[Physics] Aether theory discussion

[Ruud Loeffen]

Hello Tom.

On Thu, Dec 22, 2016 at 4:26 AM, carmam at tiscali.co.uk <carmam at tiscali.co.uk>
wrote:

> Ruud, I have now read your web page http://www.shmoop.com/forces-m
> otion/gravity-orbital.html
> <https://webmail.tiscali.co.uk/cp/ps/Mail/ExternalURLProxy?d=tiscali.co.uk&u=carmam&url=http://www.shmoop.com/forces-motion/gravity-orbital.html&urlHash=3.294858017304283E291>
>   (speed read only, a more leisurely read will follow), and find that I
> have to disagree with two parts.
>

This is not my website.  Sorry I was not clear about that.

> This is the first :-
>
> “Since we’re more familiar with weight in pounds than kilograms, we can
> use the Earth-bound conversion of 1 kg in 2.2 lb to say that this same 60
> kg person weighs 132 pounds on Earth, but this conversion doesn’t work for
> Mars: pounds aren’t the same over there.”
>
> This is not quite correct. To be sure, pounds are not the same on Mars as
> they are here, and kilograms are not the same on Mars as they are here, but
> the ratio is the same so the conversion does work on Mars the same as on
> Earth.
>

In general: I would emphasize to STICK TO THE MKS system. Perhaps you want
to share your comments with the webowner of shmoop.com?


> Then there is this statement :-
>
>  “And what’s the orbital velocity required for orbiting planet Earth? To
> answer that, we have to choose a distance from earth for the orbit. The
> closer the distance between the two, the faster the necessary velocity to
> stay in orbit. This is precisely why inner planets have shorter “years”
> around the Sun than the outer planets. Nor can the velocity change for a
> given radius: by changing the velocity, the radius of orbit changes too.”
>
> This statement is in error, but only very slightly. However, an error is
> an error. If the orbits are calculated using only the mass of the Sun and
> not the mass of the planet, then your statement that the orbital velocity
> cannot change for a given radius is correct but only as an approximation.
>

I think you are right about this. But because it would make but a very
small difference I still count on "v^2 times Distance-sun) is a constant in
the solarsystem. So if there is a "11278 m/s point" it would be on a
Distance of 8,805933E+11 meter from the sun. The everage Distance from
Jupiter to the sun is: 7,784271E+11 meter.

The calculations with v^2/c^2 velocity 12,278 km/s are related to the
Distance to the sun (the sun as reference system). So v^2/c^2 resulting in:
*1,67737912E-09 * is *in reference* to the Sun. I wonder if anybody here
around can comment on this question: How can this be related to kappa: *Kappa
times c^2   =**1,67737912E-09 **m^3/kg.s^2  *
*If the Lorentz Transformation of mass-energy can be applied here than it
seems that the velocity 12278 m/s is valuable for the visible part of our
universe and that the velocities of our planets around the sun are
“regulated” by that velocity. Please have a look at this:*
kappa ϰ <https://en.wikipedia.org/wiki/Einstein's_constant>
1,8663361230000000000000E-26  m/kg
2 x gamma -1 / c^2 1,8663359768839500000000E-26  m/kg
kappa * c2 [ m3/kg.s2 ] 1,6773791244872900000000E-09  m3/kg.s2
v^2/c^2 = 1,6773791244873000000000E-09  m2/s2
gamma -1  = 8,3868956224364400000000E-10  m3/kg.s2
0,5 * v^2/c^2 = 8,3868956224365000000000E-10  m2/s2

v^2 = 12278^2 1,507553174837990E+08
v^2 = ϰ *c^4 (kappa x lightspeed ^4) 1,507553292864810E+08
v = root (kappa x c^4) 1,227824618121340E+04

The velocity "v" is all based on the "velocity" of 12278 m/s. If its not a
coincidence what is the real meaning of this magnitude? Or is it a loop
calculation?

If you want to use my spreadsheet about these magnitudes and units have a
look at my dropbox Excel spreadsheet:
https://www.dropbox.com/s/xz8y66ha6qs8w12/calculations%20near%20kappa.xlsx?dl=0
and this nice video about the earth moving through our galaxy (just the
last few minutes of that video)
https://www.youtube.com/watch?v=IJhgZBn-LHg&feature=youtu.be&t=19m13s

Finally for Tom: the acceleration is from 0 - 3000 km/s. So the mass
increase stops if the acceleration stops and the velocity stays on 3000
km/s. For the man in the rocket there is no invrease: he is pasrt of the
mass. An independent observer would observe the mass increase as long as
the acceleration is going on (m/s^2) and also stops if the velocity is
stable (m/s). IMHO.

Time to sit down and enjoy some holidays. I am going on a round-trip. Be
back in January. Happy New Year to all of you.

Best regards.

Ruud Loeffen
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