Solar power source

I may have exaggerated slightly.
What ?!
Slightly?

Yes, slightly as explained.
There is only a slight connection to gravity.

Repeating the same nonsense will not help you.
Nothing you are providing is supporting your case.
Instead it just shows that you are either incapable of rationally defending your claims or you have no desire to.

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.
You have provided absolutely no basis for that.

Which means the formula you are using would be completely invalid as it relies upon G being a constant.
Sure

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.

Let us perform the calculation once more.

Why?
Repeatedly saying 1+1=50 wont help you.

If you want to use the calculation, you need to show it is correct.
So far all you have done is shown it isn't correct.

Stop spamming the same refuted BS and instead deal with the actual issues.

The Clayton model does not accurately describe the surface of the sun.
Repeatedly using this incorrect model, to show the model is incorrect, just shows what we already know, that this model is incorrect, and you don't care about the truth at all.
It does not show that the surface gravity of the sun is tiny.
It does not show the sun must be flat.

Yes, slightly as explained.
There is only a slight connection to gravity.

No it's not.
You have provided absolutely no basis for that.

You explained nothing at all.

There is a FULL connection to gravity.


https://openstax.org/books/astronomy/pages/8-3-earths-atmosphere

We live at the bottom of the ocean of air that envelops our planet. 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).

Now, google search "Sun’s gravity is 28 times that of the Earth’s gravity".

These are the absolute facts of science: 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.

But it is not.

In fact it is much, much lower.


The result obtained by using Clayton's equation, namely that g(sun) is much lower than the g(centrifugal acceleration) is fully corroborated by this proof as well.

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?

WRONG!!!

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.

https://image.ibb.co/fauUJy/photosph.jpg

In the chromosphere, the pressure drops to 10^-13 BAR = 0.0000000000001 BAR.

https://image.ibb.co/hkvQrJ/chromo.jpg

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

If G is a constant, as you say, then simply substituting its well-known expression, G = gr²/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 = gr²/M" but is g = GM/r².

Now we know that:
the mass of the sun = 1.989 x 10³⁰ kg, r = 700,000,000 m (you said so ^[1]) and G = 6.67 x 10^-11 m3⋅kg⁻¹⋅s⁻² (Rick Bradford used that value).

So we get g = 6.67 x 10^-11 x 1.989 x 10³⁰/700,000,000² = 270.7 m/s² but had you used r = 695,510 km, the results would have been g = 274.3 m/s², funny that!

[1]

M = 1.989 x 10³⁰ kg
Using P(700,000,000)

If the sun is spherical, you have nothing to hide/fear.

G = g²/M, a fact of science.

Then, using the Clayton equation, we should get a value/figure close to 270 m/s2.

If the sun is discoidal, then obviously g (and G) will have a very different value.

Let us find out which one it is.

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


The result obtained by using Clayton's equation, namely that g(sun) is much lower than the g(centrifugal acceleration) is fully corroborated by this proof as well.

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?

WRONG!!!

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.

https://image.ibb.co/fauUJy/photosph.jpg

In the chromosphere, the pressure drops to 10-13 BAR = 0.0000000000001 BAR.

https://image.ibb.co/hkvQrJ/chromo.jpg

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

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?

You are making no connection at all.
If you think there is one THEN PROVIDE IT!
This would go in the form of some derivation of an equation showing that the only thing determining the pressure at the surface is gravity.

That is why you have absolutely no argument.
You are just dumping out a bunch of numbers and attacking strawmen.

Now, care to stop the childish garbage and instead deal with the issues at hand?

Did you pass Physics 101?

No problem, you are going to take the final exam again right now.

Definition

"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)."

Definition

"Gravity on the surface of the earth is 9.8 m/s^2. Gravity on the surface of the sun is 274 m/s^2, or about 28 G's."

Q & A: Gravitational pull of the Sun

This makes the strength of gravity on the "surface" of the sun (that is, the photosphere, the shiny part we see), 28 times stronger than the force of gravity on the surface of the Earth.

https://van.physics.illinois.edu/qa/listing.php?id=184&t=gravitational-pull-of-the-sun


The atmospheric pressure (photosphere pressure) is 28 times stronger than the atmospheric pressure on Earth (theoretical prediction).

Q & A: How many times is the gravity of the Sun greater than the gravity of the Earth?

You may weigh 100 kilograms here on Earth. If you are on the surface of the Sun, you would feel like you weigh 2,800 kilograms. This is because the Sun’s gravity is 28 times that of the Earth’s gravity.


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?

WRONG!!!

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.

https://image.ibb.co/fauUJy/photosph.jpg

In the chromosphere, the pressure drops to 10-13 BAR = 0.0000000000001 BAR.

https://image.ibb.co/hkvQrJ/chromo.jpg

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

Quote from: rabinoz on November 13, 2019, 02:09:52 AM

Quote from: sandokhan on November 12, 2019, 09:35:13 PM

But G is not a universal constant.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The universe DOES NOT obey Newton's law of "universal" gravitation.

That's all quite irrelevant because we know that the whole "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?

Not in the slightest!

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.
Astronomers, cosmologists, particle physicists and many others are well aware that parts of the Universe differ from Newtonian Laws, somtimes very slightly and in other cases, such as near black holes and at ultrarelativistic velocities, quite drastically.

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:
In Einsteins General Relativity:
In the scaled gravitational constant, or Einstein's constant: κ = (8π/c⁴)G

That is why nobody takes your messages seriously any longer.

If the universe DOES not obey Newton's law of gravitation, then obviously G can no longer be regarded as a constant.

That does not follow at all!

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.

• I did not anywhere claim that "the universe DOES . . . obey Newton's law of gravitation" though in most regions Newtonian Mechanics is a good approximation and far simpler to use than the alternative.
  
  
• "DARK FLOW" is only a "counterexample to show that G is not constant" if it can be proven that dark flow does "show that G is not constant".
  

Statements like “The finding [that only galaxy clusters moving toward or away from a point between Centaurus and Vela] flies in the face of predictions from standard cosmological models, which describe such motions . . ." might be evidence against "standard cosmological models" but not against even Newtonian Gravitation let alone Einstein's Theory of General Relativity'

you undeniably use the Clayton equation inappropriately by forcing a value of g out of it and implicitly a value of G vastly different

Rick Bradford is assuming that the pressure in the chromosphere is the PREDICTED/THEORETICAL VALUE. That it is not.

No! Clayton's equation is an approximation to the pressure distribution inside a star, nothing more!
And you are trying to claim that you can calculate the sun's surface gravity from an extreme value - you can't rely on working backwards like that!

Moreover, G does equal g²/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 g²/M" - you made a simple mistake but G might equal g.r²/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).

G is not a constant as I have just proven.

No, you have NOT proven that "G is not a constant".

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 g²/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.

Why are you forbidding your readers to solve the equation for g?

Read my previous post! But just look what YOU have done:

Rick Bradford used G = 6.67 x 10^-11 m3⋅kg⁻¹⋅s⁻² (see p11 in his Lect 11) to calculate the central pressure.
If your calculations lead to a different value of G then you must have done something wrong.

And look at this:

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/m³, 4.1 x 10³ 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 10²⁰ protons per m³, 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.

And where have YOU chosen the get the pressure from to calculate your g and by implication, G? Past the surface of the SUn, at 700,000 km when the sun's radius is given as 695.510 km.

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!
If you uses a suitable filter and a bit of magnification the Sun is readily seen to be very slowly rotating spherical!

but not against even Newtonian Gravitation let alone Einstein's Theory of General Relativity'

Dark flow has been described as taking a hammer and beating the living tar out of Einstein’s gravitational theory of the universe.

According to the Big Bang theory, the Universe is about 13.7 billion years old; yet the gravitational attractor, tugging only on galaxy clusters, is some 32-34 billion light years away. Additionally, this gravitational force is unique and selective in its action; only affecting galaxy clusters, but not everything else. Gravity undoubtedly must affect the motion of all massive bodies and, therefore, since it is pulling the galaxy clusters, it should be pulling everything else to it, not just galaxy clusters, based on Newtonian Law.

In terms of Einstein, the identical problem exists. A massive object outside the Universe has warped space to cause galaxy clusters to move toward or away from it; that warping of space should do the same for all matter in the Universe. In terms of Dark Energy, all galaxies are supposedly moving away from each other and, therefore, would not also, at the same time, permit only galaxy clusters to not follow this expansion, but move to or away from a preferred area. If Dark Energy existed, these galaxy clusters should also be moving away from one another in different directions.
These clear-cut findings defy the Big Bang theory and, thus, have made the Dark Flow evidence very unwelcome for many cosmologists.


A total defiance of TGR/Newton's law of gravitation.

Thus G cannot be a constant, since the law of universal gravitation is defied.

‘Because the dark flow already extends so far, it likely extends across the visible universe,’ Kashlinsky says.

The entire universe does not obey anything resembling Newton's law of gravitation.


None of the Universe precisely obeys Newtonian Mechanics and Gravitation but most of it follows it very closely.

Again, you must be drunk.

If none of the Universe precisely obeys Newtonian Mechanics then obviously it won't be following it very closely.

Actually the entire Universe does not obey Newtonian Mechanics at all, since dark flow defies attractive gravitation on a grand cosmic scale.


Certainly "G is still regarded as a universal constant" and it appears in numerous places in physics such as

Can't be "universal" since you have to prove first that the Earth rotates around its axis, and that it orbits the Sun.

Since dark flow proves that the entire universe does not obey Newtonian Mechanics, then certainly you must revise your hare-brained ideas.

"G does" NOT "equal g2/M"

Your statement belong to the CN section.

Take a long look at yourself: now you are denying one of the most basic equations/formulas in physics: G = g²/M.

Certainly, if you are right, I can use g instead of G, and everything should work out fine.

But it doesn't.

That is why you are in a frenzy trying to forbid everyone to make a very simple substitution.

If your calculations lead to a different value of G then you must have done something wrong.

Not at all, it means that the shape of the Sun cannot be spherical.

Rick Bradford is counting on a much higher value of the pressure in the chromosphere for the value of G recorded on Earth.

But the value is much lower than the theoretical prediction: in fact it is 10^-13 bar.


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?

WRONG!!!

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.

https://image.ibb.co/fauUJy/photosph.jpg

In the chromosphere, the pressure drops to 10-13 BAR = 0.0000000000001 BAR.

https://image.ibb.co/hkvQrJ/chromo.jpg

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

"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.
The surface gravity of Mars is 3.72 ms⁻² and its atmospheric pressure is 0.0061 Bar
The sun has no "sea-level" nor any well-defined hard surface like the earth. so the 10^-13 BAR in the chromosphere and even the 0.16 Bar at the bottom of the photosphere is totally irrelevant.

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

No, that is quite wrong! The atmospheric pressure depends on a lot more than the surface gravity.

You are trolling this website.

Here is the recorded CHROMOSPHERE PRESSURE: 10^-13 BAR.



Here is the recorded PHOTOSPHERE PRESSURE: 8.6x10^-4 BAR.

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:




The author of the paper did perform a calculation using the Clayton model at the surface: the error is at most 40x.

NOT 1X10⁷x!!!


The result obtained by using Clayton's equation, namely that g(sun) is much lower than the g(centrifugal acceleration) is fully corroborated by this proof as well.

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?

WRONG!!!

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.

https://image.ibb.co/fauUJy/photosph.jpg

In the chromosphere, the pressure drops to 10-13 BAR = 0.0000000000001 BAR.

https://image.ibb.co/hkvQrJ/chromo.jpg

This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

The shape of the Sun cannot be spherical.

Why does the link say not to apply Clayton model to the surface of the sun?

Why did you disregard this?

Feel free to post a stellar structure equation which relates the pressure to the other variablees.

You won't find any, other than the Clayton equation.

You could attempt to integrate the polytropic equation numerically, but even then the results are not perfect.

Here is a reference which does illustrate the correctness of the Clayton model:

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 betwen 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 10⁸.

The value I derived is 1.06 x 10⁸, which is even better.

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

Did you pass Physics 101?

No. In high school it was just science and physics. No 101.
At the uni I went to the courses were 4 digits.

I did do a lot of physics courses though.

Did you realise you have no rational response so you need to keep insulting people?

The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth

BASED ON WHAT?
So far all you have been doing is repeatedly asserting that the pressure should be 24.7 times what it is on Earth.
You are yet to show that in any way.

Who cares if g for the surface of the sun is 24.7 times greater than Earth. Why should this make it the pressure 24.7 times?
You may as well say the radius of the sun is 109 times that of Earth so the pressure should be, or the volume is ~1.3 million times so the pressure should be, or the volume is ~3.3 million times so the pressure should be. You can also happily invert these. e.g. the distance from the centre is 109 times that of Earth, so the pressure should be 1/109. Or due to how so many things have an 1/r^2 relationship, you could do that as well and say that because it is 109 times as far it should be 1/11900. One which makes as much sense as yours is looking at the surface area. ie. the atmosphere is spread out over a much larger area and thus should be much lower. This gives us the 1/11900.

You have provided literally no justification for your claim at all.

As such, you have shown literally no problem.

Have you passed physics 101?
Here, I'll give you a basic rundown (rather than jumping to the final exam).
If you had, you know you can't just grab numbers like that and pretend they should have a direct connection.
You should know that such a connection would need to be established, either empirically by observing it for many different objects, or deriving it based upon the known laws of physics.

Doing the latter would mean accepting that the pressure at the surface is due to the weight of the atmosphere above, and thus as a simple approximation where g does not vary significantly it would be P=m*g/A. A more complex form where the height does very significantly and thus g does as well would be int(sigma*g*dh, h=0->inf)/A.

We know 2 of these factors. g and A. Combining them gives us the sun's atmosphere being 0.002 times the pressure.
But the m part remains "unknown". Do you know what it is?

Due to the very low density at the start of the chromosphere, that is likely going to be much smaller than for Earth, meaning the pressure should be much lower.


Again, stop using Clayton's equation.
You have already admitted it isn't valid.
Your own sources state quite clearly that it isn't valid for the surface of the sun.

So using it just shows that you have no regard for the truth.

Although I do have one question about the Clayton equation:
Is it e^(-x^2), or is it (e^(-x))^2)?


Now again, care to stop with the spam and start dealing with the topic at hand?

What evidence do you have that the sun is flat or that nuclear fusion doesn't occur at the core?
Again, attacking a specific model shows neither of those things, especially when it is already known that the model is not correct where you are applying it due the numerous assumptions made.

You are not paying attention.

Definition

"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)."

Definition

"Gravity on the surface of the earth is 9.8 m/s^2. Gravity on the surface of the sun is 274 m/s^2, or about 28 G's."

Q & A: Gravitational pull of the Sun

This makes the strength of gravity on the "surface" of the sun (that is, the photosphere, the shiny part we see), 28 times stronger than the force of gravity on the surface of the Earth.

https://van.physics.illinois.edu/qa/listing.php?id=184&t=gravitational-pull-of-the-sun


The atmospheric pressure (photosphere pressure) is 28 times stronger than the atmospheric pressure on Earth (theoretical prediction).

Q & A: How many times is the gravity of the Sun greater than the gravity of the Earth?

You may weigh 100 kilograms here on Earth. If you are on the surface of the Sun, you would feel like you weigh 2,800 kilograms. This is because the Sun’s gravity is 28 times that of the Earth’s gravity.


Here is a reference which does illustrate the correctness of the Clayton model:

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 betwen 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 108.

The value I derived is 1.06 x 108, which is even better.

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

You are not paying attention.

No! You are the one not paying attention to the simplest of matters!

Definition
"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)."

And 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.
Hence for the sun and these gas planets the surface atmospheric pressure depends where that surface is defined. Why is that simple concept so hard to understand?

You are trolling this website.

No, and you don't even know the meaning of "trolling"!
But you are continually spamming the website with the same old material over and over!

Here is the recorded CHROMOSPHERE PRESSURE: 10^-13 BAR.



Here is the recorded PHOTOSPHERE PRESSURE: 8.6x10^-4 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.

But so what?
The is only the most indirect connection between the surface gravity of the Earth or the Sun and the atmospheric pressure at a given distance above the surface.
Take a look at the surface gravity and atmospheric pressures of some planets:
  Planet        Gravity(m/s²) Pressure (bar)
Mercury:                3.7                       0
   Venus:                8.9                      92
   Earth:                 9.8                        1
    Mars:                3.7                       0.01

These just no connection because it for a planet the surface atmospheric pressure depends mainly on the mass of the atmosphere and the planet's gravity - temperature distribution etc also enter in.
But the sun has no well defined solid surface so the surface is regarded as the top of the photosphere.

Hence if that top of the photosphere is taken as the surface, the surface pressure depends largely on the mass of gas outside that radius.
It all depends on where the "sun's surface" is defined.

Which makes all of your claims about the surface gravity are simply meaningless!

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



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.



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.

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.



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”.

Rinse and repeat

Resembling...

Looks like...

Similar to...

But isnt a ripple, because ripples caused by a stone thrown into water dont accelerate. We are only beginning to understand the processes of the sun and how plasma reacts in these circumstances.

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.

These observations can only occur on a spherical sun, and do not in any way shape or form add to your position that the sun is a very young, disc, 600km away from earth.

But the Sun does have a well-defined surface.

http://www.ptep-online.com/2011/PP-26-08.PDF

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.

But the whole point is that you simply cannot calculate the surface gravity from the surface pressure.

Just face his obvious simply fact and stop this silly time wasting spamming with the same old . . . . . . .

The surface gravity of the Sun is about 274 m/s² whether you accept it or not.

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.

You haven't done your homework on the subject: no magnetic forces can cause those ripples.



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

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.




The surface gravity of the Sun is about 274 m/s2

Let's put your word to the test.

Here is a reference which does illustrate the correctness of the Clayton model:

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 betwen 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 108.

The value I derived is 1.06 x 108, which is even better.

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

The Clayton model provides us with the g value: g = 0,0000507 m/s^2 which is much lower than the centrifugal acceleration figure:

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:




There is still another way to prove the correctness of the Clayton equation.

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?

WRONG!!!

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.



In the chromosphere, the pressure drops to 10^-13 BAR = 0.0000000000001 BAR.



This is why the value for g(sun) must be MUCH LOWER than 270 m/s2.

You are not paying attention.

No, I am.
You seem to just be repeating the same nonsense and completely ignoring what I am saying.

You are doing nothing to show that the atmosphere of the sun should be 27.4 times the pressure of the atmosphere of Earth.

All you are doing is showing some numbers, and then asserting it.

The atmospheric pressure (photosphere pressure) is 28 times stronger than the atmospheric pressure on Earth (theoretical prediction).

What theory?
So far all you have done is repeatedly asserted this nonsense.

You have done absolutely nothing to substantiate it.

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?
Where? I searched for "We shall see that the Clayton model" and it did not give your result.
Searching for "reasonably correct results" gives no results.

So just where is this author saying that?

Repeating the same spam will not help you.
Falsely claiming a reference says something that it doesn't will not help you.

You have done absolutely nothing to back up your claims.

You are in no position to act here as the master of ceremonies when it comes to physics. You and your pal have always resorted to cheating, trolling, misinforming your readers, anything it takes to deny reality. Your viewers are well aware of your tactics, that you do not care at all for the truth, that you follow the path of least resistance to suit the variables and the equations to your cognitive dissonance situation.

The definitions are very clear.

Definition

"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)."

Definition

"Gravity on the surface of the earth is 9.8 m/s^2. Gravity on the surface of the sun is 274 m/s^2, or about 28 G's."

Q & A: Gravitational pull of the Sun

This makes the strength of gravity on the "surface" of the sun (that is, the photosphere, the shiny part we see), 28 times stronger than the force of gravity on the surface of the Earth.

https://van.physics.illinois.edu/qa/listing.php?id=184&t=gravitational-pull-of-the-sun


The atmospheric pressure (photosphere pressure) is 28 times stronger than the atmospheric pressure on Earth (theoretical prediction).

Q & A: How many times is the gravity of the Sun greater than the gravity of the Earth?

You may weigh 100 kilograms here on Earth. If you are on the surface of the Sun, you would feel like you weigh 2,800 kilograms. This is because the Sun’s gravity is 28 times that of the Earth’s gravity.


Where? I searched for "We shall see that the Clayton model" and it did not give your result.

Read again.


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 betwen 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 108.

The value I derived is 1.06 x 108, which is even better.

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

You are in no position to act here

And there you go with more projection, spam and insults.

Again, on what basis do you claim that the pressure at the surface of the sun should be 28 times that of the pressure at the surface of Earth?
So far all you have done is appealed to gravity being related by that factor, but you have done absolutely nothing to show that that should mean anything.
Saying your weight will increase doesn't mean pressure will.
Do you understand that?

Repeating the same spam will not help. Show an actual connection.
Until you can show an actual connection for WHY the atmospheric pressure should be linearly proportion to gravity, YOU HAVE NOTHING!

Read again.

I have, it doesn't say what you claim.
It says that such a guess can be a reasonable starting point (which clearly indicates it isn't accurate), and also directly says that it isn't valid for the sun.
It also says that such a simple model is very crude by the standards set by realistic models.
But I don't seen anywhere where it says that it can yield reasonably correct results when applied to the sun.

So again, you are using an equation which is a crude approximation and expecting it to be perfect.

Grow up.

No projection at all.

Go ahead and take your messages to some other forum, and see how long you will last: perhaps a couple of days at most. Your readers know very well what to expect of you by now: anytime you are with your back to the wall, you resort to trolling fully benefiting from the lack of moderation which would have put a stop to your shenanigans a long, long time ago.

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)."

Definition

"Gravity on the surface of the earth is 9.8 m/s^2. Gravity on the surface of the sun is 274 m/s^2, or about 28 G's."

Q & A: Gravitational pull of the Sun

This makes the strength of gravity on the "surface" of the sun (that is, the photosphere, the shiny part we see), 28 times stronger than the force of gravity on the surface of the Earth.

https://van.physics.illinois.edu/qa/listing.php?id=184&t=gravitational-pull-of-the-sun


But it is valid for the Sun and does lead to very good results.


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 betwen 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.

You haven't done your homework on the subject: no magnetic forces can cause those ripples.

Getting embarrassing.

Your source, says otherwise.

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.

You haven't done your homework on the subject: no magnetic forces can cause those ripples.

I don't know who wrote that but I'll let them answer it.

The surface gravity of the Sun is about 274 m/s2

Let's put your word to the test.

I've answered that numerous times!

Here is a reference which does illustrate the correctness of the Clayton model:

The Physics of Stars by A. C. Phillips
"We shall see that the Clayton model can yield reasonably correct results when applied to the sun."

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.

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 10⁸.
The value I derived is 1.06 x 10⁸, which is even better.

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

It might be the "correct equation" but it is not accurate enough to use as you do!

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.

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) = 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:

All 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!
Read again: "the Clayton model is . . . certainly grossly wrong at the surface of the star".

There is still another way to prove the correctness of the Clayton equation.
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 >>

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"!

But all you've done is shown that, as Rick Bradford, himself, said, "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".

So stop wasting everybody's time with this repeated spam!

"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

"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 betwen 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.


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".
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"!

Nope, it doesn't work out at all.

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.

Using the REAL VALUE for the pressure in the chromosphere, we get a(sun) = 0.0000507 m/s2.

The shape of the Sun cannot be spherical.

"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:

For clarity are you suggesting that the Sun does not have a magnetic field?

Or are you suggesting that magnetic flux tubes are not accelerating the ripples?

If its the former, so many questions!, if its the latter what is your position on the acceleration of the ripples and how does this prove the sun is a flat disc, 600km above the earth?

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.