## [geocentrism] Re: Tides and the moon and M-M

• From: Paul Deema <paul_deema@xxxxxxxxxxx>
• To: geocentrism@xxxxxxxxxxxxx
• Date: Sat, 5 Apr 2008 14:41:05 +0000 (GMT)

```Pete C
Truly eloquent sir. 'Out of the mouths of babes' so to speak!
Paul D

----- Original Message ----
From: PETER CHARLTON <peter.nambo@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Friday, 4 April, 2008 8:06:09 PM
Subject: [geocentrism] Re: Tides and the moon and M-M

As you know, I have no scientific knowledge, but it seems logical to me that,
rather than there being a "zero gravity point", that instead there is a point
where the pull of gravity from the moon, is equal to the pull of gravity from
the Earth, that is, the two forces pulling equaliy at a given point.

If you go nearer the moon, you still have the pull of gravity from the Earth,
but it is less than the increasing pull from the moon, and visa versa.

Surely, if there was a point where gravity was zero, the moon would escape from
its orbit?

Pete Charlton

----- Original Message -----
From: Bernie Brauer
To: geocentrism@xxxxxxxxxxxxx
Sent: Friday, April 04, 2008 12:04 AM
Subject: [geocentrism] Re: Tides and the moon and M-M

Statement/Question:
"It is widely accepted, although not by me, that the moon causes the tides. It
is also widely accepted, although not by me, that there exists a zero-gravity
point situated somewhere between the World and moon.
My question is this: If the ocean were situated at the zero-gravity point, then
there would be no tide. Closer to the World the pull of the World is stronger.
Closer to the moon the pull of the moon is stronger. The net effect, this side
of the zero-gravity point, is always a positive pull by the World. Since this
is equivalent to a force of gravity that produces a stronger pull as we take
the oceans further this side of the zero-gravity point, then how does the moon
produce the tides?"  Dr. Neville T. Jones
Response:
"IT DOES NOT DIRECTLY, ONLY INDIRECTLY. Hooray! I’m so glad finally someone
else sees this issue too. Further, the tides are one of the major reasons why I
model gravity as a vibration, for The Alias Effect shows that the position of
the sun and moon has a relationship to gravity on Earth but tides demonstrate
that they are not directly related due to the whole satellites issues as well
as atmosphere. However, in vibrational gravity the positions of CB's (
Celestial Bodies ) will affect the vibrational wave. In short, the tides are
caused by the squeezing effect of the gravity vibration, that is to say, that
there is no additional or absence of gravity force, only a uneven squeezing
effect that is a result in part due to sun/moon/background-stars positioning (
The Alias Effect proves this ). A vibration is the only known physical
explanation that can account for that effect while producing a non-detectable
gravity force in all of its
anomalies, which are not anomalies but rather clear indicators that gravity is
a vibration of aether waves. No other known physical construct could account
for all those things."  Allen Daves

Jack Lewis <jack.lewis@xxxxxxxxxxxx> wrote:
I haven't yet seen anyone come with an answer to something Neville, I think,
once said regarding the point, which must exist, between the Earth and the Moon
where the gravity is zero. This being the case how is it that the Moon controls
the tides? I'm sure, I think, that there must be a simple answer.

The M-M part of the subject is to ask Regner how he is getting on with the
answering the interferometer experiments wrt a non-moving Earth?

Jack

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