neuvophysics.com

 

  NOTE TO ALL THE NAYSAYERS. I wouldn't have done all this work if the formula wasn't provable by mathematically accurate results!!

 

Newton's hack 2... continuation

prerequisite study

 

PROOF THAT THE G-THEORY Nm FORMULA IS MORE ACCURATE THAN THE NEWTONIAN GRAVITY EQUATION

 

 

1/ Universal N-metric Gravitational Formula example calculations:

Fg= ( sqrt mb)4z*.ms/d2 (b... big s... small) Note *^ : The 'z' is the linear stress energy constant. -Refer to the previous tab-. I.e. momentum is not constant over vast amounts of time (by reason of gravity) and loss of velocity and energy is proportional to velocity. All the terms matter a great deal and the following form should be used. All results are correctly termed in kg and kgF. If you want Newtons   you'll have to convert.

I have also concluded that there is an apparent gravitational center in the Earth which is in reverse proportionality to height. In the beginning I estimated the first apacenter by calculus. It was thought to be at a radius of 610993 meters, so I tried that in the first calculation with less than accurate results so to be honest I have reverse engineered the true apacenter. However I have been able to derive an apacenter modifier to keep track of the changing radius. 

 

This is the complete N-metric gravity formula.  

*Fg= ( rt mb)4z.ms/d2+{}(modifier)

 

 

 Trial calculation 1

Fg --- 1kg on the ground--- assumed apacenter--- 6108993m

=9.9757e11x 39.478/4.068e13

=0.96809kgF (wrong.)

 

 

Actual apacenter radius vertically beneath. = 6255468m

Fg=9.9757e11 x 39.2262**/3.91308e13= 1kgF= 9.80665N

 

**given a slight recalculation to 4z later on.

 

Using the N-metric equation we find that it takes one million kg to see a 0.2kg reduction in the g-force at earth's surface so this formula might appear to show that the weak equivalence principle might only be a law for mere earthlings who play with insignificantly tiny feathers and hammers. Not really. The equation actually sees the density-radius-distance proportionality and knows that the object/body's N-metric solution and that if it has mass at the accepted density then it will also have the proportional 'r' and if we attempt to subjectively place the object center of gravity on the surface, naturally the result will be in error. Hint it's that little energy constant that provides the true baseline while big G which does the same thing is still in error. This will be convincingly demonstrated.

 

*Using this equation for a mass of 1kg at 10,000m alt.

Let's calculate the g-force on a 1kg object 10,000M high by Nm...

We get 0.99681... N-metric 9.77529N--- using big G... 9.76941

Lets try a 2000kg object at 10,000km...

*Fg =9.9757e11 x 2000 x 39.2262/2.6424e14 =296.17kg

2904.47N ... 'G' ( 2972.45N) 303.1kgF

 

A deviation between the two equations is beginning to show, this will be taken care of later. The apacenter has moved back towards the geocenter. The value of the force has now changed from being more than expected by G calculations for a 1kg object to being significantly less for a 2000kg object. This means that there is a value of mass for every altitude where 'G' will actually be correct.

So if the mass of a satellite changes then 'G' will be less accurate. No wonder the orbital calcs for close orbit satellites is problematic. You should notice that once we are calculating geostationary orbits the results concur to a much greater extent. However this will not be the case for much larger space stations etc. Of course this also means we need to check for WEP violation at altitude and with various masses! I did--- there's effectively none!

If we calculate the g-force on a one kg object at a vast distance then from 'G' we get... 6.7e-31N at 1e20m2... and from G-theory Nm we get 3.91e-8N so it appears that by the Nm calculation the g-force approaches infinity far less quickly than we would expect until over much more than mere millions of kilometers 'G' doesn't quite give the required result at all. How on earth are we hanging on to the sun then?

THERE IS ANOTHER MODIFICATION TO THE EQUATION WHICH TAKES INTO ACCOUNT THE OTHER DISCREPANCY REGARDING OBJECT-BODY SIZE DISPARITY WITH ALTITUDE BUT WHICH DOESN'T AFFECT THE 'WEP' BETWEEN TWO DISSIMILAR OBJECTS FALLING IN THE SAME GRAVITATIONAL FIELD.

I.e. the following modifying functor gets subtracted from the d2   in the denominator and it therefore only has effectiveness for close by and disparately sized objects such as you and the Earth for instance. Here it is... rt (ms/mb x ds) e.g... rt 9.95162e23 /1000 x 4.2e7. (Note: the 4e27 was calculated from that modifier formula)

For a start we'll check the prior example of a 1kg object at a distance of 10 billion meters; so I can just fudge this one and not add the radii in...

 

Utilizing the G-theory N metric equation

Nm-- Fg =9.9757e11 x 39.2262/1e20 = (1 divided by one heck of a lot! This denominator addition will not affect the equation. There is no change as we would suspect.)  

= 3.913e-7kgF

 

big G-- Fg =5.9736e24 x 6.67384e-11/1e20

= 4e-7kgF

 

Let's say two earth size objects were involved and in this case just 10km apart and I might add, in a great deal of trouble!.. ( ds is surface to surface distance). First we will calculate the force without the modifier...

 

**Fg =9.9757e11 x 9.95162e23 x 39.2262/1.2766e7

=3.05041e30kgF

 

Check that if we add the mass differential modifier to the denominator we get the same divisor and no derated g-force because the two masses are similar and they don't overshadow.

 

**Fg =9.9757e11 x 9.95162e23 x 39.2262/1.2766e7+ rt (1/10000)/ 39.2262 (0.01)

= 3.05041e30kgF Ditto... zero change

 

Now if we put the twin bodies out to ten million kms (10b m) apart such that once more the d2 is 1e20---

 

Nm- Fg =9.9757e11 x 9.95162e23 x 39.2262/1e20 +rt(1/1e200)/ 39.2262 (=0)

= 3.8941e17kgF --- 

 

big G-Fg =(5.9736e242 x 6.67384e-11)/ 1e20

=1.5893e29/1e20

=2.38148e19N--- / 9.80665 = 2.38148e20kg/F

 

Gravitational force is still there and significant at a distance of ten million kms. If you run through the calculation you should satisfy yourself that the mass-differential modifier has zero affect at that difference/distance. Note: This all appears to work back the front simply because it appears that the incorrect weight was initially determined on the surface of the earth.

 

#   Let's now calculate the g force at geostationary orbit height... 35786km    ( 42164000m-- o.r.) for a satellite of any mass, say 1000kg

First we'll use the centripetal force formula to find the g force per 1000kg and check that against our new equation.

 

F =m.r.4pi2/T2

=1000 x 41894987 x 39.4784/7464960000

=22.5915kgF     221.56N      0.22156N/kg

 

Utilizing the estimated 'z' of 39.2262

Nm-- Fg= ( rt mb) x ms x 4z*/d2

=9.9757e11 x 39.2262/1.7551e15

check with mod factor... 1.7551e15+ rt ( 9.95162e23/1000 x 35786000)/39.2262

1.7551e15+3.5786e10/39.2262

1.7551e15+912298412=1.7551e15 (after checking for WEP/SEP... no change)

Fg= 22.3kgF    218.64N    0.21864N/kg

 

Nm equation with the newly derived 'z'* and 1000kg...

9.9757e11 x 1000 x 39.47842/1.7551e15

for 1000kg it is 22.438kgF     220.05N compared to 221.56N from the CF equation.) 0.22N/kg This is a statistical match between the Nm and CF equation results.

 

#* However--- by the big G equation...

Fg =G m1.m2/d2

=6.67384e-11 x 5.9736e24 x1000/1.7551e15

=227.1485N 0.2271/kg

(surprised?.. a much greater deviation. Big G isn't looking all that accurate compared to the new Nm equation)  

Comparison table... 

CF---           221.56N

Nm---                    220.05N

Big G---      227.15N Big G is not very accurate at this altitude while the CF and the Nm concur.

So it appears that something's wrong with big G... not Nm! and that's only at tens of thousands of km!

Note: Gauss' gravitational equations are also variant at different altitudes but they use big G as well, so only Newtons equation has been shown.

*Refer to fundamental note.

 

Let's now try another example of a geostationary orbit and calculate the apparent altitude deviation by the N-metric formula.

 

Nm -- Orb r =9.9757e11 x 1000 x 39.47842/ 22.43866

=41894123 (actual) 41894987 m. This shows a difference of a mere 864m over forty odd thousand kms which is statistically accurate. How does that stack up to your accurate 'non re-calibrated' measurements?

 

Let ' s try big-G

Big-G --   d2=G.m1.m2/Fg

= 6.67384e-11 x 5.9736e24 x 1000/227.1485

= 41893922 That's close as well... only a little over a kilometre out! Of course that's not fair it's just the same Big G problem being exposed.




Using 'G' we should be able to calculate the altitude adjusted mass of an object orbiting at geostationary orbit altitude because the big G equation has got nothing to do with spatial displacement; just radius, mass and force.


 

 


Checking the kg mass on Earth using the centripetal force equation---

m1=Fg ( Newtons ) x r2/G x m2

=0.022438 x 9.80665 x 1.743904e15/3.98694e14 (3.98694e14 when G=6.67428e-11) which is it? It depends on who you ask!!

 

## result   0.96253kg  (0.96626kg by Nm at ##)

 

We can now use this value to assess the earth surface error because Newton assumed 1kg.

The apparent orbit radius calculated by the N-metric equation is less than 0.002% different than the CF formula result. Remember; the actual orbit value calculation by the centripetal force formula is only almost correct. It's the apparent centripetal force which has really changed but if we adjusted the kilogram then the result would truly be the same.

THIS ALL MEANS THAT THE ACTUAL MASS AFFECT (both inertial and gravitational for equivalence) OF THE OBJECT IN SPACE IS DIFFERENT THAN THAT MEASURED ON EARTH; ESPECIALLY AT MID (satellite) ORBIT ALTITUDES and it begins to come back to exactly normal at a couple of thousand odd kilometers high. This is the phenomenon which causes errant orbit calculations for close earth orbit satellites.

What happened is; that the apacenter moved back from the ground zero case by a small amount. What has really occurred is that the relative mass has been noticed as being different to the value appreciated in space* and if we were to readjust the mass in the equation then all would be correct. This applies to all the examples but big G will have to do for general calculations for around earth environs. While big G is around, the masses will always remain the same in actuality when evaluated on earth. The energy we do have a problem with though but that's no real problem on earth, and thrust calculations work well for rocket launches.

We are only looking to discover what objects would actually weigh if we tested their inertial mass and brought them back from space to our back patio and compared that with what 'G' informs us. Big G predicts they would weigh the same (in a similar gravitational force and differential) and this is not the case at all. The featured equation declares the correct centripetal forces and masses of bodies and objects out there at all altitudes, and it shows a slight notional deviation in the inverse square law curve and if scientists would stop fudging the results of noted errant orbits then I could prove it. Even geostationary orbits are very slightly anomalous.

*Refer to tab 10 -The Newton KgF enigma.

 

The return argument is going to be that there is a serious SEP violation in that statement. Yes but there wouldn't be if the inverse square law was absolutely accurate and Newton's 'G' curve was therefore not in violation. The tests for this are difficult to imagine. However what is not difficult to imagine is that Newton was making a mathematical error by applying a linear fudge to variant inverse square law values and not adjusting the starting value to suit! I guess he had no way of checking the results did he? I reckon he did brilliantly myself. He was a true genius of his day.

There mightn't be much deviation in the inverse square law geodesic but it is there, and it remains hidden by 'G' . Think about it. If you take any inverse square curve and multiply the results by anything else at all you are creating a new curve which doesn't relate to the first because the very first starting point has now been deviated from by the multiplication of the first number. I.e. 6.67384e-11 in Newton 's case.

So if we multiply the one kilogram by 'G' and add that; we find that we have adjusted the starting value of the weight by a small amount. The adjustment is small 1kg 1+6.67384e-11 but significant and therefore if the curve is taken to be the exact inverse square law from 1kg and not from the 1kg + G fudged fraction of 1kg then that curve is no longer an inverse square law curve from 1kg. It has been fudged and has therefore proven that the SEP is violated and G-rel is also wrong by consequence.

This is what I contend and that the violation noted by my theory and equations is mostly camouflaged by the inaccurate Newtonian curve. If you adhere to 'G' like Einstein has, you will never see the blind error and you will be--- the blind leading the blind!

 

 

Leaving that necessary digression; we should now apply the modifier to the previous examples and the expanded formula is now---

Fg= ( rt mb)4z.ms/d2- rt ( mbxms/*ds2)/pi  (ds is surface to surface distance)

* At zero altitude we utilize the fine constant as this denominator. (reason given in the following tab)

 

 

For the following--- The mass of the earth mb is the newly discovered 9.95162e23kg

 

(c)  **Fg with SEP violation--- 1Kg on earth surface---

 

          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.)

4.068e13 -  3.71906e12=2.8996e13

=9.806628N Note: This form can ' t be used for accurate calculations without the apacenter adjustment functor (or the accurate mass for each altitude) which would make the equation all too unwieldy when big ' G ' will do the job locally.

          *True actual mass--- 0.9629 by Nm equation & 0.9625 calculated at #* by big-G

 

 

 

 

2000kg at 1,000km...

Fg=9.9757e11 x 2000 x 39.2262/5.264e13

(rt(9.95162e23   /2e9)/19.6131=1137327   (no change- so no need for this modifier in the eq)

(aa)= 1486.7kgF   a=5264000 There is insignificant change at this altitude.

 

Now ditto by big G...

Fg=G.m.m/d2

=6.67384e-11 x 5.9736e24 x 2000/5.44363e13

(ab)= 1493.6kgF

 

now 2000kg at 10,000km

G-th Fg=9.9757e11 x 2000 x 39.2262/2.64239e14

(2.6424e14+ ( rt (9.95162e23/ 2000 x 1e7)/19.6131

=296.176kgF a=6255400. There is now no modification whatsoever because the distance between the disparate mass object-body is too great.

 

NB --- Calculated degree of deviations (not direction specific)...

On earth surface 1.9109% This is the exact 'g' deviation noted in physics.

1000km 2000kg -- 2.339%

42000km 1000kg-- 1.2935%

Moon-- 20% (Newton erred when calculating the mass of the Moon: Ref following section.)

Earth-- 17%

This inadvertent error from Cavendish-ian physics caused by the same N-metric problem is the reason for the large Earth moon mass deviations. I will try readjusting 'G' for the new earth mass and we'll run a few calcs shortly. Note: ...hats off to Cavendish too! There's always 20/20 vision in hindsight.

 

Calculating big G version 2--- if we were to envision revising the Earth's mass.

 

9.80655/9.95162e23/4.068e13

G2=4.00869e-10

 

Both the big G and the centripetal force equation are beginning to show their limitations by not being capable of taking into account the predicted and now (also to be furthermore) proven asymmetry in the Minkowski geodesic gravitational manifold.

Now it must be understood that the normal method for calculating orbital heights are fine but where they err is for calculating the actual force and energy required to put a satellite into the required orbit. We can check this by tabling different values for the N-metric equation versus the big G equation for different altitudes and weights but not just yet.

I can't adjust my equation. It won't work at all for any other data sets. The proverbial horse has bolted but the equation is still useful for gravity calculations in space. The centripetal force formula also doesn't take any variation of gravity into account. After all it too treats gravity as though the satellites are on strings attached to the geocenter of the earth, however it's all good after about three earth-diameters distance except that now G will be forever deviant with an error that never quite comes back to true.

Just to double check let's say the satellite weighs 1000kg. The figures are next to the previous calcs.

All values fit the model and all concur with the moon mass and earth moon g-force as calculated... No 'G' in sight; and as well as that we have an accurate equation for all gravitational situations. However it must be said that big G is the simplest, and best to use for 'down to earth situations' and most orbit calculations. However the equation derived here is able to be used with admirable accuracy for calculation of true masses and gravitational forces.

So it now stands that we are not required to reinvent the wheel. There is no possibility of utilizing the above results for any purposes other than for space exploration. This is because the very relationships between the kg-m-s etc will necessarily be upset and it would become necessary to readjust all the physics we have to do with and that's not a practical option at all. It should be a sobering recognition that as far as physics goes we are reliant on historical estimations and in many aspects we are bound to them by 'factuality' and not truth.

Gauss' point gravity equation g ( r)=-G.m. e r /r2 becomes modified as well whereby er is no longer a 'hat over' but a 'dot product' vector of an integral. This is because it is no longer reasonable to consider that the gravitational field from the Earth's center point of mass (traditionally assumed to be the geometric center) is spherical with respect to an object resting on or near the Earth's surface. There is obvious oblateness and the earth can only be reduced to a perfectly spherical point gravitational source when the altitude distance is infinity but practically after about three Earth diameters distance.

Including the E=mv2 debacle examined in the previous assertation; it can be understood that in consideration of the whole of physics we have an imperfect paradigm to deal with on a daily basis, and that goes a long way towards recognizing one reason why the science has met with a significant problem in all previous efforts to find unification between the various arms of physics. big G

I trust that I have convinced you; not to go out and replace big G but that Einstein's field equations by their reliance on big G are flawed, and even though they were never any proof of relativity before (as many erroneously suppose) they are even less so now. Note: if you doubt the validity of the equivalence principle violation exhibited, I must point out that any testing can only be done in deep space momentum and not in an orbital situation which is not an inertial reference frame.

 

 

EQUATORIAL CENTRIPETAL FORCE ADJUSTMENT TO 'g'.

### Let's check what the 'centrifugal' force is on a 1kg object resting on the surface at the equator.

 

F =m.r.4pi2/T2

F=1 x 6378110 x 39.4784/ 7.46496e9

F=0.03373kgF =0.33N

 

That's not much is it? No but it should still have been taken into account by Newton because it affects the weight of the kilogram! So now we know that when we weigh a kilogram it actually weighs less than it would if the earth wasn't spinning. That is just one illegal premise that  Newton  used when evaluating gravity.

##So now we have a notional 1kg mass deviation from-- 1 - 0.03373=0.96626kg. This is now looking suspiciously close to actually causing an observed acceleration rate of 9.81m/s/s if perhaps your measurement accuracy was out a bit. The acceleration rate we would expect to observe without every consideration would be 9.662m/s/s. See below.

 

 (1) THE TRUE 'g' -Earth gravitational acc rate .

 

What we now have for correct g-force and altitude calculations is a notional weight of 0.9662kg* (0.9629 actual by the Nm formula). This gives a true g force value giving an acceleration rate 'g' of 9.806628m/s/s this value is by the N-metric equation calculation at (c) . Note: There are a few more variables and slight errors which I refer to later which will account for this slight difference from the 'best measurement' acc rate. Even so, it is still statistically correct.

We also need to calculate from the height of ten meters that Newton must have dropped his object from and timed it with his ' high tech ' stop watch of the day! I did just that: It ' s almost exactly the same so we have no need to take any of that into account.

result-- 9.8696-0.03373 x 2*= 9.80214

expected-- 1N= 9.80655

 

Perhaps the 0.03373 x 2** could account for energy loss per 1kg mass proportion compared with pi(re)2 pi(r=?)2 i. That's an unknown in which case we ' ll use mass/volume (density) proportionality. The energy used and lost by the Earth ' s gravity to hold 1kg down with g-force would have been 0.03373/5.9736e24 =5.6e-27J.s.kg. So energy proportionality is 9.80214/r2 if the densities are assumed to be similar. 9.80214 is the number I maybe should have used for my later calculated ' z '*^ value instead of 9.80655. This is only a difference of 0.04496 percent.

**Twice because we have to adjust the force and the mass because both are affected.

*^ Refer to the -derivation of ' z ' tab.

 

 

True weight at earth surface might be expected to be derivable from 1kg -0.03373=0.96627kg. That ' s close to the 0.9627kg mass which was weighed as 1kg weight which has been calculated below to conform with the true values of 'g' (constant) and ' Fg ' of 9.4416N calculated at (c). (refer to (1).

 

m=F/g   9.4416/9.80665

= 0.9627kg i.e. 1 divided by 1 is 1. Mass is in kgs and not Newtons .

Compare this result with (0.9625kg by big G at #*). This is statistically correct within the margin of error and this assertation is further supported by proof from the centripetal force, N-metric and the big G equations at #. From ### we notice that both the CF and big G equations are in error (different) at low altitudes.

*Refer to the - Newton kgF enigma tab for reason why the weight is not different and that the answer to this SEP problem lies at the quantum level.

 

 

 

CONCLUSION

 

OVERALL THE NEW Nm EQUATION IS MORE ACCURATE THAN UTILIZING BIG G! However. Big G is accurate enough at real world altitudes but it becomes woefully inaccurate for solar system or deep space calcs where Einstein's field equations are used and visa versa. However the N-metric equation is truly accurate in both cases while big G is not. Relativity is no longer required.  

The slight difference noted would be caused by the apacenter shift and unknown density variations within the earth. That's an almost incalculable N-metric problem but we do know that there is some more g-force unaccounted for but the mass error (enigma) is proved and actually more serious; being 0.9627kg of notional inertial mass being measured as 1kg of weight or visa versa... 1kg mass being considered to be 1kg rather than the actual weight of 1.0373kg which would make the result low on the mass side for 1kg weight. Note: refer to the -Newton kgF enigma- tab.

This new weight must be the so called mass or 'm' term in the F=mg equation for gravity because the wrongly measured mass being stated as weight will put our energy calculations for space too low, so the Joules needed for an actual 1kg of mass will be 1.0373J. The energy on earth for the gravitational case will remain the same because that's the weight we have historically aligned with the mass and nothing changes on Earth. Note: The SEP violations in all these calculations are being used to exhibit; nothing else but that. I.e. That the traditional geodesic gravitational models are incorrect; and that includes the GTR model which by default is also incorrect.

It is important to understand that the mass kg and the weight kg as well as the joule are the units that we have 'gone' with and everything else is related to them so they can't be changed other than notionally and for true calculations, and as well as that we have to live with the Newton.

Try understanding it like this. We have a kilogram mass in space which produces one Joule of energy when we push it with any force for one meter. We bring it down to earth, only to find it weighs more than the one kilogram we would have calculated by the true N-metric equation as well as the space scales we used at 42,000km o.r. but we know it's still a kilogram of mass because of the measurements of the force it takes to accelerate it in the gravitational inertial frame on earth and we also calculate one joule of energy from E=m.d. so we are forced to conclude that the earth and space joules are different and we have a SEP violation.

Continuing with the comparisons for validation of the N-metric equation for the use it will be put to:

Was Newton's stopwatch wrong perhaps? Don't for a minute think that he derived 9.80655; that's just from supposedly more accurate experiments and perhaps a little book cooking! I would suspect that his accuracy couldn't have been much better than 9.8!

What we can most likely expect from this now is an attempt to make a change but only to 'G'. don't waste your time. I'll do it for you!

 

G=Fg x r2/m

G=9.6626 x 4.068e13/5.9736e24

G=6.5802e-11

 

Let's check in once more with the geostationary orbit satellite but now with the new G--- but first by the CF equation...

 

F =m.r.4pi2/T2

=1000 x 41894987 x 39.4784/7464960000

= 22.5915kgF    221.56N

 

Now we'll check with the new G when the mass is the nominal 1000kg.

 

Fg= Gm1.m2/d2

Fg=6.5802e-11 x 1000 x 5.9736e24/1.75518e15

= 223.95129N    mmm, still a bit wrong but better than the 227.15N calculated at #. What can be the problem? Why does the mass seem to matter with big G? First we need to check what the actual altitude deviation by the CP formula is for reference against other derived altitude answers...

Using the new G derived answer...

 

r= 22.5915  x 7464960000/39478.4 ...(1000 x 39.4784)

=42346888m which is 1.01078% more than expected.

 

Using the old big G answer... 1.0252%  even worse.

 

At first there might seem to be a violation of SEP and perhaps WEP at this altitude. (My original prediction was 1.2935% but that was calculated using the slightly incorrect initial 'z' value)*. However in spite of that; changing big G won't solve anything because we are just changing the value of the linear fudge for a different inverse square curve so we can reject that as any solution and stay with the big G form.

Using the N-metric (Nm) equation answer from #... 1.012%. There is no significant WEP violation.

 

*Refer to -the derivation of 'z' tab--- Using the new 'z' the error is now only 0.0098% but this time in the less than direction so the true deviation (which very closely agrees with the CP answer) and deviates further away from the big G answer is 1.035% and that is by using the equation modifier which turns out to give no net difference at this geostationary orbit altitude.

 

 

CONCLUSION 1 and a DOUBLE CHECK from Apollo mission data.

 

This all demonstrates that the big G value is erroneous and that the true strong equivalence deviation is only between- 0.0098% max & 0.002% min. In fact that is almost exactly correct and we can conclude that there is zero WEP or SEP violation at geostationary orbit altitudes. If there is any violation to be noticed we will have to find it at lower altitudes and/or different mass values but first we must revert back to the mass value of 1kg to avoid any possible mass disparity issues and then move to evaluating the effects on WEP and SEP with larger mass values. Note: I didn't begin by hacking this result. Go back and check if you like. This equation was totally derived from the Apollo data* and the two CF equations. The accuracy there is reflected in the accuracy here and visa versa. That's why I said you can take that moon radius value to the bank. In fact you can do that with all the masses of bodies derived herein.

*Note: The Apollo data can be trusted and it actually turns out the most astoundingly accurate results right to the correct calculation of the true 'g' on earth -as we noted in the previous two tabs. This is because there was no agenda existing at the time for any results to become "re-evaluated".

 

(a) Values of Fg for 1kg at the accepted geostationary orbital radius of 41894987m...

CF--- Fg =m.r.4pi2/T2

CF Fg = 41894987 x 39.4784/7464960000

    Fg=0.2215N

Big G --- Fg= 6.67384e-11 x 5.9736e24/1.75518e15

= 0.2271N

 

new G--- Fg= 6.5802e-11 x 5.9736e24/1.75518e15

=0.2239N

 

nM--- Fg=9.9757e11 x 39.4784/1.75518e15

=0.022438kgF = 0.22--N    

 

Compare the geostationary orbit data once again with the values from #. (and weep lol) this time with various masses.

 

          1000kg                            1kg                        10,000kg

 

CF---        221.56N 0.22N/kg          0.22N/kg               0.22N/kg

N-m---      220.05N 0.22N/k            0.22N/kg              0.22N/kg

Big-G---   227.15 0.227N/kg          0.227N/kg            0.227N/kg

 

The deviant big-G results are very telling! The accuracy of the Nm formula is also very telling.

 

 

FURTHER TECHNICAL CALCULATIONS searching for any WEP violations under many conditions. There is no need to study these if you accept that while big G does actually produce WEP violations there are no WEP violations to be found using the Nm equation under any conditions of disparate masses and distances. The Nm equation is the perfectly correct equation.

 

 

 

 

Does Newton's Big G show WEP violations? Does the new N-metric equation?

 

 

Utilizing big G---

 

 

*Earth to 1kg

G.m1m2/d2

6.67384e-11 x 5.9736e24 (x1)/4.068e13

Earth to 1kg mass gravitational force = 9.8001N on the one kg mass--- (not on the earth per kg)     

 

Earth to 1e6kg

6.67384e-11 x 5.9736e24 x 1e6/4.068e13

=9800110N

Earth to 1e6kg mass gravitational force = 9800110N

So force between Earth and a 1e6kg body is 9.8001N/kg on the body                       

 

Two earth masses

6.67384e-11 x 5.9736e24 x5.9736e24 /4.068e13

=5.854e25N

g-force on each kg--- m/F

5.854e25/5.9736e24

= 9.7997N/kg                                              pull/kg on each earth

 

When a body is larger than earth--- say 4e30kg (at a greater distance now)

6.67384e-11 x 5.9736e24 x 4e30 /4.068e20

=3.92e24N

F/kg=3.92e24/5.9736e24

=0.655N                                                        pull/kg on smaller (Earth)

3.92e24/5.9736e30

=0.000000656                   smaller  pull/kg on larger body (not Earth)

 

 

Therefore the following should be the case for one kg at that distance---

6.67384e-11 x 5.9736e24 (x1) /4.068e20

F/kg=0.000000980011N               ---there is now far less pull/kg on the smaller  object (1kg).

 

From *

6.67384e-11 x 5.9736e24/4.068e13

=9.8001N                                              ---there is far more  pull /kg on smaller 1kg from first derivation at a much closer distance.

 

 

Exercises on mass share ratios--- prepping to search for any WEP violations.

 

(a)12 x 1/2 = 6N/ to be shared by 13kg (inverse proportioned)

6/13=0.4615

5.538 N pull on the small

0.4615N/kg on the large

 

12 x 2/2= 12N to be shared by 14kg at 1:6 inverse ratio

1.166N pull on the large

6.99N on the small

total N shared = 12N  total-kgs= 14kg  total-N/kg-shared= 12/14=0.857N/kg Incorrect.

 

(b)12 + 1/2=6N/ to be shared by 13kg

ditto for (a)

 

12 + 2/2= 7N to be shared by 14kg at 1:6 inverse ratio

= 1N pull on larger

= 6N on the smaller

total N shared=7N total-kgs =14N--- total-N/kg-shared=14/7=2N/per kg correct.

 

 

 

 

 Let's say that a calculation of 1e-9 was the result required to be verified by measurements. Then like Newton we can apply the proportionality modifier Gm for the multiplication case and we'll use Ga for the addition case

 

a x12 x 1/2=1e-9

e= a.b.c/d

a=c.d/a.b

a=2/12 x 1e-9

Gm =1.66e-10

 

1.66e-10 x 12x5/2

 

=4.98e-8

 

 

Fg= Ga (m1+ m2)/2

Ga= r2 x Fg/ m1+m2

Ga=2 x 1e-8/13

Ga= 13/2x 1e-8

Ga=1.538e-9

 

1.538e-9(12 +5)/2

1.538e-9 x 17 /2

 

=1.3073e-8

 

1.538e-9 x 17 /10

 

=2.615e-9

 

Ga=9.80665 x 4.068e13/ 5.9736e24=6.675e-11

 

6.678e-11 x 5.9736e24 /4.068e13=9.8062 

 

6.678e-11 x 5.9736e24 + 1e10 /4.068e13      Ditto--- so with the Nm values we find no WEP violation.

 

(c) 6.67384e-11 x  5.9736e24 x 1e10/4.068e13 =98001107728 --- no apparent violation.

 

share ratio   5.9736e24/ 1e10 =5.9736e14 : 1

 

6.678e-11 x 5.9736e24 /4.068e20=9.8062e7--- no violation

 

6.67384e-11 x  5.9736e24 /4.068e20=8.8001e7 ---ditto

 

6.673848e-11 x 5.9736e24 x2 /1.6272e14=4.9N/kg (exactly correct)

 

 

 

Now for a look at the actual values---

 

(d) 6.67384e-11 x  5.9736e24 x 5.9736e24 /1.6272e14=1.4635e25 =1.225N/kg what the? That's an error.

 

Now we saw that the total gravitational force was apportioned to the total masses.

In the 1kg on earth case the answer was 9.81N which is shared by the ratio 5.9736e24 : 1 ---so we get---

9.81/ 5.9736e24=0.5976e-24 for the earth per kg and 9.81 for the 1kg.

 

In example (c) 6.67384e-11 x  5.9736e24 x 1e10/4.068e13 =9.8e10N apportioned inversely by the ratio 5.9736e24/ 1e10 =1:597e14N earth:m2

so 98001107728/1e10=9.8001N/kg pull on the lesser mass and zero N/kg on the earth. That can't be right. What's wrong?

 

let's go much bigger for M2 say 1e20

6.67384e-11 x  5.9736e24 x 1e20/4.068e13=9.8e20N

so9.8e20/ 1e20 = 9.81N/kg for the smaller object and once again none for the earth. When the ratio is 5976:1 so 9.8/5976=0.0016N/kg is apportioned to the earth and 9.7983N/kg apportioned to the other???

 

Let's take the earth moon case---

 

CF equation comparing N-metric derived values------

 

Fg =mmr.4 pi 2/T2 we get---

=5.97e22 x 1.5199e10/5.8525e12

1.55041e20N   1.581e19kgF*#  concurs with (Nm)

 

 

N-metric------------------------------------------

 

 (Nm) Fg =9.9757e11x 5.97e22 x 39.2262 /1.47763e17=1.58098e19kgF*#

 

1.58098e19/9.95162e23 + 5.97e22 =0.000014N/kg      

 

 

CF equation with big G values---------------

Fg =mmr.4 pi 2/T2 we get---

= 7.3477e22 x 1.5199e10/5.8525e12

1.8713e21N    1.9082e20kgF

1.8713e21/5.9736e24 + 7.3477e22=0.0003N/kg   (by big G--- 0.00003N/kg) see below^

 

Big G-----------------------------------------------

 

6.67384e-11 x 5.9736e24  x 7.3477e22/1.47763e17

=1.9824e20N---1.9082e20kgF

 

1.9824e20/5.9736e24 + 7.3477e22 =0.00003N/kg^

 

 

 

CONCLUSION

 

The Nm concurs exactly with the CF equation. Unfortunately that even when recalculating for Newton 's error the big G equation doesn't work.

THE PROOF FOR THE PROPOSED Nm EQUATION IS ASTOUNDING!

 

Let's now put the Moon right on the Earth---

 

Nm Fg =9.9757e11x 5.97e22 x 39.2262 / 6.5856e13

=3.5473e22kgF--- 3.4787e23N ---  3.5473e22/ 9.95162e23 + 5.97e22

=0.03362 --- 0.03362 x 9.80665 x 2=0.65939N. The straight altitude equation doesn't work here because there is still an N-metric problem as well as zero altitude. They are essentially one body! In this case we must use the 'one body' Nm equation---

Fg= 9.97578e11 x 5.97e22 x 9.80665 / 5.95558e34

(/6378100 +1737100 + 9.97578e11 x 5.97e22 = 5.95558e34)

=9.80665N/kg correct.                         

 

When using big G we see a problem---

Fg=6.67384e-11 x  5.9736e24 x 7.3477e22/ 6.5856e13 =4.448e23--- 4.44e23/7.3477e22 + 5.9736e24=0.07342N.kg ???

 

 

 The force per kg is decreasing. Maybe it's because we are not taking the radius of each of the bodies into account.

Try again 4.068e13 + 6378100 + 1737100 ditto. That's not the answer!

5.97e22--- Fg/kg= 1.58098e19/5.97e22 =0.0002648N/kg.

 

 

FUNDAMENTAL CONCLUSION

 

Big G is the problem because it finds WEP violations. The N-metric equation provides the correct answers. There are no WEP violations.

 

 

 

 

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