@Janus58
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:
Originally Posted by Billy T
{post 567}... 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. ...
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!