[geocentrism] Re: Tides
- From: Allen Daves <allendaves@xxxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 7 May 2008 08:14:47 -0700 (PDT)
Regner writing coments does not constitue addressing the isssue or answereing
the question...... In fact your comments tells us absolutly nothing wrt the
issue at hand..
The tides are either accelerated toward the sun moon while in free fall toward
the sun moon or not.....If they are then:
1. You can't claim a acceleration in "free fall" toward your inertial ref frame
cannot be detected..........That is the exact same condition as our
accelerometer that is in "free fall" toward its inertial fame earth...... a
mass suspended elasictly just as in the case of our accelerometer "free
falling" toward earth in the elevator or the earth falling toward the
sun....same thing "free falling" toward our inertial reference frame!.
2. You can't claim the reason our accelerometer will not detect a free fall is
because grav is pulling on all parts equaly and simoltaniously becuse the tides
are certainly not pulled equaly simoltaniously to all parts as the rest of the
accelerometer mass (earth) i.
Im going to go back now and begin in much more detail to address the error here
using Pauls diagram of an accelerometer and mass....first...... I will ask a
question to everyone.....if i am on a planet of one earth mass that is falling
toward the sun will i have any weight and how much will i weigh and why?.....
----- Original Message ----
From: Regner Trampedach <art@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Tuesday, May 6, 2008 10:18:40 PM
Subject: [geocentrism] Tides
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
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