Did Jupiter just shrink in size per its "shrunk" mass?


Are you saying (Gm) = m, or not ? :-) ...Show me.


G is in SI units: m^3/ kg/ s^2 or Nm^2/kg^2.

The second part question belongs on ATM, since it had not been observed. Any variance in G could be observable, but ask ESA if that's what they're after, in situ.

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No, we used a constant G to measure its mass so therefore, if G were different, we would have measured it's mass wrong. As long as we use the correct G, we get the correct mass.

What would be the problem with using that mass in all our familiar fomulas?

Where did you get this?
Did you actually read my post, or are you just trying to evade the point?

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papageno


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This comment seems to have been glossed over, probably because it cuts to the heart of the issue and is difficult for "nutant" to respond to.

nutant, what is your proposed definition for the kilogram?

Are your refering again to (Gm) = m, as per above? If the product of G*m remains the same no matter where G is measured, why would it affect gravity at a distance? If in proportion to the Sun's gravity, (Gm) will still respond the same whether G is changed or not (provided m is adjusted), so the same inverse square law is not violated, and all things in orbit would still follow conical sections as before. If in proportion to any other heavenly body, the same applies. The (Gm) product hasn't changed anywhere (only its internal components of G and m are mutally adjusted).

Equivalence is conserved, but that requires somebody answer my question above (re what happens to m in F = ma ?), which nobody has, or just ignored it.

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You may want to re-read this post July 5 th where I discuss this issue. Pay special attention to the BALANCE between kilograms (equal “weights” both sides) in response to G and 10G example:
Viz. “Let's say I have a basket of apples which weigh about 2.2 pounds (US) and I put them on a (European) scale in kilograms. What I find is this same basked of apples now weighs 1 kilogram. (I will address the difference between "weight" and "mass" in a minute.) Now put this 1 Kg. on a balance scale where on the other side is a cubic decimeter of water (or one kilogram platinum-irdidum rod) and what do we have? They balance, at least in Earth's gravity they balance. Now suppose that G' were different, let's say tenfold, for Earth. Would the scale still balance? Yes, the cubic decimeter of water would balance against the 1 Kg. apples (and the same for the scale using pounds), but with one difference: if you "weighed" the cubic decimeter of water in G' = 66.7E-11 Nm^2 kg^2, (meaning the Earth's gravitational acceleration was pulling on this cube with ten times the force), would it still be one kilogram compared to G where it was only a tenth, viz. G = 6.67E-11 Nm^2 kg^-2? If you put the basket of apples in 10 G (were it possible) and the cubic decimeter of water in 1 G, which would "weigh" more? Same mass.”

I'll see if I can better explain this above later, when have some free time. The kilogram is always referenced back to one cubic decimeter of water, so the mass is always the same, only how we measure this mass in terms of the G where it is. More later...

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Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter, can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.

Perhaps working through an example would help us understand what you are trying to say.

Comparing apples to oranges. Earth's surface gravity is 9.8 m/s^2, while Jupiter's "surface" gravity is 23 m/s^2, but the two are not comparable directly, since we don't know Jupiter's real surface (use top of atmosphere, not same on Earth where we use planet's crust surface).

Per the above listed examples, at Jupiter (hypothetical) 5 G, the force would not affect surface (whatever that is) acceleration since the mass of the planet is adjusted for G (in local terms): i.e., F = 5(0.2 m)a

[I think this is what papageno is referring to in his: "In your reasoning you adjusted G and m(gravitational) so that the force would not change."
This is how I see it, that the force does not change, nor how the planet interacts gravitationally per (G*m) per orbital equation, even if G and m are different (adjusted separately) so product remains same.]

Jupiter has not changed, same planet, so its gravity acceleration force has not changed. The mass of Jupiter, hypothetically, is adjusted for local G only, but it is still the same planet we figured in Earth's 1 G. It's just that Jupiter's "kilograms" are different from Earth derived kilograms. This has been my point for all of the above posts, though I am sure there would be some (all?) who would disagree.

[Edited for error in prior, viz. 5F = (0.2m)a, which was wrong. I sometimes think too fast for my fingers.] ops:

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I didn't ask you to compare anything.

Will you work through the calculations with numbers or not? I believe it should make it easier for you to explain your point.

Can you give me an example of what you have in mind? Show any math, for example?

Can you give me an equation representing what you're asking?

Gotta run, talk later. 8)

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You keep saying we need "new" kilograms to support (hypothetical, of course...) variable G. So, I'd like to see the practical application of the "variable kilogram" idea.

I'm also still waiting on your definition of "kilogram".

nutant gene 71 wrote:

I beg to differ. NPL: Frequently asked questions - mass and density

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Sorry I couldn't answer you earlier, but I was catching your posts on the fly in between other projects. Can you be more specific? Please give me an equation to which I can answer, since your question is vague.

The bottom line, if I understand what you're asking, is that 23 meters per second squared is still Jupiter's gravity (at its atmospheric surface), same as now have it. Remember the planet did not change, it's still the same planet, even if its G is different affecting how its mass is measured. Measured in our 1 G postulate for the universe, or measured for its local G, the end result is the same. Please provide me with an equation to address this better. Thanks.

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Correct, cubic decimeter of water is not longer used, which has been replaced by a high controlled environment platinum-irridium rod, where the accumulated molecules are carefully cleaned off periodically. My reference was to the original idea behind what represented the mass of one kilogram. If you look up historical references to the kilogram, you'll see what I mean.

A "kilogram" is a scalar, arbitrarily chosen unit of mass. Here's a reference: http://physics.nist.gov/cuu/Units/kilogram.html

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WHAT HAPPENS TO THE KILOGRAM IN A VARIABLE G?

I'd like to add this, how I see it. I don't know if any of this is true or not, and will not know for certain until such time that we find G varies, so for now this is all hypothetical.

I suspect, per my above reasonings, that the kilogram, which is an arbitrary unit of mass (and with which we also calculate weight), is a defined unit in its own right. Taking the same representative mass and putting it elsewhere where G might be different does not change the mass itself, except in how it interacts with other mass at that locale. This means the same gravitational parameters exist as before, but the size of the material mass may be different. If we use 5G as an example, in that locale, the material size of a representative unit of kilogram would be five times smaller, but still exhibit the same characteristics we know for one kilogram (on Earth). The effect should be that smaller size bonds with other equivalent smaller size, what we can also call "weight" for gravitational, or molecular attraction for chemical, as if the smaller mass weighed more. Not knowing how this will affect chemical bonding, I am forced to leave that to some future study. And ditto for how it affects metal springs, or rocket propulsion, both unknowns. I assume that "heavier" gravitational mass will somehow affect these, but can't guess how. Obviously the Cassini-Huygens springs worked, though operating in extreme cold space (which might make them brittle?) and launched the separated craft away from the mother craft without mishap. But I do suspect that this different "kilogram", which is heavier than our kilogram per volume, does affect gases and matter. There should be more compactness to the planet cores for the gas giants, if such a core exists, and there should be more abundant gas retention per mass. This means that even if Jupiter's core planet is only two Earth masses (in size and volume) it may retain an atmosphere that is substantially greater than if Earth was twice its own size here. I can't calculate this, since I don't know how, so treat this only as a conceptual idea worth examining. When I learn how to calculate it, I will. Separately, though I am not yet ready to release it, I did calculate what the gravitational mass is for the hydrogen atoms in 99.9% of deep space (I used one atom of hydrogen per cubic centimeter of space), away from those tiny galaxy islands of electromagnetic radiant energy, and came up with a startling number, where G is 100,000 times what it is on Earth. When I applied this greater G against a gravitational-redshift value (measured on Earth), I came up with close to Hubble's constant. But that's all I can say for now, not ready to show this yet.

Overall, I am intrigued by the idea that maybe we don't have the G "constant" right, and that perhaps its variance had been (very well) hidden from us all this time. Do I know this for sure? Hardly! I'm fascinated by it. And if it's self delusion, then it's self delusion, and no loss to anybody. But if it's right, how exciting it will be to know that the isotropic universe is still homogenous, but at a substantially higher G, and that the galaxies and stars within them are but low G islands within the cosmic scale of things. If we find that G is variable, our universe is a very different place from what we had thought before, and very exciting too.

Well, gotta walk the dogs on the beach and as usual look up at the sky, except today it's more of the same, cloudy. I hope I added something of value to you all in considering these ideas, as I know for sure that you had added something of value to me by questioning them. Cheers.

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You mean: the weight changes.


No.
The kg-mass is the same on Earth, on Mars, on Jupiter or on your 5G-Earth.
1 kg-mass will still be accelerated by 1 m/s^2 if 1 N of force is applied, whether this force is gravitational or not.

The difference between the Earth and your 5G-Earth is the gravtitational force at the surface, hence the wieght is different.
And since the force on the 1 kg-mass at the surface, changes, the gravitational acceleration (downwards) changes.
But if you go ice-skating on 5G-Earth, you still need the same force as on Earth to accelerate horizontally, even if your weight is different.

This is what happened to the Apollo astronauts on the Moon.


No need to change the size, because the (inertial) mass does not change.
You can take take a sample kilogram from and use it on 5G-Earth.

That would work at most in the vertical direction, but not horizontally.
Since the inertial mass of atoms does not change, I do not expect significant changes in molecular bonding.
If chemical bonds were affected by the weight of atmos and molecules, we would not be able to recognize spectra from the Sun or the gas giants.
All the spectroscopy in astrophysics would be useless, because we would not be able to compare spectra taken in labs on Earth with spectra from stars.
The electronics of the probes we sent throughout the Solar System and of the artificial satellites around Earth would not work as designed, becuase chemical bonds determine the band-structure of the semiconductors used (which includes CCDs), and hence it determines their electric properties.

Since spectroscopy works fine, and the probes we sent out work fine, it is safe to say that weight does not significantly affect chemical bonding.

Why?
Spectra from the Sun look like spectra taken on Earth, hence atoms in gases have still the same electronic structure, which determines chemical bonds.

Do you still think that the mass of gas of Jupiter does not exert gravitational force?
WHy do you treat gases as passive objects in gravitational interaction?