[geocentrism] Re: Moon Rotation
- From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Thu, 4 Dec 2008 09:31:48 +1000
Please clarify, the highlighted question.
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. 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.. (I am not comfortable with orthogonal.) 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.
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.
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.
What you have done is tipped the axis of the real moon over by 90 degrees. and
left it in the same orbit. 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) will appear to the earth bound observer to
rotate one complete revolution per orbit.
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.
I hope I got it right this time.. but come again..
Philip.
----- Original Message -----
From: Paul Deema
To: geocentrism@xxxxxxxxxxxxx
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
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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
To: geocentrism@xxxxxxxxxxxxx
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
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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
To: Geocentrism@xxxxxxxxxxxxx
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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