You know that Bible won't allow Sun to be older than 6000 years.
Plus, FE won't allow the existence of Moon rocks and meteorites, they come from space and space doesn't exist.
Are you saying it is a nuclear furnace?How?
Then, you are wrong.
The Carbon-Nitrogen-Oxygen (CNO) cycle defies the solar nuclear furnace hypothesis.
It is actually some 600 meters in diameter, in the shape of a disk:If that was the case it would vary in size dramatically throughout the day, or it would have to circle around Earth. But at just 600 meters, in order to have an angular diameter of 0.5 degrees it would need to be 68 km above Earth.
The Sun is powered by laevorotatory subquarks emitted by the Black Sun.Just more empty words.
How?
Don't just link to some random page with no real explanation, explain here in your own words just how the CNO cycle defies the fact that the core of the sun is undergoing nuclear fusion?
You have no rational basis for your claim that from nothing nothing comes.
wow you can use the quotes properly keep up the good work.How?
Don't just link to some random page with no real explanation, explain here in your own words just how the CNO cycle defies the fact that the core of the sun is undergoing nuclear fusion?
You are in no position to issue demands around here.
Not while you are the author of something like this:You have no rational basis for your claim that from nothing nothing comes.
You are philosophically illiterate.
Since you are unable to face reality, you are forced, by default, to troll this forum: this is what you have been doing here for the past three years.
Rest assured, this is going to stop.
An allowance must be made for ether refraction.Not until you show it exists, and that is quite irrelavent to the topic, unless you are saying ether refraction is magically powering the sun.
Here is the most precise proof that the shape of the Sun cannot be spherical at all:You mean here is even more spam that proves nothing.
Observational Confirmation of the Sun's CNO CycleWould mean an observation confirmation that the atoms in the sun undergo nuclear fusion.
You are philosophically illiterate.If you weren't so philosophically illiterate, you would know just how invalid that claim of clickjamas is.
Rest assured, this is going to stop.I take it that means you are reporting me for hurting your feelings and not just accepting your unsubstantiated claims?
Here is the most precise proof that the shape of the Sun cannot be spherical at all:
You have no rational basis for your claim that from nothing nothing comes.
Observational Confirmation of the Sun's CNO Cycle
https://arxiv.org/ftp/astro-ph/papers/0512/0512633.pdf
Measurements on gamma-rays from a solar flare in Active Region 10039 on 23 July 2002 with the RHESSI spacecraft spectrometer indicate that the CNO cycle occurs at the solar surface, in electrical discharges along closed magnetic loops.
"But the nuclear furnace theory assumes that these nuclear events are separated from surface events by hundreds of thousands of years as the heat from the core slowly percolates through the Sun’s hypothetical “radiative zone”."
What us the relevance of any of that to the topic, "Solar power source", might I suggest none?You have no rational basis for your claim that from nothing nothing comes.
You are philosophically illiterate.
Only someone who has inherited the IQ of an ape, such as yourself, could make such a statement.
Since you have no arguments to make, you are trolling this forum. Everything you said has been debunked.
You are not here to debate, but only to sabotage: it takes less than ten seconds to prove you are wrong, since you cannot accept defeat, you are forced to resort to basic trolling.
You are also scientifically illiterate: you are denying Stokes' theorem.
You are philosophically illiterate.That is from a different thread, which you had plenty of opportunity to chime in and object/refute it.
it takes less than ten seconds to prove you are wrongYet instead of spending those 10 seconds to prove I am wrong, you spend far more on insulting me and spamming and derailing the thread.
Fair enough.
Professor P.M. Robitaille (Ohio State University) has put the information into a form that all the students can understand.
He is not FE/GE, in fact he is one of the top heliocentrists in the world.
RADIUS OF THE SOLID SURFACE SUN
Within the context of modern solar theory, the Sun cannot have a distinct surface. Gases are incapable of supporting such structures. Modern theory maintains the absence of this vital structural element. Conversely, experimental evidence firmly supports that the Sun does indeed possess a surface. For nearly 150 years, astronomy has chosen to disregard direct observational evidence in favor of theoretical models.
Dr. P.M. Robitaille
http://www.ptep-online.com/2011/PP-26-08.PDF
On the Presence of a Distinct Solar Surface: A Reply to Herve Faye
(https://image.ibb.co/nnigj8/rds.jpg)
Spectacular images of the solar surface have been acquired in recent years, all of which manifest phenomenal structural elements on or near the solar surface. High resolution images acquired by the Swedish Solar Telescope reveal a solar surface in three dimensions filled with structural elements.
Beyond the evidence provided by the Swedish Solar Telescope and countless other observations, there was clear Doppler confirmation that the photosphere of the Sun was behaving as a distinct surface. In 1998, Kosovichev and Zharkova published their Nature paper X-ray flare sparks quake inside the Sun. Doppler imaging revealed transverse waves on the surface of the Sun, as reproduced in Figure 2: “We have also detected flare ripples, circular wave packets propagating from the flare and resembling ripples from a pebble, thrown into a pond”. In these images, the “optical illusion” was now acting as a real surface. The ripples were clearly transverse in nature, a phenomenon difficult to explain using a gaseous solar model. Ripples on a pond are characteristic of the liquid or solid state.
http://vixra.org/pdf/1310.0159v1.pdf
Commentary on the Radius of the Sun:
Optical Illusion or Manifestation of a Real Surface?
Observational astronomy continues to report increasingly precise measures of solar radius and diameter. Even the smallest temporal variations in these parameters would have profound implications relative to modeling the Sun and understanding climate fluctuations on Earth. A review of the literature convincingly demonstrates that the solar body does indeed possess a measurable radius which provides, along with previous discussions (Robitaille P.M. On the Presence of a Distinct Solar Surface: A Reply to Herve Faye. Progr. Phys., 2011, v. 3, 75–78.), the twenty-first line of evidence that the Sun is comprised of condensed-matter.
On the Temperature of the Photosphere: Energy Partition in the Sun
http://vixra.org/pdf/1310.0140v1.pdf
If the local thermal equilibrium and its extension of Kirchhoff’s formulation fails to guarantee that a blackbody spectrum is produced at the center of the Sun, then the gaseous models have no mechanism to generate its continuous emission. In part, this forms the basis of the solar opacity problem.
Stellar Opacity: The Achilles’ Heel of the Gaseous Sun
http://vixra.org/pdf/1310.0139v1.pdf
Given the problems which surround solar opacity, it remains difficult to understand how the gaseous models of the Sun have survived over much of the twentieth century. Local
thermal equilibrium does not exist at the center of the Sun. Both Kirchhoff and Planck require rigid enclosure which is not found in the Sun. Planck has also warned that the Sun fails to meet the requirements for being treated as a blackbody.
On the validity of Kirchhoff's law of thermal emission
https://ieeexplore.ieee.org/document/1265348/
https://www.libertariannews.org/2014/04/04/kirchhoffs-law-proven-invalid-the-implications-are-enormous/
Further, all blackbodies are limited to solids, since only they can be perfect absorbers, and unlike liquids, they cannot sustain convection. Prof. Robitaille also explains why gases do not follow these laws because they do not emit radiation in a continuous manner, further discrediting the standard model of stars. The emissivity of a real gas drops with temperature. Planck’s equation remains the only fundamental equation that has yet to be linked to physical reality, which is a direct result of Kirchhoff’s error.
Prof. Robitaille notes that the standard gaseous Sun model uses equations of radiative transfer, and those equations all have, at their source, KLTE. The invalidity of KLTE means there cannot be blackbody radiation at the center of the Sun, which means the entire standard model of the gaseous Sun is invalid.
https://principia-scientific.org/new-study-invalidates-kirchhoff-s-law-of-thermal-emission/
https://web.archive.org/web/20160211150839/http://www.ptep-online.com/index_files/2015/PP-41-04.PDF
“The Theory of Heat Radiation” Revisited:
A Commentary on the Validity of Kirchhoff’s Law of Thermal Emission
and Max Planck’s Claim of Universality
"Since the corona must be excessively hot to produce such
ions in a gaseous context, the continuous spectrum of the K-corona
has been dismissed as a strange artifact, produced
by electronic scattering of photospheric light. Otherwise,
the coronal continuous spectrum would be indicating
that apparent coronal temperatures are no warmer than those
of the photosphere. It would be impossible for the gaseous
models to account for the presence of highly ionized
species within the outer solar atmosphere.
Current temperature estimates are
flirting with violations of both the first and second laws of
thermodynamics: it is difficult to conceive that localized temperatures
within flares and the corona could greatly exceed
the temperature of the solar core."
P.M. Robitaille
http://vixra.org/pdf/1310.0134v1.pdf
Commentary Relative to the Distribution of Gamma-Ray Flares on the Sun:
Further Evidence for a Distinct Solar Surface
http://vixra.org/pdf/1310.0108v1.pdf
The Solar Photosphere: Evidence for Condensed Matter
http://vixra.org/pdf/1310.0110v1.pdf
Forty Lines of Evidence for Condensed Matter — The Sun
Dr. P.M. Robitaille is a Professor at Ohio State University.
His detailed explanations are very well spelled out.
Here is the proof that the information contained in his paper is correct:
(https://image.ibb.co/nnigj8/rds.jpg)
Dr. P.M. Robitaille
http://www.ptep-online.com/2011/PP-26-08.PDF
On the Presence of a Distinct Solar Surface: A Reply to Herve Faye
Spectacular images of the solar surface have been acquired in recent years, all of which manifest phenomenal structural elements on or near the solar surface. High resolution images acquired by the Swedish Solar Telescope reveal a solar surface in three dimensions filled with structural elements.
Beyond the evidence provided by the Swedish Solar Telescope and countless other observations, there was clear Doppler confirmation that the photosphere of the Sun was behaving as a distinct surface. In 1998, Kosovichev and Zharkova published their Nature paper X-ray flare sparks quake inside the Sun. Doppler imaging revealed transverse waves on the surface of the Sun, as reproduced in Figure 2: “We have also detected flare ripples, circular wave packets propagating from the flare and resembling ripples from a pebble, thrown into a pond”. In these images, the “optical illusion” was now acting as a real surface. The ripples were clearly transverse in nature, a phenomenon difficult to explain using a gaseous solar model. Ripples on a pond are characteristic of the liquid or solid state.
It is actually some 600 meters in diameter, in the shape of a disk:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765#msg1939765
in fact he is one of the top heliocentrists in the world.You sure do love making these pointless appeals to authority.
RADIUS OF THE SOLID SURFACE SUNWhere does this say the sun is flat or doesn't undergo nuclear fusion in its core?
The gaseous modelWho cares?
it isn't undergoing nuclear fusion in the core.That sure looks like far more than the 10 seconds you claimed before.
Fair enough.
Here is the most devastating proof against the nuclear furnace model.
Actually, much, much younger.Yet not much much flatter, or much much not a nuclear furnace.
I have done so from the very start.
Clayton model/equation, which is more accurate than the Lane-Emden equation, and comparable to the integrated hydrostatic equation:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765#msg1939765
"Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun."
I have done so from the very start.Have you tried looking at the sun with the proper solar filters?
Clayton model/equation, which is more accurate than the Lane-Emden equation, and comparable to the integrated hydrostatic equation:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765#msg1939765
"Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun."
The RE have computed the centrifugal acceleration:No it won't! And you've given no valid reason why it should.
http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html
a_{c} = 0.0063 m/s^{2}
Let us now use the Clayton equation to calculate the actual acceleration of gravity.
(https://image.ibb.co/hkvQrJ/chromo.jpg)
PRESSURE: 10^{-13} BAR = 0.0000000000001 BAR
The entire chromosphere will then be subjected to the full centrifugal force of rotation, as will the photosphere itself of course.
It cannot be 274 m/s2.
That is the ASSUMED figure.
However, the pressure in the chromosphere is extremely small.
(https://image.ibb.co/hkvQrJ/chromo.jpg)
The Clayton model is the standard, accepted stellar model.
It is not my equation, the heliocentrists have been using it for decades to obtain the necessary results.
Here is the final result using the Clayton equation:
g = 0,0000507 m/s2
That is, this figure is much smaller than the value for the computed centrifugal acceleration: the Sun cannot be a sphere.
Here is the pressure diagram using the Clayton model vs. the integrated equation:
(https://image.ibb.co/nsZDdy/chro1.jpg)
Not my equation, not my diagram.
This is the standard accepted today in heliocentrism.
It cannot be 274 m/s2.Why not? I work it out as 274.35 m/s^{2} from the Sun's GM product and its radius.
That is the ASSUMED figure.So what?
However, the pressure in the chromosphere is extremely small.
Please show exactly where the Clayton equation claims "g = 0,0000507 m/s2".
The Clayton model is the standard, accepted stellar model.
It is not my equation, the heliocentrists have been using it for decades to obtain the necessary results.
Here is the final result using the Clayton equation:
g = 0,0000507 m/s2
Flat Earth Believers / Re: Advanced Flat Earth Theory « Message by sandokhan on August 09, 2017, 03:24:06 AM » (https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765;topicseen#msg1939765)You say "the gravity is opposed by the internal pressure of the stellar gas" but that has no effect on the gravitation outside that region.
"However, the gravity is opposed by the internal pressure of the stellar gas which normally results from heat produced by nuclear reactions. This balance between the forces of gravity and the pressure forces is called hydrostatic equilibrium, and the balance must be exact or the star will quickly respond by expanding or contracting in size. So powerful are the separate forces of gravity and pressure that should such an imbalance occur in the sun, it would be resolved within half an hour."
The RE have already checked the equation, the figures, the graphs years ago.
They know that it is not my equation: it is the standard formula used in stellar structure physics.
I just plugged in the numbers and obtained a final result.
If you do not like this equation, then please write to your nearest university and let them know of your concern.
I have done so from the very start.No, you have avoided the issue from the start.
"Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun.No it explains it quite well.
Let us now use the Clayton equation to calculate the actual acceleration of gravity.What equation?
P(r) = 2πgr^{2}a^{2}ρ^{2}_{c}e^{-x2}/3MWhat is this equation? Where is it from?
where a = (3^{1/2}M/2^{1/2}4πρ_{c})^{1/3}
The Clayton equation was requested, by the RE, years ago.
Not my equation, not my graphics.
It is the standard RE (modern stellar structure physics) formula used in such calculations.
All I did is to plug in the numbers, that is all.
Do not complain to me about the equation.
Dr. D. Clayton, one of 20th century's best astrophysicists, based his formula exactly on the hydrostatic equilibrium equation, and its parameters on the known laws of physics such as the Boltzmann constant, the polytropic conditions, and much more.
It is based exactly on the principles of physics as they are known to modern science.
The Clayton model is an accepted fact of astrophysics, it is the RE's own model, not mine.
A few words about Dr. D. Clayton.
He was awarded the NASA Exceptional Scientific Achievement Medal (1992).
He has done research at CalTech, Rice University, Cambridge University, Max-Plank Institute for Nuclear Physics, Durham University and Clemson University during an academic career spanning six decades.
No, it's my link from three years ago.
Chapter 18 from the same work contains the graphics as well.
You cannot complain about the Clayton equation since it is YOUR very own formula, derived by one of the best astrophysicists of the 20th century.
It is the standard in stellar structure physics.
Thus, whilst hydrostatic equilibrium is respected by the Clayton model, Equs.[2,3] are not. Hence the Clayton model is not really a correct solution, and this shortcoming becomes apparent at larger radii. In particular, the Clayton model is completely wrong as regards the temperature predicted near the surface of the star.
The Clayton equationStill not providing any details, as such it is entirely useless.
Am I supposed to do your homework for you?No. You are supposed to do YOUR homework.
http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdfHave you read this source?
We consider only the spherically symmetric problem (hence a star which is not rotating an unrealistic simplification in truth)
. In general this will be a combination of gas pressure and radiation pressure. In this Section we shall assume that the gas pressure dominates.
In some circumstances this can be approximated as a polytropic equation of the form
Hence, the use of Equ.(9) is a convenient approximation only
The Clayton model consists of assuming a specific analytic form for the pressure distribution
This differs from the original Clayton model which imposed zero pressure at the surface, r = R, of the starSo it is making simplifications.
This would seem to imply that the Clayton model pressure might be quite seriously in error except near the centre
However, at other locations the pressure is not given in terms of the density by a power-law like Equ.(9), i.e. Equs.(18) and (20) are not related by a power law. In as far as Equ.(9) is a good approximation to the equation of state, this exposes the approximation inherent in the Clayton models
Am I supposed to do your homework for you?That equation! I've seen and downloaded Chapter 11: Stellar Structure Part 1: Pressure, Density and Temperature Distributions in Spherically Symmetric, Main Sequence Stars, The Clayton Model (http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf).
You, the RE, do not know the form of the Clayton equation, one of the most famous equations in astrophysics?
http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf (http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf)
equation (18)
P(r) = 2πgr^{2}a^{2}ρ^{2}_{c}e^{-x2}/3M, where a = (3^{1/2}M/2^{1/2}4πρ_{c})^{1/3}
" See: NIST RReference on Constants, Units and Uncertainty: Fundamental Physical Constants. (https://physics.nist.gov/cgi-bin/cuu/Value?bg)
There is no more connection between the "pressure in the chromosphere" and the surface gravity of the Sun than between the pressure at, say, 400 km and the surface gravity of Earth.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sure, "It is not your equation" but that does not mean that you've been using it correctly.
No, it's my link from three years ago.But we can complain about YOUR equation since it is not the Clayton equation nor is it Eqn (18) in Rick Bradford's Chapter 11!
Chapter 18 from the same work contains the graphics as well.
You cannot complain about the Clayton equation since it is YOUR very own formula, derived by one of the best astrophysicists of the 20th century.
It is the standard in stellar structure physics.
I don't believe he is Donald D. Clayton.
You cannot complain about the Clayton equation since it is YOUR very own formula
And I haven't even mentioned the Nelson effect.
The Nelson effect is very real:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1645824#msg1645824 (synchronized relationship between the Sun’s periodic peak sunspot cycles and the orbital positions of the Jovian planets -- Jupiter and Saturn)
You already have at your disposal the equation, the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct.Really?
jackblack, you cannot dismiss the Clayton equation.Why can't I? Your own source says it only accurate at the core.
Go ahead and plug in the numbers, the registered figure for the pressure of the chromosphere, you will see that my calculations are totally correct.Your "calculations" being correct is completely irrelevant.
You already have at your disposal the equation, the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct.I am not questioning any of that, just your use of the equation that ends up in changing a fundamental constant, the Newtonian gravitational constant G.
rabinoz... please take your trolling to the CN. It is simply pathetic.I am not and never have been "trolling"! Get used to it.
G = gr^{2}/m(r); Are you saying that this equation is false?Not at all but that form is only useful in calculating G if you have very accurately measured values of g, r and m and you have not got these.
M = 1.989 x 10^{30} kgThe "mechanics" of you calculations are correct but you simply cannot change a fundamental constant, G, like that.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
G = gr^{2}/m(r)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Using P(700,000,000) = 1.0197 x 10^{-9} kg/m^{2} value, we get: g = 0,0000507 m/s^{2}
But G is not a universal constant.And more baseless claims for you.
Surely I can substitute gr^{2}/M to check out the equationAt which point you are then rejecting the values provided. In order to use the equation with the accepted RE figures, you need to accept G.
As such, my calculations are very correct.Again, your calculation are pure nonsense as you are not correctly applying the calculation.
The shape of the Sun cannot be spherical.Again, you have absolutely no basis for this claim.
The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth drops to one-thousandth of the pressure on the earth in the photosphere.And more pure nonsense.
It has been tested here on Earth, not anywhere else.
The entire universe does not obey at all Newton's laws.
Therefore, G is a constant only here on Earth, not anywhere else.
The shape of the Sun cannot be spherical.So what shape do you propose the Sun might be. Based on observations alone it is obviously circular because it presents a disk on the sky and the fact that we have been watching sunspots traverse that disk since Galileos time is evidence enough that the Sun rotates. Larger sunspots disappear on one side only to return a couple of weeks later on the other side.
QuoteThe shape of the Sun cannot be spherical.So what shape do you propose the Sun might be. Based on observations alone it is obviously circular because it presents a disk on the sky and the fact that we have been watching sunspots traverse that disk since Galileos time is evidence enough that the Sun rotates. Larger sunspots disappear on one side only to return a couple of weeks later on the other side.
You don't need any of your complex mathematics or theory to figure out that the Sun is obviously spherical. Just observe the Sun (safely of course). Common sense shows it is spherical. Read the bit at the top of your little embedded graphic where it says at the start "The Sun is a ball of gas". The last time I checked a ball was spherical in shape or do you deny that as well? You seem to be contradicting your own attachments! The graphic is entirely true and what I've known all along. So to state explicitly that "the shape of the Sun cannot be spherical" is just being silly to be quite frank with you.
g being 27.4 times as large doesn't mean the pressure will be 27.4 times as large.Is that why instead of you trying to rationally respond you have to resort to insults?
There is literally no connection there.
You are trolling the flat earth debate section.
So you want to claim that G is a variable and can vary in the sun?Which means the formula you are using would be completely invalid as it relies upon G being a constant. That means you are completely wrong regarding the Clayton equation.
It is absolutely variable.
In order to use the equation with the accepted RE figures, you need to accept G.Yes, exactly. Not making a ridiculous substitution, but just accepting it.
Exactly.
Surely, if what you say is true, then the equation will work JUST AS FINE with g, isn't it?If by "just as fine" you mean still very wrong, then yes.
Let's put your word to the test.Sticking the numbers in again and showing it produces a nonsense value just further supports the claim of your source, that this equation does not apply to the surface of the sun.
Accuracy of the Clayton model:Yes, amazingly 0 is quite similar to 0.
(https://image.ibb.co/eHYH5d/chro2.jpg)
The author of the paper did perform a calculation using the Clayton model at the surface: the error is at most 40x.The author of what paper, and what error?
There is still another way to prove the correctness of the Clayton equation.You mean there is still another way you can try and escape from reality?
I have the equationsYou have equations which do not apply.
https://arxiv.org/abs/1402.1534
Is this supposed to be a joke on your part?
They are using the redshift distance relation!
The Picture that Won’t Go Away
"Only in the rarest instances has a single picture altered the direction of a scientific discipline. But in the case of the galaxy NGC 7319 and the "misplaced" quasar in front of it, the message is inescapable: its presence threatened to shatter one of the most cherished themes of mainstream astronomy, the Big Bang.
The rationale for the Big Bang rests substantially on an interpretation of a well-known phenomenon called “redshift”. The term refers to the shift of light from distant galaxies toward red on the light spectrum.
Many years ago, astronomers decided that redshifted objects must be moving away from the observer, stretching out their lightwaves. This “Doppler interpretation” of redshift enabled astronomers, based on the degree of redshift, to calculate both the distances and velocities of the objects. From these calculations, certain conclusions were inescapable. If all redshifted objects are moving farther away, the universe must be expanding. If the universe is expanding, the expansion must have had a starting point—an unimaginable explosion producing a universe of galaxies receding in every direction from the observer.
Then came the Hubble photograph, taken on October 3, 2003. The picture showed a galaxy (NGC 7319) known for its dense clouds that obstruct all objects behind its core. In front of the galaxy's core is a strongly redshifted quasar. In fact, under the prevailing assumptions, the redshift of the quasar would put it more than 90 times farther away from us than the big galaxy behind it."
(http://electric-cosmos.org/NGC7319quasarLabeled.jpg)
A higher magnification image of the quasar shows a "jet" of matter extending out from the center of NGC 7319 toward the quasar:
(http://electric-cosmos.org/NGC7319quasar2.jpg)
http://ucsdnews.ucsd.edu/archive/newsrel/science/mcquasar.asp
The Discovery of a High Redshift X-Ray Emitting QSO Very Close to the Nucleus of NGC 7319:
https://arxiv.org/pdf/astro-ph/0409215.pdf
Published in the Astrophysical Journal
Geoffrey Burbidge, a professor of physics and astronomer at the University of California at San Diego’s Center for Astrophysics and Space Sciences
"The quasar was found embedded in the galaxy NGC 7319 only 8 arc sec from its centre. According to the Hubble law the galaxy NGC 7319, with a redshift of 0.022, is at a distance of about 360 million light-years. Therefore these objects could not be physically connected to each other if this was true."
At the meeting of the American Astronomical Society held in Texas in 2004, Professor Margaret Burbidge presented a paper that she had co-authored with Arp and several other leading astronomers, including her husband [subsequently published in the Astrophysical Journal]. It detailed the discovery of a high redshift quasar close to a low redshift galaxy. This time, though, the alignment was different in every significant way.
This time, no one could argue. You see, the high redshift [more distant] quasar lay in front of the [less distant redshift] galaxy NGC 7319! There was no longer occasion to debate the veracity of [Arp’s] matter bridge [connecting galaxies with quasars]. The quasar was in the foreground [the galaxy in the background]. In that impressive gathering of astronomy’s who’s who, you could have heard a pin drop. It was a deafening silence.”
“The significance of this discovery is huge. We have direct, irrefutable, empirical evidence that the Hubble law stands on feet of clay, that the observational justification of an expanding Universe is fatally flawed.”
Hilton Ratcliffe
Based on observations alone it is obviously circular because it presents a disk on the sky
Yes, the shape of the Sun is discoidal.
we have been watching sunspots
In the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.
g(sunspot) = g(earth)/10,000 = 0.00098 m/s2.
The shape of the Sun cannot be spherical.
We have clearly demonstrated that the ULX lying 8 arc sec from the nucleus of NGC 7319 is a high redshift QSO. This is to be added to a list of more than 20 ULX candidates which have all turned out to be genuine QSOs (cf. Burbidge, Burbidge & Arp 2003; Arp, L´opez-Corredoira and Guti´errez 2004). Since all of these objects lie within a few arc minutes or less of the centers of these galaxies, the probability that any of them are QSOs at cosmological distance, observed through the disk of the galaxy, is negligibly small. Thus this is further direct evidence that high redshift QSOs are generated and ejected in low redshift active galaxies.
mak3m, you are expected to be posting while sober.
Thus this is further direct evidence that high redshift QSOs are generated and ejected in low redshift active galaxies.
Exactly my point.
The redshift distance relation is completely useless.
Your link made use exactly of this relation.
.
You don't stand a chance with me here.
But G is not a universal constant.That's all quite irrelevant because we know that the whole "universe DOES NOT obey Newton's law of "universal" gravitation.".
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The universe DOES NOT obey Newton's law of "universal" gravitation.
DARK FLOW:I'll omit any reference to that because it is quite irrelevant here.
Ricks Cosmology Tutorial: Chapter 11 Stellar Structure Part 1 (http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf)
The central pressure from Equ.(22) is also given above (using G = 6.67 x 10-^{11} in MKSA units). Since 1 atm = 10^{5} Pa, the larger of the above pressures is a huge 1.9 x 10^{11} atmospheres pressure. Standard solar models give the central pressure as 1.65 x 10^{16} Pa, so the Clayton model with R/a ~ 5.4 is reasonably close.
Take a look at yourselves: now you are denying Clayton's equation.Nothing of the sort! You are the one misusing Clayton's equation.
Surely I can substitute gr^{2}/M to check out the equation, since YOU SAY that G is indeed a universal constant.But you cannot use that to find the surface g on the sun when Rick Bradford has already used "G = 6.67 x 10^{-11} in MKSA units" to find the parameters of the Clayton's equation.
But G is not a universal constant.That's all quite irrelevant because we know that the whole "universe DOES NOT obey Newton's law of "universal" gravitation.".
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The universe DOES NOT obey Newton's law of "universal" gravitation.
But G is still regarded as "universal constant" and is used with the same value everywhere, including in Clayton's equation.
I may have exaggerated slightly.Yes, slightly as explained.
What ?!
Slightly?
the theoretical/predicted/assumed value is that the atmospheric pressure of the Sun should be 27.47 times greater than the atmospheric pressure of the Earth.No it's not.
Which means the formula you are using would be completely invalid as it relies upon G being a constant.And thanks for admitting that you know the formula is invalid and thus your application of it is invalid, and all the claims based upon it are invalid.
Sure
Let us perform the calculation once more.Why?
I've nothing to hide but the "well-known expression" is not "G = gr^{2}/M" but is g = GM/r^{2}.
If G is a constant, as you say, then simply substituting its well-known expression, G = gr^{2}/M, will modify nothing at all, unless you have something to hide.
M = 1.989 x 10^{30} kg
Using P(700,000,000)
You explained nothing at all.There you go projecting again.
There is a FULL connection to gravity.Is that why you are completley unable to provide it and instead just appeal to the changes in gravity to then claim that the pressure should change?
Not in the slightest!But G is not a universal constant.That's all quite irrelevant because we know that the whole "universe DOES NOT obey Newton's law of "universal" gravitation.".
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The universe DOES NOT obey Newton's law of "universal" gravitation.
But G is still regarded as "universal constant" and is used with the same value everywhere, including in Clayton's equation.
Are you high on something?
You have just made TWO CONTRADICTORY STATEMENTS.No, I did NOT make "TWO CONTRADICTORY STATEMENTS"!
because we know that the whole "universe DOES NOT obey Newton's law of "universal" gravitation."None of the Universe precisely obeys Newtonian Mechanics and Gravitation but most of it follows it very closely.
But G is still regarded as "universal constant" [/i]Certainly "G is still regarded as a universal constant" and it appears in numerous places in physics such as:
That is why nobody takes your messages seriously any longer.That does not follow at all!
If the universe DOES not obey Newton's law of gravitation, then obviously G can no longer be regarded as a constant.
It takes a single counterexample to show that G is not constant.
DARK FLOW:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1936995#msg1936995
‘Because the dark flow already extends so far, it likely extends across the visible universe,’ Kashlinsky says.
you undeniably use the Clayton equation inappropriately by forcing a value of g out of it and implicitly a value of G vastly differentNo! Clayton's equation is an approximation to the pressure distribution inside a star, nothing more!
Rick Bradford is assuming that the pressure in the chromosphere is the PREDICTED/THEORETICAL VALUE. That it is not.
Moreover, G does equal g^{2}/M. Surely if you what you say is right, there should be no problem to use the formula with g instead of G.No, "G does" NOT "equal g^{2}/M" - you made a simple mistake but G might equal g.r^{2}/M IF you had precisely correct values of g, r and M.
We can even keep G as it is and solve for it in terms of P(700,000).No, you have NOT proven that "G is not a constant".
G is not a constant as I have just proven.
Certainly you have something to hide if you are so troubled by a simple substitution in the equation.It is a "simple substitution" but using values taken from at region that Rick Bradford says "the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star" and so you answer is likewise "expected to be seriously in error".
Are you saying that G does not equal g^{2}/M?No, you made a mistake as noted before.
If the Sun is a sphere, you should have nothing to fear.I've nothing to hide and I'm hiding nothing just saying that YOU are wrong and misusing Clayton's equation,
If the Sun is a disk, then obviously the values of g and G will be markedly different.Read my previous post! But just look what YOU have done:
Why are you forbidding your readers to solve the equation for g?
From page 11:
We have seen above that the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star. Nevertheless, it is of interest to see what numerical values are predicted at the surface. For this we use R/a = 5.4, which gives the density, pressure and temperature at the surface to be 1.3 x 10^{-6} kg/m^{3}, 4.1 x 10^{3} Pa (=0.041 atm) and 233,000 K respectively.
The density and pressure are at least small - though whether they are small enough is not obvious without comparison with the correct result. The density is equivalent to about 7 x 10^{20} protons per m^{3}, and a pressure of 0.041 atm is a very poor vacuum. . . . . . Hence, for reasonable predictions near the surface of the star we need to solve the full structure equation, (10), for which a polytopic equation of state is required.
The shape of the Sun cannot be spherical.Of course, the shape of the Sun is spherical. Just poke your head out the door and have a look!
"The atmosphere, weighing down upon Earth’s surface under the force of gravity, exerts a pressure at sea level that scientists define as 1 bar (a term that comes from the same root as barometer, an instrument used to measure atmospheric pressure)."Try again! If you want to be accurate it is 1.01325 Bar!
Standard Atmospheric Pressure
The Standard Atmospheric Pressure (atm) is normally used as the reference when listing gas densities and volumes. The Standard Atmospheric Pressure is defined at sea-level at 273K (0°C) and is 1.01325 bar
The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth drops to one-thousandth of the pressure on the earth in the photosphere.Why is that even relevant? The pressure at the Karman Line (100 km above earth) is only 1.0 x 10^{-5} bar.
This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.No, that is quite wrong! The atmospheric pressure depends on a lot more than the surface gravity.
The shape of the Sun cannot be spherical.
Did you pass Physics 101?No. In high school it was just science and physics. No 101.
The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earthBASED ON WHAT?
You are not paying attention.No! You are the one not paying attention to the simplest of matters!
DefinitionAnd all that depends where you chose to define the surface! On "Earth-like" planets it is obvious but the Sun and the gas planets have no such well-defined surface.
"An atmosphere is a layer or a set of layers of gases surrounding a planet or other material body, that is held in place by the gravity of that body."
Definition
"The atmosphere, weighing down upon Earth’s surface under the force of gravity, exerts a pressure at sea level that scientists define as 1 bar (a term that comes from the same root as barometer, an instrument used to measure atmospheric pressure)."
You are trolling this website.No, and you don't even know the meaning of "trolling"!
Here is the recorded CHROMOSPHERE PRESSURE: 10^{-13} BAR.Thre is no "recorded CHROMOSPHERE PRESSURE" nor any single "recorded PHOTOSPHERE PRESSURE" because neither have been directly measured but inferred from other measurements and calculations.
(https://image.ibb.co/hkvQrJ/chromo.jpg)
Here is the recorded PHOTOSPHERE PRESSURE: 8.6x10^{-4} BAR.
But the Sun does have a well-defined surface.
http://www.ptep-online.com/2011/PP-26-08.PDF
On the Presence of a Distinct Solar Surface: A Reply to Herve Faye
(https://image.ibb.co/nnigj8/rds.jpg)
Spectacular images of the solar surface have been acquired in recent years, all of which manifest phenomenal structural elements on or near the solar surface. High resolution images acquired by the Swedish Solar Telescope reveal a solar surface in three dimensions filled with structural elements.
Beyond the evidence provided by the Swedish Solar Telescope and countless other observations, there was clear Doppler confirmation that the photosphere of the Sun was behaving as a distinct surface. In 1998, Kosovichev and Zharkova published their Nature paper X-ray flare sparks quake inside the Sun. Doppler imaging revealed transverse waves on the surface of the Sun, as reproduced in Figure 2: “We have also detected flare ripples, circular wave packets propagating from the flare and resembling ripples from a pebble, thrown into a pond”. In these images, the “optical illusion” was now acting as a real surface. The ripples were clearly transverse in nature, a phenomenon difficult to explain using a gaseous solar model. Ripples on a pond are characteristic of the liquid or solid state.
(https://apod.nasa.gov/apod/image/9806/sunquake_soho_big.jpg)
In 1998, Kosovichev and Zharkova published their Nature paper X-ray flare sparks quake inside the Sun. Doppler imaging revealed transverse waves on the surface of the Sun, as reproduced in the above figure: “We have also detected flare ripples, circular wave packets propagating from the flare and resembling ripples from a pebble, thrown into a pond”.
But the Sun does have a well-defined surface.As I said the Sun's surface is defined as the surface of the photosphere but that is just the visible surface, not a solid surface.
http://www.ptep-online.com/2011/PP-26-08.PDF
You are not paying attention.No, I am.
The atmospheric pressure (photosphere pressure) is 28 times stronger than the atmospheric pressure on Earth (theoretical prediction).What theory?
Here is a reference which does illustrate the correctness of the Clayton model:You mean where it says it is not appropriate for today's sun?
"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."Really?
You are in no position to act hereAnd there you go with more projection, spam and insults.
Read again.I have, it doesn't say what you claim.
You haven't done your homework on the subject: no magnetic forces can cause those ripples.
The evidence you provide was part of a theory that solar flares cause these ripples, that are carried by the magnetic force, produce Xrays, microwaves and a shockwave that further heats that layer of the sun. These observations appear to confirm this theory.I don't know who wrote that but I'll let them answer it.
You haven't done your homework on the subject: no magnetic forces can cause those ripples.
The surface gravity of the Sun is about 274 m/s2I've answered that numerous times!
Let's put your word to the test.
Please note the "can yield reasonably correct results"! The Clayton model is an approximation this is not accurate enough at the surface to use the way you do.
Here is a reference which does illustrate the correctness of the Clayton model:
The Physics of Stars by A. C. Phillips (https://books.google.ro/books?id=ue2D__e06XkC&pg=PT146&lpg=PT146&dq=clayton+model+accuracy+stellar+pressure&source=bl&ots=nw7jNgMv4i&sig=ACfU3U1JJ5IALZvJlJw3avQmR0XQXHjnnQ&hl=en&sa=X&ved=2ahUKEwicyeGX0uflAhURPFAKHWMuBi0Q6AEwBnoECAkQAg#v=onepage&q=clayton%20model%20accuracy%20stellar%20pressure&f=false)
"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."
The author continues:It might be the "correct equation" but it is not accurate enough to use as you do!
"To find the variation in the pressure inside the sun we adopt the appropriate value for the length parameter."
"This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."
Now, for the most important part:
"However, the necessary small pressure gradient near the surface of the star will be reproduced if the value of the length parameter a is small with respect to the radius R of the star."
The value for a is: 1.29 x 10^{8}.
The value I derived is 1.06 x 10^{8}, which is even better.
So, I am using the correct equation, with the correct parameter a.
Ricks Cosmology Tutorial: Chapter 11' Stellar Structure Part1 (http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf)
We have seen above that the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star. Nevertheless, it is of interest to see what numerical values are predicted at the surface. For this we use R/a = 5.4, which gives the density, pressure and temperature at the surface to be 1.3 x 10^{-6} kg/m^{3}, 4.1 x 10^{3} Pa (=0.041 atm) and 233,000K respectively. The density and pressure are at least small - though whether they are small enough is not obvious without comparison with the correct result. The density is equivalent to about 7 x 10^{20} protons per m^{3}, and a pressure of 0.041 atm is a very poor vacuum. The temperature is certainly wrong by a large factor, the correct value being about 6,000 K. Hence, for reasonable predictions near the surface of the star we need to solve the full structure equation, (10), for which a polytopic equation of state is required.
The Clayton model provides us with the g value: g = 0,0000507 m/s^2 which is much lower than the centrifugal acceleration figure:Only if you use the Clayton model in a range where Rick Bradford, himself, says, "the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star".
P(r) = 2πgr^{2}a^{2}ρ^{2}_{c}e^{-x2}/3MAll the above is a total waste of time because your are assuming that the Clayton model is completely accurate outside the surface of the sun and it is NOT!
where a = (3^{1/2}M/2^{1/2}4πρ_{c})^{1/3}
a = 106,165,932.3; x = r/a
M = 1.989 x 10^{30} kg
central density = 1.62 x 10^{5} kg/m^{3}
G = gr^{2}/m(r)
m(r) = M(r/R)^{3}(4 - 3r/R); if r = R, then M = m(r)
Using P(700,000,000) = 1.0197 x 10^{-9} kg/m^{2} value, we get:
g = 0,0000507 m/s^{2}
RATIO
a_{c}/g = 0.0063/0.0000507 = 124.26
Accuracy of the Clayton model:
(https://image.ibb.co/eHYH5d/chro2.jpg)
There is still another way to prove the correctness of the Clayton equation.Yes, if you do it correctly "Everything works out fine".
Modern science tells us that one gram at the surface of the earth would weigh 27.47 grams on the surface of the sun (as expected because of the gravitational pull of the large solar mass; atmospheric pressure of the sun being 27.47 times greater than the atmospheric pressure of the earth, according to the presumed/theoretical value).
g(sun) = 27.47 x g(earth)
g(sun) = 270 m/s2
Everything works out fine, right?
<< Ignored >>
"But the most pressing complication lies in the reality that gases are unable to generate powerful magnetic fields. They can respond to fields, but have no inherent mechanism to produce these phenomena. Along these lines, how can magnetic fields be simultaneously produced by gases while at the same time prevent them from rising into the sunspot umbra? On Earth, the production of powerful magnets involves the use of condensed matter and the flow of electrons within conduction bands, not isolated gaseous ions or atoms.
Astrophysics cannot hope that magnetic fields impart ‘illusionary’ details and emissive properties to photospheric objects (e.g. sunspots and faculae), while at the same time requiring that real structure exists in a gaseous Sun. This structure must somehow enable the formation of powerful magnetic fields and the buildup of a solar dynamo. The fact remains that the generation of strong magnetic fields on Earth always requires the action of condensed matter. As they have no structure, gases are unable to generate magnetic fields on a macroscopic level. They are simply subject to their action. It
is improper to confer upon gases behavior which cannot even be approached in the laboratory."
See also:
The Clayton model is an approximation this is not accurate enough at the surface to use the way you do.
But it is.
https://books.google.ro/books?id=ue2D__e06XkC&pg=PT146&lpg=PT146&dq=clayton+model+accuracy+stellar+pressure&source=bl&ots=nw7jNgMv4i&sig=ACfU3U1JJ5IALZvJlJw3avQmR0XQXHjnnQ&hl=en&sa=X&ved=2ahUKEwicyeGX0uflAhURPFAKHWMuBi0Q6AEwBnoECAkQAg#v=onepage&q=clayton%20model%20accuracy%20stellar%20pressure&f=false
No projection at all.You are accusing others of what is wrong with you. That is projection.
Go ahead and take your messages to some other forum, and see how long you will lastAnd how long do you last on other forums before they ban you for your persistent, pathetic spam and refusal to even attempt to debate?
"But the most pressing complication lies in the reality that gases are unable to generate powerful magnetic fields.Only someone completely ignorant of what the sun is, or with no interest in the truth would make such a claim regarding the sun.
It takes less than 10 seconds to defeat you. Very fast. Then, you resort to trolling techniques to get through the debate. If there was any moderation here, as it should be, you'd be banned in less than a couple of days.
The admin and mods do not seem to understand what is happening here: the drop in new users/new FE members/# of messages is due directly to the sabotaging efforts of jackblack and rabinoz.
Here is the paradox: on one hand, the admin wish to have more debates, more new members joining this forum, more messages. On the other hand, they permit two amateurish arm-chair physicists, jackblack and rabinoz, to troll this forum, so that no one can debate anything at all.
Magnetic fields cannot generate the physical structures which can be observed in the Sun's atmosphere.
(https://apod.nasa.gov/apod/image/9806/sunquake_soho_big.jpg)
http://www.ptep-online.com/2011/PP-26-08.PDF
Beyond the evidence provided by the Swedish Solar Telescope and countless other observations, there was clear Doppler confirmation that the photosphere of the Sun was behaving as a distinct surface. In 1998, Kosovichev and Zharkova published their Nature paper X-ray flare sparks quake inside the Sun. Doppler imaging revealed transverse waves on the surface of the Sun, as reproduced in Figure 2: “We have also detected flare ripples, circular wave packets propagating from the flare and resembling ripples from a pebble, thrown into a pond”. In these images, the “optical illusion” was now acting as a real surface. The ripples were clearly transverse in nature, a phenomenon difficult to explain using a gaseous solar model. Ripples on a pond are characteristic of the liquid or solid state.
Nope the author clearly states its a simple approximation, that gives enough detail to permit a general insight into some of the general features of a stellar structure.
Let's see.
https://books.google.ro/books?id=ue2D__e06XkC&pg=PT146&lpg=PT146&dq=clayton+model+accuracy+stellar+pressure&source=bl&ots=nw7jNgMv4i&sig=ACfU3U1JJ5IALZvJlJw3avQmR0XQXHjnnQ&hl=en&sa=X&ved=2ahUKEwicyeGX0uflAhURPFAKHWMuBi0Q6AEwBnoECAkQAg#v=onepage&q=clayton%20model%20accuracy%20stellar%20pressure&f=false
"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."
The author continues:
"To find the variation in the pressure inside the sun we adopt the appropriate value for the length parameter."
"This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."
Now, for the most important part:
"However, the necessary small pressure gradient near the surface of the star will be reproduced if the value of the length parameter a is small with respect to the radius R of the star."
The value for a is: 1.29 x 10^8.
The value I derived is 1.06 x 10^8, which is even better.
So, I am using the correct equation, with the correct parameter a.
It takes less than 10 seconds to defeat you.If that was the case you would have done so already rather than repeatedly spamming the same refuted nonsense.
The Clayton model is an approximation this is not accurate enough at the surface to use the way you do.There is nothing in that to say that "it is"!
But it is.
"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."
"This figure shows impressive agreement betwen the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."Maybe so, but There is nothing in that to say that "it is" accurate enough to use the way you did!
Now, for the most important part:I'm not disputing that!
"However, the necessary small pressure gradient near the surface of the star will be reproduced if the value of the length parameter a is small with respect to the radius R of the star."
The value for a is: 1.29 x 10^8.
The value I derived is 1.06 x 10^8, which is even better.
So, I am using the correct equation, with the correct parameter a.
Obviously, Rick Bradford did not fully investigate/research the Clayton model, as did Dr. A.C. Phillips (University of Manchester).
Remember that "The Physics of the Stars" is published by Wiley, one of the top publishers of scientific textbooks, and at such it is heavily peer-reviewed at every step.
The statements/graphs used by Dr. A.C. Phillips, pertaining to the Clayton model, passed this review process with flying colors.
Yes, if you do it correctly "Everything works out fine".No they do not get "flushed down the toilet"! Your misuse of Clayton's equation gets "flushed down the toilet"!
You wrote that
M = 1.989 x 1030,
G = gr2/m(r) or g = G.m(r)/r2,
R = 700,000,000 m (695,510,000 m is more accurate)
and Rick Bradford used G = 6.67 x 10-11 N.m2/kg2.
So simply solve those for g = 6.67 x 10-11 x 1.989 x 1030/695,510,0002 = 274.25 m/s2 using R = 695,510,000 m.
And "Everything works out fine"!
You see, you and R. Bradford are using the THEORETICAL/ASSUMED value for the pressure of the chromosphere.
Once we plug in the real figure, 10^{-13} BAR, your calculations are flushed down the toilet, that is how useless they are.
"This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."I have!
The Clayton model is an approximation this is not accurate enough at the surface to use the way you do.
But it is.
https://books.google.ro/books?id=ue2D__e06XkC&pg=PT146&lpg=PT146&dq=clayton+model+accuracy+stellar+pressure&source=bl&ots=nw7jNgMv4i&sig=ACfU3U1JJ5IALZvJlJw3avQmR0XQXHjnnQ&hl=en&sa=X&ved=2ahUKEwicyeGX0uflAhURPFAKHWMuBi0Q6AEwBnoECAkQAg#v=onepage&q=clayton%20model%20accuracy%20stellar%20pressure&f=false
Nope the author clearly states its a simple approximation, that gives enough detail to permit a general insight into some of the general features of a stellar structure.
You are trolling, yet again, this forum.There is nothing in "This figure shows impressive agreement" to support agreement of exceedingly low pressures like 10^{-13} BAR when the central pressure is 1.65 x 10^{11} BAR.
But your expecting agreement of 10-13 BAR when the central pressure range is 1.65 x 1011 BAR is quite outside anything envisaged by Stomgren's "impressive agreement".
Clayton's model takes care of both situations: core and surface.
You are being ridiculous to claim that "This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."
covers anything like 10-13 BAR when the total pressure range is zero to 1.65 x 1011 BAR.
Clayton's model approximates the pressure at the surface of the Sun very well.
"This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."
The Clayton model is an approximation this is not accurate enough at the surface to use the way you do.
But it is.
https://books.google.ro/books?id=ue2D__e06XkC&pg=PT146&lpg=PT146&dq=clayton+model+accuracy+stellar+pressure&source=bl&ots=nw7jNgMv4i&sig=ACfU3U1JJ5IALZvJlJw3avQmR0XQXHjnnQ&hl=en&sa=X&ved=2ahUKEwicyeGX0uflAhURPFAKHWMuBi0Q6AEwBnoECAkQAg#v=onepage&q=clayton%20model%20accuracy%20stellar%20pressure&f=false
Nope the author clearly states its a simple approximation, that gives enough detail to permit a general insight into some of the general features of a stellar structure.
Shall we all just start repeating ourselves ::)
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
The Clayton model is valid: textbook published by Wiley.Pretending your source says something that it doesn't will not help.
Please read:Why?
"This figure shows impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:No, it also matches observations of other planets, not just Earth.
I'd rather say that I've "LINKED THE CORRECTNESS OF THE VALUE OF 274 m/s2 TO THE" well verified "ORBITAL MOTION OF THE EARTH".Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Do you, rabinoz, understand what you have just done?
You have just LINKED THE CORRECTNESS OF THE VALUE OF 274 m/s2 TO THE HYPOTHESIZED ORBITAL MOTION OF THE EARTH!!!
M = 1.989 x 10^{30} kgAnd the easily checked average angular diameter of the Sun viewed from Earth is 0.533°.
central density = 1.62 x 10^{5} kg/m^{3}
Using P(700,000,000) = 1.0197 x 10^{-9} kg/m^{2} value, we get:
Not very bright of you.Oh, I should have realised that the Sun is only 600 m in diameter and only 15 km from the Earth ;D ;D ;D!
Thanks to you now, we have at our disposal the very best proof that indeed the 274 m/s2 figure is pure bonkers.
GPS satellites do not register the SOLAR GRAVITATIONAL POTENTIAL EFFECT:That is possibly because any changes due to the solar gravitational potential effect are too small to register.
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706
GPS satellites do not register the ORBITAL SAGNAC/CORIOLIS EFFECTS: https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1917978#msg1917978It's NOT Coriolis effect! The Coriolis acceleration is given by, a = 2.v.Ω - that's nothing like the Sagnac effect! See The Coriolis acceleration by E. Linacre and B. Geerts (http://www-das.uwyo.edu/~geerts/cwx/notes/chap11/coriolis_acc.html).
The Earth is not orbiting the Sun.Well, that does seem a little hard to swallow because:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.
C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .I am not casting doubt on any of that! Just on your attempt to use those publications to support your hypotheses!
LISA Space Antenna
(https://image.ibb.co/ivHjjS/lisa2.jpg)
The LISA interferometer rotates both around its own axis and around the Sun as well, at the same time.
I have another theorem that I would like to present to Sandy. Its called Virial theorem and I would like to know what he knows about that. I have some specific questions I would like to address which would help my own research.
Welcome diversion
https://www.scienceforums.net/
Is a good forum, great physics sections, if you dont already know it. If you look close enough you can see Sandy get his ass handed to him there too.
On this comparison graph, even an error of 1 BAR (105 Pa) would be completely lost.OK, then you show a comparison graph that expands the pressure scale enough to observe 10^{-13} bar.
Not at all; simply the range of the graph has to be extended to better cover the 0.8 - 1 solar radius axis.
You are continuously trying to fool/trick/obfuscate your readers.No I'm trying to save them from your obvious misuse of the Clayton model.
So you expecting agreement down around 10^{-13} BAR is ludicrous!I never disputed that. All I'm disputing is that it is ludicrous to expect the Clayton pressure equation to fit to that precision.
That is the REAL VALUE of the pressure of the chromosphere! 10^{-13} BAR.
Obviously, your 274 m/s2 figure for g(sun) is a piece of garbage.Is not just my figure! It seems to be everybody's figure except yours! That says more about your ideas than mine.
The REAL DATA ON THE FIELD says otherwise: that the value of g(sun) must be MUCH LOWER than 274 m/s2.No, it does not!
Ricks Cosmology Tutorial: Chapter 11' Stellar Structure Part1 (http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf)Don't you understand simply words like "the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star."?
We have seen above that the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star. Nevertheless, it is of interest to see what numerical values are predicted at the surface. For this we use R/a = 5.4, which gives the density, pressure and temperature at the surface to be 1.3 x 10^{-6} kg/m^{3}, 4.1 x 10^{3} Pa (=0.041 atm) and 233,000K respectively. The density and pressure are at least small - though whether they are small enough is not obvious without comparison with the correct result. The density is equivalent to about 7 x 10^{20} protons per m^{3}, and a pressure of 0.041 atm is a very poor vacuum. The temperature is certainly wrong by a large factor, the correct value being about 6,000 K. Hence, for reasonable predictions near the surface of the star we need to solve the full structure equation, (10), for which a polytopic equation of state is required.
How much lower?Sure but you are "using the correct equation, with the correct parameter a" where it certainly grossly wrong!
<< Repeated calculations deleted! >>
So, I am using the correct equation, with the correct parameter a.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
You are an embarrassment to the RE.The embarrassment here is you.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
You are an embarrassment to the RE.
They will never forgive you for this.
You should have stayed in AR, here you are more than useless.
That is possibly because any changes due to the solar gravitational potential effect are too small to register.
You must be dreaming.
http://www.tuks.nl/pdf/Reference_Material/Ronald_Hatch/Hatch-Relativity_and_GPS-II_1995.pdf
It is very important to note that the GPS satellites' clock rate and the
receiver's clock rate are not adjusted as a function of their velocity relative to one
another. Instead, they are adjusted as a function of their velocity with respect to the
chosen frame of reference—in this case the earth-centered, non- rotating, (quasi) inertial
frame.
N. Ashby tried to make a similar claim.
Ashby’s claim is equivalent to the claim
found elsewhere [22] that the local frame rotates with the
orbit and that the sun’s differential gravitational potential
is canceled by “centripetal acceleration,” i.e. by the
differential velocity with respect to the sun. In other
words, it is claimed that the inertial frame indeed rotates
once per year. However, the GPS clocks clearly show
this argument is not valid. The orientation of the GPS
orbital planes does not rotate to maintain the same angle
with respect to the sun, so there is no differential velocity
orthogonal to the orbital plane. And there can be no
differential velocity within the orbital plane or else
Kepler’s laws would be violated. Thus, GPS clocks do not
suffer centripetal acceleration. Furthermore, if this
argument were correct, the differential gravitational
potential would be canceled in the sun’s frame as well.
The JPL reference document [7] and the Hill pulsar
document [19] clearly show that such a cancellation does
not occur.
http://www.tuks.nl/pdf/Reference_Material/Ronald_Hatch/Hatch-Clock_Behavior_and_theSearch_for_an_Underlying_Mechanism_for_Relativistic_Phenomena_2002.pdf
YOU HAVE FAILED TO ADDRESS THE GRAVITATIONAL POTENTIAL ANOMALY INHERENT IN GPS TECHNOLOGY:
Many people believe that GR accounts for all the observed
effects caused by gravitational fields. However, in
reality GR is unable to explain an increasing number of
clear observational facts, several of them discovered recently
with the help of the GPS. For instance, GR
predicts the gravitational time dilation and the slowing of
the rate of clocks by the gravitational potential of Earth,
of the Sun, of the galaxy etc. Due to the gravitational
time dilation of the solar gravitational potential, clocks in
the GPS satellites having their orbital plane nearly parallel
to the Earth-Sun axis should undergo a 12 hour period
harmonic variation in their rate so that the difference
between the delay accumulated along the half of the orbit
closest to the Sun amounts up to about 24 ns in the time
display, which would be recovered along the half of the
orbit farthest from the Sun. Such an oscillation exceeds
the resolution of the measurements by more than two
orders of magnitude and, if present, would be very easily
observed. Nevertheless, contradicting the predictions of
GR, no sign of such oscillation is observed. This is the
well known and so long unsolved non-midnight problem.
In fact observations show that the rate of the
atomic clocks on Earth and in the 24 GPS satellites is
ruled by only and exclusively the Earth’s gravitational
field and that effects of the solar gravitational potential
are completely absent. Surprisingly and happily the GPS
works better than expected from the TR.
Obviously the gravitational
slowing of the atomic clocks on Earth cannot be due to
relative velocity because these clocks rest with respect to
the laboratory observer. What is immediately disturbing
here is that two completely distinct physical causes produce
identical effects, which by it alone is highly suspicious.
GR gives only a geometrical interpretation to the
gravitational time dilation. However, if motions cause
time dilation, why then does the orbital motion of Earth
suppress the time dilation caused by the solar gravitational
potential on the earthbased and GPS clocks? Absurdly
in one case motion causes time dilation and in the
other case it suppresses it. This contradiction lets evident
that what causes the gravitational time dilation is not the
gravitational potential and that moreover this time dilation
cannot be caused by a scalar quantity. If the time dilation
shown by the atomic clocks within the earthbased
laboratories is not due to the gravitational potential and
cannot be due to relative velocity too then it is necessarily
due to some other cause. This impasse once more
puts in check the central idea of the TR, according to
which the relative velocity with respect to the observer is
the physical parameter that rules the effects of motions.
The above facts show that the parameter that rules the
effects of motions is not relative velocity but a velocity
of a more fundamental nature.
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1784780#msg1784780
but who accepts the local ether model?
Every scientist who is faced with the fact that the GPS satellites do not record/register the orbital Sagnac effect must accept eventually the Lorentz ether model.
C.C. Su certainly accepts the reality of the Earth rotating and orbiting the Sun, so completely ignore I'll your attempt to use his work to disprove those things!
Dr. Su just presented ABSOLUTE PROOFS that the GPS satellites do not record the ORBITAL SAGNAC EFFECT.
His papers are published in the most respected journals.
However, if the GPS satellites do not record the ORBITAL SAGNAC, then obviously the Earth is not orbiting the Sun.
Very simple to understand.
That is why HE IS FORCED TO ACCEPT THE LOCAL-AETHER MODEL.
But this contradicts each and every statement ever made by Newton or by Einstein.
EVERYONE ACCEPTS THE EXISTENCE OF THE MISSING ORBITAL SAGNAC EFFECT.
LISA Space Antenna
(https://image.ibb.co/ivHjjS/lisa2.jpg)
The LISA interferometer rotates both around its own axis and around the Sun as well, at the same time.
That is, the interferometer will be subjected to BOTH the rotational Sagnac (equivalent to the Coriolis effect) and the orbital Sagnac effects.
Given the huge cost of the entire project, the best experts in the field (CalTech, ESA) were called upon to provide the necessary theoretical calculations for the total phase shift of the interferometer. To everyone's surprise, and for the first time since Sagnac and Michelson and Gale, it was found that the ORBITAL SAGNAC EFFECT is much greater than the CORIOLIS EFFECT.
The factor of proportionality is R/L (R = radius of rotation, L = length of the side of the interferometer).
(https://image.ibb.co/iMSdB7/lisa3.jpg)
Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet
https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe
This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:
http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta
In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.
This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.
For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.
The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.
The MISSING ORBITAL SAGNAC EFFECT IS A FACT OF SCIENCE, ACCEPTED BY BOTH NASA AND ESA AND CALTECH.
Now, we are back to your catastrophic derivation and comparison.Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
If the Earth is not orbiting the Sun, a(sun) DOES NOT equal 274.35m/s2: IN FACT IT IS EQUAL TO ZERO.
Then, we are left with the centrifugal acceleration: a_{c} = 0.0063 m/s^{2}.
Thus, the Clayton model is fully vindicated, as is all of the information I have provided in this thread.
GPS satellites DO NOT REGISTER/RECORD THE ORBITAL SAGNAC EFFECT.
This is a fact of science.
Then, the Earth is not orbiting the Sun at all.
As for the scienceforums link, I WAS THE ONE WHO MENTIONED IT, remember?
Let's go to page 1 of that link.
https://www.scienceforums.net/topic/118524-globalgeneralized-sagnac-effect-formula/#comments
A total disaster for the "physicists" at scienceforums.
They are unable to mount any kind of a defense.
Their star, swansontea, cannot explain anything at all.
Page 2
https://www.scienceforums.net/topic/118524-globalgeneralized-sagnac-effect-formula/page/2/#comments
A huge disaster for scienceforums: they resort to trolling to escape the unavoidable conclusions.
Page 3
https://www.scienceforums.net/topic/118524-globalgeneralized-sagnac-effect-formula/page/3/#comments
A total disaster for scienceforums: they cannot explain the fact that the SAGNAC EFFECT does not feature any area at all.
So, they are forced to close the thread.
As for the comments of the mods, they do this with every thread they close, in order to convey a positive image for themselves, but the thread speaks for itself: they were unable to explain the formulas I presented.
rabinoz linked the MISSING ORBITAL SAGNAC EFFECT WITH THE 274 M/S2 FIGURE.
Since the GPS satellites do not register the MISSING ORBITAL SAGNAC, the Earth is not orbiting the Sun at all.
Therefore a(sun) = ZERO.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
You are an embarrassment to the RE.You're really getting worked up now aren't you?
They will never forgive you for this.
You should have stayed in AR, here you are more than useless.
That is possibly because any changes due to the solar gravitational potential effect are too small to register.Nope! As Neil Ashby correctly shows "any changes due to the solar gravitational potential effect are too small to register".
You must be dreaming.
RELATIVITY AND GPS, Ronald R. Hatch (Section I: Special Relativity) (http://www.tuks.nl/pdf/Reference_Material/Ronald_Hatch/Hatch-Relativity_and_GPS-II_1995.pdf)You mean the Ronald R. Hatch of the Electric Universe?
<< Hence ignored >>SO you name ONE!
but who accepts the local ether model?
Every scientist who is faced with the fact that the GPS satellites do not record/register the orbital Sagnac effect must accept eventually the Lorentz ether model.
C.C. Su certainly accepts the reality of the Earth rotating and orbiting the Sun, so completely ignore I'll your attempt to use his work to disprove those things!He did? But who's arguing?
Dr. Su just presented ABSOLUTE PROOFS that the GPS satellites do not record the ORBITAL SAGNAC EFFECT.
His papers are published in the most respected journals.Where did C.C. Su ever even hint at that? Chapter and verse please!
However, if the GPS satellites do not record the ORBITAL SAGNAC, then obviously the Earth is not orbiting the Sun.
Very simple to understand.The LISA geometry is totally different but I'll look into it later.
That is why HE IS FORCED TO ACCEPT THE LOCAL-AETHER MODEL.
But this contradicts each and every statement ever made by Newton or by Einstein.
EVERYONE ACCEPTS THE EXISTENCE OF THE MISSING ORBITAL SAGNAC EFFECT.
LISA Space Antenna
(https://image.ibb.co/ivHjjS/lisa2.jpg)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Now, we are back to your catastrophic derivation and comparison.Catastrophic?
That method might but I could use values from you and your own first reference and do it like this:Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
If G is a constant, as you say, then simply substituting its well-known expression, G = gr^{2}/M, will modify nothing at all, unless you have something to hide.I've nothing to hide but the "well-known expression" is not "G = g.r^{2}/M" but is g = G M/r^{2}.
Now we know that:
the mass of the sun = 1.989 x 10^{30} kg, r = 700,000,000 m (you said so ^{[1]}) and G = 6.67 x 10^{-11} m^{3}⋅kg^{−1}⋅s^{−2} (Rick Bradford used that value in evaluating the parameters).
So we get g = 6.67 x 10^{-11} x 1.989 x 10^{30}/700,000,000^{2} = 270.7 m/s^{2} but had you used r = 695,510 km, the results would have been g = 274.3 m/s^{2}, funny that!
[1] M = 1.989 x 10^{30} kg
Using P(700,000,000)
If the Earth is not orbiting the Sun, a(sun) DOES NOT equal 274.35m/s2: IN FACT IT IS EQUAL TO ZERO.No, we are NOT, see above.
Then, we are left with the centrifugal acceleration: a_{c} = 0.0063 m/s^{2}.
Thus, the Clayton model is fully vindicated, as is all of the information I have provided in this thread.Sure, "the Clayton model" is "fully vindicated" as an approximation but it is "certainly grossly wrong at the surface of the star".
From page 11:
We have seen above that the Clayton model is expected to be seriously in error for values of x greater than unity, and certainly grossly wrong at the surface of the star. Nevertheless, it is of interest to see what numerical values are predicted at the surface. For this we use R/a = 5.4, which gives the density, pressure and temperature at the surface to be 1.3 x 10^{-6} kg/m^{3}, 4.1 x 10^{3} Pa (=0.041 atm) and 233,000 K respectively.
The density and pressure are at least small - though whether they are small enough is not obvious without comparison with the correct result. The density is equivalent to about 7 x 10^{20} protons per m^{3}, and a pressure of 0.041 atm is a very poor vacuum. . . . . . Hence, for reasonable predictions near the surface of the star we need to solve the full structure equation, (10), for which a polytopic equation of state is required.
GPS satellites DO NOT REGISTER/RECORD THE ORBITAL SAGNAC EFFECT.What a magnificient leap of logic that is! Congratulations, but does:
This is a fact of science.
Then, the Earth is not orbiting the Sun at all.
As for the scienceforums link, I WAS THE ONE WHO MENTIONED IT, remember?I see a classic case of the Dunning-Kruger Syndrome coming up!
Let's go to page 1 of that link.Projecting again, I see! Don't you mean that you were "unable to mount any kind of" retional "defense"?
https://www.scienceforums.net/topic/118524-globalgeneralized-sagnac-effect-formula/#comments
A total disaster for the "physicists" at scienceforums.
They are unable to mount any kind of a defense.
Their star, swansontea, cannot explain anything at all.Let's igmore the rest of that debacle!
A huge disaster for scienceforums: they resort to trolling to escape the unavoidable conclusions.
rabinoz linked the MISSING ORBITAL SAGNAC EFFECT WITH THE 274 M/S2 FIGURE.How do you work that out? Exact quote please.
Since the GPS satellites do not register the MISSING ORBITAL SAGNAC, the Earth is not orbiting the Sun at all.Wow, what a leap!
Therefore a(sun) = ZERO.
Do you understand what we are discussing here?Please desist from this continued spamming with exactly the same material over and over again.
In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.
This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.
For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Now, here is an in-depth analysis of N. Ashby's failed claims regarding the solar gravitational potential.The topic is "Solar power source" and I've had a gutful or your continual off-topic spamming of quite irrelevant material.
rabinoz linked the MISSING ORBITAL SAGNAC EFFECT WITH THE 274 M/S2 FIGURE.Stop shouting it makes it look as though you're losing your cool.
Since the GPS satellites do not register the MISSING ORBITAL SAGNAC, the Earth is not orbiting the Sun at all.Only some devoid of any logical thought could make such a claim.
Therefore a(sun) = ZERO.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
GPS satellites DO NOT REGISTER/RECORD THE ORBITAL SAGNAC EFFECT.
Of course "the value of 274 m/s^{2}" does not rest entirely on that statement!Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
that means that the Earth is not orbiting the Sun.
Do you understand what we are discussing here?Yes.
Unfortunately for you, my equations and formulas really do work.Except as shown by me and others repeatedly, they don't.
These are just my own thoughts from reading through this discussion but I suspect (with quite high confidence) that many others share them. Over to the others now...They are definitely not just your own.
Unfortunately for you, my equations and formulas really do work
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
I told you that your trolling was going to stop.Are you hoping to try and troll so much I leave?
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
The ORBITAL SAGNAC EFFECT is not being registered by GPS satellites.
QuoteThe ORBITAL SAGNAC EFFECT is not being registered by GPS satellites.
I think I can say without too much doubt in my mind that GPS satellites and whether they register the orbital sagnac effect or not has got absolutely nothing to do with the power source in the [/b] flat Earth[/b] version of the Suns power source.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Since the GPS satellites ARE NOT registering/recording the missing ORBITAL SAGNAC, that means that the Earth is not orbiting the Sun.
The factYou mean your blatant lie?
Of course "the value of 274 m/s2" does not rest entirely on that statement!Yes, and I was correct because who is going the question that a sidereal year is 365.256363004 Ephemeris days or 31,558,150 SI seconds.
You have linked the correctness of the 274 m/s2 value to the Earth's orbital angular velocity figure.
He also kindly shows the equation to use (though we knew that didn't we?): (https://www.dropbox.com/s/m52xrm9w8m7mo0w/Phillips%20A.C.%20The%20physics%20of%20stars%20%28Wiley%2C1994%29%20g%28r%29%20vs%20r.png?dl=1)And not only that but all references that I can find shoe the same result - just possibly the 274 m/s^{2} is really pretty close!
So g_{sun} = (6.67430 x 10^{-11} x 1.99 x 10^{30})/(6.96 x 10^{8})^{2} = 274.2 m/s^{2} - funny that!
Your doing.No, not just me! It seems that everybody except YOU is doing it - claiming that g_{sun} = 274 m/s^{2}.
And now you are going to have to answer for it.What do YOU plan to do about it? Post more spam totally irrelevant to the topic, "Solar power source".
Therefore, the value of 274 m/s2No, it does not rest entirely on that quite accurate statement! Can't YOU read plain English?RESTS ENTIRELY ON THIS STATEMENT<< Don't shout so much, it's bad manners! >> rests entirely on this statement:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Now, everyone can see that you are have a problem.No, everyone can see that you have a problem.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
https://web.archive.org/web/20170808104846/http://qem.ee.nthu.edu.tw/f1b.pdfIt is you who are trolling this thread and your message belongs to the CN.
This is an IOP article.
The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.
Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.
C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).
[ur=https://web.archive.org/web/20170808104846/http://qem.ee.nthu.edu.tw/f1b.pdfl]Reinterpretation of the Michelson-Morley experiment based on the GPS Sagnac correction[/url]Many statements in Su's paper would be quite meaningless if the Earth were not a rotating Globe that orbited the Sun yet you deny both of these.. . . . . .
Abstract. – By examining the effects of rotational and orbital motions of the Earth on wave propagation in the global positioning system and an intercontinental microwave link, it is pointed out that the Earth’s orbital motion has no influence on these earthbound wave propagations, while the Earth’s rotation does contribute to the Sagnac effect. As the propagation mechanism in the Michelson-Morley experiment cannot be different from that in the aforementioned ones, it is concluded that due to the Earth’s rotation, the shift in interference fringe in this famous experiment is not exactly zero. However, by virtue of the round-trip propagation path, this shift becomes second order and hence is too small to observe within the present precision.
GPS propagation model and Sagnac correction. – It is well known that GPS provides a high accuracy in positioning. The NAVSTAR GPS employs about 24 half-synchronous satellites carrying highly precise and synchronized atomic clocks around six nearly circular
orbits of radius of about 26600 km. Each GPS satellite repeatedly broadcasts microwaves carrying a sequence of its own unique codes which can be used by a receiver to determine the propagation delay time from the satellite to the receiver and then the instant of signal emission.
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/sYou claim that "the Earth is not orbiting the Sun" but Su many times states that the earth does indeed orbit the Sun and his paper relies on it.
If the Earth is not orbiting the Sun, g(sun) DOES NOT equal 274.35m/s2: IN FACT IT IS EQUAL TO ZERO.
Since the GPS satellites ARE NOT registering/recording the missing ORBITAL SAGNAC, that means that the Earth is not orbiting the Sun.
you agree that the 24 GPS satellites exist, and they are in motion, around Earth.Again you deny what Su writes in "the Earth’s rotation rate . . . . This range formula is practiced numerously everyday around the globe" and in "GPS employs about 24 half-synchronous satellites carrying highly precise and synchronized atomic clocks around six nearly circular orbits of radius of about 26600 km".
Satellites orbit above the flat surface of the Earth, at a much lower altitude.
I have just debunked your failed claims and beliefs.No, you have just repeated the same, already refuted, off topic spam.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
The GPS satellites DO NOT RECORD THE ORBITAL SAGNAC EFFECT.
You are trolling this thread.
The ORBITAL SAGNAC EFFECT is proportional to the LINEAR VELOCITY (thus to the RADIUS OF ROTATION).
As such, it is much greater than the effect currently registered by GPS satellites.
Are you able, scientifically, to understand the difference between 4Aω/c^{2} and 2VL/c^{2}?
https://web.archive.org/web/20170808104846/http://qem.ee.nthu.edu.tw/f1b.pdf
This is an IOP article.
The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.
Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.
C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).
(https://image.ibb.co/cEyxQd/sa3.jpg)
By comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.
Dr. Massimo Tinto, Principal Scientist at CALTECH, Jet Propulsion Laboratory
https://web.archive.org/web/20161019095630/http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf
In the SSB frame, the differences between back-forth delay times are very much larger than has been previously recognized. The reason is in the aberration due to motion and changes of orientation in the SSB frame. With a velocity V=30 km/s, the light-transit times of light signals in opposing directions (Li, and L’i) will differ by as much as 2VL (a few thousands km).
SSB = solar system barycenter
Published in the Physical Review D
http://tycho.usno.navy.mil/ is the U.S. Naval Observatory website
https://arxiv.org/pdf/gr-qc/0310017.pdf
Within this frame, which we can assume to be Solar System Barycentric (SSB), the differences between back-forth delay times that occur are in fact thousands of kilometers, very much larger than has been previously recognized by us or others. The problem is not rotation per se, but rather aberration due to motion and changes of orientation in the SSB frame.
The kinematics of the LISA orbit brings in the effects of motion at several orders of magnitude larger than any previous papers on TDI have addressed. The instantaneous rotation axis of LISA swings about the Sun at 30 km/sec, and on any leg the transit times of light signals in opposing directions can differ by as much as 1000 km.
Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.
The ORBITAL SAGNAC calculated at the Jet Propulsion Laboratory amounts to an admitted difference in path lengths of 1,000 kilometers.
The difference in path lengths for the rotational Sagnac is 14.4 kilometers:
https://arxiv.org/pdf/gr-qc/0306125.pdf (Dr. Daniel Shaddock, Jet Propulsion Laboratory)
https://gwic.ligo.org/thesisprize/2011/yu_thesis.pdf (pg. 63)
Formula for the orbital SAGNAC EFFECT derived by Dr. Massimo Tinto, from CALTECH: 2VL
Formula for the rotational SAGNAC EFFECT derived by Dr. Daniel Shaddock, from CALTECH: 4Aω/c^{2}
The factor of proportionality is R/L (R = radius of rotation, L = length of the side of the interferometer).
Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet
https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe
This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:
http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta
In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.
This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.
For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.
(https://image.ibb.co/iMSdB7/lisa3.jpg)
You are trolling this thread.Are you able to explain the derivation of 2VL/c^{2}?
The ORBITAL SAGNAC EFFECT is proportional to the LINEAR VELOCITY (thus to the RADIUS OF ROTATION).
As such, it is much greater than the effect currently registered by GPS satellites.
Are you able, scientifically, to understand the difference between 4Aω/c^{2} and 2VL/c^{2}?
https://web.archive.org/web/20170808104846/http://qem.ee.nthu.edu.tw/f1b.pdfAnd the author recognises that the Earth is a rotating Globe that orbits the Sun once per year and you deny all those so how dare you try to prop up your fiction with Su's paper. .
This is an IOP article.
The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.
Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.
C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).
HYPER, sun-synchronous satellite, SAGNAC EFFECT interferometer aboard:If you accept the truth of NASA's papers I assume that you also accept that this demonstrates that the Earth must be a rotating Globe.
https://bhi.gsfc.nasa.gov/documents/Mission_Concept_Work/ISAL_January_2002_SST/Astrometry/hyper-atomic-gyro.pdf
https://books.google.ro/books?id=-9zD3WSO4DIC&pg=PA154&lpg=PA154&dq=sun+synchronous+satellite+sagnac+effect&source=bl&ots=CVwfjMgK7C&sig=ACfU3U38nv7shvMYY9ZP1BEds0o-PvNNIw&hl=en&sa=X&ved=2ahUKEwjF3fP1ze7lAhWCxosKHa6sDag4ChDoATAIegQIBxAC#v=onepage&q=sun%20synchronous%20satellite%20sagnac%20effect&f=false
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
HYPER, sun-synchronous satellite, SAGNAC EFFECT interferometer aboard:
https://bhi.gsfc.nasa.gov/documents/Mission_Concept_Work/ISAL_January_2002_SST/Astrometry/hyper-atomic-gyro.pdf
https://books.google.ro/books?id=-9zD3WSO4DIC&pg=PA154&lpg=PA154&dq=sun+synchronous+satellite+sagnac+effect&source=bl&ots=CVwfjMgK7C&sig=ACfU3U38nv7shvMYY9ZP1BEds0o-PvNNIw&hl=en&sa=X&ved=2ahUKEwjF3fP1ze7lAhWCxosKHa6sDag4ChDoATAIegQIBxAC#v=onepage&q=sun%20synchronous%20satellite%20sagnac%20effect&f=false
Dr. C.C. Su proves that the GPS satellites do not register the orbital Sagnac effect.
You can't have the Earth orbit the Sun and at the same time have no recording/registering of the ORBITAL SAGNAC EFFECT.Like I said, take your ignorance regarding the Sagnac effect where it belongs.
means that there are elements lighter than hydrogen.There can be no elements lighter than hydrogen.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Satellites orbit above the flat surface of the Earth, at a much lower altitude.What sort of altitude would you be talking here just out of interest?
The diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).And that is based upon what?
As for the distance, here is the real cruising altitude of aircrafts:And how is rejecting the areal cruising altitudes of aircraft going to magically make the sun lower?
The diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).Has FE Theory always been based on such bizarre and ridiculous ideas or suggestions like this? If so no wonder it has never managed to establish itself on the main stage! What does this even mean?
QuoteThe diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).Has FE Theory always been based on such bizarre and ridiculous ideas or suggestions like this? If so no wonder it has never managed to establish itself on the main stage!
to back up your hypothesis of a 600m d sun 15km away.
But I have.
The diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).
to back up your hypothesis of a 600m d sun 15km away.Why is there the slightest connection between the diameter of the Sun and "the total westward precessional movement in the course of a single year"?
But I have.
The diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).
As for the distance, here is the real cruising altitude of aircrafts:Oh come off it! You say:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2044464#msg2044464
An altimeter actually includes an aneroid barometer which measures the atmospheric pressure (actually it measures the effect of the dextrorotatory ether waves).What hogwash! An aneroid barometer measures the atmospheric pressure due to the easily measured g = 9.8 m/s^{2}!
Hahaha cosmic ray device did itNikola Tesla firmly believed in the Globe earth, the Heliocentric Solar System and the Cosmology of his time.
Sandy you do know that Teslas cosmic ray device patent was a passive detector. If he had managed to build it he would have been able to detect ionized particles, effectively inventing a Gieger counter about 100 years before Gieger.
HOW COSMIC FORCES SHAPE OUR DESTINIES, ("Did the War Cause the Italian Earthquake") by Nikola Tesla (http://www.tfcbooks.com/tesla/1915-02-07.htm)I have also read, though I cannot verify it right now, that one reason Tesla disliked Einstein so much is that he believed that Einstein destroyed "Newton's gravitation".NATURAL FORCES INFLUENCE US. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Accepting all this as true let us consider some of the forces and influences which act on such a wonderfully complex automatic engine with organs inconceivably sensitive and delicate, as it is carried by the spinning terrestrial globe in lightning flight through space. For the sake of simplicity we may assume that the earth's axis is perpendicular to the ecliptic and that the human automaton is at the equator. Let his weight be one hundred and sixty pounds then, at the rotational velocity of about 1,520 feet per second with which he is whirled around, the mechanical energy stored in his body will be nearly 5,780,000 foot pounds, which is about the energy of a hundred-pound cannon ball.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The sun, having a mass 332,000 times that of the earth, but being 23,000 times farther, will attract the automaton with a force of about one-tenth of one pound, alternately increasing and diminishing his normal weight by that amount
Though not conscious of these periodic changes, he is surely affected by them.
The earth in its rotation around the sun carries him with the prodigious speed of nineteen miles per second . . . . .
Tesla universe INVENTIONS OF TESLA (https://teslauniverse.com/nikola-tesla/articles/inventions-tesla)It sure looks as though Tesla believed the earth to be a rotating Globe and the Heliocentric Solar System.
Mother Earth Put to Work.
“By this invention every live part of Mother Earth's body would be brought into action. Energy will be collected all over the globe in amounts small or large, as it may exist, ranging from a fraction of one to a few horse power or more. Every water fall can be utilized, every coal field made to produce energy to be transmitted to vast distances, and every place on earth can have power at small cost. One of the minor uses might be the illumination of isolated homes. We could light houses all over the country by means of vacuum tubes operated by high frequency currents. We could keep the clocks of the United States going and give everyone exact time; we could turn factories, machine shops and mills, small or large, anywhere, and I believe could also navigate the air."
Transmission of Intelligence.
One of the most important features of this invention,” said Mr. Tesla, “will be the transmission of intelligence. It will convert the entire earth into a huge brain, capable of responding in every one of its parts. By the employment of a number of plants, each of which can transmit signals to all parts of the world, the news of the globe will be flashed to all points. A cheap and simple receiving device, which might be carried in one's pocket, can be set up anywhere on sea or land, and it will record the world's news as it occurs, or take such special messages as are intended for it.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
There are no strikes, just the local-aether model which is a fact of science since the GPS satellites DO NOT RECORD/REGISTER THE ORBITAL SAGNAC EFFECT.Incorrect! I asked you to:
to back up your hypothesis of a 600m d sun 15km away.Why is there the slightest connection between the diameter of the Sun and "the total westward precessional movement in the course of a single year"?
But I have.
The diameter is based on the subdivision of the distance between the two tropics and the fact that the diameter of the Sun MUST BE LESS than the total westward precessional movement in the course of a single year (1.5 km).
And how does that back up your hypothesis of a 600 m diameter sun 15 km high?
Dr. C.C. Su explains:
"It is supposed that in the region under sufficient influence of the gravity due to the Earth, the Sun, or another celestial body, there forms a local ether which in turn moves with the gravitational potential of the respective body. For earthbound waves, the medium is the earth local ether which as well as earth’s gravitational potential is stationary in an ECI (earth-centered inertial) frame, while the sun local ether for interplanetary waves is stationary in a heliocentric inertial frame."
Each layer of the ether field has a different index of refraction. The angular size of the Sun is a DIRECT MEASURE of this ether field.
Make sure you understand the fact that the GPS satellites do not register the ORBITAL SAGNAC EFFECT.Who said that you can't?
You cannot have a globe orbiting the Sun if there is no ORBITAL SAGNAC EFFECT.
That is why each and every physicist who discovers this fact is practically forced to accept the local-aether model.Really? Please prove that claim, thank you!
Algebraic Approach to Time-Delay Data Analysis for Orbiting LISA by K. Rajesh Nayak and J-Y. Vinet (https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe)That presentation is based on LISA orbiting the Sun about 20° behind the Earth so K. Rajesh Nayak and J-Y. Vinet both must accept that the Earth orbits the Sun. Look at the diagram of LISA's orbit that they use!
This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:Again all of Masimo Tinto's calculations are based on LISA rotating on its own axis and orbiting the Sun about 20° behind the Earth. Look at the diagrams he uses:
LISA: THE LASER INTERFEROMETER SPACE ANTENNA by Masimo Tinto of JPL, CalTech (http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta)
In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.
(https://www.dropbox.com/s/1zoe8pzqaonlmvk/LISA-%20THE%20LASER%20INTERFEROMETER%20SPACE%20ANTENNA%20by%20Masimo%20Tinto%20of%20JPL%2C%20CalTech%20-%20paper34%20page%20384%20%232%20LISA%20orbit.png?dl=1) | (https://www.dropbox.com/s/akdq9lu2h8m7x5f/LISA-%20THE%20LASER%20INTERFEROMETER%20SPACE%20ANTENNA%20by%20Masimo%20Tinto%20of%20JPL%2C%20CalTech%20-%20paper34%20page%20384%20%233%20LISA%20orbit.png?dl=1) |
The reason is in the aberration due to motion and changes of orientation in the SSB frame. With a velocity V=30 km/s, the light-transit times of light signals in opposing directions (L and L’_{i}) will differ by as much as 2VL (a few thousands km).Masimo Tinto states that LISA and hence the Earth are orbiting the Sun at about 30 km/s and this is consistent with the Sun being about 15,000,000 km from Earth but NOT with the Sun being 15 km above the Earth.
Well, it is just as well that your own references by K. Rajesh Nayak, J-Y. Vinet[/color] and Masimo Tinto support my claim that the Earth does orbit the Sun once a year and not your claim that "the Earth is not orbiting the Sun".Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
If the Earth is not orbiting the Sun, a(sun) DOES NOT equal 274.35m/s2: IN FACT IT IS EQUAL TO ZERO.
Then, we are left with the centrifugal acceleration: a_{c} = 0.0063 m/s^{2}.
<< There's no point dealing with this part because it presents no evidence supporting you claim "that the Earth is not orbiting the Sun". >>But you own references are based entirely on the Earth's orbiting the Sun once per year so you have presented nothing to support your claim.
Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
If the Earth is not orbiting the Sun, a(sun) DOES NOT equal 274.35m/s2: IN FACT IT IS EQUAL TO ZERO.
Since the GPS satellites ARE NOT registering/recording the missing ORBITAL SAGNAC, that means that the Earth is not orbiting the Sun.No, your logic is completely false and your none of your support that claim.
There are no strikes, just the local-aether model which is a fact of scienceNo, it is an idea, not a fact.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
All of you here still do not understand what is going on, so it seems I have to repeat these messages until you do.Possibly, but I'd be far more certain that the rotating Globe Earth orbiting the Sun are undeniable facts of science.
Who said that you can't?
Dr. C.C. Su obviously understands that the earth is a "Globe orbiting the Sun".
The GPS satellites do not record/register the ORBITAL SAGNAC EFFECT.
This is an undeniable fact of science.
Now, the Earth is stationary.You have never proven that and none of your own references support that.
All of you here still do not understand what is going on, so it seems I have to repeat these messages until you do.No, that would still be you who doesn't seem to understand, especially given that you want to repeat the same spam.
Of course, the surface gravity of the Sun is roughly 274 m/s^{2}!
And here is another way to check that 274 m/s^{2} value for the Sun's surface gravity.
Average distance from earth to Sun: 149,597,870,000 m.
Radius of Sun: 695,510,000 m
Sidereal year: 31,558,150 secs
Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s
Hence Earth's centripetal Acceleration about Sun = (1.99099 x 10^{-7})^{2} x (149,597,870,000) = 0.005930 m/s^{2}.
But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) = 0.005930 m/s^{2}.
Now the gravity due to the Sun decreases as 1/(distance from the sun)^{2}.
The Earth is 149,597,870,000 m from the Sun's centre and the Sun's surface is 695,510,000 m from the Sun's centre.
Therefore the Sun's gravity at its surface = 0.005930 x (149,597,870,000/695,510,000)^{2} = 274.35 m/s^{2} - QED.
So that agrees quite well with the surface g of the Sun as calculated from its mass, radius and the Universal Gravitational Constant - funny that!
Just answer that. Ideally in just a couple of sentences.So you are admitting that you are a paid shill refusing to understand anything because you are paid not to?
(https://image.slidesharecdn.com/dialogkonferansenstromstad26082014novideo-140826081635-phpapp02/95/how-to-sell-through-social-media-and-why-b2cs-have-a-lot-to-learn-from-b2bs-17-638.jpg?cb=1409291927)
Show us the line of reasoning which leads you to conclude that the sun is the size you claim.Your ignorance on the Sagnac effect and the gravity of the sun has nothing at all to do with your number.
The accuracy of the GPS system with or without it is enough to do the job.But none of that is evidence that the Earth is not a Globe orbiting the sun and it is quite irrelevant to the topic, "Solar power source".
What?
The GPS on your mobile phone is functioning BECAUSE the signal does not register/record the MISSING ORBITAL SAGNAC EFFECT.
Just answer that. Ideally in just a couple of sentences.There's nothing to answer unless you are accusing all the authors of the papers that YOU quoted from of lying simply because they do not supports your ridiculous hypotheses.
(https://image.slidesharecdn.com/dialogkonferansenstromstad26082014novideo-140826081635-phpapp02/95/how-to-sell-through-social-media-and-why-b2cs-have-a-lot-to-learn-from-b2bs-17-638.jpg?cb=1409291927)
Show us the line of reasoning which leads you to conclude that the sun is the size you claim.Nothing you posted is relevant to to my request!
You have to accept reality: CALTECH/NASA/ESA is telling you that THE ORBITAL SAGNAC EFFECT IS MISSING. Then, the Earth is stationary.CALTECH/NASA/ESA are not telling us that any orbital Sagnac effect in the GNSS is missing.
Or you have to accept the local-ether model.Dr C. C. Su does not accept relativity so he bases his work on an aether model.
So, it's pretty obvious who the paid shills are.I would say the more telling part is the ability to debate, i.e. respond to what has been said and back up arguments, vs repeatedly copying and pasting off topic nonsense.
All of you here still do not understand what is going on, so it seems I have to repeat these messages until you do.
CALTECH/NASA/ESA are not telling us that any orbital Sagnac effect in the GNSS is missing.That is irrelevant to the shape of the Earth, the movement of the Earth and to the topic, "Solar power source".
For a start the LISA analysis was just for uncancelled noise and the geometry of the LISA satellites is quite different from that of the GNSS.
Both LISA and GPS satellites do not record or register the missing ORBITAL SAGNAC EFFECT.
If the Sun is just . . . 600 metres across, then what energy source has been able to sustain its luminosity for over 4.6 billion years?
The GPS on your mobile phone is functioning BECAUSE the signal does not register/record the MISSING ORBITAL SAGNAC EFFECT.
If it did, the errors would be measured in kilometers.
[spam spam spam and more spam]Again, none of that spam is relavent to the topic.
On the other hand, the orbital
motion of the earth around the sun has a linear speed of
about 30 km/s which is about 100 times that of earth’s
rotation.
<< Irrelevant to the shape of the Earth, its movement or the topic, "Solar power source" so bug out! >>
Your article describes what Dr. C.C Su calls the ROTATIONAL SAGNAC EFFECT.
Not the ORBITAL SAGNAC EFFECT, which is much greater.