[geocentrism] Re: Tides
- From: Allen Daves <allendaves@xxxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Fri, 9 May 2008 15:25:13 -0700 (PDT)
Hi Martian,
Yes, I can see your point...But I was trying to point out that you can't
explain one while ignoring the other, particularly when those two explanations
would be mutually exclusive as it pertains to a detectable and yet not
detectable accelerations in "free fall" to the sun on one hand and the stars in
the other in the same "frame of Reference".....It is my belief that this
"slight of hand" by MS is the cause of much deception and difficulty for the
other side to see clear through. Although, I believe Paul sees the paradox.
Otherwise, it is difficult to imagine why he would have such opposition to
those terms (acceleration/grav/inertia) in my diagram.
----- Original Message ----
From: Martin G. Selbrede <mselbrede@xxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Friday, May 9, 2008 2:50:14 PM
Subject: [geocentrism] Re: Tides
Hi Allen,
Of course you know that we'd both agree that the equatorial bulge effect is
"with respect to the distant stars." Superimposed on top of that are the ocean
tidal cycles. If you're talking about the former, Regner's point is indeed
"kinda irrelevant," but since this thread started focusing on the ocean tides,
I limited my remarks to the superimposed effect.
Martin
On May 9, 2008, at 4:39 PM, Allen Daves wrote:
I would like to point out however, that the tidal bulges are on both sides of
the earth & earth's oceans at the same time, the earth is flatened around it's
whole cercumfrence supposedly due to the centrifical force that is cused by
inertia wrt the stars not just the tidal actions of the sun/moon.....Even that
explination would not reslove the the fact that the acceleration is detectable
but it does not address the overall inertial paradox that exist for both the
tides as well as any given orbiting body in a eliptical orbit where the
accelerations advance and slow down wrt the very same causes of inertial
reaction when not in orbit. (the whole point of the diagram)Therfore that (sun
only) explination is not only too simplistic but kinda irrelevant to the issue
as a whole that I am trying to point out.
----- Original Message ----
From: Martin G. Selbrede <mselbrede@xxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Friday, May 9, 2008 2:25:19 PM
Subject: [geocentrism] Re: Tides
I understood Regner to be mounting an argument based on relative differential
force (the entire premise of tidal forces in the first place). It is this
differential force that supposedly gave rise to the rings of Saturn, when a
sufficiently large chunk of material was disrupted as it passed within the
Roche limit (meaning the force between the near-side and far-side parts of the
body exceeded the tensile strength of the material comprising that body). This
tidal disruption causes the alleged shattering of the body into a myriad of
small chunks, each finding a unique orbit around Saturn due to inter-body
collisions over time.
I believe Regner mounted a Sun-only explanation for the sake of didactic
simplicity, and he even went so far as to point this out. I saw nothing faulty
in his analysis given the limits he himself provided to frame his discussion.
Martin
On May 9, 2008, at 4:14 PM, Neville Jones wrote:
Regner,
Your analysis of the cause of the tides is invalid, because gravitational
attraction between parts of extended objects being exactly equivalent to all
mass of each object being concentrated into mathematical points is valid for
these objects being rigid, not flexible. You cannot assume the World to be
rigid on the one hand and then subtract the theoretical force on a rigid World
from both sides of a flexible World in order to 'explain' the tides.
Where would these forces on the oceans be acting? At the centre of mass of the
ocean, or on individual water molecules, or at the centre of mass of the World?
The pull would vary as the inverse-square of the distance, which would result
in very serious tidal waves arising as the Moon-side ocean was pulled more than
the other was 'left behind'.
Rather than being
<- | -> as
depicted, the force would be proportional to
x*x^(-1) (x*x+2xR+R*R)^(-1) (x*x+4xR+4R*R)^(-1)
where x is the distance from the centre of mass of the Moon to the closest tip
of the ocean nearest to the Moon and R is the radius of the World, and x >> R
in the heliocentric model.
Ocean swell would also vary with the Moon's orbit.
The analysis is thus far too simplistic. As with other aspects of the so-called
'solar system', such forces would lead very quickly to disorder and destruction.
Neville.
-----Original Message-----
From: art@xxxxxxxxxxxxxx
Sent: Wed, 07 May 2008 15:18:40 +1000
There has been a couple of references to tides, lately, and much confusion
on the subject. I'll do what I can to clear it up a bit.
Please take the time to read the whole post before commenting.
The force of gravity moves objects
The Moon and the Earth orbit each other due to their gravitational
interaction.
It can be shown that you can simplify this problem:
Every single bit of the Moon interacting gravitationally with every single
bit of the Earth.
is exactly equivalent to
All the mass of the Moon being located at the centre-of-mass of the Moon,
and all the mass of the Earth being located at the centre-of-mass of the
Earth.
From such an analysis we find that the two objects will orbit their common
centre-of-mass in elliptical orbits.
Tidal forces deform objects
But the Earth and the Moon are extended objects so what effect does that
have? Well, the Moon is still close enough to the Earth, that the gravity
from the Moon will change over the diameter of the Earth:
Moon Near Centre Far
side of Earth side
O <-------- <------- <------
Okay, if we now subtract the part that actually moves the Earth, that is
we subtract <------- which is the same as adding -------> to all three
vectors (arrows) above, and we get the tidal forces:
Moon Near Centre Far
side of Earth side
O <- | ->
So looking at the Moon-Earth system from the outside we see that the
Moon-side of Earth experience a larger gravitational pull from the Moon
than the centre of the Earth, and the opposite side of Earth experience
experience even less gravitational pull from the Moon. Seeing it from the
Earth point of view, we have a Moon-ward force on the Moon-side of Earth
and an opposite force on the opposite side of Earth. This is what causes
the tidal bulge, and this is why there are (approximately) two high- and two
low-tides a day. The oceans, being a liquid, obviously responds much stronger
than the bedrock, but the Earth's crust is flexing too, as has been measured.
The Earth rotates under this tidal bulge, which makes it a dynamical phenomenon.
The gravitational force is proportional to 1/r² where r is the distance
between the Earth and the Moon. The tidal force is the difference in
gravitational force over a certain distance, which means it can be expressed
as the r-derivative of the gravitational force. The tidal force is therefore
proportional to 1/r³. That means the tidal force falls off more rapidly with
distance than the gravitational force. This is the reason that the gravitational
force of the Sun dominates the orbit of the Earth, but the Moon's tidal forces
on Earth are larger than the Sun's.
The Sun also contributes a significant tide (about half that of the Moon)
and when the two are aligned at new Moon or full Moon, we get more
powerful tides called spring tides. At either half Moon we get neap tides.
The ellipticity (eccentricity) of the Moon-Earth and the Sun-Earth orbits
also causes variations in the strength of tides on Earth.
The Moon experience (about 22 times) stronger tides from the Earth
because of the Earth/Moon mass ratio of 81 and the size ratio of 3.7.
The Lunar tide from the Sun is 3.7 times weaker than here on Earth, due
to the 3.7 times smaller diameter of the Moon.
The tides would of course be exactly the same in a geocentric Universe.
Allen's concerns in the "Acceleration thread"
To account for tidal influences, the centre-of-mass of the test-mass of
the accelerometer, should coincide with the centre-of-mass of the free-
falling laboratory. Tides would have NO effect on such a set-up.
But then the tidal effects on such a lab are very small in the first place:
Less than a billionth per meter of the gravitational force keeping the lab
AND the test-mass in orbit.
That's all for now (sorry for such a long post), and I hope it clarified a
few things.
- Regner
________________________________
Receive Notifications of Incoming Messages
Easily monitor multiple email accounts & access them with a click.
Visit www.inbox.com/notifier and check it out!
--------
Martin G. Selbrede
Chief Scientist
Uni-Pixel Displays, Inc.
8708 Technology Forest Place, Suite 100
The Woodlands, TX 77381
281-825-4500 main line (281) 825-4507 direct line (281) 825-4599 fax (512)
422-4919 cell
mselbrede@xxxxxxxxxxxxx / martin.selbrede@xxxxxxxxxxxx
--------
Martin G. Selbrede
Chief Scientist
Uni-Pixel Displays, Inc.
8708 Technology Forest Place, Suite 100
The Woodlands, TX 77381
281-825-4500 main line (281) 825-4507 direct line (281) 825-4599 fax (512)
422-4919 cell
mselbrede@xxxxxxxxxxxxx/ martin.selbrede@xxxxxxxxxxxx
Other related posts:
