[geocentrism] Re: acceleration calcs
- From: Regner Trampedach <art@xxxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Thu, 24 Jan 2008 10:52:06 +1100
allendaves@xxxxxxxxxxxxxx wrote:
**
*Regner......This is the problem with the equivalence principle argument.*
*1. It can only be proven to be false it has never been positivly
proven...*
That, Allen, is the problem with all theories in physics, as I have said
repeatedly
in the past. But all physical theories are also disprovable - i.e., you
can measure
things that does not neccesarily agree with the theory, thereby proving
it wrong.
This is in contrast to, e.g., religion, where an almighty God can do
anything she
pleases. All the physical laws that we see enacted in nature, could of
course just
be staged by such a God to look like that but in fact be something quite
different.
In science, Okham's razor comes in and says that the physical laws are a
much
simpler explanation for what we see, compared to having an omnipotent
sentient
being stage everything to look the same. Where would this God come from?
What does she consist of? Does this God exist within our outside our
Universe?
Is this God restricted by any laws of (other) physics? - are there any
laws of
physics? - and why do they seem to work anyway?
Since the laws of physics seems to work, and science has no way of
addressing
the question of God, we use Okham's razor to say that the laws of
physics are real.
That is the working assumption of scientists - and it works because that
is what we
observe.
*Notice* that this does not exclude the existence of a God.
*
*
*...only assumed.*
That is not correct. That their has been no proof against a particular
physical theory,
through a century of experiments and observations, really makes it a
very solid and
robust theory - far from a mere assumption.
*2. The solar system itself is in free fall around the galactic center
and thus the solar system is a inertial ref fame that we should be
able to measure within just as the earth is with the solarsytem and we
can measure inertia in earths inertail reference system.....*
Correct, the Solar system's orbit in the Galaxy, and the Earth's orbit
in the Solar system, are
both cases of free fall - which means we cannot measure them dynamically
(by means of
extra fictitious forces showing up in our experiments). We can measure
them kinematically,
however, that is by means of observing motions.
*So you cant hid behind the equivalence principle for a lack of
detection of inertia any more then you could with a car or a airplane
in free fall on earth.......inertia is still measured against all free
falling objects in every case. If you could not detect the earths
inertia then you could not detect any inertia from any object in space
or on the earth......*
Inertia is easy to measure: You measure how much force, /F/, is needed
to obtain a particular
acceleration, a, of a specific object: F= m*a, m is the inertia or mass.
This can be measured (and is unchanged) whether the object is in free
fall or not.
This is obviously a bit harder to do with the Earth, so instead we need
to rely on Newton's
law of gravity a_G = G*M/R^2
and the radius, R, of Earth coupled with measurements of the local
acceleration
of gravity, a_G... The gravitational constant, G, can be measured in the
lab, leaving just the
Earth's mass, M, as unknown.
*3. All ref frames are equivalent in relativity...you can have any
number or pick any sections for your reference frame....*
True.
* individual atoms are inertial reference frames!?*
A reference frame is a 3D coordinate system - so you need both a
position and two directions
in order to specify a reference frame.
*Position:* Most atoms would be close to other atoms, and would
constantly collide with
(feel the electric forces) of these other atoms - our atom would
therefore be accelerated
constantly and randomly. In a gas or a liquid, this is know as Brownian
motion.
Not an inertial rest frame at all.
A lone, free-falling atom could be an inertial rest frame, but read on.
*Orientation:* Some atoms are spherical (no directional indicators are
avaiable) and others
are non-spherical in various ways (some would be able to indicate two
directions). But
atoms are governed by quantum mechanics, and the orientation of an atom
can therefore
only be known in some (well-defined) statistical sense. The same goes
for it's position.
*Conclusion:* You cannot use single atoms as inertial reference frames.
**
*Philip...........What we are measuring is not the velocity of the
earth in its orbit but its change in velocity, in the same way that a
change in velocity is felt/ measured when you either put the brakes on
a car or you give the car some gas...*
...or you turn a corner.
I have the distinct impression that Philip understands all of that.
*you may not feel like your moving 100 mph but if you either put on
the brakes or give it some gas you will feel the change even as small
as it is.....*
Excellent Allen.
A very good summary of Newton's 1st law.
You don't feel velocity, but you do feel accelerations = forces, relative to
an inertial rest frame.
That's the reason we don't feel the 30 km/s speed of Earth in it's orbit
around the Sun.
However, there is that acceleration of 0.3776 cm/s2 to keep Earth in
it's elliptic orbit.
If Earth was a cart traveling on a track that kept it in orbit (that is,
no gravity from
the Sun), then we would all feel a centrifugal force (we would weigh
more at noon
than at midnight), because the force of the tracks would work on the
Earth and only
through it, on us. This is analogous to your example of a car.
In the actual case, where the centripetal force is caused not by a
track, but by
gravity, the centripetal force (gravity from the Sun) works on every atom of
everything, and exactly balances the fictitious centrifugal force. The
result is
that we do not experience any net force from the Earth travelling in
it's orbit.
Exactly as Philip wrote.
Regards,
Regner
----- Original Message ----
From: Regner Trampedach <art@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Tuesday, January 22, 2008 8:10:40 PM
Subject: [geocentrism] Re: acceleration calcs
Exactly, Philip.
The Earth, and we with it, are in free fall around the Sun, with the
gravitational acceleration
by the Sun (and towards the Sun) keeping us in our elliptic orbit.
Without careful analysis, I actually thought that you might be able
to detect it, but you
are right, Philip. This is also stated in Einsteins equivalence
principle which states that
a /free-falling/ reference frame is an /inertial /reference frame, and
there will therefore be no
fictitious forces (centrifugal-, Coriolis- and Euler-forces). The
equivalence principle
means that the orbit of Earth can just as well be seen as the Earth
traveling along a straight
line in a curved space - the two are equivalent - and the latter is
described by /general relativity/.
As a partial reply to your (much) earlier post on pseudo forces, I
will note a few facts
on them here - and there is nothing dubious about them.
Pseudo forces, more often called fictitious forces, arise when your
reference frame is
being accelerated. Let's say you set up a laboratory inside a
container on a trailer truck.
* The truck drives along a turn in the road.
* A ball is dropped from the ceiling of the container.
Imagine the container turning transparent, so that your colleague can
record the
trajectory of the ball, as seen from the roadside
* your colleague will see the ball follow a parabola determined by the
speed of
the truck when the ball was released, and the local acceleration of
gravity.
Only one force, gravity, acts on the ball: F_obs = F_grav.
* You, however, will see the ball being acted upon by another force,
since the
ball (and you...) will be accelerated towards the side of the
container:
F_obs = F_grav + F_fict
This force is entirely due to the truck accelerating iin the
opposite direction,
towards the inside of the bend in the road, and we call it a fictive
force.
Fictive forces are trivial (but often cumbersome) to derive as the
opposite of
the acceleration of your (non-interial) reference frame.
Regards,
Regner
philip madsen wrote:
re Alan and Regners figures.
On this business of "feeling" acceleration, whilst I do not pretend
to having had enough interest in checking the figures, I still reason
that its a matter of how forces are applied, as to whether you feel
anything.
In a suddenly braking car you get flung forward... because the force
is at the wheels.. But if the breaking force was applied to every
molecule of the vehicle including you, then I concieve no effect to
be "felt"
If I take the orbiting space station as an example, the people inside
and even ouside are all exposed to the same accelerating forces..
They follow the orbit of the vehicle.. when the man steps outside,
he does not get flung off on a merry plunge towards the sun or the
earth for that matter. He would not "feel" any movement. Yet he is
circling the earth every few hours. Thats travelling a fast corner.
Philip.
----- Original Message -----
*From:* Regner Trampedach <mailto:art@xxxxxxxxxxxxxx>
*To:* geocentrism@xxxxxxxxxxxxx <mailto:geocentrism@xxxxxxxxxxxxx>
*Sent:* Tuesday, January 22, 2008 4:59 PM
*Subject:* [geocentrism] Re: acceleration calcs
Alan,
Thanks for your calculation, but I'm afraid you made a mistake -
it's easy to do with all those
crazy units Americans juggle with. You forget that your
velocities are still per hour, while
you have the change of velocity per second, so your result is
actually:
(1) 11.43 cm/(hour*s) = 0.003175 cm/s2 (cm per
second squared)
The actual change is:
(2) (30.29e5 cm/s - 29.29e5 cm/s) / (year/2d0) =
0.006338 cm/s2
We agree on the velocities and the difference in velocity - I
just use centimeter-gram-second (cgs)
units. One year is 365.26 days * 24 hours/day * 3600 s/hour =
3.155693e7 s.
The change happens during half a year (I divide year by 2, in Eq.
[2]) so you would actually
have underestimated the change (as you can see from my
correction, Eq. [1]).
It is always a good idea to put your result in perspective by
comparing with another relevant
quantity - the gravitational acceleration at the surface of Earth
is about g=9.8 m/s2 on average,
which means that the acceleration along Earth's orbit is
(3) (0.006338 cm/s2) / (980 cm/s2) = 0.000006467
times the average gravitational acceleration at the surface of
Earth, g.
or conversely, the acceleration along Earth's orbit is 154600
times smaller than g. I don't
think you would notice that!
But that is obviously a tiny component of the accelerations
actually involved.
Remember that (in HC) the direction of the velocity has also
changed over the 6 months
and the velocities in the two instances will be exactly opposite.
We can get a rough estimate
of that acceleration by just adding the two velocities in Eq.
(2), since a-(-b) = a+b, to get:
(4) (30.29e5 cm/s + 29.29e5 cm/s) / (year/2d0) =
0.3776 cm/s2
which is then 2595 times smaller than g. Absolutely measurable,
but it wouldn't exactly
knock you over.
Regner
allendaves@xxxxxxxxxxxxxx wrote:
Try to move _*4.5 inches within one sec*_ without feeling/
/(being abel to detect that using current technology)/ it. This
demonstrates the crux of the problem with earths inertial
motion. Appealing to some imaginary reason why you could not
detect it in the earth but you could with anything and
everything else is not going to work untill you can _*first*_/
/prove that your imaginary reason exist in reality. NO one
isarguing it could be, but if we are to arive at a conclusion
and proclaim it logical we have to prove the variables along the
way not make them up as we go along. that is the fundimental
difference between GC & HC. GC accepts as proof only the
effidence presented as it goes along through the discovery
process.....HC makes it up as it goes along to save it's
conclusions.....
----- Original Message ----
From: Allen Daves <allendaves@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Monday, January 21, 2008 8:38:51 AM
Subject: [geocentrism] Re: acceleration calcs
Here it is ..quick & rough....
18.5 miles per second *_average speed_* * 60 sec for min * 60
min for MPH = 66600MPH or
The ~ avg change over the course of a year is 3.4%* 66600=
2264.4 / 365.4 days=6.2038 MPH per day /24 hours = .25840166
MPH change per hour /60min=.00430819444 MPH change per min
……There are 5,280 *feet* *in* *a* *mile* .0043081944 MPH =
22.747266432 feet /( or 6.933366807864/
/meters)/ per min /60 to convert to seconds = .3791 feet per
sec/ per second change /( or .11554968 meters per sec per sec)./
_*This is a change in velocity of ~4.5 inches per sec/ per
sec*_ Or _*11.43 centimeters per sec per second*_
There is now way to consider this amount to be inertial change
negligible. The effect rate of change regardless of how fast the
earth is supposed to be traveling because only the rate of any
change from the effective inertail 0 is measured.
This means that the velocity change of the earth going around
the sun is not just moving 4.5 inches ever second but changing
by 4.5 inches per sec. During the earths closest approach to the
sun /(such as traveling in a moving car)/if we experience 0
velocity change because we are traveling with the earth then
whatever the current velocity is would be felt as 0. However,
the rate of change just as in a moving vehicle would be changed
if we "give it some gas" and in this case the rate of change
would be a increase ~4.5 inches every sec every sec. This is to
say we on second one we increase by 4.5 sec on second two we
have increased to 9 by second three we have increased to
13.5….the rate makes for a exponential distance traveled curve.
In any case this is the rate of change. Assume for the sake of
argument that your body could not detect that change rate,
current instrumentation however ( acelerometers) are able to
detect that amount of inertial change to almost infinite
amounts, and they are not "aetheraly" depemdent).
*/Neville Jones <njones@xxxxxxxxx>/* wrote:
Not me.
-----Original Message-----
*From:* allendaves@xxxxxxxxxxxxxx
*Sent:* Fri, 18 Jan 2008 07:55:05 -0800 (PST)
Since the earth changes its speed throught it's orbit,
has anyone out there ever calculated the actual
acceleration force changes to the earth as it moves back
and fourth through its apogee and perigee elliptical
orbit around the sun?
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