Dark energy (appendice 3)

The nex web
address is for the first part of this paper:

The second part
(annex 1) is for show that énergy from the zone of contraction(clusters of
galaxies) is enough for explain the Univers expansion accélération,the web page
is:

The
third part(annex 2) of this paper ,show a total accélération expansion that
include accélération from dark énergy ans accélération gravity,the
web page is :

Before begin,i have found a french paper that show that the current
accélération expansion begin there is about 6 billions years et before it

was a décélération expansion, i am going try to
give it to you, first do the first next web page and after go on the forum
message on the top ;

In this part ,we compare constant find ( in the preceding part) with the
Hubble constant and we wil studied more ma version.

Bégin
with a équation that allow to get the Hubble constant :

Be Vmoy. the avérage speed, H the Hubble constant,
R the lenght that go from the avérage contraction zone center,with this we can
get

the équation that allow to get the Hubble
constant:

Vmoy.=
HR Vmoy. = R/T = HR H =
1/T T is the nécessary time for a avérage expansion
speed Vmoy. for get a lenght R,

This time T is a estimation of the Universe
age. With use of accélération from dark énergy(dea),we get
:

(1/2)(dea)T = HR = R/T =
Vmoy. dea =
(constant)R

(dea) = 2R//T^2 = (constant)R

2/T^2 = (constant)

2(H)^2 = (constant) this 2
équations give us the 2 nexts équation:

[(4/3)(pi)G]d = (constant)

GM/(Rcritique)^3 = (constant)

d is the libération speed for a avérage contraction
zone. For this 2 équations , i have consider the next 2 équations :

T = [{3/2(pi)G}^(1/2)](1/d)^(1/2)

T is the nécessary time for do R and if we consider
the avérage speed (Vmoy.) is équal to (1/2)VL ou VL is the libération speed,then
this time

come from the équarion : (1/2)VL = Vmoy. =
R/T

T = R/(Vmoy.) = R/[(1/2)VL]

VL = [2GM/R]^(1/2)

We have find that: 2/(T)^2 = (constant) =
2(H)^2 this
give :

[(4/3)(pi)G]d = (constant) =
2(H)^2 this give :

GM/(Rctitical)^3 = (constant)

Try a exemple for a estimatate M :

Take a Hubble constant équal to (75 km/s) per
millions parsec, as there is (3.086)(10)^19 km in a millions parsecs,then
:

H = 75 km/s per (3.086)(10)^19 km , la
corresponding value give :

(2.043)(10)^(-18) per second ou T =
(4.11)(10)^11 second (or 13 billions years).

With our 2 following équations, we are ready
to estimate M and d :

[(4/3)(pi)G]d = (constant) = 2(H)^2 =
2(1/T)^2

GM/(Rcritical)^3 = (constant) = 2(H)^2 =
2(1/T)^2

If the Universe is 30 billions light years lenght
for its Ray, with 100 billions of galaxy as Milky Way, with a mass (10)^13 solar
mass ,and that

Rcritical for a avérage contraction zone is
estimated to (1/2) billions lignt years, this we give :

[30/(.5)]^3 = 216 000 (216 thousands
contraction zone) and 100 billions divided by 216 thousands give about
:

463 thousands galaxy (for a avérage contraction
zone) and see if that is the same as M :

For M (included in a avérage contraction zone) : M
= (1.64)(10)^49 KG = 824 535 galaxy mass as our, or (824.535)thousands mass as
Milky Way,

this is (1.78) time the number that we have
estimated witout our constant, this is a little leass that the
first estimated value, and that give a good look

of the error of my estimations, this can not
cancel my version of dark énergy and a wished estimated this version with the
informations that we have.

We get the densitée d =
(4.23)(10)^(-26) kg/(m)^3

I hope that
i have not do too much errors, what i wish with this last calculus was to
compare (constant) with Hubble constant and give

a estimation for M ,
Rcritical , and the dnsity d.