Solar power source

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mak3m

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Re: Solar power source
« Reply #60 on: November 12, 2019, 04:25:03 AM »
No, it's my link from three years ago.

Chapter 18 from the same work contains the graphics as well.

You cannot complain about the Clayton equation since it is YOUR very own formula, derived by one of the best astrophysicists of the 20th century.

It is the standard in stellar structure physics.

Nearly there

The link you meant to post was

http://www.rickbradford.co.uk/Chapter18_StellarStructurePart2.pdf

Your graphs are there so you are comparing a Clayton calculation to a Polytropic-Hydrostatic equation while following neither.

Quote
Thus, whilst hydrostatic equilibrium is respected by the Clayton model, Equs.[2,3] are not. Hence the Clayton model is not really a correct solution, and this shortcoming becomes apparent at larger radii. In particular, the Clayton model is completely wrong as regards the temperature predicted near the surface of the star.

So show how you made your assumptions, as they dont match your cited source



You have to learn to reply without quoting a long previous answer.

Re: Solar power source
« Reply #61 on: November 12, 2019, 04:34:03 AM »
The Clayton equation
Still not providing any details, as such it is entirely useless.

Still not dealing with the actual issues and instead just attacking models and thus entirely useless.

Am I supposed to do your homework for you?
No. You are supposed to do YOUR homework.
Instead you seem to be trying to get us to do it for you.
No thanks.

http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf
Have you read this source?

Quote
We consider only the spherically symmetric problem (hence a star which is not rotating an unrealistic simplification in truth)
Quote
. In general this will be a combination of gas pressure and radiation pressure. In this Section we shall assume that the gas pressure dominates.
Quote
In some circumstances this can be approximated as a polytropic equation of the form
Quote
Hence, the use of Equ.(9) is a convenient approximation only
Quote
The Clayton model consists of assuming a specific analytic form for the pressure distribution
Quote
This differs from the original Clayton model which imposed zero pressure at the surface, r = R, of the star
So it is making simplifications.
As such, it is not based EXACTLY on the principles of physics as known to modern science.
Instead they are based upon approximations of those principles and thus will have a limited validity.

In fact, your own source has this to say:
Quote
This would seem to imply that the Clayton model pressure might be quite seriously in error except near the centre
Quote
However, at other locations the pressure is not given in terms of the density by a power-law like Equ.(9), i.e. Equs.(18) and (20) are not related by a power law. In as far as Equ.(9) is a good approximation to the equation of state, this exposes the approximation inherent in the Clayton models

So your own source says that what you are doing is complete nonsense, that the formula you are using does not hold where you are trying to use it.

Now again, stop attacking specific models and deal with a round sun in general and a sun undergoing nuclear fusion in its core in general.
You are yet to do either.

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rabinoz

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Re: Solar power source
« Reply #62 on: November 12, 2019, 04:39:16 AM »
Am I supposed to do your homework for you?

You, the RE, do not know the form of the Clayton equation, one of the most famous equations in astrophysics?

http://www.rickbradford.co.uk/Chapter11_StellarStructurePart1.pdf

equation (18)
That equation! I've seen and downloaded Chapter 11: Stellar Structure Part 1: Pressure, Density and Temperature Distributions in Spherically Symmetric, Main Sequence Stars, The Clayton Model.

But this equation is NOT your equation:
P(r) = 2πgr2a2ρ2ce-x2/3M, where a = (31/2M/21/24πρc)1/3

Look carefully and note that Rick Bradford has G, often called big G, the "Newtonian constant of gravitation" but you have g that you interpret as the sun's gravity.

There is a vast difference between g and the "Newtonian constant of gravitation, G = 6.67430 x 10-11 m3 kg-1 s-2[/quote]

Maybe you should have taken notice when I wrote:
" See: NIST RReference on Constants, Units and Uncertainty: Fundamental Physical Constants.
There is no more connection between the "pressure in the chromosphere" and the surface gravity of the  Sun than between the pressure at, say, 400 km and the surface gravity of Earth.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sure, "It is not your equation" but that does not mean that you've been using it correctly.

And still you go on with:
No, it's my link from three years ago.
Chapter 18 from the same work contains the graphics as well.
You cannot complain about the Clayton equation since it is YOUR very own formula, derived by one of the best astrophysicists of the 20th century.
It is the standard in stellar structure physics.
But we can complain about YOUR equation since it is not the Clayton equation nor is it Eqn (18) in Rick Bradford's Chapter 11!

So now do you accept that the surface gravitation of the Sun is about 274 m/s2?
« Last Edit: November 12, 2019, 12:48:30 PM by rabinoz »

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sandokhan

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Re: Solar power source
« Reply #63 on: November 12, 2019, 04:43:23 AM »
Your graphs are there so you are comparing a Clayton calculation to a Polytropic-Hydrostatic equation

Until you read my messages, you had no idea what a polytropic stellar equation is; I was the first to bring these concepts here to the FES.

You already have at your disposal the equation, the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct.


rabinoz... please take your trolling to the CN. It is simply pathetic.

G = gr2/m(r)

Are you saying that this equation is false?


jackblack, you cannot dismiss the Clayton equation.

It is the ONLY explicit equation modern astrophysics has to offer.

Go ahead and plug in the numbers, the registered figure for the pressure of the chromosphere, you will see that my calculations are totally correct.

Of course I used the equation correctly, check it out yourself.

Even if we assume an unimaginable 100x error factor, you are still left with a figure which does not help your cause at all.

Using Clayton's equation we get for the g acceleration of the Sun:

g = 0.0000507 m/s^2 which is much lower than the centrifugal acceleration figure

Let us now assume an unheard of 100x error figure, even then with g = 0.00507 m/s2, you are nowhere near to explain the hypothesized spherical shape of the Sun.

And I haven't even mentioned the Nelson effect.

Re: Solar power source
« Reply #64 on: November 12, 2019, 05:13:33 AM »

You cannot complain about the Clayton equation since it is YOUR very own formula
I don't believe he is Donald D. Clayton.
Quote from: mikeman7918
a single photon can pass through two sluts

Quote from: Chicken Fried Clucker
if Donald Trump stuck his penis in me after trying on clothes I would have that date and time burned in my head.

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mak3m

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Re: Solar power source
« Reply #65 on: November 12, 2019, 05:46:39 AM »

And I haven't even mentioned the Nelson effect.

Do you mean the Mandela effect, where you thought you remembered to show your workings and assumptions, but never did...ever.

I've lost count of the threads where you obfuscate, deflect and runaway when word salad doesn't get rid of questioning. 

You have to learn to reply without quoting a long previous answer.

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sandokhan

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Re: Solar power source
« Reply #66 on: November 12, 2019, 06:41:24 AM »
The Nelson effect is very real:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1645824#msg1645824 (synchronized relationship between the Sun’s periodic peak sunspot cycles and the orbital positions of the Jovian planets -- Jupiter and Saturn)

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mak3m

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Re: Solar power source
« Reply #67 on: November 12, 2019, 07:08:31 AM »
The Nelson effect is very real:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1645824#msg1645824 (synchronized relationship between the Sun’s periodic peak sunspot cycles and the orbital positions of the Jovian planets -- Jupiter and Saturn)

So again following your citations you are postulating now that the Sun is a rotating sphere and at the very least Earth, Jupiter and Saturn orbit it? And, that the moon orbits a rotating earth causing solar and lunar eclipses, and that the Alias effect is inconclusive.

Truly a victory for FE
You have to learn to reply without quoting a long previous answer.

Re: Solar power source
« Reply #68 on: November 12, 2019, 12:25:53 PM »
You already have at your disposal the equation, the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct.
Really?
Is that why the paper you quoted literally says that it isn't correct, that it is quite seriously in error except near the centre?

That sure doesn't sound like accepting that they are indeed correct.

jackblack, you cannot dismiss the Clayton equation.
Why can't I? Your own source says it only accurate at the core.
The graph's you have shown yourself, also show it is accurate at the core, but not elsewhere.

Go ahead and plug in the numbers, the registered figure for the pressure of the chromosphere, you will see that my calculations are totally correct.
Your "calculations" being correct is completely irrelevant.
You are using an equation which is not correct for this location.
That is the problem.
That means you are not using it correctly.

You may as well use the equation P=rho*g*h.
It would be use as useless, as it does not apply in this context.

Then there is also the fact that you tried to solve for a constant.
Why not solve for the variables, such as a, or rho_c, or M?

And again, all you are doing is attacking a particular model, a model which RE already acknowledges has significant limitations due to the assumptions and simplifications used and which does NOT work accurately where you are trying to use it.

So this in no way proves anything about the sun.

Now how about you try and actually back up your claims?

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rabinoz

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Re: Solar power source
« Reply #69 on: November 12, 2019, 02:26:06 PM »
You already have at your disposal the equation, the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct.
I am not questioning any of that, just your use of the equation that ends up in changing a fundamental constant, the Newtonian gravitational constant G.

Sure, "the registered pressure and the RE have already checked the figures three years ago, they had to accept that they are indeed correct" but "Clayton's equation" is still only an approximation.

You have not been using the exact "Clayton's equation" but you own re-arranged version where you replaced G by gr2/m(r)?
Now that technically is correct but G is a fundamental constant that YOU cannot change just because you feel like it.

Quote from: sandokhan
rabinoz... please take your trolling to the CN. It is simply pathetic.
I am not and never have been "trolling"! Get used to it.

Quote from: sandokhan
G = gr2/m(r); Are you saying that this equation is false?
Not at all but that form is only useful in calculating G if you have very accurately measured values of g, r and m and you have not got these.

You cannot arbitrarily use that equation to re-calculate some new value of G. You use your calculations to find the surface gravity of the Sun as in:
M = 1.989 x 1030 kg
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
G = gr2/m(r)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Using P(700,000,000) = 1.0197 x 10-9 kg/m2 value, we get: g = 0,0000507 m/s2
The "mechanics" of you calculations are correct but you simply cannot change a fundamental constant, G, like that.

You have GSand = gr2/m(r)  or  GSand = gr2/M.

So your value of GSand is 0.0000507 x 700,000,0002/1.989 x 1030 = 1.24902 x 10-17 N⋅m2/kg2 (or m3⋅kg−1⋅s−2) and you simply cannot do that!

To illustrate what you have done let's find the resulting value of G and use that to find the surface gravity of Earth.
gEarth = G x massEarth/(rEarth^2) where massEarth = 5.972 × 1024 kg and rEarth = 6.371 x 106 m.

So gEarth = 0.00000184 m/s2 and you must admit that is a tad low!

Where you have gone wrong is to forget that the Clayton's equation is just an approximation but have used the pressure vs distance from the centre for a star to re-calculate the fundamental constant, G.

Summarising: Your calculations are technically correct but you have grossly misused the "Clayton's equation".



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sandokhan

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Re: Solar power source
« Reply #70 on: November 12, 2019, 09:35:13 PM »
But G is not a universal constant.

Newton had to first prove that the Earth does indeed rotate around its axis, and that it does indeed orbit the Sun.

He never provided anything on the matter.

The universe DOES NOT obey Newton's law of "universal" gravitation.

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.

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.


The entire universe does not obey at all Newton's laws.

Therefore, G is a constant only here on Earth, not anywhere else.


Take a look at yourselves: now you are denying Clayton's equation.

Surely I can substitute gr2/M to check out the equation, since YOU SAY that G is indeed a universal constant.

Everything should work out fine: an answer in the range of 200 - 300 m/s2.

Yet the answer we get is 0.0000507 m/s2.

An error of the order of 1x107!!!

The author of the paper does perform a calculation and observes that the error is of the order of 40x at the surface.

Even if we allow an error of 100x (unheard of in such formulas), we still get 0.00507 m/s2, which is not anywhere near the value you need.


Dark Flow proves that G is NOT a universal constant. It does work here on Earth, not anywhere else.

As such, my calculations are very correct.

Modern science gives us the Clayton model.

You do the math: substitute the REGISTERED/RECORDED value of the pressure for the chromosphere and you get 0.0000507 m/s2.

The shape of the Sun cannot be spherical.

« Last Edit: November 12, 2019, 09:42:47 PM by sandokhan »

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mak3m

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Re: Solar power source
« Reply #71 on: November 12, 2019, 09:51:01 PM »
Big G is a fundamental constant in physics, the hypothesis has been tested numerous times, here and far out into the universe.

Can you provide evidence it isnt constant?

Also asking again for the nth time can you show your assumptions for Clayton
You have to learn to reply without quoting a long previous answer.

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sandokhan

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Re: Solar power source
« Reply #72 on: November 12, 2019, 10:16:30 PM »
It has been tested here on Earth, not anywhere else.

First, you have to prove assumed the orbital motions of the Earth.

DARK FLOW defies the G constant on a universal scale.


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.

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.


The entire universe does not obey at all Newton's laws.

Therefore, G is a constant only here on Earth, not anywhere else.


Also asking again for the nth time can you show your assumptions for Clayton

You are trolling this thread.

You have at your disposal the link/reference.

Plug in the numbers yourself.


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.

Use the Clayton equation provided by modern science.

Final answer: g(sun) = 0.0000507 m/s2.

The shape of the Sun cannot be spherical.

« Last Edit: November 12, 2019, 10:20:19 PM by sandokhan »

Re: Solar power source
« Reply #73 on: November 12, 2019, 11:06:44 PM »
But G is not a universal constant.
And more baseless claims for you.
So you want to claim that G is a variable and can vary in the sun?
I guess that means the Clayton model needs to be updated as well.

And of course, now you are running off on a tangent.

Surely I can substitute gr2/M to check out the equation
At which point you are then rejecting the values provided. In order to use the equation with the accepted RE figures, you need to accept G.

As such, my calculations are very correct.
Again, your calculation are pure nonsense as you are not correctly applying the calculation.
You are completely ignoring the approximations of the model and the FACT that it does not work at the surface of the sun.

The shape of the Sun cannot be spherical.
Again, you have absolutely no basis for this claim.
All you have is you completely misapplying a calculation to try and attack one specific model, which is already known to not work at the surface of the sun. The Clayton model is wrong on the surface of the sun. So what?
That in no way means the sun is not spherical.

The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth drops to one-thousandth of the pressure on the earth in the photosphere.
And more pure nonsense.
g being 27.4 times as large doesn't mean the pressure will be 27.4 times as large.
There is literally no connection there.

Now, stop using an equation WHICH DOES NOT APPLY TO WHERE YOU ARE USING IT!
Stop attacking specific models.

Start actually trying to prove that the sun is flat as you haven't even attempted to do so yet.

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sandokhan

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Re: Solar power source
« Reply #74 on: November 12, 2019, 11:44:02 PM »
g being 27.4 times as large doesn't mean the pressure will be 27.4 times as large.
There is literally no connection there.


You are trolling the flat earth debate section.

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".

You see, you are too desperate to cover up the fact that the shape of the sun cannot be spherical at all.


So you want to claim that G is a variable and can vary in the sun?

It is absolutely variable.

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.

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.

The entire universe does not obey at all Newton's laws.

Therefore, G is a constant only here on Earth, not anywhere else.

In order to use the equation with the accepted RE figures, you need to accept G.

Exactly.

I substitute the known formula for G: G = gr2/M.

Surely, if what you say is true, then the equation will work JUST AS FINE with g, isn't it?

Again, your calculation are pure nonsense as you are not correctly applying the calculation.
You are completely ignoring the approximations of the model and the FACT that it does not work at the surface of the sun.


Let's put your word to the test.

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) = 2πgr2a2ρ2ce-x2/3M

where a = (31/2M/21/24πρc)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


ac/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!!!


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.

The shape of the Sun cannot be spherical.

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mak3m

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Re: Solar power source
« Reply #75 on: November 12, 2019, 11:51:55 PM »
It has been tested here on Earth, not anywhere else.


The entire universe does not obey at all Newton's laws.

Therefore, G is a constant only here on Earth, not anywhere else.


You are trolling this site.

https://www.sciencedaily.com/releases/2014/03/140324230254.htm

Put this thread down to another example of you not being able to back your assumptions, not wasting another day chasing around your gish

Put up or shush
You have to learn to reply without quoting a long previous answer.

Re: Solar power source
« Reply #76 on: November 12, 2019, 11:52:28 PM »
Quote
The shape of the Sun cannot be spherical.
So what shape do you propose the Sun might be.  Based on observations alone it is obviously circular because it presents a disk on the sky and the fact that we have been watching sunspots traverse that disk since Galileos time is evidence enough that the Sun rotates. Larger sunspots disappear on one side only to return a couple of weeks later on the other side.

You don't need any of your complex mathematics or theory to figure out that the Sun is obviously spherical.  Just observe the Sun (safely of course). Common sense shows it is spherical.  Read the bit at the top of your little embedded graphic where it says at the start "The Sun is a ball of gas".  The last time I checked a ball was spherical in shape or do you deny that as well? You seem to be contradicting your own attachments!  The graphic is entirely true and what I've known all along.  So to state explicitly that "the shape of the Sun cannot be spherical" is just being silly to be quite frank with you.
« Last Edit: November 12, 2019, 11:59:27 PM by Nucleosynthesis »

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sandokhan

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Re: Solar power source
« Reply #77 on: November 13, 2019, 12:00:45 AM »
https://arxiv.org/abs/1402.1534

Is this supposed to be a joke on your part?

They are using the redshift distance relation!

The Picture that Won’t Go Away

"Only in the rarest instances has a single picture altered the direction of a scientific discipline. But in the case of the galaxy NGC 7319 and the "misplaced" quasar in front of it, the message is inescapable: its presence threatened to shatter one of the most cherished themes of mainstream astronomy, the Big Bang.

The rationale for the Big Bang rests substantially on an interpretation of a well-known phenomenon called “redshift”. The term refers to the shift of light from distant galaxies toward red on the light spectrum.

Many years ago, astronomers decided that redshifted objects must be moving away from the observer, stretching out their lightwaves. This “Doppler interpretation” of redshift enabled astronomers, based on the degree of redshift, to calculate both the distances and velocities of the objects. From these calculations, certain conclusions were inescapable. If all redshifted objects are moving farther away, the universe must be expanding. If the universe is expanding, the expansion must have had a starting point—an unimaginable explosion producing a universe of galaxies receding in every direction from the observer.

Then came the Hubble photograph, taken on October 3, 2003. The picture showed a galaxy (NGC 7319) known for its dense clouds that obstruct all objects behind its core. In front of the galaxy's core is a strongly redshifted quasar. In fact, under the prevailing assumptions, the redshift of the quasar would put it more than 90 times farther away from us than the big galaxy behind it."



A higher magnification image of the quasar shows a "jet" of matter extending out from the center of NGC 7319 toward the quasar:




http://ucsdnews.ucsd.edu/archive/newsrel/science/mcquasar.asp


The Discovery of a High Redshift X-Ray Emitting QSO Very Close to the Nucleus of NGC 7319:

https://arxiv.org/pdf/astro-ph/0409215.pdf


Published in the Astrophysical Journal

Geoffrey Burbidge, a professor of physics and astronomer at the University of California at San Diego’s Center for Astrophysics and Space Sciences

"The quasar was found embedded in the galaxy NGC 7319 only 8 arc sec from its centre. According to the Hubble law the galaxy NGC 7319, with a redshift of 0.022, is at a distance of about 360 million light-years. Therefore these objects could not be physically connected to each other if this was true."


At the meeting of the American Astronomical Society held in Texas in 2004, Professor Margaret Burbidge presented a paper that she had co-authored with Arp and several other leading astronomers, including her husband [subsequently published in the Astrophysical Journal]. It detailed the discovery of a high redshift quasar close to a low redshift galaxy. This time, though, the alignment was different in every significant way.

This time, no one could argue. You see, the high redshift [more distant] quasar lay in front of the [less distant redshift] galaxy NGC 7319! There was no longer occasion to debate the veracity of [Arp’s] matter bridge [connecting galaxies with quasars]. The quasar was in the foreground [the galaxy in the background]. In that impressive gathering of astronomy’s who’s who, you could have heard a pin drop. It was a deafening silence.”

“The significance of this discovery is huge. We have direct, irrefutable, empirical evidence that the Hubble law stands on feet of clay, that the observational justification of an expanding Universe is fatally flawed.”

Hilton Ratcliffe


Based on observations alone it is obviously circular because it presents a disk on the sky

Yes, the shape of the Sun is discoidal.

we have been watching sunspots

In the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

g(sunspot) = g(earth)/10,000 = 0.00098 m/s2.

The shape of the Sun cannot be spherical.

Re: Solar power source
« Reply #78 on: November 13, 2019, 12:25:22 AM »
Forget all this stuff you keep on going on about pressure, temperature and whatever other physical properties of the Sun you wish to involve.  The Sun is obviously spherical.  Your own eyes can tell you that. 

Sunspots even look flattened as the approach the limb on either side because the Suns spherical surface is turning them away from the observers direct line of sight. If you look at a sunspot in the centre of the solar disk it looks regular and circular. If you see it on the edge it looks elliptical.  No science required to realise that.  It is just common sense.

https://en.wikipedia.org/wiki/Wilson_effect

If your theories and equations tell you something different to that my friend, then I can only suggest they are wrong. Sometimes you can be blinded by the maths and fail to see what is clearly there in front of you.

« Last Edit: November 13, 2019, 12:27:09 AM by Nucleosynthesis »

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sandokhan

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Re: Solar power source
« Reply #79 on: November 13, 2019, 12:27:42 AM »
The Sun is obviously spherical. 

But it can't be.

I have the equations, the numbers, everything.

You, on the other hand, have NOTHING.

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.

The shape of the Sun cannot be spherical.

Re: Solar power source
« Reply #80 on: November 13, 2019, 12:58:48 AM »
I have my own eyes and my own common sense which is all I need to tell me that the Sun is spherical. You don't need numbers, equations or anything else. You can make numbers and equations tell you anything you wish but sometimes plain old direct observation will tell you the truth.

Your argument that "I have nothing" is based on nothing more than I am saying something that you can't or won't accept. I also have two dedicated solar telescopes to show me the Sun is spherical as well.

I have a very nice image taken in the wavelenth of CaK (near ultra violet) of the Mercury transit that happened on Monday. Happy to share it if you are interested. I used a Lunt LS152 with a dedicated CaK filter.

Have you counted up how many times you have posted that graphic about the photosphere of the Sun now?  I am familiar thank you with the properties of the photosphere as well as the other layers of the Suns atmosphere through the university module I did last year about solar astrophysics.

If your theories, numbers and equations tell you that the Sun is not spherical then there is something clearly wrong with them. As Richard Feynman famously said...

"Now I’m going to discuss how we would look for a new law. In general, we look for a new law by the following process. First, we guess it (audience laughter), no, don’t laugh, that’s the truth. Then we compute the consequences of the guess, to see what, if this is right, if this law we guess is right, to see what it would imply and then we compare the computation results to nature or we say compare to experiment or experience, compare it directly with observations to see if it works.

If it disagrees with experiment, it’s wrong. In that simple statement is the key to science. It doesn’t make any difference how beautiful your guess is, it doesn’t matter how smart you are who made the guess, or what his name is … If it disagrees with experiment, it’s wrong. That’s all there is to it."

For experiment of course we can also use the word observation.

« Last Edit: November 13, 2019, 01:14:21 AM by Nucleosynthesis »

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mak3m

  • 737
Re: Solar power source
« Reply #81 on: November 13, 2019, 01:06:49 AM »
Quote
The shape of the Sun cannot be spherical.
So what shape do you propose the Sun might be.  Based on observations alone it is obviously circular because it presents a disk on the sky and the fact that we have been watching sunspots traverse that disk since Galileos time is evidence enough that the Sun rotates. Larger sunspots disappear on one side only to return a couple of weeks later on the other side.

You don't need any of your complex mathematics or theory to figure out that the Sun is obviously spherical.  Just observe the Sun (safely of course). Common sense shows it is spherical.  Read the bit at the top of your little embedded graphic where it says at the start "The Sun is a ball of gas".  The last time I checked a ball was spherical in shape or do you deny that as well? You seem to be contradicting your own attachments!  The graphic is entirely true and what I've known all along.  So to state explicitly that "the shape of the Sun cannot be spherical" is just being silly to be quite frank with you.

Sandy has not answered a question in years.

He diverts, he posts proofs that match up with standard models then concludes something completely left field, he copy pastas even within the same thread and posts links without even reading them.
You have to learn to reply without quoting a long previous answer.

Re: Solar power source
« Reply #82 on: November 13, 2019, 01:12:37 AM »
g being 27.4 times as large doesn't mean the pressure will be 27.4 times as large.
There is literally no connection there.


You are trolling the flat earth debate section.
Is that why instead of you trying to rationally respond you have to resort to insults?

I may have exaggerated slightly. There is a very slight connection in that gravity does contribute, but it is not a simple linear relationship.
Instead it depends on the strength of gravity and the amount of gas above it.

Just because a planet has x times the gravity, that doesn't mean the pressure will be x times.

If you would like to try justifying such nonsense, go ahead.
But insults wont help you.

So you want to claim that G is a variable and can vary in the sun?
It is absolutely variable.
Which means the formula you are using would be completely invalid as it relies upon G being a constant. That means you are completely wrong regarding the Clayton equation.

Do you understand that?
Is that why you had to completely ignore me saying that very thing and instead just skip over it and post more spam?
Because you don't want to admit you are wrong and have absolutely no rational basis for your insane claims?

In order to use the equation with the accepted RE figures, you need to accept G.
Exactly.
Yes, exactly. Not making a ridiculous substitution, but just accepting it.
If you do that, the pressure you calculate is wrong.
Again, this shows the Clayton model doesn't work at that altitude.
Why do you keep ignoring this fact?

Surely, if what you say is true, then the equation will work JUST AS FINE with g, isn't it?
If by "just as fine" you mean still very wrong, then yes.
But I would phrase it in a much more meaningful way, i.e. it will still not work.

Let's put your word to the test.
Sticking the numbers in again and showing it produces a nonsense value just further supports the claim of your source, that this equation does not apply to the surface of the sun.

Accuracy of the Clayton model:

Yes, amazingly 0 is quite similar to 0.
You are dealing with basically no pressure.
Underestimating or overestimate

The author of the paper did perform a calculation using the Clayton model at the surface: the error is at most 40x.
The author of what paper, and what error?
There are quite a lot of variables in there. Just which one has the 40x error?

There is still another way to prove the correctness of the Clayton equation.
You mean there is still another way you can try and escape from reality?

Repeating the same nonsense will not help you.

You are yet to prove anything except that you no concern for the truth.

I have the equations
You have equations which do not apply.
You have a pathetic attack on a single model, a model which relies upon the sun being spherical.

So you have nothing.

Meanwhile we have mountains of evidence.

All you seem to be capable of doing is repeating the same refuted nonsense.

Now how about you stop discussing the flawed Clayton model and instead try to prove your insane claims.

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mak3m

  • 737
Re: Solar power source
« Reply #83 on: November 13, 2019, 01:15:45 AM »
https://arxiv.org/abs/1402.1534

Is this supposed to be a joke on your part?

They are using the redshift distance relation!

The Picture that Won’t Go Away

"Only in the rarest instances has a single picture altered the direction of a scientific discipline. But in the case of the galaxy NGC 7319 and the "misplaced" quasar in front of it, the message is inescapable: its presence threatened to shatter one of the most cherished themes of mainstream astronomy, the Big Bang.

The rationale for the Big Bang rests substantially on an interpretation of a well-known phenomenon called “redshift”. The term refers to the shift of light from distant galaxies toward red on the light spectrum.

Many years ago, astronomers decided that redshifted objects must be moving away from the observer, stretching out their lightwaves. This “Doppler interpretation” of redshift enabled astronomers, based on the degree of redshift, to calculate both the distances and velocities of the objects. From these calculations, certain conclusions were inescapable. If all redshifted objects are moving farther away, the universe must be expanding. If the universe is expanding, the expansion must have had a starting point—an unimaginable explosion producing a universe of galaxies receding in every direction from the observer.

Then came the Hubble photograph, taken on October 3, 2003. The picture showed a galaxy (NGC 7319) known for its dense clouds that obstruct all objects behind its core. In front of the galaxy's core is a strongly redshifted quasar. In fact, under the prevailing assumptions, the redshift of the quasar would put it more than 90 times farther away from us than the big galaxy behind it."



A higher magnification image of the quasar shows a "jet" of matter extending out from the center of NGC 7319 toward the quasar:




http://ucsdnews.ucsd.edu/archive/newsrel/science/mcquasar.asp


The Discovery of a High Redshift X-Ray Emitting QSO Very Close to the Nucleus of NGC 7319:

https://arxiv.org/pdf/astro-ph/0409215.pdf


Published in the Astrophysical Journal

Geoffrey Burbidge, a professor of physics and astronomer at the University of California at San Diego’s Center for Astrophysics and Space Sciences

"The quasar was found embedded in the galaxy NGC 7319 only 8 arc sec from its centre. According to the Hubble law the galaxy NGC 7319, with a redshift of 0.022, is at a distance of about 360 million light-years. Therefore these objects could not be physically connected to each other if this was true."


At the meeting of the American Astronomical Society held in Texas in 2004, Professor Margaret Burbidge presented a paper that she had co-authored with Arp and several other leading astronomers, including her husband [subsequently published in the Astrophysical Journal]. It detailed the discovery of a high redshift quasar close to a low redshift galaxy. This time, though, the alignment was different in every significant way.

This time, no one could argue. You see, the high redshift [more distant] quasar lay in front of the [less distant redshift] galaxy NGC 7319! There was no longer occasion to debate the veracity of [Arp’s] matter bridge [connecting galaxies with quasars]. The quasar was in the foreground [the galaxy in the background]. In that impressive gathering of astronomy’s who’s who, you could have heard a pin drop. It was a deafening silence.”

“The significance of this discovery is huge. We have direct, irrefutable, empirical evidence that the Hubble law stands on feet of clay, that the observational justification of an expanding Universe is fatally flawed.”

Hilton Ratcliffe


Based on observations alone it is obviously circular because it presents a disk on the sky

Yes, the shape of the Sun is discoidal.

we have been watching sunspots

In the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

g(sunspot) = g(earth)/10,000 = 0.00098 m/s2.

The shape of the Sun cannot be spherical.

Lol read your links half wit

Quote
We have clearly demonstrated that the ULX lying 8 arc sec from the nucleus of NGC 7319 is a high redshift QSO. This is to be added to a list of more than 20 ULX candidates which have all turned out to be genuine QSOs (cf. Burbidge, Burbidge & Arp 2003; Arp, L´opez-Corredoira and Guti´errez 2004). Since all of these objects lie within a few arc minutes or less of the centers of these galaxies, the probability that any of them are QSOs at cosmological distance, observed through the disk of the galaxy, is negligibly small. Thus this is further direct evidence that high redshift QSOs are generated and ejected in low redshift active galaxies.

 ::)
You have to learn to reply without quoting a long previous answer.

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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 4871
Re: Solar power source
« Reply #84 on: November 13, 2019, 01:23:27 AM »
I may have exaggerated slightly.

What ?!

Slightly?


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.

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

Sure, which means you have nothing against substituting the correct formula G = gr2/M, since it works just as well with g.

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.

Let us perform the calculation once more.

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) = 2πgr2a2ρ2ce-x2/3M

where a = (31/2M/21/24πρc)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


ac/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!!!


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.

The shape of the Sun cannot be spherical.


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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 4871
Re: Solar power source
« Reply #85 on: November 13, 2019, 01:26:50 AM »
mak3m, you are expected to be posting while sober.

Thus this is further direct evidence that high redshift QSOs are generated and ejected in low redshift active galaxies.

Exactly my point.

The redshift distance relation is completely useless.

Your link made use exactly of this relation.

You don't stand a chance with me here.

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mak3m

  • 737
Re: Solar power source
« Reply #86 on: November 13, 2019, 02:02:05 AM »
mak3m, you are expected to be posting while sober.

Thus this is further direct evidence that high redshift QSOs are generated and ejected in low redshift active galaxies.

Exactly my point.

The redshift distance relation is completely useless.

Your link made use exactly of this relation.
.

Resorting to insults again chap, well it is easier than actually sticking to the point and answering questions.

Take some time read your papers than try and make a hypothesis on how a phenomena documented dozens of times has broken G and BBT, you are the only person on the planet to jump to that conclusion

You don't stand a chance with me here.


You beat yourself either with sloppy math, failing to read your own citations, or jumping to wild unsubstantiated claims.


You have to learn to reply without quoting a long previous answer.

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rabinoz

  • 23800
  • Real Earth Believer
Re: Solar power source
« Reply #87 on: November 13, 2019, 02:09:52 AM »
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.
Quote from: sandokhan
DARK FLOW:
I'll omit any reference to that because it is quite irrelevant here.

You have claimed that "G is not a universal constant" so, I have to ask why the source of your equation, Ricks Cosmology Tutorial, uses "G = 6.67 x 10-11 in MKSA units".
Then go and "calculate", via your g calculations, a vastly different value of G.
Quote from: Rick Bradford
Ricks Cosmology Tutorial: Chapter 11 Stellar Structure Part 1
The central pressure from Equ.(22) is also given above (using G = 6.67 x 10-11 in MKSA units). Since 1 atm = 105 Pa, the larger of the above pressures is a huge 1.9 x 1011 atmospheres pressure. Standard solar models give the central pressure as 1.65 x 1016 Pa, so the Clayton model with R/a ~ 5.4 is reasonably close.

So whether or not you think that G is a universal constant you undeniably use the Clayton equation inappropriately by forcing a value of g out of it and implicitly a value of G vastly different from the one Rick Bradford used.

Quote from: sandokhan
Take a look at yourselves: now you are denying Clayton's equation.
Nothing of the sort! You are the one misusing Clayton's equation.

Quote from: sandokhan
Surely I can substitute gr2/M to check out the equation, since YOU SAY that G is indeed a universal constant.
But you cannot use that to find the surface g on the sun when Rick Bradford has already used "G = 6.67 x 10-11 in MKSA units" to find the parameters of the Clayton's equation.

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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 4871
Re: Solar power source
« Reply #88 on: November 13, 2019, 02:26:42 AM »
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?

You have just made TWO CONTRADICTORY STATEMENTS.

because we know that the whole "universe DOES NOT obey Newton's law of "universal" gravitation."

But G is still regarded as "universal constant"


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.

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.

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.


The entire universe does not obey at all Newton's laws.

Therefore, G is a constant only here on Earth, not anywhere else.


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.

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.

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.

Certainly you have something to hide if you are so troubled by a simple substitution in the equation.

Are you saying that G does not equal g2/M?

If the Sun is a sphere, you should have nothing to fear.

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?

It is a valid substitution: we don't know if g will be different, we simply want to find out for ourselves.

Yet you are telling everyone here that they cannot solve for g.

Please come to your senses.

The calculation is very straightforward.

g(sun) = 0.0000507 m/s2.

The shape of the Sun cannot be spherical.


My calculation is proven correct again:

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.

The shape of the Sun cannot be spherical.


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mak3m

  • 737
Re: Solar power source
« Reply #89 on: November 13, 2019, 02:43:24 AM »
And there it is the 10th copy pasta of same post in a single thread

We salute you Sandokhant
You have to learn to reply without quoting a long previous answer.