neuvophysics.com
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Newton's hack 3 continuation
Derivation of the constant 'z'
So now we can see that the Nmetric
equation concurs completely with the centripetal force (CF) equation while big
G is slightly erroneous; which is why the Nmetric equation is exactly accurate
for altitudes beyond around 42,000kms into space. Wherever the Nm equation can
be defined for use; it is accurate. In fact if we hadn't begun with the wrong
weight in the first place, it would be accurate everywhere.
Note: the true reason big G is
in error is because it is following it's own different inverse square curve
from an erroneous starting point so it's deviation from reality will be
conditional but sure.
Also note that both the CF
equation and big G are happily ignorant of reality for lower altitudes because
they both fail to take the Nmetric anomaly into account.
(b)
Earth surface
values of Fg for 1kg at an
orbital radius of 6378100m (meaning on the surface of the earth)
0.0m

using Big G
Fg= 6.67384e11 x 5.9736e24/4.068e13
=
9.8001N
With the new G we calculate
Fg= 6.5802e11 x
5.9736e24/4.068e13
=
9.6626N
because we are looking for the
real SEP violation which is masked by the Newtonian big G
this will be used to convert to
Newtons in the equations which follow.
(c) Nm
Fg=9.9757e11 x 0.9629* x 38.4784
x 9.80665/3.69609e13
4.068e13 
rt
(9.95162e23
/
0.00729)/ 3.1416 (0.00729 is
the fine constant in earth gravity read the end of this tab to discover why
the fine constant is qualified to be used here)
4.068e13 
3.71906e12=2.8996e13
=9.806628N
*corrected mass
N.B. The following equivalence
modifier results are NOT CHANGING THE GRAVITATIONAL FORCE OR WEIGHT; they are
computing the inertialgravitational inertialequivalence deviation. This is
because the original mass and weight values were derived by comparisons made on
the surface of the earth instead of in space.
The weight remains as previously
calculated but now we have a gravitational inertial force deviation from the
classical physics inertial force relative to the same conditions in the
sideways direction (space case). If you contemplate the relative inertial
(retroactive) force at ten meters you can see why Newton observed an
acceleration rate of only 9.81m/s/s.
Einstein's GR is now in the
toaster!
Here are the SEP violation
calculation results for plotting:
Using
F_{g}= (
rt
m_{b}) m_{s}.4z
/
d^{2}

t
(
m_{b}x 2z/
rt
h)
(assuming the small mass is unity
kg) This is a truncated version of
F_{g}= (
rt
m_{b})m_{s}.4z/(
d^{2} (
(
rt
m_{b})
x
4pi
/
rt
h) x 1 + 2
pi
h/m) This addition tests for WEP
violation and its omission should have no affect.
begin with
the weight (gforce) on the surface of the earth
Fg= 9.97578e11 x 9.80665 / 9.9758e11
=9.80665N
Only kidding! That's all just
because the radius is too small to have any affect.
So because there is effectively
zero height; the object is considered to be part of the Earth; so we use the
equation for depth of zero at (y) from the following section and we get that
same result.
Next we calculate the actual
gravitational force at 1m without any apacenter adjustment in order to discover
the apacenter caused SEP violation. (Note: these results are very slightly
incorrect because I was winging it at the time of calculation and I hadn't yet
discovered the true derivation of 'z' 4z should actually be 4pi^{2}
because 'z'=pi^{2})
See below
F_{g}= (
rt
m_{b}) m_{s}.4z
/
d^{2}

t
(
m_{b}x 2z/
rt
h)
Fg= 9.9757e11 x
1.03373
x 38.6504 x 9.6626* / 4.06851e13
4.068e13 +
rt
(9.95162e23/1)/
19.3252
4.068e13 + 5.16205e9=4.06851e13
=9.4659
*Newton conversion from the new
big G
10m
Fg=9.9757e11 x 1.07215 x 38.6504
x 9.6626/4.06853e13
4.06802e13 +
rt
(9.95162e23/100) /
19.3252
4.06802e13+5.16205e9=4.06853e13
=9.8176
50m
Fg=9.9757e11 x 1.07215 x 38.6504
x 9.6626/4.06817e13
=4.06807e13 +
rt
(9.95162e23/2500)/
19.3252
4.06807e13 +1.03241e9=4.06817e13
=
9.8185
100m
Fg=9.9757e11 x 1.07215 x 38.6504
x 9.6626/4.06805e13
=4.068e13 +
rt
(9.95162e23/10000) /
19.3252
4.068e13 + 5.16205e8=4.06805e13
=
9.8188
500m
Gth Fg=9.9757e11 x 1.07215 x 38.6504x
9.6626/4.06861e13
4.0686e13 +
rt
(9.95162e23/250000)/
19.3252
4.0686e13 + 103241163=4.06861e13
9.81749 (no significant change)
Now for 1kg at
1000m
Gth Fg=9.9757e11 x 1.07215 x
38.6504 x 9.6626/4.0712e13
F_{g}=9.8112 (The
modifier function ceases to have any affect beyond this altitude so we can
ignore it for the following calculations and I will simply give calculated
values which I have added to the list below.)
0m
Fg=9.80653N
(fudged but fair
invalid for use) (
9.80665 is calculated by utilizing the larger
underground equation, because it only cares about the size of the object at the
center of the Earth and visa versa for the altitude equation.
1m Fg=9.81774
10m Fg=9.8176
50m Fg=9.8185
100m Fg=9.8188
500m Fg=9.81749
1000m Fg=9.8112
5000m Fg=9.8035
10,000m Fg=9.7882 (big G 9.7696)
Still SEP violation here at 10km
20,000m Fg=9.7576
50,000m Fg=9.6667
100,000m Fg=9.5181
(G 9.4998)
SEP violation beginning to
vanish.
1,000,000m Fg=7.3376 (G 7.3235)
effectively
Nil SEP violation
10,000,000m Fg=1.489 (G 1.4862)
ditto
41,894,987m Fg=0.22757 geostationary
orbit altitude. There is no SEP violation at this altitude so
this answer
compares exactly with the CF formula and only almost with big G.
100,000,000m Fg=0.0329
Graph SEP violation (sorry no
graph on website)
I also carried out computations
for 1000kg at the same altitudes and the results are exactly the same per kg. So
there is
no
(zip zero)
WEP violation. However the graphs above
depicts the strong equivalence energy deviation blip.
Plotting Fg values inside the
Earth.
The
relationships in the modifier are now changed to suit the completely reversed
conditions inside the earth. There is still a SEP violation and you will
notices how the curve transitions smoothly through the Earth's surface without
any weird or sharp curve deviations.
(y)
F_{g}= (
rt
m_{b})/(r.h/2
pi
)
+((
rt
m_{b})
+
(h/2)^{2 }
rt
(
m_{b}x
2z/
rt
h)
r=63771000m. At rest on earth
surface.
Fg= 9.97578e11 x 9.80665 /
9.9758e11
6378090 +
9.97578e11
=9.97641e11
=9.80665N
r=6377100 1m
below earth surface
Fg= 9.97578e11 x 9.80665 /
9.9764e11
6378100 +
9.97578e11
=9.97641e11
9.8065N
10m
below earth surface
Fg= 9.97578e11 x 9.80665 /
9.98215e11
6378100 + NA
+
9.97578e11=
9.8065N
100m
below earth surface
Fg= 9.97578e11 x 9.80665 / 9.9821e11
637810000 + 98696 +
9.97578e11=
9.8004N
500m
below earth surface
9.782N
1000m
below earth surface
9.744N
378100/2pi 1000000 below
Fg= 9.97578e11 x 9.80665 /
1.63138e12
3.8381e11 +
2.5e11+
9.97578e11
=
5.999N
2000000 down
=2.881N
r=4378100/2pi x 6378100
(5000000/2)squared
m down
Fg= 9.97578e11 x 9.80665 /
2.4674e14
4.444e12+
24674e14 + 9.97578e11
=
0.0396N
Earth geocenter 6378100/
2pi x 6378100 + (6378100/2)^{2}
Fg= 9.97578e11 x 9.80665 /
6.337e16
6.4744e12 +
6.337e16 +
9.97578e11
0.0001N
geocenter
=0
table of theoretical gforces at
depth
0m=9.80665N
1m=9.8065N
10m=ditto
100m=9.8004N
1000m=9.744N
1000km=5.999N
2000km=2.881N
5000km=0.0396N
Earth center 6378.1km=0
Note: There can only be exactly
zero gravitational force if the small object is infinitely small but the
equation knows that the 1kg mass has a radius.
The SEP violation in space:
The 1kg is actually heavier by )
0.00373 from CF. So the spacecase mass will be 1+0.00373Kg =1.00373kg.
Also because of the apacenter
shift phenomenon the weight is actually more on earth than the mass in space
also.
So from (c) we derived a gforce
value of 9.4416N (0.94416kgF) from m=a/F we derive a weight value
9.80665/9.4416
=1.03866kg
So we need to add the two correctforspace
values and divide by two and we get
1.021195kg.
This is the actual space mass of
a 1kg earth weight. That shows a permanent SEP violation. We don't need to
adjust the kg to suit but we do need to recognize that the energy requirements
to move a 1kg object in space will be greater than expected but not by much.
The GTR geodesic can never account for or predict this! The GTR is incorrect
science.
*Note: Because of the
adjustments required when dropping weights of different density simultaneously
I have constrained the argument to the objects being of exactly similar
density. In the real world it is highly unlikely that 1000kg will ever be
experimentally compared against a 1kg object.
N.B.
I am not declaring that the
inverse square law of gravity is being violated; it's just that the energy
relationship between gravity and the energy expected from the same action in
space is violated (SEP). This is why the energy realized in a free fall to the
ground is noticeably greater than by E=2mv, and that has in part been
responsible for the E=mv or E=mv^{2} controversy. The SEP violation
reduces dramatically by inverse square law over the first few meters of free
fall to be completely removed beyond about ten thousand meters of altitude.
This concept needs to be
understood. It affects E=mv especially close to earths surface but not F=ma!
The energy is proportional to
the amount of gravity actually being used. This 'z' modification accounts for
the energy conservation violation conditionally implicit in the Newtonian and
Einsteinian equations. Consider this first and then I'll explain why 'z'
appears to be pi squared.
(2) How about a quick estimation
of the gravitational energy used by the earth every second.
with true mass
F_{g}= (
rt
m_{b}) 2z/4/3pi.r^{3}
=
9.97578e11 x 6.28318/1.08683e21
E=F.t
0.00000000057/J.kg.s
Note: I left this in reductive
terms of pi to try and show the relationships. If you can't see that then the
real denominator is actually the calculated volume of the earth.
The energy output of the sun
times the gravity difference between GD and GS (: z) is almost 10 million times
that of the earth. That might look like a vast energy for the earth to have to
dissipate but not all that much of the sun's energy reaches earth but even that
is greater than the infinitesimal energy loss per kg of gravity in the earth.
That's only a very insignificant value for each kg in the earth. The thermal
conductivity of the earth had better be greater than that or I'm in trouble
with the facts!
Note: Ten million times seems to be a little lower than intuition would
suggest; however the solar energy component of gravity is just the energy of
gravity which gets trapped inside the vast volume of the much less dense sun.
This helps keep the fusion going which is basically gobbling gravity, burning
hydrogen and turning it into helium and that accounts for the vast energy which
is emitted by the sun compared to earth which is (a) not fusing (b) much denser
and so has a better thermal conductivity coefficient.
WHY 'z' APPEARS TO BE PI
SQUARED.
Yes it does appear to be pi
related and here's why
Fundamental mass is the
reference frame independent measurement of force.
1eV in Gtheory is not energy;
it's a
force
and herein lies the proof. That
force is recognizable in the kinetic energy of one electron theoretically
travelling at velocity 'c'. Electrons, protons and neutrons are the known
standards of force and mass relationships, not by measuring lumps of lead being
dropped in clay. Not these days anyway. That stuff is so retro.
Moving on: That electron force
is proportionally related to the electron joule of energy, which is related to
the volt, the Joule, the kilogram and the second by one coulomb number of
electrons blah blah. One coulomb is 6.241509e18 electrons per second.
Now if an electron travelling at
'c' has mass 'm' and it bumps into a neutron which is bound to an almost
infinite number of other nuclei then the neutron would be considered to
absorb/stop the electron and the energy would be lost to the surrounds and that
resulting Joule of energy is vector/parity shared by every atom in the mass.
(assuming a theoretical closed system.)
If the neutron was free to move
then we could calculate that its resulting velocity would be proportional to
its mass divided by an electron mass. However we are dealing with atoms here
and not neutrons but we know that a perfect atom or 1H isotope still has the
mass of one neutron because it has unused binding energy or else very close.
We can derive 'z' if we find the
precursor value by multiplying a coulomb number of electrons producing the
energy of one Joule in one second multiplied by the speed of light at which
speed that coulomb will have that kinetic energy as the energy being 1J.kg.sec.
Then divide by the number of atoms that each coulomb is colliding with over one
second in the making of that Joule; being is related to a mass of perfect atoms
with just one neutron value of mass per atom and that is 1kg of hydrogen.
Note:
1kg is not an infinite number of atoms but we know what it is and its close
enough; but the real connection is that the Joule and the kg have been
fundamentally related. We still divide by the calculated quantity of protons in
a kg of hydrogen 1H and not infinity.
Z=coulomb x 'c'/n
z= 6.241509e18 x
3e8
/5.96e26=
3.1416991610738eJ/s
Does that look like pi to you?
Here's the real pi
3.141592653589 more unbelievable? That's the 'z1'
energy loss constant relationship. This has to be one of the most
remarkable coincidences in physics. Or is it a coincidence? Looks like just one
more fingerprint of God to me. Feynman's week might not be far away! Atheism is
starting to look so boring.
So precursor
Z1=pi
This is a constant at the macro
level which is relatable to Planck's constant at the quantum level. How? Because
the gravitons trace a wave and the wave has a wavelength and a frequency as
well as an amplitude.
If you look at a sine wave and
consider the radius of the circle that prescribes the sine wave to be one
quantum in a full photon or in this case graviton then Pi is the relationship
between the wavelength and the amplitude and we also have the relationship
between a quantum of sub photon bosons by E=hf. So the quantum of bosons is
energy proportional to frequency.
It appears obvious that the
amplitude of a particle stream is fully relatable to the frequency as well as
the quantum energy. The frequency is determined at emission but the amplitude
takes over with ENERGY LOSSES. Code: Sorry no symbols on this cruddy website
Aamplitude Wwavelength. The rest you
know.
In other words; after emission
A:f:E so by reason A:W:E. In the last paragraph we noted that Pi was relatable
to A and W so by deduction pi is also totally relatable to particle stream
energy; and in our case gravitational energy.
None of this is big deal really.
If you go through Gauss, Coulomb, Ampere and Maxwell's law equations you will
notice that pi is involved because pi is relatable to energy at the spherical
quantum field. Therefore it doesn't take much intelligence to see how pi is
related to the coulomb and PEP and ohms law and it becomes a circular
interrelationship. The reason pi becomes derived in the kilogram is because
the cross sectional area of a conductor is proportional to the ampere etc and
that the kilogram is directly related to the Joule, which is related to the
coulomb, which is related to the quantum, which is related to Planck's
constant, which is related to the energy and we now have 'z1'. It's still
pretty amazing that it just came out to be pi like that.
Apart from the units such as the
ampere which are defined constants, we know we have the fine constant which by
the way is only a constant in the same gravitational field which is pretty
upsetting for atomic clocks and derived further from that we now have the
gravitational energy constant 'pi^{2}'. Squared; because the gravity
equation demands it for reasons to be forthcoming. This is likely the
relationship between the speed of light and the speed of gravity because
Planck's constant is really to do with light. I must admit to being at a bit of
a loss to explain this; but it works. We might understand later.
This brings me to a point of
conjecture regarding Maxwell who studied the work of Weber and Kohirausch not
long after the speed of light had been determined to be 3.14e8m/s and was most
likely still the talk of 'physicstown'. Maxwell's assumption was that this
must mean that the proportions of 'u_{o}
e_{o}' measured by previous
physicists had to be close to the new speed of light because he had already
pre decided that emr propagated at the speed of light so he added two and
two to make three and came up with the idea that emr fields build at velocity
'c' also but even worse that charge could be applied to a plate with
instantaneity and that's not true. This was in spite of the fact that the
proportioning wasn't velocity related at all but force/energy
related. I contend that they had in a clumsy way (by measuring the emptying of
Compare with the similar tags below:
E1=6.2833983221476 x 9.8696 for a coulomb of electrons per atom
bombarding 1kg of hydrogen for one second compared with the 'vg' of 9.81m/s.
E=mv
E2 supposedly used=9.81J/kg
E3 =9.81417 from 4/ add in the energy loss and deviation
components from 3a/ 3b/ and 3c/ below=9.8341 (pi sq=9.8696). you can see from
the following that there are some other losses and reasons for the incorrect
weight to be assumed, but the general and I hope valid assumption is that the energy
constant is related to the energy of a joule by the square of pi. This means
that all joules (mechanical) are anomalous and the Joule should be exactly
relatable to pi squared which has been done in the fundamental derivatives of
electrodynamics laws and we should now be able to see why this isn't the case
with the mechanical Joule.
The use of the energy constant in my calculations is therefore
valid and accurate and if the weight had been true the Nmetric equation would
be true for all gravitational calculations and there would be no SEP violation
but a variant inverse square law curve for gravity per individual case which
the equation would declare.
From (2)
(3) Gravitational energy used for the earth is
0.000000018J/kg/sec.
that is
(E1) That is
the energy lost by gravity for every kilogram in the earth per second.
(E2) 3.33% energy violation per kg due to the equatorial CF
problemâ€¦ is 1.5e11 which isn't much at all and for most intents and purposes
we can ignore it. But we are left with actual energy used/lost by gravity and
the CF error
3a/ by E1+E2=1.8e8J/kg/s.
Now this electron beam induced velocity and energy has nothing to
do with our gravitational motion case except that E=2mv as well because there
is a constant flux of force acting in both cases. However the energy constant
does in every case. In both cases we have different energies though. Reason;
the forces were not acting similarly.
From (1) above we found that the original kilogram
weight
was less than 1kg by 0.03373 so the extra energy attributed to 1kg
mass
has really been the energy used by moving 1kg0.03373kg over one meter. Our kg
of mass is now 0.96627kg. If we subtract that from the
3b/ measured 'g' (for proportionality) we get
0.014395
1.4395% which is the proportional energy violation component of the
gravitational
component of energy 1J only because the
inertial energy of 0.96627kg
has been fully accounted for in the derivation of the joule. From the SEP
energy violation section above; we found that the total energy violation in
space was
2.1195%. however from the graph you can see that the Fg
violation is a whopping 4% between 1000 and 10000 meters. That's airline
territory*. With reference to inertial energy it takes more energy to move an
object sideways in space.
*Unfortunately
this won't help them much because this SEP violation is not a reality changer.
That would be ridiculous. But it does explain one reason why it takes less
energy to fly at altitude. By my calculations the most efficient range is from
six to ten thousand meters, after which the benefit begins to erode. This curve
can be utilized with a drag curve to derive this which is what I have done. It
is interesting and profound for model support that airlines discovered this
experimentally.
So our inertial kg problem has been taken care of but on top of
that we note that half of that
remaining gravitational energy was lost
in the lift and half in the fall so if we only derive results from the fall we
have a loss during that action which will again have to be divided by 2 to
account for the half lost during the lift
0.014395/2=0.00719 so this relates to one E=mv of the fall
E=0.00719 x 6.2833983221476 =0.045
E loss=0.045J
E loss remaining in 1kgF 0.045 0.03373/2
=0.0056
3c/
=0.0056 (extra
gravitational energy loss
component per J during the free fall)
At (3) I calculated 1.5e11J/s energy loss but that was the
overall energy loss of the total gravitational energy in the earth per
second and 0.0056J is the energy loss component of the free fall.
There is no relationship but energy is always being lost.
Newton must have got his measurements and calcs wrong
because of equatorial oblateness due to centrifugal force etc. I've just used
an average earth radius because I'm not setting out to calculate everything. So
there are other phenomenon which can cause 'g' to deviate from the expected
value of 9.83135N to be the experimentally measured 9.80665N which is variable
but 0.2512% deviation at earth surface due to the weight anomaly and if we add in
the centrifugal anomaly of 0.373%. As you can see from the graph the deviation
becomes greater at certain altitudes.
There are a several points to make here: 1/ the acceleration rate
is less from any given height above the earth. 2/ The earth actually falls
towards a falling object so the distance of the fall is shorter. 3/ The gravity
change with height is greater than ever imagined close to the surface of the
earth. I.e. actual Gforce change
9.4416 to 9.4659 over just the first
meter. That's an increase of 0.0243N in one meter by inverse square law
which keeps that difference for about 0.707 of the distance which will give an
increase in the observed acceleration rate over one meter of 0.017m/s/s
The first problem will cause the terminal velocity to be
less. The
second will cause the terminal velocity to also be
less. So a
measurement of 9.80665m/s/s is less that it would be if all were as we suspect.
But they're not! This is why I have suggested elsewhere that the acceleration
rate should be termed in meters per second per meter and not meters per second
per second.
It would get too involved to go any further than this, the
accepted acceleration rate for 'g' will suffice.
Energy constant value I used originally 9.80655 for plotting
purposes 9.6626
4/ Energy constant
'4z'
calculated by utilizing
'z'=pi^{2}

9.81417â€¦ Pi squared=9.8696
This shows that Newton's measurements were out by 1.057% from
the predicted value. That's close enough for most purposes but we need to
remove his error to find the true percentage of loss. We can do that by
utilizing the precise gravitational equation which features Pi even though the volumes or radii of the
bodies/objects are not evaluated by represented by their mass. The equation
doesn't assume equal densities.
F_{g}= (
rt
m_{b})
4pi^{2}
.m_{s}/
d^{2 }
+
rt
(
m_{b} / m_{s}
x d_{s}^{2})
/
2pi^{2}
I'm not going back to adjust everything now because that's not the
purpose of the exercise but you can!
There is actual concurrence with the inverse square law under
Gtheory gravity in the earth's gravitational field at different altitudes;
however the SEP is violated. This was not appreciated because the mass was
weighed on the surface of a spinning globe.
Tests have proven that the inverse square law is accurate at less
than one millimetre from the earth surface. The Gtheory metric needs to concur
with this.
*^Fundamental note 2: I'm going to forestall the argument
regarding the earth being observed from an external stationary reference frame
and the idea that the falling object is actually falling over a longer curvy
path.
That's purely observational relativity. In that case the actual
force (if we could observe it) would be notionally following that curve as well
so it all becomes a zero sum game and we are able to consider the whole
argument from the reference frame of the surface of the earth and a vertically
falling object. Yes you can probably derive a mathematical relationship to
describe that warp in the geodesic. However it must be pointed out that the
math still would not be able to cause the force we call gravity.
Please note that I am not calling for a change. The force remains
as it is because it 'thinks' the mass is one kilogram. For all earth bound
applications that will suffice.
Toast anyone?
PS you can plot these results
yourself and you will find a very small but significant SEP deviation beginning
from the surface to around a couple of thousan
d meters from where it comes back into
close agreement with the accepted (but wrong) inverse square curve before three
or four thousand meters. In reality the SEP violation is strongest on the
surface but Newton's errors and fudge have covered that up. Once again it is
very important to note that there is no WEP violation whatsoever at any
altitude. That baby is a law! the feather and the hammer fall together! It's a
shame about the SEP!.. Oh no Einstein's toast is burnt!
ASSERTATION CONCLUSION
The gravitational law is not an
inverse square law after all it becomes modified close to bodies and especially
when large bodies are in close proximity (e.g. binary pulsars). Big G is a
perfect inverse square law equation which fails to predict
such anomalous interactions. The featured equation does, because as
the masses become greater in comparison to distance the softening of the square
root
of mass
function gradually
bites
deeper into the inverse distance squared function and this relates to reality
whereby the law is closer to a 1/r function. For further support of this
assertation visit
http://metaresearch.org/cosmology/PhysicsHasItsPrinciples.asp
PS. If you wonder if G is a
useful number which should have been able to be derived somehow Are you
ready?
The approximate
speed of light divided by the approximate speed of gravity (derived in the
thesis) comes to 6.7e11. Look familiar! Now you know how to find out what
the speed of gravity is. You just can't make this stuff up! I would suggest
that the mathematical patterns that are just dropping out of Gtheory
should give reason for pause!
neuvophysics.com