Hypothetical variable mass in hypo variable G?

[Luna2uno]

Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?

Asking it this way may be illustrated as follows:

We are in a region of space where G is much higher than here, say beyond the solar system somewhere. Let's say it's 10G. Then, per equivalence, what would the kilograms measuring mass, or inertia, be in that region? Would kilograms be 10 times greater than here, 10kg? Or perhaps 100 times greater?

Think of this, and why I am bringing up this hypothetical question:

If G is 10 times what we know as the universal Newton's G, and the equivalence requires that inertial mass measured in kg is also 10 times, what happens to the kg in terms of what we know as measurement of mass here? So per equivalence, 10G gives us 10kg, but this may be only a local pehonomenon, meaning that 10kg in our kilograms may be 10 times that, viz. 100kg.

Is there an issue here, or kill the thread now? :roll:

[papageno]

This should be in the "Against The Mainstream" forum.

Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?

The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass.
It has nothing to do with the value of G.

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).

Asking it this way may be illustrated as follows:

We are in a region of space where G is much higher than here, say beyond the solar system somewhere. Let's say it's 10G. Then, per equivalence, what would the kilograms measuring mass, or inertia, be in that region? Would kilograms be 10 times greater than here, 10kg? Or perhaps 100 times greater?

As I said, 1 kg gravitational mass is 1 kg inertial mass, whatever the value of G.
What would change is the gravitational force on the mass.

Think of this, and why I am bringing up this hypothetical question:

If G is 10 times what we know as the universal Newton's G, and the equivalence requires that inertial mass measured in kg is also 10 times, what happens to the kg in terms of what we know as measurement of mass here? So per equivalence, 10G gives us 10kg, but this may be only a local pehonomenon, meaning that 10kg in our kilograms may be 10 times that, viz. 100kg.

Is there an issue here, or kill the thread now? :roll:

Well, I suggest you look up the threads in the "Against The Mainstream" forum where Lunatik and Jerry try to argue in favour of a variable G.

[Luna2uno]

Thanks papageno for your response. My purpose for entering this hypothetical question on the Equivalence Principle here (rather than Against the Mainstream, per Lunatik & Jerry) is not to argue for a variable G, which would be speculative, but to consider how such a (hypothetical future) discovery would affect our measure of mass in kilograms. Which kilograms would we use, and how would they be affected? Your response addresses how kilograms work at 1 G, which is known, but how would this change if we found a variable G, at 10G for example? Or would it not change at all, and still preserve equivalence?

To my thinking (and I must admit I really do not know the answer to this hypothetical question on measuring mass under a variable G scenario), the kilograms we developed in our 1G universe are in part a function of gravity, mainly Earth's gravity, so we can weigh things in kilograms. The Equivalence is that this same kilograms applies to F = Ma, as you pointed out, so we can measure inertial mass with the same unit. I believe it was Einstein who thus resolved that gravity and inertial mass are linked, which we know as the Equivalence Principle. So the question remains, in a hypothetical variable G, would the kilogram units remain the same, or forced to change?

I would think this is a valid astronomy-physics question, in anticipation of some point in the future that our distant space probes, or other observations, yield a variable G. To date, this has not been observed, to my knowledge. Perhaps this question of measurement in kilograms (at this point a merely philosophical question since we have not confirmed any change in Newton's G from its universal constant) should be explored in the event we find the universal G is something else. We must allow for nature to be a tricky place, so she might throw us a surprise. Would we know what to do with our units of measure of mass at that point if she did? :-?

[papageno]

Thanks papageno for your response. My purpose for entering this hypothetical question on the Equivalence Principle here (rather than Against the Mainstream, per Lunatik & Jerry) is not to argue for a variable G, which would be speculative, but to consider how such a (hypothetical future) discovery would affect our measure of mass in kilograms. Which kilograms would we use, and how would they be affected? Your response addresses how kilograms work at 1 G, which is known, but how would this change if we found a variable G, at 10G for example? Or would it not change at all, and still preserve equivalence?

Equivalence principle and the value of G are independent from each other.
Do not confound mass and weight.

To my thinking (and I must admit I really do not know the answer to this hypothetical question on measuring mass under a variable G scenario), the kilograms we developed in our 1G universe are in part a function of gravity, mainly Earth's gravity, so we can weigh things in kilograms. The Equivalence is that this same kilograms applies to F = Ma, as you pointed out, so we can measure inertial mass with the same unit. I believe it was Einstein who thus resolved that gravity and inertial mass are linked, which we know as the Equivalence Principle. So the question remains, in a hypothetical variable G, would the kilogram units remain the same, or forced to change?

It would be the same.
The Equivalence principle is based on experimental results. Einstein decided to elevate to the status of postulate.

I would think this is a valid astronomy-physics question, in anticipation of some point in the future that our distant space probes, or other observations, yield a variable G. To date, this has not been observed, to my knowledge.

I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).

Perhaps this question of measurement in kilograms (at this point a merely philosophical question since we have not confirmed any change in Newton's G from its universal constant) should be explored in the event we find the universal G is something else. We must allow for nature to be a tricky place, so she might throw us a surprise. Would we know what to do with our units of measure of mass at that point if she did? :-?

Yes, because the experiments used for the units would not change their outcome if we found something new.

[Luna2uno]

I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).

I see this really as a question referring to our units of measure, what we call kilograms. Can the same kilograms be used if G is different from what we know it to be as a universal constant?

In yours you said: "The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass. It has nothing to do with the value of G."

Granted, given that G is universally the same, it has nothing to do with it, though G is part of the function describing Newton's formula for gravitation, as per yours above:

F = G * (m*M) / r^2 , which is related to Newton's second law:

F = M * a

Now, this equivalence can be also shown as:

F = M * a = M * (G*m) / r^2, where by default a = (G*m) / r^2

which also means: G = (r^2 * a) / m

Now assume that both a and r^2 are fixed, same values, but G is greater, viz. G1 = 10G. So we have:

G1 = (r^2 * a) / m1, except now of necessity, m1 = 1/10th of m, if G1 = 10G.

However the mass had not changed, same mass (same atomic composition and volume), so the mass did not suddenly shrink to a tenth of its original form. What changed instead was that the measures in kilograms had changed, to where now the kilograms are 10 times greater than the kilograms used earlier, to match up with G ten times Newton's G.

Can you see how this could be a problem? Though for now, given that G is universal, we don't have a problem. But if it were discovered that G is different, something might have to be adjusted in the measure of our (Earth derived) kilograms.

(That said, I still think that the answer above, kg1 = 10kg is wrong, but I'm not sure of what the right answer is. I suspect a is in fact not fixed as assumed, for a variable G. Hypothetically, the real answer may be more like kg1 = 100 kg, if G1 = 10G, or its squared. It may take 10 times as much acceleration to move the same mass in 10G, so a is not fixed, but rather a1 = 10X a. But I don't know this.)

So you can see why I am frustrated, and I don't like my own answers! There must be a better way to see this.

Interesting if this might not apply as well to a " variable dielectric constant in electromagnetism", since it might impact how light bends around stars, which would impact gravitational lensing. :-?

Actually, now that I re-read this, I can almost begin to appreciate the frustration Galileo must have had trying to prove why the Earth is not standing still with the heavens going around, but instead it is spinning.

[papageno]

I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).

I see this really as a question referring to our units of measure, what we call kilograms. Can the same kilograms be used if G is different from what we know it to be as a universal constant?

The definition of the unit kilogram has nothing to do with G.

In yours you said: "The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass. It has nothing to do with the value of G."

Granted, given that G is universally the same, it has nothing to do with it,...

No, it has nothing to do with G, whether it is constant or not.

... though G is part of the function describing Newton's formula for gravitation, as per yours above:

F = G * (m*M) / r^2 , which is related to Newton's second law:

F = M * a

Now, this equivalence can be also shown as:

F = M * a = M * (G*m) / r^2, where by default a = (G*m) / r^2

which also means: G = (r^2 * a) / m

As you can see, G does not affect M, which is the mass the Equivalence has been applied to, but affects the acceleration a the mass M is subjected to.

Now assume that both a and r^2 are fixed, same values, but G is greater, viz. G1 = 10G. So we have:

G1 = (r^2 * a) / m1, except now of necessity, m1 = 1/10th of m, if G1 = 10G.

But we applied the Equivalence principle to M.
By changing the value of G, you changed the force M and m are subjected to.

If the only mean we had to measure the mass m, was from the acceleration of M due to its gravitational interaction with m, then changing G would affect our measured m because the acceleration is different (assuming that we did not know that G has a different value).

However the mass had not changed, same mass (same atomic composition and volume), so the mass did not suddenly shrink to a tenth of its original form. What changed instead was that the measures in kilograms had changed,...

No. What changed is the gravitational force between the two masses, and hence the acceleration changed.

...to where now the kilograms are 10 times greater than the kilograms used earlier, to match up with G ten times Newton's G.

This has nothing to do with the units.

Can you see how this could be a problem? Though for now, given that G is universal, we don't have a problem. But if it were discovered that G is different, something might have to be adjusted in the measure of our (Earth derived) kilograms.

No. The units would not need to be adjusted.

We don't need to change the unit of electric charge because the dielectric constant is not universal in materials.

(That said, I still think that the answer above, kg1 = 10kg is wrong, but I'm not sure of what the right answer is. I suspect a is in fact not fixed as assumed, for a variable G.

This is the point.

Hypothetically, the real answer may be more like kg1 = 100 kg, if G1 = 10G, or its squared. It may take 10 times as much acceleration to move the same mass in 10G, so a is not fixed, but rather a1 = 10X a. But I don't know this.)

So you can see why I am frustrated, and I don't like my own answers! There must be a better way to see this.

You just need to realize that the value of the constant G does not affect the mass of an object.

Interesting if this might not apply as well to a " variable dielectric constant in electromagnetism", since it might impact how light bends around stars, which would impact gravitational lensing. :-?

Refraction is very common phenomenon, which is due to a change in dielectric constant: this is why lenses work.
A G dependent on position would not be more exotic than refraction.

[Luna2uno]

Let's see if this anecdotal illustration better explains how I see it:

I live on planet X (not a real planet we know) where gravity's proportional G is ten times what we know here as G, so Xian's gravity is 10G (in Earth terms). I very carefully measure this Gx, set up my weights of measure in kilograms per this Gx, then work out the Equivalence Principle per F = Ma = G*(Mm)/ r^2 (everyone knows gravity and inertial mass are related), so my mass Mx (and mx) is measured in the kilograms I developed. Now I am content, since I worked it all out, where my units of measure for weight on X are measured in kilograms, for which I then established an equivalence with F = Ma, where Mx is measured in kgx. Confident, I now teach at a prominent Xian university and (since I never traveled off world) merrily accept that my Gx and kilograms kgx are universal.

Four hundred years go by and in a very fancy space ship arrive people (to every Xian's surprise) who say they're from some far off place called Earth. Now these Earthians (all descendents of a prominent university where physics had been taught with confidence for the past 400 years) are very eager to impress their newfound Xians, so they too go and measure G and kilograms. To their surprise, they discover that the Xians are using a different unit of measure for kilograms than the Earthian measure. So they carefully explain to the Xians that G is not what they thought it is (since it is universal), but it is 10 times less, and that only the "acceleration" derived from the greater gravity of their planet is 10 times greater. Kilograms cannot change. They further explain that what they had done wrong was make a tenfold mistake (or hundredfold?) in estimating their kilograms. They made the error of thinking that their Gx (which is 10G in Earthian terms) is the correct G, so the kilograms they developed was based on this error. Since, as your Earthian student descendents take great pains to explain, only "aceleration" is ten times what it should be, so mass measured in kilograms has to be the same, so their Xian kilograms are obviously wrong. The Xians challenge this, saying no, that the Earthian kilograms are wrong, because they are only a tenth of what they should be for G, as everyone who studied at the Xian universsity can tell you, and that their kilograms are correct, since the acceleration works out exactly for their equivalence principle F = Ma. In fact, they (barely) tolerantly explain, the Earthians had got it wrong. In thinking that G is only a tenth of Gx (everyone knows this is a universal constant), it is the Earthians who should adjust their kilograms to reflect the correct G. And that correct measure of mass is tenfold Earthian kilograms, so obviously Xian kilograms are the correct measure for mass.

Well, this heated discussion goes on for some time, and as their appears no solution to this problem, with both Earthians and Xians convinced their measures for mass are correct, it appears the two worlds are in danger of declaring war. Knowing that we really don't know, my (long descendent) student politely (diplomatically) reminds his friends at the Xian academy that it was smart Earthians who came to Xian, and not the otherway around, so perhaps Earthian measured kilograms of mass should be adopted (though they are different from Xian kilograms), just to keep the peace. But he gets shouted down because they say the Earthians are on X now, so must use Xian kilograms instead. :-?

Who is right?

[nutant gene 71]

DISCLAIMER

Let it be know to all who post here and read, that "Lunatik" had been retired, put into permanent "safing" with post # 555, and that my lame attempt to revive him with "Luna2uno" had been in violation of BA rules (prior unbeknownst to me, but Phil made the point), so neither name shall henceforth be shown. All future posts will now default to my other (unwitting alias), from here on in my legitimate handle: "nutant gene 71".

I fully accept any and all criticisms, scorn, ridicule, shunning, or wisecracks, for I am truly repentant.

And I would not be here were it not the high level with which I hold the participants of this board, myself excepted.

Mea culpa. ops:

[Tobin Dax]

Luna2uno, I still don't understand your question. I'm pretty confused about what you're asking right now. I've only skimmed the last half of the thread, but let me attempt a response.

As papageno said, don't confuse mass and weight. Mass is an intrinsic property to matter, where as weight is force caused by gravity and determined by mass, distance, and the value of G. (The equivalence principle does not apply here, as a = G*M/R^2 [your mass, m, remains the same].) The amount of mass here where G=G would be the same amount where G'=2G.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.

[papageno]

So, Luna2uno, you were Lunatik.
And you still don't get the distinction between mass as physical quantity and the unit of measurement to express that quantity in numbers.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.

Exactly my point.

Whether G is a universal constant or not, it has absolutely no bearing on the unit kilogram.

[nutant gene 71]

So, Luna2uno, you were Lunatik.
And you still don't get the distinction between mass as physical quantity and the unit of measurement to express that quantity in numbers.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.

Exactly my point.

Whether G is a universal constant or not, it has absolutely no bearing on the unit kilogram.

Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same. Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m. But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know. 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.

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours. Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs. I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density. Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be. 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).

Can you see where this is taking me? If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere? Unless the G is much greater than supposed, it is virtually impossible as a gas. (This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow. This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body. If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments. In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G? That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)? 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.

And I still do not know what the Xian's kilograms should be, ten fold or a hundred fold.

Sorry about the identity mixups, it may be due to a "multiple personality" syndrome. ops: I never like "Lunatik" anyway.

[Tassel]

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?

[nutant gene 71]

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?

Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" page for now, only very indirectly related. This ESA page on trajectory corrections is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now...

[Tassel]

Alas, I cannot at this time, so take my statement under advisement.

Since you cannot provide any evidence for "numerous inflight adjustments" or "adjustment tables", I will take your statements as if you just made them up for your own benefit.

I can only refer you to this "gravity anomaly" page for now, only very indirectly related.

The subject matter on that page is in no way related to your claims.

This ESA page on trajectory corrections is better, but still not it.

Agreed. This page also has no bearing on your claims.

If I find different, I'll report. I read something somewhere, but can't put my fingers on it now...

Yes, cerainly, if you can find any evidence that your claims have any validity whatsoever, I'd be happy to see it. Until then, I'm not sure why you would make statements about "adjustment tables" and "numerous inflight adjustments".

[papageno]

Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.

So what is the problem?

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.

Not defined, but measured.
The formula you just wrote is a way to measure G.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.

You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?

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.

And how is that different from using another system of units?

Why is this important, since it appears a rather mundane pro
blem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.

If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.

In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.

Europeans do not think that the US are 2.5 times taller than Europe!

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.

The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.

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).

Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.

Can you see where this is taking me?

You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?

The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!

Unless the G is much greater than supposed, it is virtually impossible as a gas.

Why?
On what are your expectations based?

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.

You still assume that a gas does not interact gravitationally.
On what is this "idea" based?

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.

Weird... Huygens landed as planned, didn't it?

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

What makes you think that we did not get that right?

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.

Dynamics does not need "local" kilograms, because the kilogram is the same.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?

You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?

I already gave you references to speculations about variable G.
None seem to have given us a groundbreaking new understanding of the Universe.

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.

The postulate is based on a wealth of experimental evidence, which you simply ignored.

[Maksutov]

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?

Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" page for now, only very indirectly related. This ESA page on trajectory corrections is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now...

There will always be course corrections. Among other real factors, this is a result of metrology. All instruments used to measure something have intrinsic errors. These errors start with accuracy and precision.

In the case of measurements of a body's location in space, typically the farther away from the measuring instrument the body is, the greater the error factor that subtracts from the results reported by the measuring instrument. In the case of dynamic systems, the persons responsible for keeping the object on the intended path will use the data based on nominal values or values corrected for known or estimated errors. There is no absolute way to perform these adjustments precisely when based on data which have error factors.

The only feasible method is to spread the measurements and adjustments out over time such that the results of the corrections can be seen as a cumulative change either toward the planned path or away from it. This is fundamental to navigation.

Just as when the body was near the measuring instruments, when the body is approaching the target, the target affords a calibration standard that allows for finer and finer adjustments, since the adjustments can be readily compared to a known standard, in this case the target.

Of course the "burns" performed to achieve course corrections are variable too, and become another of the many factors which, if considered here, would overextend this post.

That is a simple overview of how course corrections work, and why they are necessary.

Course corrections have nothing to do with such non-factors as variable G, and other imaginary concepts.

[worzel]

I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?

[nutant gene 71]

I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?

Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2, what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before and still maintain the equivalence principle?

That essentially is the real question of this "hypothetical variable mass in a hypothetical variable G". I realize the new artefact for one kilograms is now a platinum-iridium rod kept under highly controlled conditions in France at the International Bureau of Weights and Measures, and that there had been proposals to replace it with something more scientific than a matter prototype, such as a count of atoms in a perfect crystal. At present, nothing else replaces the definition of one Kilogram of mass, a scalar, dimensionless, and represented by this artefact.

[nutant gene 71]

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?

Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" page for now, only very indirectly related. This ESA page on trajectory corrections is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now...

There will always be course corrections. Among other real factors, this is a result of metrology. All instruments used to measure something have intrinsic errors. These errors start with accuracy and precision.

In the case of measurements of a body's location in space, typically the farther away from the measuring instrument the body is, the greater the error factor that subtracts from the results reported by the measuring instrument. In the case of dynamic systems, the persons responsible for keeping the object on the intended path will use the data based on nominal values or values corrected for known or estimated errors. There is no absolute way to perform these adjustments precisely when based on data which have error factors.

The only feasible method is to spread the measurements and adjustments out over time such that the results of the corrections can be seen as a cumulative change either toward the planned path or away from it. This is fundamental to navigation.

Just as when the body was near the measuring instruments, when the body is approaching the target, the target affords a calibration standard that allows for finer and finer adjustments, since the adjustments can be readily compared to a known standard, in this case the target.

Of course the "burns" performed to achieve course corrections are variable too, and become another of the many factors which, if considered here, would overextend this post.

That is a simple overview of how course corrections work, and why they are necessary.

Course corrections have nothing to do with such non-factors as variable G, and other imaginary concepts.

I believe what you are describing is a "chaotic effect" of computations over time and distance. Totally valid, and I take it for granted that it is so. What I had mentioned in my above regarding trajectoral corrections would have more to do with variability of the inertial mass itself.

Again, as I said to worzel above, there is no hard evidence this is so, and any anecdotal evidence of possible variable G, such as gravity anomalies, dark matter, or the Pioneers Anomaly, does not count as hard evidence, only possible reasons to research this further. The hypothetical question raised in this thread is whether or not our measurement of kilograms would be affected if G were something other than what we now know as a universal constant: G = 6.67E-11 Nm^2 kg^-2. Also bear in mind that should inertial mass be different (such as our space probes going into a higher G region) the overwhelming mass of solar system bodies, planets and moons, would still totally dominate per the G*M effect (meaning if G were different, M would be different too), so we would not necessarily see the variable inertial mass effect (of the probes), since this would be like comparing a speck of dust on the back of an elephant. The elephant controls the spec of dust, not the other way around.

The more pressing issue would then follow that if G were different, hypothetically, would it explain gravity anomalies such as neutron stars, giant planet atmospheres, dark matter, and possibly the Pioneers anamolous constant acceleration towards the Sun as well?

[nutant gene 71]

Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.

So what is the problem?

See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.

What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.

Not defined, but measured.
The formula you just wrote is a way to measure G.

Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.

You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?

Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.

Since this is such a challenging question, let's spend some time here, and compare apples to apples. 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.

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. 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.

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!

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.

And how is that different from using another system of units?

Who is right, the Xians or the Earthians?

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.

If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.

See my answer above regarding how 1 kg does not equal 10 kg.

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.

In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.

Your "height" would be equivalent to "r" in G = r^2 * a/ m, different issue.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.

Europeans do not think that the US are 2.5 times taller than Europe!

Rightly so.

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.

The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.

Of course.

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).

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. 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). 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.)

Can you see where this is taking me?

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. 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.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?

The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!

I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition. This new way of seeing mass no longer requires anything so exotic. The Sun's mass is very great, though not totally understood, in my opinion. Neither is Jupiter's composition well understood. Much of what we think we know on these matters are only one step removed from speculation. Steven Hawking at one time said the Sun's interior has a mini-black hole, for example. I realize the mass of the atmosphere plays into the equation of the planet's total mass, but I question how a "bubble" of atmosphere can hold itself together in its orbit around the Sun. I suspect, my spec, that Jupiter has a "small" rocky core.

Seismically, we're not even sure what the Earth's core is made of, though for now we have it as "solidified" iron due to the planet's gravitational pressure, but seismic waves don't go there (which gives the "hollow Earth" people cause to cheer, wrongly in my opinion). Why are seismic waves deflected from it? How much gravity is there at the center of the Earth, if mass is totally ambient and distributed evenly in all directions away from the center? Anyway, it looks like we really don't know what's at the center, so to explain the magnetic field, we gave it an iron core. Okay, I'm cool with that, though in the future we may discover otherwise. Ditto for the Sun and Jupiter, etc. For now, I don't think we really know these things, so only speculative theories.

Unless the G is much greater than supposed, it is virtually impossible as a gas.

Why?
On what are your expectations based?

Small rocky core with greater G would be one possibility, but I don't know either.

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.

You still assume that a gas does not interact gravitationally.
On what is this "idea" based?

A "bubble" of gas would consolidate gravitationally into something at its center, much like our Earth's molten core consolidates into a solid core, if this is true. 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!

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.

Weird... Huygens landed as planned, didn't it?

Yes. Though, if I may add, Jerry has reasons to think it landed with a clang, but that thread's been closed, so no further comment.

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

What makes you think that we did not get that right?

It was just an idea, not too attached to it. We may have it right, with gravitation assist, but we may not know why we have it right. I'm actually rather intrigued by how we maneuver these things to get to where we want them. I think Huygens is a great success, and so is Deep Impact. =D>

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.

Dynamics does not need "local" kilograms, because the kilogram is the same.

Ah, "kilograms are the same", so we're back to square one! "Who was right?" The Xians or the Earthians?

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?

You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"

This revives the "why are comets not showing orbital anomalies" question. I don't know if they are or not, since we don't have two way communications with them. By all "appearances" they seem to be where they're supposed to be, around the Sun in a very large elliptical orbit. Remember, comets are only specs of dust on the Sun's back.

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?

I already gave you references to speculations about variable G.
None seem to have given us a groundbreaking new understanding of the Universe.

What are you refering to? MOND? Dark Matter? Testing for Einstein's Relativity from Jupiter using the Sun's gravitational lensing? Like I said before, there is no "hard" evidence for a variable G, only anecdotal soft evidence, and even that is contested. Ibid. We don't know.

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.

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.

So, who was right? The Xians or the Earthians?

[TheOncomingStorm]

Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often. In metric or SI the unit of weight is the newton in imperial system it is the lb so if you did change the G the weight of the objects may change but the amount of material inside the objects has not changed. and by the way since I am in canada my 2 cents is worth about a penny us currency. sorry for any grammer or spelling mistakes and if I had scientific I could spit the numbers for up above I am at work so I dont have the calculator and hopefully i do not have re edit again.

[Celestial Mechanic]

A 10 kg mass weighs 98 Newtons on Earth. It weighs about 16 Newtons on the Moon, but still has a mass of 10 kg.

If somehow G were to be made 10 times bigger than it is now and if the Earth does not compress and get smaller then that 10 kg mass will still be 10 kg but it will weigh 980 Newtons.

This is elementary physics. Please review a good elementary physics text on the distinction between mass and weight.

[worzel]

I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?

Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2,

Well simply quoting the number over and over now doesn't change my point. You haven't used the value in any of your arguments.

what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before

Depends on what you mean by kilogram. If you mean the amount of mass then yes, if you mean how much does it weigh then no. The former is the correct usage of the term kilogram, the latter is a convenience because we live in an almost constant gravitational field where a 1kg "weight" is the amount of mass that happens to weigh 9.8N.

and still maintain the equivalence principle?

In Newtonian mechanics it was a bit of mystery why the inertial mass equals the gravitional mass - but given that they're equal, I don't see any reason to suppose they wouldn't be if G changed because that would require a big coincidence: that they are equal now!

Einstein's resolution was that spacetime is warped so that free falling masses are following Newton's first law (they keep going in a straight line) as best they can in warped spacetime (they follow geodesics). For the equivalence to break there would have to be a different warping of spacetime for different masses starting out on the same geodesic.

As I see it, a more interesting question would be: if intertia is resistance to the gravitional field of the universe as a whole (Mach, Lense Thirring), then is there even any meaning to the quesion "what if G changed?" ?

[papageno]

Again, as I said to worzel above, there is no hard evidence this is so, and any anecdotal evidence of possible variable G, such as gravity anomalies, dark matter, or the Pioneers Anomaly, does not count as hard evidence, only possible reasons to research this further. The hypothetical question raised in this thread is whether or not our measurement of kilograms would be affected if G were something other than what we now know as a universal constant: G = 6.67E-11 Nm^2 kg^-2.

And this question has been thoroughly answered with no.

Also bear in mind that should inertial mass be different...

It is not.

...(such as our space probes going into a higher G region) the overwhelming mass of solar system bodies, planets and moons, would still totally dominate per the G*M effect (meaning if G were different, M would be different too), so we would not necessarily see the variable inertial mass effect (of the probes), since this would be like comparing a speck of dust on the back of an elephant. The elephant controls the spec of dust, not the other way around.

So, why do you think anomalies like the Pioneer anomlies, have anything to do with G?

[papageno]

Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.

So what is the problem?

See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.

What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?

What part of the Equivalence Principle does not depend on the value of G is not clear?
To the best of our knowledge, the actual value of G, in any units, has no effect whatsoever on the equivalence of inertail and gravitational mass:
m(inertial) = m(gravit.), whatever the units and whatever the value of G.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.

Not defined, but measured.
The formula you just wrote is a way to measure G.

Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.

Wrong.
The definition of G as physical quantity is in Newton's law for gravitation.
The formula above is a way to measure the value of G.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.

You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?

Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.

No.
You still do not understand that the Equivalence Principle is independent of the value of G.
How many times do I have to refer you to the equations?

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).

M(1) = M(2): where does G enter?

Since this is such a challenging question, let's spend some time here, and compare apples to apples. 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.

Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.

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.

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.

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.

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.

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!

I see only a confusion about the unit kilogram.
It is not for weight, but for mass.

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.

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.

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.

If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.

See my answer above regarding how 1 kg does not equal 10 kg.

Your "answer" is absed on a confusion between kg(mass) and kg(weight).

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.

In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.

Your "height" would be equivalent to "r" in G = r^2 * a/ m, different issue.

Your issue is a misunderstanding of the conventional units for wieght and mass.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.

Europeans do not think that the US are 2.5 times taller than Europe!

Rightly so.

So, you are worng if you think that mass changes if G is different.

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.

The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.

Of course.

So, why do you say that the unit kg for mass is different on your planet?

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).

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.

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).

Only if you use kg for weight, which is wrong.
You should use N, the unit for force.

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.)

And you see wrong.
Becuase the inertial mass has not changed.

Can you see where this is taking me?

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.

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.

Wrong.
I already explaiend to you uncountable times that a variable G is not an exotic concept, but unsupported by observations.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?

The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!

I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition.

You were talking about Jupiter.

This new way of seeing mass no longer requires anything so exotic. The Sun's mass is very great, though not totally understood, in my opinion. Neither is Jupiter's composition well understood. Much of what we think we know on these matters are only one step removed from speculation. Steven Hawking at one time said the Sun's interior has a mini-black hole, for example. I realize the mass of the atmosphere plays into the equation of the planet's total mass, but I question how a "bubble" of atmosphere can hold itself together in its orbit around the Sun. I suspect, my spec, that Jupiter has a "small" rocky core.

Seismically, we're not even sure what the Earth's core is made of, though for now we have it as "solidified" iron due to the planet's gravitational pressure, but seismic waves don't go there (which gives the "hollow Earth" people cause to cheer, wrongly in my opinion). Why are seismic waves deflected from it? How much gravity is there at the center of the Earth, if mass is totally ambient and distributed evenly in all directions away from the center? Anyway, it looks like we really don't know what's at the center, so to explain the magnetic field, we gave it an iron core. Okay, I'm cool with that, though in the future we may discover otherwise. Ditto for the Sun and Jupiter, etc. For now, I don't think we really know these things, so only speculative theories.

Your opinion does not seem to be based on a real understanding.

Unless the G is much greater than supposed, it is virtually impossible as a gas.

Why?
On what are your expectations based?

Small rocky core with greater G would be one possibility, but I don't know either.

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.

You still assume that a gas does not interact gravitationally.
On what is this "idea" based?

A "bubble" of gas would consolidate gravitationally into something at its center, much like our Earth's molten core consolidates into a solid core, if this is true.[/quote]
What makes you think that the material composing Jupiter should behave exactly as the material composing Earth?

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?

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.

Weird... Huygens landed as planned, didn't it?

Yes. Though, if I may add, Jerry has reasons to think it landed with a clang, but that thread's been closed, so no further comment.

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.

What makes you think that we did not get that right?

It was just an idea, not too attached to it. We may have it right, with gravitation assist, but we may not know why we have it right. I'm actually rather intrigued by how we maneuver these things to get to where we want them. I think Huygens is a great success, and so is Deep Impact. =D>

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.

Dynamics does not need "local" kilograms, because the kilogram is the same.

Ah, "kilograms are the same", so we're back to square one! "Who was right?" The Xians or the Earthians?

Consider that your questions are all based on your misunderstanding of unit of kilograms for mass and for weight.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?

You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"

This revives the "why are comets not showing orbital anomalies" question. I don't know if they are or not, since we don't have two way communications with them. By all "appearances" they seem to be where they're supposed to be, around the Sun in a very large elliptical orbit. Remember, comets are only specs of dust on the Sun's back.

So, you have no idea why we get it right.
Maybe because we are right?

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?

I already gave you references to speculations about variable G.
None seem to have given us a groundbreaking new understanding of the Universe.

What are you refering to? MOND? Dark Matter? Testing for Einstein's Relativity from Jupiter using the Sun's gravitational lensing? Like I said before, there is no "hard" evidence for a variable G, only anecdotal soft evidence, and even that is contested. Ibid. We don't know.

Actually, you do not know.

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.

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?

[Tassel]

Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often.

Particularly because this thread is in the "General Astronomy" forum, I think that it should be pointed out that you're right on the money, David. All evidence points to a constant G and it seems to me that you correctly understand the difference between weight and mass.

[nutant gene 71]

Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often. In metric or SI the unit of weight is the newton in imperial system it is the lb so if you did change the G the weight of the objects may change but the amount of material inside the objects has not changed. and by the way since I am in canada my 2 cents is worth about a penny us currency. sorry for any grammer or spelling mistakes and if I had scientific I could spit the numbers for up above I am at work so I dont have the calculator and hopefully i do not have re edit again.

(bold mine)
Thanks for your input davidlpf, 2 cents is good, as indeed I should earnestly thank ALL who had contributed to this debate on hypothetical variable mass. Turning over these ideas like this is an important form of conceptual examination. It is not my wish to be proven right on this issue, but rather to find a right conceptual way to cope with a universe if our concept of G should prove wrong.

Yes, we know this is true if "G is a constant in the universe", and if this is so, then there is no debate here. This hypothetical question stems from a possibility that at some future point we may discover than the 1 G we know is not everywhere the same, and that would change things. In fact, I suspect it might even change how we model cosmology in general, which could be a significant change. But not there yet, since at present there is no firm reason to doubt the 1 G universe.

[nutant gene 71]

A 10 kg mass weighs 98 Newtons on Earth. It weighs about 16 Newtons on the Moon, but still has a mass of 10 kg.

If somehow G were to be made 10 times bigger than it is now and if the Earth does not compress and get smaller then that 10 kg mass will still be 10 kg but it will weigh 980 Newtons.

This is elementary physics. Please review a good elementary physics text on the distinction between mass and weight.

Thanks C.M., that is how I see it too, in a 1 G universe scenario. I will show below the fundamental difference between that and a variable G universe scenario, should there be a different G elsewhere. It may have a profound effect on lots of things we now cannot seem to explain. Remember that the "kilogram" is dimensionless, a scalar, unlike G, which is a cubic meter per kilogram per second squared (m^3 kg^-1 s^-2) dimensional. Take out the kilograms, and you still have the cubic meters per second squared, so this is something not contingent on the kilogram itself, but a "proportional" force in its own right which affects the mass. The mass is in kilograms. Now, the current existing physics books do not address the possibility that these kilograms, really an arbitrary unit developed from the (metric system) cubic decimeter of water, which we use to measure mass are capable of being anything else. That's what's being explored here.

[nutant gene 71]

I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?

Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2,

Well simply quoting the number over and over now doesn't change my point. You haven't used the value in any of your arguments.

what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before

Depends on what you mean by kilogram. If you mean the amount of mass then yes, if you mean how much does it weigh then no. The former is the correct usage of the term kilogram, the latter is a convenience because we live in an almost constant gravitational field where a 1kg "weight" is the amount of mass that happens to weigh 9.8N.

Look at what Wiki has to say about this:

"Mass is the amount of matter in a body. The mass of an object is the same everywhere."

That's the long and short of it, as papageno and others have explained here. This is how we now see it, period. The question I am raising is whether or not this is still true if G is different elsewhere.

and still maintain the equivalence principle?

In Newtonian mechanics it was a bit of mystery why the inertial mass equals the gravitional mass - but given that they're equal, I don't see any reason to suppose they wouldn't be if G changed because that would require a big coincidence: that they are equal now!

Exactly, there seems no apparent reason to doubt the Equivalence Principle if G changed, but that would be in a 1 G universe. What if, let's say where exists Dark Matter, G were different? If inertial mass and gravitational mass are the same, what are they in a different G?

Einstein's resolution was that spacetime is warped so that free falling masses are following Newton's first law (they keep going in a straight line) as best they can in warped spacetime (they follow geodesics). For the equivalence to break there would have to be a different warping of spacetime for different masses starting out on the same geodesic.

As I see it, a more interesting question would be: if intertia is resistance to the gravitional field of the universe as a whole (Mach, Lense Thirring), then is there even any meaning to the quesion "what if G changed?" ?

That depends on whether Machian background gravity is 1 G or X G. If the universal space vacuum, at say past the Oort Cloud, is some G much greater than here, then the Machian principle would not apply directly, unless it was figured out in a different G. (What would the Machian gravity background be in a Dark Matter Galaxy, for example?) The same would be true of the geodesic, where its curvature would be different. Again, we can't calculate any of this because we have no reason to assume G is different anywhere else, and thus we operate conceptually in a 1 G universe. But if gravitational mass and inertial mass are the same, the resulting values for a different G should be different.

[nutant gene 71]

Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.

So what is the problem?

See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.

What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?

What part of the Equivalence Principle does not depend on the value of G is not clear?
To the best of our knowledge, the actual value of G, in any units, has no effect whatsoever on the equivalence of inertail and gravitational mass:
m(inertial) = m(gravit.), whatever the units and whatever the value of G.

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). 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. 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. 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.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.

Not defined, but measured.
The formula you just wrote is a way to measure G.

Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.

Wrong.
The definition of G as physical quantity is in Newton's law for gravitation.
The formula above is a way to measure the value of G.

Why would you say G is a "physical quantity"? Is it not merely a "proportional" quantity between gravitationally attracted masses? 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. But if each mass is now tenfold in terms of its (equivalent) inertial mass, then the attraction is ten times ten (two bodies interacting), so the interaction between them is what? 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.

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, 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.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.

You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?

Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.

No.
You still do not understand that the Equivalence Principle is independent of the value of G.
How many times do I have to refer you to the equations?

No. The Equivalence Principle is NOT independent of the value of G. We know the inertial mass and gravitational mass are the same. Gravity acceleration is a function of its "proportional" G. Acceleration of mass is inertially "proportional" to G. Therefore, they are BOTH a function of the G proportional. Change G and you change the "proportional" to how mass interacts. 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!

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).

M(1) = M(2): where does G enter?

a = G * m/ r^2.

Since this is such a challenging question, let's spend some time here, and compare apples to apples. 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.

Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.

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?

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.

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". 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.

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. 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.

Are we conceptually prepared to think this way? In my opinion, we are not. 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.

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.

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.

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.

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.

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!

I see only a confusion about the unit kilogram.
It is not for weight, but for mass.

ibid.

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.

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...

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).

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.

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).

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.

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.)

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.

Can you see where this is taking me?

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.

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.

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.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?

The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!

I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition.

You were talking about Jupiter.

See this Giant Planets article

..snip...

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.

...snip...

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.

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.