[Physics] Shadow Gravity and Spiral Galaxies

Arend Lammertink lamare at gmail.com
Thu Apr 30 08:06:06 CEST 2020


Hi all,

Paul sent me some interesting (draft) stuff about shadow gravity and
spiral galaxies in august/september last year:

-:-
I am including below the progress I've made in the recent past.  First
I've found an expression the given us the gravitational constant.

G = ([4π]c/q)(2a^2c/A)^2 = 6.665E-11         m^3/kg-sec^2
Where a = Sqrt(3)8pi^2
      A = Avogadro Number  = 6.02214129E+23
      q = Elemental Charge = 1.60403E-19    kg/sec
      c = Light Speed      = 2.9979E+08      m/sec
G = h'c(8π/mₑN')² = 6.673274E-11  Verses CODATA Concise form  6.67384E-11
h' = h-bar          (1.05457E-34)
c = Light Speed     (2.99792E+08)
mₑ= Electron Mass   (9.10938E-31)
N'= Avrogadro Prime (5.97864E+23) = 6.02214E+23((1.66054 / 1.67262)

Not sure how it all fits in yet but the Coulomb constant can also be
expressed in terms of h'c so yes, gravity is a second order effect of
EM.  I also realized that Shadow Gravity must weaken as it passes
through multiple bodies.  I am in the early stages of writing this up
an am attaching the very first draft writeup and would be interested
in your take & comments.
-:-

-:-
I am including an analysis of the cumulative shadowing effects of a
flat spiral galaxy and what the Shadow (LeSagian) Gravity orbital
dynamics process produces as a result.  I will also include a graphic
plot of actual observations.
-:-

Appears to be helpful in your discussion.

All the best,

Arend.



On Wed, Apr 29, 2020 at 3:17 PM Tom Hollings <carmam at tiscali.co.uk> wrote:
>
> Got it that time Mike, thanks. The slow rotation rate makes it difficult to estimate the speed in any case, but point taken.
> Tom.
>
>
> > On 29 April 2020 at 12:54 mikelawr at freenetname.co.uk wrote:
> >
> >
> > Tom
> >
> > I said the rotational rates would not be dfferent. What will be
> > different is the actual velocities of the arms because you need to
> > separate out the viscosity red shift of the centre and arms, which will
> > be the same because they are the same distance from us, from the local
> > reference frame where the centre and arms will be different.
> > Unfortunately we don't know what the value of any viscosity red shift is
> > yet. Because the galaxy as a whole may be closer to us, then the
> > distance from centre to arms will be different to that which we
> > currently think it is. So with a different radius, but same rotational
> > frequency, then the estimaed velocity of the arms will be lower.
> > Cheers
> > Mike
>
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