendulums don't spin 360 since they aren't set up on the poles (except a crude one once), so they spin less times per day at lessor latitudes, and not at all at the equator.

Set up a Foucault Pendulum at 45˚ latitude at any spot on the Earth and it will show rotation. Set up a Foucault Pendulum at 45˚ latitude at any spot on the moon and I seriously doubt that the pendulum will be affected by the moon's orbital revolution since a Foucault Pendulum wasn't designed to do that.

Unless you have proof otherwise, we can agree to disagree on that side issue.

At issue here are the effects of tidal braking, and in regards to tidal braking, the moon's gravity is OVER TWICE as strong at affecting the Earth's tides than the sun's gravity is.

Unintelligible, but see my above response.

Since the Earth and moon together orbit the sun at a far greater speed than the moon orbits the Earth, you appear to be referring to this concept ("The Orbit of the Moon around the Sun is Convex!"):

(Add the http & www): math.nus.edu.sg/aslaksen/teaching/convex.html

Here's a good graph of the concept (Add only the HTTP):

wikipedia.org/wiki/Image:Moon_trajectory1.svg

That DOES NOT mean that the moon doesn't orbit the Earth too!

I'm sure you've been told that many times, eh?

Ken

 

Explain lunar libration with this model. Hint: You can't. Lunar libration results precisely because the moon's rotation rate (and rotational axis) is not exactly equal to the moon's orbital rate (and orbital axis). The moon's rotational rate is nearly constant; it's orbital rate is not because the moon's orbit is slightly elliptical.

Think of it this way: An observer see would see the Sun and the stars rise and set. The Moon is rotating.

Umm, turning about a polar axis is precisely the definition of rotation.

Your problem is that you are thinking too geocentrically, as if that is the only "true" way of looking at things. The same is happening here:

We do have 366.24 sidereal days per year. What constitutes a "day" depends on one's frame of reference. You are assuming one and one only frame of reference is "true", and that viewpoint is simply wrong. All reference frames are equally valid, including the geocentric frame in which the Moon's motion appears to go a rather complex longitudinal and latitudinal libration and an inertial frame in which the Moon appears to undergoing nearly uniform rotation about an axis passing through its center of mass.

Ancients (and modern crackpots) viewed the Moon as wobbling side-to-side and up and down by some mystical force. Modern physicists and astronomers view the Moon as rotating about its own axis because that is by far the much simpler way to explain the Moon's motion. No mystical forces are needed.

 

Your graphics #1 thru #3 are NOT posted anywhere close to your graphic #4 – it would have been easier to post one of the #1 thru #3 graphics near the #4 graphic than it is to bitch about it. What are you afraid of?

Soon I'll have full posting rights and I can do it for you.

YES, you are correct, Venus' 243 rotational period is as viewed from the sidereal perspective. HOWEVER, Venus' rotation has everything to do with the discussion at hand since its retrograde rotation is a prime example of how a sidereal perspective often confuses people!

E.g., google and you'll find that many astronomy websites claim Venus' day is longer than Venus' 224.65 day orbit. When IN FACT, Venus has TWO (2) complete days per each 243 day orbit – this graphic on this page clearly shows that Venus has NEARLY two full days per each orbit:

(Add the HTTP & www): geocities.com/kfuller2001/tVenus.html

If Venus' rotation were a tad 19 days faster, then Venus would have a sidereal spin rate of (REMARKABLY) 1:1!!! If Venus' rotation continues to slow down from the sun's tidal braking, then eventually Venus will reach the 0:1 (zero sidereal spin) that was on this page:

(Add the HTTP & www): geocities.com/kfuller2001/tVenus.html

CLEARLY, Venus would still have one internal polar rotation at a sidereal spin rate of 0:1! And after Venus is fully braked by the sun, then Venus will once again have another 1:1 sidereal spin - i.e., there will be TWO 1:1 sidereal spin rates as Venus winds down to ZERO actual axial rotations!

Of course, Venus may actually be speeding up and not slowing down:

---snip---
"Research suggests that {Venus'} "backward" rotation is caused by tides which are raised in the thick atmosphere by the Sun, and with friction interaction between the atmosphere and planet itself. It is hypothesized that these interactions caused Venus's rotation to slow, stop, and then reverse. This is somewhat similar to what is happening here on Earth, as our Moon's pull on our oceans causes tides whose subsequent friction is gradually slowing Earth's rotation"

(Add only the HTTP): chuckayoub.googlepages.com/venus_information__the_planets.htm
--/--

If that's the case, then Venus will speed up to a perfect 1:1 (w/ 2 solar days), then 2:1 (w/ 3 solar days.)

As that theory holds, Venus first spun down from a normal prograde rotation, stopped, and then slowly started turning in the opposite direction – and, as viewed from the sidereal perspective, that would be ("+" is prograde and "-" is retrograde):

+4:1 (w/ 3 solar days)
+3:1 (w/ 2 solar days)
+2:1 (w/ 1 solar day)
1:1 (w/ 0 solar days) – i.e., one long day on the near-side and total darkness on the far-side)
-0:1 (w/ 1 solar day)
-1:1 (w/ 2 solar days)
-2:1 (w/ 3 solar days)
-3:1 (w/ 4 solar days)

Spin it any way you like, but whether an astronomical body spins down to 1:1 from the retrograde or prograde direction, when it finally stops at 1:1, then it has LOST all of its polar axial rotations!!!

An astronomical body spinning down from the retrograde direction will also have TWO 1:1 spin rates on either side of the 0:1 sidereal spin rate - so much for the God's-eye sidereal perspective!

No one knows if our own moon spun down from a normal prograde direction or a retrograde direction, but either way, the moon stopped spinning around its polar axis over 3 billion years ago!

1:1 equals ZERO!

Many websites do list Venus correctly:

---snip---
Sidereal Rotation 243 Earth days
Length of Day 116.75 Earth days
Sidereal Revolution 225 Earth days

(Add the HTTP & www):crh.noaa.gov/fsd/astro/venus.php

Many other websites list Venus' day as 243 Earth days – the sidereal perspective can be confusing if you don't understand it.

That's because you have poor spatial reasoning!

Whether a large astronomical body is spinning or not can be determined by its shape. No matter your reference frame, the Earth rotates and its rotation forces it into the shape of an oblate sphere.

The moon has stopped rotating and the Earth's gravity has distorted the moon's shape into a prolate spheroid (i.e., a football shape.)

Venus still slightly rotates, just enough to keep its sun-caused tidal bulges in constant motion, so of all our solar system's astronomical bodies, Venus is the closest in shape to a perfect sphere.

Yes and no. If the Earth slowly ground to a halt, then when it finally stopped rotating on its polar axis, then the Earth would switch over to rotating around the Earth-moon barycenter. The Earth would then have a near-side and a far-side that faced the moon, and the Earth's new day would be identical to the moon's length of day.

Billions of years ago when the Earth finally stopped the moon's rotation (at 1:1), ONLY then did the moon's tidal bulges stop and become fixed in place. HOWEVER, prior to fully stopping at 1:1, the moon was likely a perfect sphere, just as round as Venus is today (with its slow .93:1 spin rate.)

SIMPLE EXPERIMENT:

Spin 360 around with an orange in your out-stretched hand, and as you spin around looking at the orange (which isn't rotating around its own center of gravity), and then tell me the orange was spinning on its internal axis, or not?

Try it again, but this time, use your free hand to give the orange one 360 axial rotation as you spin your body around 360 – so, what happens if you rotate both the orange and your body at the same time? Did the orange keep one face pointed towards you?

NO!

Do the simple orange experiment and get back to me.

That, too, depends upon one's perspective!

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Ken

 

Yes, the orange spun once on its axis for each turn of me. If it hadn't, it would have appeared to rotate as I did, but it showed no apparent rotation (this is logical since it was stationary wrt my hand), therefore it must have rotated about its own axis.