By the way, do YOU understand the difference between a pinned-axis graph showing rotation even when representing forward momentum around another axis ... versus a graph showing what the axis relative to the body is doing, when the pinned axis is representing NO forward movement of the axis?
The Earth would continue to show spin; the Moon would not.
That is because in real space, the Earth takes more of a turn than its path (orbit) requires, whereas the Moon follows its orbit face forward. These people are graphing the axial movement, directing the body at each point, but not the body's movement relative to its axis on the path -- with or without a path.
So you are defining lunar rotation relative to the Earth, despite the fact that the Moon primary orbit is always falling toward the Sun.
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link to www.mathpages.com]
The Moon takes more of a turn than its path (orbit) requires in its motion around the Sun, after all. Is it spinning now?
And I covered that too: it's still optical illusion and sloppy definitions, but yes, it does seem to spin over space as it also seems just to "wobble" ... around the Earth's barycentre.
It is why the astronomers can call THAT feature "spin on its axis" -- which is not the thing anyone here so far understood, in the sense that it was not the point they were making and in the sense that though I made it, everyone called that gobbledigook.
But even so, the way to tell that even around the Sun it's an illusion of "spin on its axis" in proper motion, is that for every forward movement on its wobble, it turns relative to the point outside it, not more.
It's a tricky optical illusion, but the Earth, if it also orbited another body nearer to it as they passed around the Sun, would turn MORE than that body's forward momentum would indicate.
So it's of course a tricky illusion that the Moon goes around its axis, but it's observationally accurate enough, and there's the confusion from graphs of its total axial momentum turn (which people here got hung up on).
True axial spin in proper motion is when all axial momentum (path) is removed from the thought experiment AND where the graphic representation is not graphing that momentum AT ALL (unlike the graphic representation which is OF the momentum).
By the way, over space, the problem is more what Kepler was getting at when he pointed out that orbits maintain constant velocity over the AREA of the distance, not the linear total distance.
In this case, the analogy is this: the illusion of turn of the Moon on its axis proper, over the rotation around the Sun, is less about total area vs. velocity, than turn (of Moon) vs. velocity (of barycentre with Earth).
So, the distance the Earth (its barycentre) travels forward around the Sun vs. TURN of the Moon is the Keplerian likeness. How so:
The Moon will be seen to turn more at some points in its progression around the Sun as the Earth-Moon barycentre moves forward the same distance, because it does not have an exact same rotation momentum as the Earth's spin. It creates a long "sine-wave" of turning, in space, instead of a shorter one.
Hence, not a true rotation spin around its axis in its own right. But we know this without the 2ndary movement around the Sun. That is again, an illusion impression, or if you will, a mapping through space impression.
The Moon would have to rotate IN RELATION TO ITS ANGULAR MOMENTUM (or if it were physicaly not moving at all, still rotate), in order to be actually rotating on its own axis as the Earth does: "proper" (self-same) spin/rotation motion.
It was THIS which Nancy was referring to; and it is THIS why she said if you stopped the Earth's rotation around the Sun AND its spin ...
But that the Moon would only have to be stopped (in relation to the Earth), not stopped from spinning.
HOWEVER, as I acknowledged all along,
Nancy was wrong that the Moon does not have ANY ORBITAL DIFFERENCE than the Earth. It does: its own angular momentum.
