[geocentrism] Re: Moon Rotation

  • From: Paul Deema <paul_deema@xxxxxxxxxxx>
  • To: geocentrism@xxxxxxxxxxxxx
  • Date: Thu, 4 Dec 2008 07:14:44 -0800 (PST)

Philip M
I'll add some <colour>

Paul D



From: philip madsen <pma15027@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Wednesday, 3 December, 2008 11:31:48 PM
Subject: [geocentrism] Re: Moon Rotation


 
Please clarify, the highlighted question. <They all seem to be highlighted -- I'll do them all!>
Object A. Looking down on the two body system shown, the Black Spot needs to keep facing the primary and to remain in the centre of the disk. This can ONLY occur if a line joining the centre of the primary and passing through the Black Spot AND the centre of the satellite rotates around the plane of the orbit. To do this, the satellite MUST rotate at the same rate as the line. There can be no other rotation of the primary because this would cause precession preventing the Black Spot from remaining at the centre of the seen disk. Because it is rotating, it has angular momentum which has energy.
Cant follow why rotation of the primary has anything to do with things. <You're absolutely right - it doesn't. And for the reasons you've outlined in you comments below.> The radian from the centre of the orbit to the satellite is a "vector" (an imaginary line)  of the angular movement of the orbiting satellite. It is fixed to the satellite, not the primary. I see two lines or vectors/  One is moving with respect to the centre of the orbit. And the other is moving with respect to the centre of the satellite. Because they are synchronised they remain locked together. Your last line is self evident.
 
Paul I am having difficulty with B  yes it is trickier Meaning I am not comprehending the motions involved in your text.  Let me guess, by putting it in my brains visualisation.
 
Object B. This is trickier as the satellite is in fact not rotating. We could still give it a gemetrical axis with poles in the normal sense as being at right angles to the orbital plane..<There is nothing says right angles (or small angles eg ~25 deg) are 'normal' except that coalescence from a primordial cloud would favour this. But that is only to do with the manner of the forming of the Solar System. We are here considering only mechanical possibilities ie would it be stable if it were different from 'normal' or if not, how long would it last. The example of Uranus is a beacon. Yes, you're correct, we could have put the poles and equator in the 'normal' position but then it would have been useless for my illustration.>  (I am not comfortable with orthogonal.) <It took me a while too but I think I have it right.> As such, you could argue that it therefore has no poles and no equator but if I choose to define the location of poles and equator and then describe the view of the Black Spot from the primary, you would have to conclude that its radial velocity is zero. As such it has no angular momentum and therefore no energy from such a motion.. Absolutely.  And my experiment with the water glass is the example proof. If the satellite presents one complete revolution of its face to the observer on the primary per one orbit period, then it has in reality not turned in any way, and as you say has no angular momentum. Its mass is undisturbed, has no centrifugal forces, and thus no stored energy due to rotation..  It simply is not rotating. <Good -- but no cigar! It could meet your description if it rotated twice per revolution and it would then certainly be rotating but of course in the opposite direction.>
 
I am sending another experiment out today which will disturb minds I hope, showing this principle you seem to be showing, and make one wonder about the complex energy formula re flywheels. <I have to admit that, sadly, applied maths is not a friend of mine.>
 
As regards C  I have to guess again, but within your text you give me clues.. 
Object C.
The satellite begins its orbit and the Black Spot would -- if the satellite is not rotating -- begin to drift from the centre of the seen disk towards one limb still in the plane of the orbit. However, if it is rotating synchronously (one rotation per one orbit) it will drift towards the limb at an apparent angle of about 45 deg. After 180 deg of revolution (the satellite is now on the left hand side of the illustration) and 180 deg of rotation, the Black Spot will again occupy the centre of the seen disk having drifted in from the limb at an angle of about 45 deg. This can ONLY happen if there is synchronous 1:1 rotation per revolution. Since the satellite is rotating, it has angular momentum and therefore energy from this rotation.
Well this was confusing till you explained below that the spin of the satellite is horizontal to the orbital plane. Of course once again, what is observed on the primary has nothing to do with reality.. Its a normal visualisation due to two apparent relative motions, one of which is not real. <I'm not really sure what you're getting at here so I won't comment (except for this comment!) Is that a recursive statement? I've never gotten my mind to comprehend recursion.> <[Later] I think I have just tumbled to what you said. The satellite appears to be rotating backwards in the plane of its axis because it is not rotating in the plane of its axis but is revolving (or orbiting if you prefer) this being your 'not real' motion..>
What you have done is tipped the axis of the real moon over by 90 degrees. and left it in the same orbit. <Quite so. It was also so for Object B of course.> It will still spin synchronously with the orbit period, have angular momentum with respect to its new axis, and the energy therefrom. But because of its orbit this axis though remaining unmoved with respect to itself (stationary) <Well it points to a fixed point in space (ignoring parallax).> will appear to the earth bound observer to rotate one complete revolution per orbit. <I didn't think of this aspect but of course you are correct. This is again your 'not real' motion as above.>
This is not an idle exercise. The object was to retain the relationships but by shifting the axis of rotation from perpendicular to the orbital plain, to parallel to the orbital plain, the appearance of the motions is more easily visualised.
Just how many times have I said to Bernie, Appearances are not what you get.  Like we see a moving picture on TV, entirely due to an inherent weakness of the human eye..  the dog sees the reality, a sequence of separate images. <When things are slow you might talk to me about this -- it's new to me (the dog's eye view I mean).>
 
I hope I got it right this time..   but come again..  <Come back if I've confused you here.>
Philip. 
----- Original Message -----
From: Paul Deema
Sent: Thursday, December 04, 2008 3:08 AM
Subject: [geocentrism] Re: Moon Rotation

Philip M

OK -- I'll go through my reasoning with you. (you'll need to make reference to ThreeObjects.png)
 
Object A. Looking down on the two body system shown, the Black Spot needs to keep facing the primary and to remain in the centre of the disk. This can ONLY occur if a line joining the centre of the primary and passing through the Black Spot AND the centre of the satellite rotates around the plane of the orbit. To do this, the satellite MUST rotate at the same rate as the line. There can be no other rotation of the primary because this would cause precession preventing the Black Spot from remaining at the centre of the seen disk. Because it is rotating, it has angular momentum which has energy.
 
Object B. This is trickier as the satellite is in fact not rotating. As such, you could argue that it therefore has no poles and no equator but if I choose to define the location of poles and equator and then describe the view of the Black Spot from the primary, you would have to conclude that its radial velocity is zero. As such it has no angular momentum and therefore no energy from such a motion.
 
Object C. Here I have to make an embarrassing admission and seek your collective forgiveness. For this exercise, the Black Spot is shown on the wrong side of the satellite. (It was all clear in my mind when I put this together!) With the Black Spot moved 180 deg around the primary, it will now be in the centre of the seen disk with the North Pole to one side and the South Pole to the other, on the plane of the orbit.
 
The satellite begins its orbit and the Black Spot would -- if the satellite is not rotating -- begin to drift from the centre of the seen disk towards one limb still in the plane of the orbit. However, if it is rotating synchronously (one rotation per one orbit) it will drift towards the limb at an apparent angle of about 45 deg. After 180 deg of revolution (the satellite is now on the left hand side of the illustration) and 180 deg of rotation, the Black Spot will again occupy the centre of the seen disk having drifted in from the limb at an angle of about 45 deg. This can ONLY happen if there is synchronous 1:1 rotation per revolution. Since the satellite is rotating, it has angular momentum and therefore energy from this rotation.
 
This is not an idle exercise. The object was to retain the relationships but by shifting the axis of rotation from perpendicular to the orbital plain, to parallel to the orbital plain, the appearance of the motions is more easily visualised.
 
And if Allen is listening, I'd ask him -- where is his "... common point," to which this satellite is "...progressively radially oriented."
 
Paul D


From: philip madsen <pma15027@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Tuesday, 2 December, 2008 11:20:36 PM
Subject: [geocentrism] Re: Moon Rotation


The angular momentum -- I think I may have the wrong term -- which I am referring to is that energy which exists in a body rotating -- note, NOT revolving or otherwise translating. The energy which would be present if it were the only body in the universe. The energy which would be demonstrable without reference to any other body.
 
If you re-examine my illustration in that light, you may wish to modify your response.  Paul.
 
You ask a complex question Paul. But the only reference in my physics is the flywheel ..  Flywheel math is simplified by haveing all the mass concentrated at the rim.   to establish the energy of a moon or planet is more difficult for my math, but it has been done for the moon and the earth somewhere about how long the tides take to slow down our moons orbit and planet rotation .
 
In any case, in all cases my answer remains..  There will be an energy called kinetic I hazard, inherent to the rotations of the moon object you show. It is essentially a flywheel operation.
 
Philip. 
----- Original Message -----
From: Paul Deema
Sent: Wednesday, December 03, 2008 4:50 AM
Subject: [geocentrism] Re: Moon Rotation

Philip M

Thank you for your response. It indicates among other things that I have something (many things?) to learn about terminology. I've found a new and interesting site which I think might help me -- assuming I can get my rapidly 'addling' (present tense of addled) brain to cooperate. It is http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html.
 
The angular momentum -- I think I may have the wrong term -- which I am referring to is that energy which exists in a body rotating -- note, NOT revolving or otherwise translating. The energy which would be present if it were the only body in the universe. The energy which would be demonstrable without reference to any other body.
 
If you re-examine my illustration in that light, you may wish to modify your response.
 
Regarding the poles as depicted in that part of my illustration which you have included here. I know it's a stretch but I meant literally what I depicted. The poles are indeed the 'ends' of the axis of rotation and the axis of rotation is indeed in the plane of the orbit. It closely resembles the orientation of Uranus in its orbit. This puts the equatorial plane orthogonal to its ecliptic ie the plane of its orbit.
 
Please do re-evaluate -- I need your understanding.
 

Paul D



From: philip madsen <pma15027@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Monday, 1 December, 2008 9:41:54 PM
Subject: [geocentrism] Re: Moon Rotation


 

well yes it has two elements of angular momentum..  One due to the orbital motion, and one due to its angular rotation around its own axial centre..   As my first answer was.. 

 

 

the answer is the same for diagram B..  The angular momentum of an object still has two elements...  one with respect to its own centre axis, and one with respect to the primary centre.

 

However I have difficulty still with your terminology..  Poles generally refers to the opposite ends of the axis of rotation. In B you seem to have the poles shifted to the plane of the orbit, which now makes these the equator. and the rotation is still vertical to plane of the orbit, which make for new poles, to and bottom. and a new equator.. 

 

I did tell you I have difficulty reading static diagrams of a dynamic system. 

 

Phil

----- Original Message -----
From: Paul Deema
Sent: Monday, December 01, 2008 6:44 PM
Subject: [geocentrism] Re: Moon Rotation

Philip M
re Moon Rotation -- From Paul Deema Thu Nov 27 01:37:59 2008 (Attachment ThreeObjects.png) addressed to Allen D.
I recall your oft stated difficulty visualising physical motions, moving mechanisms et al, but regardless, I am interested in your take on the questions included in the illustration. Allen of course has a vested interest in simply pronouncing my offerings as "Your post is nonsense!" but I believe that you may well be able to see what I am getting at.
Would you comment please?

Paul D


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