Solar power source

No projection at all.

You are accusing others of what is wrong with you. That is projection.
The below is a wonderful example of that.

Go ahead and take your messages to some other forum, and see how long you will last

And how long do you last on other forums before they ban you for your persistent, pathetic spam and refusal to even attempt to debate?

You are completely incapable of defending any of your claims, so you just repeat the same copy-pasted spam.
It is pathetic.

Repeating the same spam and never backing up your arguments will not help you.
Making the same baseless claims will not help you.
Repeating the same false claims about your source will not help you.

Your own sources show you are wrong.

If you want to use the Clayton model to make any such wild claims you will need to justify it, and no, repeating the same lies about your source wont help you.

"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.
The sun is not gas, it is plasma.
Do you know the difference?

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.




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. 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.




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.

So what force is accelerating the ripples?

Do any of these links contain any evidence that the sun is a 600m wise disk?

Have any of the scientists cited come to the conclusion that the sun is anything but a bloody great fusion powered ball in the middle of the solar system?

Is there any mention of a “black sun” in any of these?

If not, then what the hell is the point of all this?

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.

You have provided absolutely no justification for your claim that the pressure at the surface of the surface of the sun should be 28 times that of Earth.
Instead you repeatedly say the gravity is 28 times and then just jump straight to the atmosphere, claiming the pressure should be 28 times.

You repeatedly use the Clayton model, even though your own sources say that it is an approximation and isn't valid.
You are quote mining your sources and blatantly misrepresenting them.

They are not saying the Clayton model is accurate.
They are saying it is a crude approximation which is not valid.

The Clayton model is valid: textbook published by Wiley.


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:

"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.


Please read:

"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.

"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."

There is nothing in that to say that "it is"!

"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!
Look at the scale of the graph and look at the pressure you are using!
"Once we plug in the real figure, 10^-13 BAR, your calculations are flushed down the toilet".

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.
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.

I'm not disputing that!

Yes, if you do it correctly "Everything works out fine".
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.

No they do not get "flushed down the toilet"! Your misuse of Clayton's equation gets "flushed down the toilet"!
From Ricky Bradford: "Standard solar models give the central pressure as 1.65 x 10¹⁶ Pa" or 1.65 x 10¹¹ BAR.
You are being ridiculous to claim that "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."
covers anything like 10^-13 BAR when the total pressure range is zero to 1.65 x 10¹¹ BAR.

"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!

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".

You seem to have no concept of relative values.

You are trolling, yet again, this forum.

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 betwen 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 betwen the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution of the equations of stellar structure."

This includes the photosphere, chromosphere, corona.

If you are desperate enough to use these kinds of tactics it means you really know you lost the argument.

By now, your calculations are well into the sewer system.

Mine are fully corroborated by this impressive textbook.

Just watch.

P(r) = 2πgr²a²ρ²_ce^-x2/3M

where a = (3^1/2M/2^1/24πρ_c)^1/3

a = 106,165,932.3

x = r/a

M = 1.989 x 10³⁰ kg
central density = 1.62 x 10⁵ kg/m³

G = gr²/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/m² value, we get:


g = 0,0000507 m/s²


RATIO


a_c/g = 0.0063/0.0000507 = 124.26

Accuracy of the Clayton model:

https://image.ibb.co/eHYH5d/chro2.jpg



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.

Quote from: sandokhan on November 14, 2019, 01:56:59 AM

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 

You are trolling, yet again, this forum.

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."

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 10¹¹ BAR.

On this comparison graph, even an error of 1 BAR (10⁵ Pa) would be completely lost.
It would not only be classed as "impressive agreement between the results obtained from Clayton's simple model and obtained by Stomgren's results numerical solution" but it would be classed as absolute perfect agreement!
So you expecting agreement down around 10^-13 BAR is ludicrous! Have you no concept of relative errors etc?



So stop being totally ridiculous! Of course, the surface gravity of the Sun is roughly 274 m/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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!

Hence Earth's orbital Angular Velocity = 2 x π / (Sidereal year) = 1.99099E-07 rad/s

You are assuming that the Earth is orbiting the Sun.

However, the GPS satellites do not register the orbital SAGNAC/CORIOLIS effects at all. Nor do they record the solar gravitational potential effect.

See, that is why everyone here is laughing at you: you are creating your own fantasy world, where the Earth is orbiting the Sun.

Here, in the real world, GPS satellites fail to register the orbital SAGNAC/CORIOLIS/SOLAR GRAVITATIONAL POTENTIAL effects.


Your calculations, therefore, are now deposited at the bottom of the sea.


On this comparison graph, even an error of 1 BAR (105 Pa) would be completely lost.

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.


So you expecting agreement down around 10-13 BAR is ludicrous!

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.

The REAL DATA ON THE FIELD says otherwise: that the value of g(sun) must be MUCH LOWER than 274 m/s2.

How much lower?

P(r) = 2πgr²a²ρ²_ce^-x2/3M

where a = (3^1/2M/2^1/24πρ_c)^1/3

a = 106,165,932.3

x = r/a

M = 1.989 x 10³⁰ kg
central density = 1.62 x 10⁵ kg/m³

G = gr²/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/m² value, we get:


g = 0,0000507 m/s²


RATIO


a_c/g = 0.0063/0.0000507 = 124.26

Accuracy of the Clayton model:

https://image.ibb.co/eHYH5d/chro2.jpg



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.

Quote from: mak3m on November 14, 2019, 02:05:03 AM

Quote from: sandokhan on November 14, 2019, 01:56:59 AM

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 

Sandy do you only read the parts of your link that you think agrees with you?

Of course, the surface gravity of the Sun is roughly 274 m/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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!!!

Not very bright of you.

Thanks to you now, we have at our disposal the very best proof that indeed the 274 m/s2 figure is pure bonkers.

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

GPS satellites do not register the SOLAR GRAVITATIONAL POTENTIAL EFFECT:

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#msg1917978

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1985230#msg1985230

The Earth is not orbiting the Sun.

Therefore your value of 274 m/s2 is catastrophically wrong.

The Clayton model is valid: textbook published by Wiley.

Pretending your source says something that it doesn't will not help.

This is what it actually says:

Notice how it is saying it is a crude approximation?

Find a single valid source that says that the Clayton model produces the correct pressure in the chromosphere.

Please read:
"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."

Why?
It is just showing "reasonable agreement" between 2 models.
It isn't saying it is correct.

Now again, ENOUGH OF YOUR NONSENSE REGARDING THE CLAYTON MODEL!
It does not help your case at all.
All it does is show that the Clayton model is wrong.
It does not show the sun is flat.
It does not show the sun does not have nuclear fusion occurring in the core.


If you want to continuing asserting such nonsense, then justify the Clayton model from first principles.
If you can't do that, then you shouldn't be using it.

OK I have been into astronomy for over 40 years and yet I have never (up to now) heard of this Clayton model. I also know several career astrophysicists who are authorities in the field of stellar astrophysics.  I have asked around a few of them if the Clayton Model means anything to them and how valid it is a working model of star evolution. So far the replies have rather vague and along the lines of 'huh... what is that about then?'

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.

You are really frightened this time around.

Nothing is going to work for you anymore, not the trolling, nor the continuous spamming.

Of course, the surface gravity of the Sun is roughly 274 m/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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/s².

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.

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://web.archive.org/web/20050217023926/https://www.ee.nthu.edu.tw/ccsu/










Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. 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.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


https://pdfs.semanticscholar.org/f606/87008dd7b3e872c67770eaa9ada9128bbf8b.pdf

Journal of Electromagnetic Waves and Applications:

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.

The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.


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.

He uses GPS and a link between Japan and the US to prove this.

In GPS the actual magnitude of the Sagnac correction
due to earth’s rotation depends on the positions of
satellites and receiver and a typical value is 30 m, as the
propagation time is about 0.1s and the linear speed due
to earth’s rotation is about 464 m/s at the equator. The
GPS provides an accuracy of about 10 m or better in positioning.
Thus the precision of GPS will be degraded significantly,
if the Sagnac correction due to earth’s rotation
is not taken into account. 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. Thus the present high-precision GPS would be
entirely impossible if the omitted correction due to orbital
motion is really necessary.


In an intercontinental microwave link between Japan and
the USA via a geostationary satellite as relay, the influence
of earth’s rotation is also demonstrated in a high-precision
time comparison between the atomic clocks at two remote
ground stations.
In this transpacific-link experiment, a synchronization
error of as large as about 0.3 µs was observed unexpectedly.


Meanwhile, as in GPS, no effects of earth’s orbital motion
are reported in these links, although they would be
easier to observe if they are in existence. Thereby, it is evident
that the wave propagation in GPS or the intercontinental
microwave link depends on the earth’s rotation, but
is entirely independent of earth’s orbital motion around
the sun or whatever. As a consequence, the propagation
mechanism in GPS or intercontinental link can be viewed
as classical in conjunction with an ECI frame, rather than
the ECEF or any other frame, being selected as the unique
propagation frame. In other words, the wave in GPS or the
intercontinental microwave link can be viewed as propagating
via a classical medium stationary in a geocentric
inertial frame.


LISA Space Antenna



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).



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.

Therefore, the value of 274 m/s2 RESTS ENTIRELY ON THIS STATEMENT:

No, it also matches observations of other planets, not just Earth.
But if you are rejecting that you are no longer using real physics and instead are running off into fantasy.

The simple fact is the only thing you have to back up your nonsense is a manipulation of a formula from a model which is admitted to be in invalid for the surface of the sun and in some cases the sun in general.

Take your Sagnac BS back to the threads where you have already been repeatedly refuted.

Your claims based upon the Clayton model have been repeatedly refuted.
Bringing them up anymore is just spamming.

If you have something else, present it.
Otherwise you have no case.

Now then, do you have anything at all to show the sun is flat or the sun does not have nuclear reactions occur at the core?

Quote from: rabinoz on November 14, 2019, 04:49:53 AM

Of course, the surface gravity of the Sun is roughly 274 m/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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!!!

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".
You used R_sun = 700,000,000 m in

M = 1.989 x 10³⁰ kg
central density = 1.62 x 10⁵ kg/m³
Using P(700,000,000) = 1.0197 x 10^-9 kg/m² value, we get:

And the easily checked average angular diameter of the Sun viewed from Earth is 0.533°.
Your 700,000,000 m Sun's radius would make the distance to the Sun roughly 2 x 700,000,000 / radians(0.533°) = 150,440,598,461 m - looks close to waht I used.
But had I used the more accepted value of 695,510,000 m the distance to the Sun would have been 2 x 695,510,000  / radians(0.533°) = 149,475,629,479  m.

I'll stick to the distance from the Earth to the Sun that I used, thank you!

Not very bright of you.

Thanks to you now, we have at our disposal the very best proof that indeed the 274 m/s2 figure is pure bonkers.

Oh, I should have realised that the Sun is only 600 m in diameter and only 15 km from the Earth !

Do you suggest that I re-run the calculations based on YOUR values ?

No, I don't think I'll bother. Now let's look at the rest:

GPS satellites do not register the SOLAR GRAVITATIONAL POTENTIAL EFFECT:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846706#msg1846706

That is possibly because any changes due to the solar gravitational potential effect are too small to register.
see: Why there is no noon-midnight red shift in the GPS by Neil Ashby and Marc Weiss

GPS satellites do not register the ORBITAL SAGNAC/CORIOLIS EFFECTS: https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1917978#msg1917978

It'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.

And, just possibly, when the angular velocity of the Earth is taken as (2 x π)/(seconds in a sidereal day) there is no orbital Sagnac effect.
I'm sure that the physicists at ScienceForums.net could sort this out for us in: Global/Generalized Sagnac Effect Formula by sandokhan.
Should we check what they think of your "Coriolis/Sagnac" confusion?

Then you go on to say:

The Earth is not orbiting the Sun.

Well, that does seem a little hard to swallow because:

• The angular size of the Sun only changes slightly throughout the year.
  So, assuming that the Sun does not magically shrink and expand, the distance to the Sun only changes slightly throughout the year.
  
  
• The position of the Sun relative to the background stars changes over a year and then seems to magically repeat after one year.

The simplest explanation seems to be that the Earth orbits the Sun with a period of one year and at a distance of about 149,000,000 km.

That seems far more logical than a 600 m diameter Sun only 15 km from the Earth - that could explain none of those things!

<< Added a bit >>

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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).

• Sure, but who accepts the local 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!
  Do YOU pretend to know more about the motion of the Earth than C.C. Su?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
LISA Space Antenna



The LISA interferometer rotates both around its own axis and around the Sun as well, at the same time.

I am not casting doubt on any of that! Just on your attempt to use those publications to support your hypotheses!

Are you really claiming that your knowledge of astronomy etc exceeds that of C.C. Su and all those working on the  LISA interferometer?

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.

Ooo nice little diversion

Big stress on Astrophysics ain't my bag, but it was a good read.

So unless I missed the mark Virial Theorum in its purest form describes the effects of gravity on particles or groups of particles.  But current work has seen it scaled up, albeit with some caveats, to galactic clusters particularly looking at dark energy / matter estimations.

So small and big. Is there any research going on in the middle on single bodies  like a star, particularly in light of this thread our nearest one?

Yes it basically deals with the distribution of thermal KE and GPE within a system of particles.  Those particles could be the molecules making up a star, the stellar members and gas associated with a star cluster or even a cluster of galaxies.

I am doing a University module (2nd yr U/G) as part of my BSc degree in astronomy about galaxy formation and structure. My concentration currently relates to star cluster ansd specifically the total mass of stars and total mass of gas within them.  I first of all need to calculate the total kinetic energy within the cluster assuming it is static (unrotating) and in a state of equilibrium.  After that we introduce some additional destabilising forces such as the influence from young stars being born within the cluster and supernova explosions external to the cluster.  The latter send out pressure waves across space and when those pressure waves interact with the cluster they destabilise the gas causing dense regions of it to collapse and ultimately form new stars. We may make the assumption that the gas and stars are distributed as spheres of constant density and some finite radius r sub c.

Appreciating that this is not a forum about astrophysics specifically and Virial theory is a little bit of a diversion (and perhaps a welcomed diversion by some) from the ongoing bickering which I refuse to have any part in, I just wondered if Sandys encyclopaedic knowledge of advanced astrophysics included anything on virial theory. In which case I welcome his thoughts... otherwise as you were.

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.

Grey haired engineer this end, did my first bit of amateur astronomy this year, Northumberland Dark park then onto Kielder Observatory in July to see Saturn in opposition fairly melted my brain.

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.

One thread ends with this:

"Moderator Note
Since the OP appears impervious to reason and genuine scientific rebuttal, this thread is closed."
https://www.scienceforums.net/topic/118524-globalgeneralized-sagnac-effect-formula/#comments

On this comparison graph, even an error of 1 BAR (105 Pa) would be completely lost.
Not at all; simply the range of the graph has to be extended to better cover the 0.8 - 1 solar radius axis.

OK, then you show a comparison graph that expands the pressure scale enough to observe 10^-13 bar.

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!

That is the REAL VALUE of the pressure of the chromosphere! 10^-13 BAR.

I never disputed that. All I'm disputing is that it is ludicrous to expect the Clayton pressure equation to fit to that precision.
The 10^-13 BAR might be "the REAL VALUE of the pressure of the chromosphere" but the predictions of the Clayton model are NOT real life values - they are estimates based on one model!

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!
That is because you are expecting  perfect predictions from the the Clayton pressure equation in a region where it is "certainly grossly wrong.

Read again!
I do wish that you would read and believe your own references, such as:

Ricks Cosmology Tutorial: Chapter 11'  Stellar Structure Part1
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/m³, 4.1 x 10³ 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 10²⁰ protons per m³, 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.

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."?

Yet your are expecting perfect accuracy in the Chromosphere, outside the surface of the star.

How much lower?

<< Repeated calculations deleted!  >>

So, I am using the correct equation, with the correct parameter a.

Sure but you are "using the correct equation, with the correct parameter a" where it certainly grossly wrong!

I simply cannot comprehend how you are unable to grasp this.

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



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).



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/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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/s².

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.

Ronald R Hatch is an interesting character, and another fine example of you using heliocentric 100% RE theory, because you spotted a couple of words that appear in your copy pasta.

Hatch had an interesting theory, he made a hypothesis and the whole thing fell down at the first hurdle when it came to prediction and observation.

Hatch's theory wasn't fully fleshed out  but his main hypothesis was that LIGO would not be able to detect gravitational waves...

Oh wait

You are trolling this thread.

Ron Hatch is an internationally renowned expert on GPS satellite technology.

Director of Navigation Systems Engineering and Principal and co-founder of NavCom Technology, Inc.
Institute of Navigation (ION), including Chair of the Satellite Division, President and Fellow.
https://www.gps.gov/governance/advisory/members/hatch/

Ronald R. Hatch is a recipient of the Johannes Kepler Award from
the Institute of Navigation because he was the most significant
contributor to the advancement of satellite navigation. He has
over 30 years experience in designing navigation systems and has
been consulted by government agencies and companies.

The RUDERFER EXPERIMENT completely proves the statement made by Ron Hatch.

Ruderfer, Martin (1960) “First-Order Ether Drift
Experiment Using the Mössbauer Radiation,”
Physical Review Letters, Vol. 5, No. 3, Sept. 1, pp
191-192

Ruderfer, Martin (1961) “Errata—First-Order Ether
Drift Experiment Using the Mössbauer Radiation,”
Physical Review Letters, Vol. 7, No. 9, Nov. 1, p 361

In 1961, M. Ruderfer proved mathematically and experimentally, using the spinning Mossbauer effect, the FIRST NULL RESULT in ether drift theory.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1846721#msg1846721

You are an embarrassment to the RE.

The embarrassment here is you.
You are an embarrassment to the entire human race.


All you are doing is repeating the same refuted nonsense.

Linking to elsewhere where you have spouted this nonsense and been refuted wont help your case.
Instead it just makes it worse and shows just how little you care about the truth, reality being correct and so on.
Are you trying to show everyone that you are a troll, or that you are deluded enough to think you are correct?

Now again, enough of the spam.
Justify your claims that the sun is a flat disk that does not undergo nuclear fusion.

You have provided literally nothing to justify it so far.

Of course, the surface gravity of the Sun is roughly 274 m/s²!

And here is another way to check that 274 m/s² 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/s².

But the (Sun's gravity at the Earth) = (Earth's centripetal Acceleration about Sun) =  0.005930 m/s².
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/s² - 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/s².


EVERYONE ACCEPTS THE EXISTENCE OF THE MISSING ORBITAL SAGNAC EFFECT.

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.


LISA Space Antenna



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).



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.

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.