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 g

^{2}/M. Surely if you what you say is right, there should be no problem to use the formula with g instead of G.

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 g

^{2}/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.