[geocentrism] Re: Tides
- From: Allen Daves <allendaves@xxxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 7 May 2008 14:43:18 -0700 (PDT)
ok ....see attached diagram......
A. HOw can i be pulled away differently then the earth in a "free fall"?...I
thought grav in f a "freefall "pulls centers of mass at the same rate?.....if
that is true about a person on the planet then why is it not true for the
accelerometer.....well could say the accelerometer is at the center of
mass...ok....so then what we are saying is all we have to do to detect our
acceleration is move the mass or person in our accelerometer/ planet offset wrt
the center of totlal mass by puting it on the sun side (side of the direction
of our "free fall") ?.......ummmmm that is not very sporting of the sun if it's
gravity is pulling all centers of mass at the same rate wrt each other....where
is the additional energy comming from.....ummmm....perhaps we are saying the
further we fall the larger the deviation between any two centers of masses are
because of the rate is determined by the square of the distance............. in
a orbit the square of that distance changes by how
much?.......oh the tides thats right.....so can we or can we not detect the
acceleration of the earth in "freefall"...?
B. If there was no sun would we weigh anything at all?..we would say
yes....because the earth has it's own grav feild..ok... so how does the sun
negate the earth's grav in free fall?..even more to the point if it can why
does it not do that when we are on the sun side during the day?. if it
accelerates the center of mass of the oceans then why not the center of my mass
? ....why not?... We are said to "freefall" toward the sun even now! ( on the
sun side of the earth) But we dont appear to fly off toward the sun....why not,
if we are "freefalling" and the sun's grav feild would negate the pull of the
earth's grav feild?.if it does not negate/overcome the pull of the earth's
feild then how would we fall away toward the sun?!......If the pull is equal
from the sun and the earth then how can we weigh anything....if the earths pull
is greater at our location then the sun's then how do the tides get accelerated
differently then we do by the sun?...why does our weight not
fluctuate with the tides with the same magnitude of force that accelerates all
that sea water as the sun passes over us....
C. on the night side of that earth, there is the difference between a person
on that earth "freefalling" to the center of the sun's center of mass and a
person on the elevator "freefalling" to the earth's center of mass ....? is the
earth's grav over power the suns...then how come there are tides on the night
side as well?....what is accelerating them in a new moon where the sun and moon
are on the same side wrt the earth....
Summary: The earth still has a gravity that attracts mass with or without the
sun.....There is no way to negate the earth's pull on person by claiming a
"freefalling" toward the sun even if we say the earths is stronger we still
have a detectable acceleration wrt the sun in free fall the tides!?.....or more
precisely the whole idea of "free falling" to the sun is not really
"freefalling" in the same way that a apple falling toward the earth. The
falling apple is not the same as a satellite in orbit...using the term
"freefall" is just a free and loose use of terminology/cop out for the fact
that those two circumstances are not even remotely the same and yet they are
explained away the exact same way....?
Neville Jones <njones@xxxxxxxxx> 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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