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Newton's hack 3--- continuation

 

Derivation of the constant 'z'

 

So now we can see that the N-metric equation concurs completely with the centripetal force (CF) equation while big G is slightly erroneous; which is why the N-metric 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 N-metric 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= 6.67384e-11 x 5.9736e24/4.068e13

= 9.8001N

 

With the new G we calculate---

Fg= 6.5802e-11 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 inertial-gravitational inertial-equivalence 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---  

Fg= ( rt mb) ms.4z / d2 - t ( mbx 2z/ rt h) (assuming the small mass is unity kg)    This is a truncated version of---

Fg= ( rt mb)ms.4z/( d2- ( ( rt mb) 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 (g-force) 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 4pi2 because 'z'=pi2) See below

 

Fg= ( rt mb) ms.4z / d2 - t ( mbx 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---

G-th 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---

G-th Fg=9.9757e11 x 1.07215 x 38.6504 x 9.6626/4.0712e13

 

Fg=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)--- Fg= ( rt mb)/(r.h/2 pi ) +(( rt mb) + (h/2)- rt ( mbx 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 g-forces 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 space-case 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 g-force 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 correct-for-space 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=mv2 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

Fg= ( rt mb) 2z/4/3pi.r3

= 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 G-theory 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- A-amplitude   W-wavelength. 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 inter-relationship. 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 'pi2'. 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 'physics-town'. Maxwell's assumption was that this must mean that the proportions of 'uo eo' 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 Leyden jars by their unseen radiation of electromagnetic fields into the vacuum) found the energy constant 'z1' or 3.14--- and nothing else at all. Just a thought. We now know that the speed of light is even less than 3e8m/s. Refer to the other profound errors tab re Maxwell. *^  

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 N-metric 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 problem… is 1.5e-11 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=1.8e-8J/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 1kg-0.03373kg 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 1.5e-11J/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 G-force 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'=pi2 ---    9.81417…  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.

 

Fg= ( rt mb) 4pi2 .ms/ d2 + rt ( mb / ms x ds2) / 2pi2

 

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 G-theory 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 G-theory 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--- 6.7e-11. 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 G-theory should give reason for pause!

 

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