[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
*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....*


* 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.
----- 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.



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

     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
    (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.


    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
----- 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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