[geocentrism] Re: Tides
- From: "Martin G. Selbrede" <mselbrede@xxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 7 May 2008 15:48:47 -0500
Neville
Of course, the principle of superposition must be applied. The world
and you are in identical free-fall (setting aside the negligible
difference between your center of mass and the Earth's in respect to
astronomical distances), and for all intents and purposes can be
neglected in the analysis while free fall is in effect. Technically,
as Newton pointed out, the Earth is currently in free fall within a
Copernican frame of reference. It's not until the "extended object"
effect begins to dominate that there'd be any significant difference
in the rates of the Earth's fall into the Sun and your own fall into
the sun. For all practical purposes, they're identical. In short,
your analysis appears to omit the g' acceleration on the Earth
incurred by the Sun, which is essentially identical to the g'
acceleration upon you. (As to extended object effects, one could,
for the sake of completeness, note that IF you are positioned on the
Sunward side of the Earth, there may be a microscopic reduction in
weight due to the difference in the value of r for your center of
mass and the Earth's center of mass with respect to the Sun's center
of mass, it is also true that if you were positioned 90 degrees out
from the Sunward side (such that your center of mass and the Earth's
center of mass are equidistant to the Sun's center of mass), there'd
be a microscopic INCREASE in your weight as you approach the Sun,
insofar as your geodesic and the Earth's geodesic are not exactly
parallel but are pointed to a common center (the Sun's), meaning the
lines of fall are slowly converging, the more so as you get closer to
the Sun and configurational considerations begin to dominate in the
Gedankenexperiment's geometric framework.
Martin
--------
Martin G. Selbrede
Chief Scientist
Uni-Pixel Displays, Inc.
8708 Technology Forest Place, Suite 100
The Woodlands, TX 77381
On May 7, 2008, at 3:17 PM, Neville Jones wrote:
-----Original Message-----
From: allendaves@xxxxxxxxxxxxxx
Sent: Wed, 7 May 2008 08:14:47 -0700 (PDT)
... 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?.....
Your weight is a force and is therefore given by Newton's F=ma,
where m is your mass (in kg for example) and a is the acceleration
(m/s/s) produced by this force.
There will be a force acting on you directed towards the centre of
the World, where the acceleration resulting from this force at the
World's surface is g m/s/s. And there will be a force acting on you
and directed towards the centre of mass of the Sun, where the
acceleration due to this at the World's surface is g' m/s/s. These
will be acting in opposite senses.
Your weight, W, will therefore be
W = mg - mg' = m(g - g') N
and will act towards the centre of the World until such time (since
the World is in 'free fall' towards the Sun) as g' = g, when you
will become weightless. Then, as g' > g, you will be pulled off the
World and fall into the Sun before the World does. I.e., your
weight would be constantly changing.
The above assumes that you are positioned on the same side of the
World as the Sun and that you are not an 'extended' object, at
least not in relation to the strength (size) of the gravitational
field of either the World or the Sun.
Neville.
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