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Originally Posted by nutant gene 71
Per the Equivalence Principle, gravitational mass equals inertial mass, always. "Gravitational" mass is a function of G, as my basket of apples illustration shows above, which is equivalent to its "inertial" mass. The same basket, or cubic decimeter of water, can be either one kilograms (in 1 G), or ten kilograms (in 10 G).
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:roll:
I see that you still do not grasp the difference between
kilogram-mass and
kilogram-weight.
You basket has a
weight of
1 kg-weight = 1 kg-mass * g on Earth, and of
10 kg-weight = 1 kg-mass * (10*g) on your planet.
The inertial mass of
1 kg-mass has not changed.
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Originally Posted by nutant gene 71
If we want the whole universe to be figured in Earth's arbitrary measure for mass, our kilograms, then 10 G means the inertial-gravitational masses are always 10 kilograms.
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Wrong.
The force on them is ten time higher: their masses are not.
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Originally Posted by nutant gene 71
But this causes a problem with how masses interact locally, because if kilograms for their local measure of G are different from ours, then each kilogram will need to be "locally" different from ours.
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Of course you are referring to the weight, measured in
kg-weight, and not to the mass, measured in
kg-mass.
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Originally Posted by nutant gene 71
The ramifications of this is that local mass interacts differently than here on Earth. If so, then using Earth's kilograms becomes a poor, and ingenuous choice, since it fails to explain how in a different gravitational G "proportional" masses may hold together in ways that our 1 G (1 kg) cannot explain. Let me explain this further in your next.
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The masses are not proportional to
G.
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Originally Posted by nutant gene 71
Why would you say G is a "physical quantity"?
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Because it can be measured.
Newton's law for gravitation gives at least an operative definition of
G, which allows researchers to measure it.
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Originally Posted by nutant gene 71
Is it not merely a "proportional" quantity between gravitationally attracted masses?
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Is it not a quantity that can be
measured?
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Originally Posted by nutant gene 71
If G attracts at 1 G, that is the proportional attraction between the masses, which are measured in kilograms. If G's proportional attracts at ten times our 1 G, then the proportional attraction between masses increases by tenfold.
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The
force increases, but not the masses.
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Originally Posted by nutant gene 71
But if each mass is now tenfold in terms of its (equivalent) inertial mass,....
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It is not: the gravitational force increased because you increased
G not the masses.
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Originally Posted by nutant gene 71
... then the attraction is ten times ten (two bodies interacting), so the interaction between them is what?
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Why?
You already increased
G:
G' = 10 G.
Then you assume out of the blue an increase in mass:
m' = 10 m,
M' = 10 M.
That would make the force:
F' = 10*10*10*F = 1000 F.
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Originally Posted by nutant gene 71
If each side has an "eqivalent" mass that is ten times greater, and the G "proportional" between them is ten times greater, saying merely that the mass is now 10 kilograms is not enough, because they attract by a larger proportional. And that, really, is why it is important to redefine our (arbitrary) kilogram in a different G scenario.
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It is not, because the definition of
kilogram-mass does not depend on the value of
G.
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Originally Posted by nutant gene 71
This is also why I brought up the question in the first place, because I don't know if the answer is tenfold or one hundredfold. This question cannot be raised in a 1 G universe,...
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Your question can be raised with
G as universal constant.
Your problem is the distinction between
kg-mass and
kg-weight: you just need to go to the Moon to raise that question, and find the answer.
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Originally Posted by nutant gene 71
....but it can be raised in a (hypothetical) variable G universe. So, yes, I understand what you are saying, if the universe is only 1 G throughout, but I am forced to disagree with you, because it does not apply in a variable G universe.
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No, you do not understand.
You still confuse weight with mass.
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Originally Posted by nutant gene 71
No. The Equivalence Principle is NOT independent of the value of G.
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So, how come it does not show up in the relevant equation?
M(grav) = M(inertial).
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Originally Posted by nutant gene 71
We know the inertial mass and gravitational mass are the same. Gravity acceleration is a function of its "proportional" G.
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And of the distance and mass of the other mass (for example, Earth).
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Originally Posted by nutant gene 71
Acceleration of mass is inertially "proportional" to G.
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Because the gravitational force is proportional to
G.
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Originally Posted by nutant gene 71
Therefore, they are BOTH a function of the G proportional.
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:roll:
If the acceleration of a mass is due to gravity, then your "gravity acceleration" and "acceleration of mass" are one and the same!
And sinc this acceleration depends on the garvitational force exerted on the mass, of course it depends on
G.
But this does not imply that the mass that is accelerated depends on
G.
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Originally Posted by nutant gene 71
Change G and you change the "proportional" to how mass interacts.
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Because the force changes.
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Originally Posted by nutant gene 71
Why is this so difficult to understand? Or is it because you think ONLY of G as a "universal constant", and cannot imagine it being something different? Okay, for now, that is how the world of physics sees it. But change G, hypothetically, and what have you got? The same "proportinal"? No!
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Change
G and you change the force.
Change the force and you change the acceleration.
G mM/r^2 = F = a M, and
G' mM/r^2 = F' = a' M
If
G' = 10 G,
G mM/r^2 = F = a M, and
(10 G) mM/r^2 = (10 F) = (10 a) M.
You see?
M is subjected to ten times the acceleration, becasue the force is ten times stronger.
But you, with no justification whatsoever, assume a change in
M.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by papageno
Newtons' formula for gravitation:
F = G * (m*M) / r^2 (1)
Newton's second law:
F = M * a (2)
Equivalence principle: M in (1) is the same as M in (2).
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M(1) = M(2): where does G enter? |
a = G * m/ r^2. |
So, you admit that
G, even through
a, does
not enter
M(1) = M(2).
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Originally Posted by nutant gene 71
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Originally Posted by papageno
Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.
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You're really stubbornly holding on to this notion that our 1 G is it. Let's go back to this:
F = Ma
If the gravity F is ten times, then (as an either or case) either 10 F = M * 10 a; or, 10 F = 10 M * a. Which would you choose? |
Experimental results show that it is (
10 a).
Why are you attributing the change to
M?
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Originally Posted by nutant gene 71
They are not the same: If you choose the prior, mass is calculated in Earth's 1 G kilograms, and acceleration is tenfold (gravity acts ten times on mass). This has been your argument all along, I believe. On Jupiter, a much greater mass than Earth's, the acceleration is increased by its greater (1 G) gravity.
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And on Jupiter the weight of
1 kg-mass is higher!
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Originally Posted by nutant gene 71
But if you choose the latter, you're in a 10 G universe, then mass is calculated in 10 G "kilograms" (where each kilogram is tenfold ours, same cubic decimeter of water but "weighs" ten times ours, and ten times per equivalence), but acceleration remains the "same".
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Unfortunately for you, we observe the first case, where the acceleration is larger.
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Originally Posted by nutant gene 71
Is this the same acceleration we had in our 1 G universe? I don't think so, since it is already tenfold ( a = 10 G * m/ r^2 ), so that it pulls ten times as hard on the (tenfold kilograms) of mass. The end result is that in 10 G universe, tenfold acceleration pulls on tenfold mass.
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You forgot to justify where the change in mass comes from.
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Originally Posted by nutant gene 71
Whether we are pushing or pulling on this mass, it should remain equivalent. If 10 F = 10 M * a, and the "a" is already tenfold because G is tenfold, then gravity acts ten times on a mass that is ten times greater.
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You cannot even be consistent.
You had mutually exclusive options: either
10 M or
10 a.
You chose
10 M, so you cannot change
a:
a stay the same (not the "same").
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Originally Posted by nutant gene 71
Therefore, in 10 G universe, the 10 F (gravitational equivalence) acting on mass is tremendous (a square of 10), and that means matter interacts there differently from our 1 G universe. Conversely, per equivalence, the nertial mass will now take a much greater (1 G) force to move the 10 G mass (10 squared). If the inertial mass is now ten times (10 kg) what it was in our 1 G universe, the force needed to move it will be 100 times our 1 G force. but only tenfold in local 10 G "kilograms". And THAT is why a variable G universe is different from our known 1 G universe.
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You forgot that the inertial mass can be measure using non-gravitational forces.
And by doing such experiments, you would see that the inertial mass has not changed, even if
G changes.
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Originally Posted by nutant gene 71
Are we conceptually prepared to think this way? In my opinion, we are not.
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Your opinion on serious misconceptions and deep misunderstandings.
I explained to you uncountable times that a variable
G is not an exotic concept, nor mathematically challenging.
It would exactly like a variable dielectric constant in electrostatics, which is commonplace, and easily dealt with.
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Originally Posted by nutant gene 71
Equivalence is still preserved, but it takes a different set of rules for a (hypothetical) universe where G is variable:
Mass has not changed, only how we measure it changed.
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:roll:
You still do not understand how mass is measured.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.
The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs.
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kilogram is a unit of mass, not weight.
The weight you get is (1 kg)*(local g) N, where "local g" is the local gravitational acceleration on the surface of the planet; on Earth local g = 9.8 m/s^2, so 1 kg -> 9.8 N of weight.
On your planet, local g = 98 m/s^2, so 1 kg -> 98 N of weight.
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Originally Posted by nutant gene 71
Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.
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I see that you are confusing weight and mass.
1 kg(weight on Earth) = 1 kg(mass)*g = 9.8 N.
On your planet, g is different: g' = 10 g, hence
1 kg(weight on planet) = 1 kg(mass)*g' = 98 N. |
ibid. |
So, you still do not understand.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
Now can you see why I find this hypothetical question so challenging? I realize this is merely a "what if" question, but what if we find that other worlds or regions of space have a different G? If we find this, then it is truly exciting!
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I see only a confusion about the unit kilogram.
It is not for weight, but for mass. |
ibid. |
You persist in your misconception.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
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And how is that different from using another system of units? |
Who is right, the Xians or the Earthians? |
They are both right: they are just using different units.
Europeans are right when they say that I am 180 cm tall;
US citizens are right when they say that I am 71 inches tall. |
Yes! It's all relative to where you measure.
...snip... |
The numbers attributed to physical quantities depends on the system of units.
You have shown that this concept is beyond your grasp.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
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Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass. |
Wrong. Yes, only "the gravitational force would be different", but so also would the effective "kilograms" for mass. |
In this case, there is no such thing as "effective kilograms".
The kg unit for mass odes not change.
What changed is the gravitational acceleration at the surface, hence the gravitational force is difeerent = different weight. |
Mass does not change, same basket of apples. But the inertial mass changed. |
You are contradicting yourself: either mass does not change, or it does.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
Same mass, but in 10 G, equivalence is now 10 kg (for the 1 kg basket of apples). Remember that it is G we're talking about for a planet that had not changed in size or volume, only the G changed. The cubic decimeter of water had not changed, only its effective "weight" had changed. Where m = 1 kg before (at 1 G), the mass (same mass, still cubic decimeter) is now m' = 10 kg (at 10 G).
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Only if you use kg for weight, which is wrong.
You should use N, the unit for force. |
Either or. Kilograms are derived from Earth's gravitational force on one cubic centimeter of water (plantinum-irridium artefact) and also a standard of measure for weight through most of the world. |
:roll:
Why can't you understand the difference between
kg-mass and
kg-weight?
Why is it so hard for you?
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
I see that as a change in the mass's "dynamic measurement", where a new "kilogram" defines (measures) the cubic decimeter of water. (Remember, 1 kg does not equal 10 kg for the same mass.)
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And you see wrong.
Becuase the inertial mass has not changed. |
In your 1 G universe, inertial mass has not changed. In a variable G universe, it has changed, as per above. |
The "above" is misguided and wrong.
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
Can you see where this is taking me?
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You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere. |
I am beginning to see you have a serious conceptual disconnect with what is being discussed here. |
You do not grasp the difference between weight and mass, confusing the unit kg(mass) with the unit kg(weight on Earth) = 1 kg(mass)*9.8 m/s^2. |
ibid. |
Translated: "I have no clue." Is this what you mean?
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
It looks to me like you're still thinking 1 G, but ten times more powerful, in a 1 G universe. The conceptual adjustment necessary is to think in terms of a 10 G (or any variable G) universe, and then measure the new "kilograms" from there.
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Wrong.
I already explaiend to you uncountable times that a variable G is not an exotic concept, but unsupported by observations. |
Correct, a variable G is unsupported by current observations. That's why this exercise in reason is only hypothetical. |
But you clearly do not grasp the basics for this kind of speculations.
Hence you are reaching wrong conclusions.
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Originally Posted by nutant gene 71
[snip!]
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Originally Posted by papageno
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Originally Posted by nutant gene 71
How would such a buble of gas exist otherwise? We're not talking about soap bubbles here (where the outer surface is held together by water tension), but vast collections of molecules put into immense spin, around what? More gas? Since I obviously don't know, I'd be curious as to your idea of what's inside Jupiter, really!  |
I see no numbers to support your idea.
Can you show that the mass of Jupiter cannot be held together by gravity? |
I don't see your answer. |
You made claim: the burden of proof is yours.
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Originally Posted by nutant gene 71
...snip...
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Originally Posted by papageno
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Originally Posted by nutant gene 71
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Originally Posted by papageno
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Originally Posted by nutant gene 71
These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
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The postulate is based on a wealth of experimental evidence, which you simply ignored. |
No, not ignored, but only considering as possible theoretical explanations without final verdict. |
Where did you consider it? |
Experiments are not final judgements forever. They need to be periodically and critically reviewed, so new models emerge. Otherwise, your skating dangerously close to the thin edges of dogma. |
:roll:
Galileo's experimental results are still valid today.