Hypothetical variable mass in hypo variable G? - Page 4

nutant gene 71 · #91 (**permalink**) 15-July-2005, 12:48 AM

Quote:

Originally Posted by papageno

Quote:

Originally Posted by nutant gene 71

All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".

So, you are arguing that the gravitational mass changes, but the inertial mass does not.
You are claiming that the Equivalence Principle must be violated in your variable G "speculation".

Here is once again the kernel of the argument, why you and I see it differently.

Inertial-mass is always equivalent to its gravitational-mass. Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference. But Jupiter is not 5 times the mass it was before, same mass. If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram.. Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G. Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down. This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.

Here is something by way of illustration that maybe will give you a better sense of what I am talking about: Enceladus, especially where they say:

"Likewise, with the atmosphere, Enceladus does not possess the gravitational attraction necessary to hold on to a cloud of water ions, so this must be being replenished also."

Only problem, Enceladus being about the size of Arizona, is too small gravitationally in 1 G; but it may not be too small in 5-10 G (I think Saturn has ~10 G). The scientists who are puzzling over this (they know nothing of a greater G) are saying that there must be a source replenishing these water ions, and atmosphere. But there is no evidence of this since we began looking, so they're mystified, naturally.

This is why mass behaves differently in 1 G or 5 G: in 5 G it can do things that in 1 G it cannot, like hold onto more ambient mass, gases, water ions, etc. BTW, the current thought is that Saturn's core planet must be much larger than Jupiter's small core, but that again is a function of not knowing that G on Saturn may be twice that of Jupiter, so to compensate theory they give Saturn a bigger core, which has not been found to date due to Saturn's thicker (bigger G, hypothetically) atmosphere.

I think where you run into trouble here, conceptually (and why this whole idea rubs you the wrong way), is that by referring back to all our spaceprobes that successfully maneuvered their way around Jupiter and Saturn, there was no variable G ever detected. And this is positively true, but we may have the planet's mass-volume figured in 1 G, when it fact it should be substantially smaller in 5 G. We use gravity assist, which is a function of (G*m) to navigate the spacecrafts, and the spacecraft mass is infinitesimal in comparison to the mass of the planet, so everything more or less works fine (after a few inflight adjustments). You see, it is well hidden, even if wrong, because we modeled the universe after our planet's 1 G environment, so even if the G is different elsewhere, we can't tell... well, except for all the puzzling things like an atmosphere on Enceladus or Pluto, where we explain it away saying that they are "gassing out". End result? We get a very confused view of cosmological events from our solar system out to deep space. Is the Pioneers Anomaly giving us a clue? Most people, including yourself, don't think any of these puzzling phenomena are gravitational. I do...* But I do not wish to convince, only to illustrate how I see it, and remember I may be wrong.

*(I think it is at least possible, but we can't know until we find a way to measure for this, so only hypothetical for now.)

nutant gene 71 · #92 (**permalink**) 15-July-2005, 01:20 AM

I'd like to make an editorial comment here.

I know all you guys, and gals, are smart here. Yes, Tassel, I think you're smart too. I wouldn't be going through the trouble to explain this if I didn't think so. I had been on other boards where I did not find this the case (won't mention names) and left after some time. But in the same vein I do not want you to accept my word on this, because until we have evidence to bear this out, we cannot assume any of it is right. My goal is for you to see that there is another way to understand what gravity is doing out there, and to file it away into your little grey cells until such time that something gets revealed that maybe it is so. Until then, what you learn in school, or had always accepted professionally, will have to serve. Three hundred years of science cannot be dismissed as wrong so easily, and it should not be, but if there should be evidence in the future that we did not get gravity right, I ask only this, at least be prepared conceptually to deal with it. But for now, all we know is what we know.

Metricyard · #93 (**permalink**) 15-July-2005, 02:50 AM

Quote:

Originally Posted by nutant gene 71

Inertial-mass is always equivalent to its gravitational-mass. Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference. But Jupiter is not 5 times the mass it was before, same mass. If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram.. Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G. Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down. This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.

Ah, were back to changing variable mass.
When does the variable light speed come back to your theory?

Quote:

I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference.

What are you showing us here??? 5F=5ma -> 5*F=5*m*a #-o
This is basic algebra. Of course they're going to balance. Moving an '*' isn't going to revolutionize physics anytime soon.

Edit to add. Sorry, it must be way past my bed time. Actually, your formula will never balance. Why do you have F multiplied by 5?

If I make m=2 and a=3 (no units needed here) then F should equal 30 using you formula.

But in your formula, F would equal 150, hence, they would never balance.

Sleep on this I must.

papageno · #94 (**permalink**) 15-July-2005, 11:23 AM

Quote:

Originally Posted by nutant gene 71

Quote:

Originally Posted by papageno

Quote:

Originally Posted by nutant gene 71

All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".

So, you are arguing that the gravitational mass changes, but the inertial mass does not.
You are claiming that the Equivalence Principle must be violated in your variable G "speculation".

Here is once again the kernel of the argument, why you and I see it differently.

Inertial-mass is always equivalent to its gravitational-mass.

So, since you claim that the mass changes, it wouuld observable using non-gravitational forces.
These experiments show that the inertial mass does not change.

Quote:

Originally Posted by nutant gene 71

Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference.

And the "5a" interpretation is consistent with the observations.
And we know that, because we have observe known masses in situations where the gravitational force is different.
And because the spectra from other planets and stars are recognizable from spectra taken on Earth, and because the electonics of our porbes worked fine (if it did not, we would not have any image taken with CCD-based cameras).

Quote:

Originally Posted by nutant gene 71

But Jupiter is not 5 times the mass it was before, same mass.

But the gravitationally force Jupiter exerts is 5 times larger.
This would be observable, unless you invoke a violation of the Equivalence Principle.

Quote:

Originally Posted by nutant gene 71

If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram..

The weight of 1 kg-mass would be 5 times larger on Jupiter.
But Jupiter's kg-mass would be exactly the same as Earth's.

Quote:

Originally Posted by nutant gene 71

Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G.

Quote:

Originally Posted by nutant gene 71

Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down.

Of course, because 1 kg > 0.2 kg.
But 0.2 kg is not Juptier's kg, it is 0.2 Jupiter's kg.

You are still confused about weight and mass.
Why don't you express the wieght in Newtons?

Quote:

Originally Posted by nutant gene 71

This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.

You still forget that Jupiter's atmosphere itself has a significant mass.

Quote:

Originally Posted by nutant gene 71

Here is something by way of illustration that maybe will give you a better sense of what I am talking about: Enceladus, especially where they say:

"Likewise, with the atmosphere, Enceladus does not possess the gravitational attraction necessary to hold on to a cloud of water ions, so this must be being replenished also."

Only problem, Enceladus being about the size of Arizona, is too small gravitationally in 1 G; but it may not be too small in 5-10 G (I think Saturn has ~10 G). The scientists who are puzzling over this (they know nothing of a greater G) are saying that there must be a source replenishing these water ions, and atmosphere. But there is no evidence of this since we began looking, so they're mystified, naturally.

This is why mass behaves differently in 1 G or 5 G: in 5 G it can do things that in 1 G it cannot, like hold onto more ambient mass, gases, water ions, etc. BTW, the current thought is that Saturn's core planet must be much larger than Jupiter's small core, but that again is a function of not knowing that G on Saturn may be twice that of Jupiter, so to compensate theory they give Saturn a bigger core, which has not been found to date due to Saturn's thicker (bigger G, hypothetically) atmosphere.

I think where you run into trouble here, conceptually (and why this whole idea rubs you the wrong way), is that by referring back to all our spaceprobes that successfully maneuvered their way around Jupiter and Saturn, there was no variable G ever detected. And this is positively true, but we may have the planet's mass-volume figured in 1 G, when it fact it should be substantially smaller in 5 G. We use gravity assist, which is a function of (G*m) to navigate the spacecrafts, and the spacecraft mass is infinitesimal in comparison to the mass of the planet, so everything more or less works fine (after a few inflight adjustments).

5 times the expected force is not small.

Quote:

Originally Posted by nutant gene 71

You see, it is well hidden,...

To hide it, you need a violation of the Equivalence Principle.

Quote:

Originally Posted by nutant gene 71

...even if wrong, because we modeled the universe after our planet's 1 G environment, so even if the G is different elsewhere, we can't tell... well, except for all the puzzling things like an atmosphere on Enceladus or Pluto, where we explain it away saying that they are "gassing out". End result? We get a very confused view of cosmological events from our solar system out to deep space. Is the Pioneers Anomaly giving us a clue? Most people, including yourself, don't think any of these puzzling phenomena are gravitational. I do...* But I do not wish to convince, only to illustrate how I see it, and remember I may be wrong.

You just claimed that the variable G is well hidden.
Why would it show up only in the Pionerr probes?

Quote:

Originally Posted by nutant gene 71

*(I think it is at least possible, but we can't know until we find a way to measure for this, so only hypothetical for now.)

nutant gene 71 · #95 (**permalink**) 15-July-2005, 06:36 PM

Quote:

Originally Posted by papageno

Quote:

Originally Posted by nutant gene

Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G.

Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down.

Of course, because 1 kg > 0.2 kg.
But 0.2 kg is not Juptier's kg, it is 0.2 Jupiter's kg.

You are still confused about weight and mass.
Why don't you express the wieght in Newtons?

I’ve avoided the “weight” discussion because it may confuse. Your weight on Jupiter will be the same as we now estimated, since your material mass had not changed when you got there.

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earth’s N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesn’t work, which is why I avoided this analogy.

How would you calculate it to show your analogy better?

Tassel · #96 (**permalink**) 15-July-2005, 06:41 PM

The definition of Kilogram:

Quote:

The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. A gram is defined as one thousandth of a kilogram. Conversion of units describes equivalent units of mass in other systems.

Since 1889, the SI system defines the unit to be equal to the mass of the international prototype of the kilogram, which is made from an alloy of platinum and iridium of 39 mm height and diameter, and kept at the Bureau International des Poids et Mesures (International Bureau of Weights and Measures).

The definition of Mass:

Quote:

Mass is a property of physical objects that, roughly speaking, measures the amount of matter they contain. It is a central concept of classical mechanics and related subjects.

Based on the definitions above, kilograms are a numeric representation of the amount of matter that an object is composed of. 1 kilogram is defined as the amount of matter that composes a specific platinum and iridium rod.

We could take that rod anywhere in the universe, even in a universe where G varies greatly (which we know isn't the case in our universe). Since the amount of matter in the rod does not change, then by definition the mass of the rod is still one kilogram.

The amount of matter that makes up an object is a physical property that is measured in kilograms. What 1 kilogram means is not up for debate. It is a defined value. That definition does not include the value of G. Don't bother: it doesn't include the value of G not because "we've always assumed constant G", but because G has no bearing on how much matter an object is composed of.

Kilograms are a measure of mass. Mass is a physical property of an object that represents the amount of matter that makes up the object. The amount of matter in an object does not depend on the value G. Therefore, kilograms, which are the unit we use to represent quantities of matter, would not be different if G were different.

If you want to debate what happens to the unit "kilogram" if G were somehow different, you are not debating anyone in this forum. You are debating the definition of "kilogram".

Quote:

Originally Posted by nutant gene 71

Yes, Tassel, I think you're smart too.

You launched an ad hominem attack on my intelligence while simultaneously demonstrating that you do not have any kind of intuitive understanding about how orbits are determined. With that, and your history of intellectual dishonesty in mind, I couldn't possibly care any less if you think I'm smart or not.

nutant gene 71 · #97 (**permalink**) 15-July-2005, 06:59 PM

Quote:

Originally Posted by Tassel

...snip...

If you want to debate what happens to the unit "kilogram" if G were somehow different, you are not debating anyone in this forum. You are debating the definition of "kilogram".

Quote:

Originally Posted by nutant gene 71

Yes, Tassel, I think you're smart too.

You launched an ad hominem attack on my intelligence while simultaneously demonstrating that you do not have any kind of intuitive understanding about how orbits are determined. With that, and your history of intellectual dishonesty in mind, I couldn't possibly care any less if you think I'm smart or not.

You're absolutely right, it is not a debate against anyone's person but a debate against understanding, or lack of it. Since this is becoming nasty, I'm going to move on from this "1 kilogram" discussion and turn my attention to an earlier question raised about the mass of atoms in space at 1 z distance. Later.

Tassel · #98 (**permalink**) 15-July-2005, 07:40 PM

Quote:

Originally Posted by nutant gene 71

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earth’s N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesn’t work, which is why I avoided this analogy.

You are so confused it is becoming painful to read this thread.

W = mg

The mass of an object is 100kg.
The weight on Earth is: 100kg * 9.8m/s^2 = 980N
The weight on Jupiter is: 100kg * 23m/s^2 = 2300N

G is required to calculate the mass of Jupiter. If we discovered G were different at Jupiter, then we'd realize we had always had Jupiter's mass wrong. We'd recalculate Jupiter's mass and find it was much lower than we thought. You've pointed this out seemingly hundreds of times: GM remains constant. Therefore, acceleration due to gravity remains constant. The mass of the object in question remains constant. If m (mass of the object) and g (acceleration due to gravity) are the same before and after the "discovery" that we were wrong about G, guess what? W is the same before and after the "discovery". If you'd just take 30 seconds and walk through the calculations step-by-step instead of running around randomly multiplying everything by 5, you'd see this.

nutant gene 71 · #99 (**permalink**) 15-July-2005, 09:45 PM

Quote:

Originally Posted by Tassel

Quote:

Originally Posted by nutant gene 71

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earth’s N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesn’t work, which is why I avoided this analogy.

You are so confused it is becoming painful to read this thread.

W = mg

The mass of an object is 100kg.
The weight on Earth is: 100kg * 9.8m/s^2 = 980N
The weight on Jupiter is: 100kg * 23m/s^2 = 2300N

G is required to calculate the mass of Jupiter. If we discovered G were different at Jupiter, then we'd realize we had always had Jupiter's mass wrong. We'd recalculate Jupiter's mass and find it was much lower than we thought. You've pointed this out seemingly hundreds of times: GM remains constant. Therefore, acceleration due to gravity remains constant. The mass of the object in question remains constant. If m (mass of the object) and g (acceleration due to gravity) are the same before and after the "discovery" that we were wrong about G, guess what? W is the same before and after the "discovery". If you'd just take 30 seconds and walk through the calculations step-by-step instead of running around randomly multiplying everything by 5, you'd see this.

-- bold mine

Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G? Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G? Don't you see that in Jupiter's greater graivtational G "proportional" each kilogram is now five times greater? Of course the end result is the same mass with 100 kg, but can you see the qualitative difference here, or not?

Don't get frustrated about it, but think of what is happpening if a body's gravitational G is different: to arrive at the same kilograms of mass we use on Earth, the "kilograms" of a different G world have to be adjusted for the higher G. There is no getting around this, and this is why the mass acts differently there, because (per above example) each kg.-Earth is now equivalent to 0.2 kg-Jupiter. This is not so confusing, and I'm sure if you can understand Relativity, you can understand a variable G, in fact it's easier!

Metricyard · #**100** (**permalink**) 15-July-2005, 10:20 PM

Quote:

Originally Posted by nutant gene 71

Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G? Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G? Don't you see that in Jupiter's greater graivtational G "proportional" each kilogram is now five times greater? Of course the end result is the same mass with 100 kg, but can you see the qualitative difference here, or not?

What happens to the other 80kgs? And how does having a larger G replace the missing mass? This is why I think you didn't want to do my little thought experiment. To prove you theory without using any math, you're going to have to remove mass. All you have to do is explain why a larger G would have anything to do with replacing mass.

Quote:

Don't get frustrated about it, but think of what is happpening if a body's gravitational G is different: to arrive at the same kilograms of mass we use on Earth, the "kilograms" of a different G world have to be adjusted for the higher G. There is no getting around this, and this is why the mass acts differently there, because (per above example) each kg.-Earth is now equivalent to 0.2 kg-Jupiter. This is not so confusing, and I'm sure if you can understand Relativity, you can understand a variable G, in fact it's easier!

Why is this easier? You're creating a formula that is way overcomplicated. Why spend so much time to try and prove a variable G and m that's trying to give the same results as a standard formula would?

Tassel · #**101** (**permalink**) 15-July-2005, 10:39 PM

Quote:

Originally Posted by nutant gene 71

Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G?

Yes. Kilograms are a unit of measure indicating the amount of matter that an object is composed of. The amount of matter making up the object has not changed.

Quote:

Originally Posted by nutant gene 71

Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G?

No.

W = 20kg * 23m/s^2 = 460N which is NOT what we would observe, if we were wrong about G at Jupiter.

Do the math, and post the calculations step-by-step (you won't do this). Start with Jupiter's mass based on the currently accepted value for G and calculate "g" for Jupiter. Then, replace the currently accepted value of G with any value you choose. Recalculate Jupiter's mass utilizing this new value of G and then recalculate "g". Since GM remains the same, you will find that "g" is the same.

"g" and the mass of the object you are weighing are all you need to determine weight, since "g" is calculated using G.

You do not need to change the definition of kilograms to support variable G. You will never realize this until you actually sit down, take a deep breath, and work through the calculations step by step. You are trying to take shortcuts, and you are making mistakes. For example:

Quote:

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earth’s N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N.

You are saying that W(Jupiter) is W(Earth) * 5 * g(Jupiter). This is just horrendously wrong. In the real, constant G universe, you don't multiply W(earth) by g(Jupiter) to get W(Jupiter). You multiply the mass of the object by the "g" of the planet you are weighing it on. You further mangle the issue by multiplying by the increase in G, which also makes no sense, since g(Jupiter) already includes G (old or "new").

Again, I suggest you work out the calculations with real numbers, starting with the currently accepted values for G and Jupiter's mass. If you just do this, you will see why variable G works just fine (mathematically...) in all regards with the current definition of "kilogram".

Edit to add: if you do the calculations the value for "g" at Jupiter should be closer to 25m/s^2, which is Jupiter's "surface gravity". On the fact sheet, "surface acceleration" (which for Jupiter is listed at around 23m/s^2) includes the effects of the rotation of the planet.

nutant gene 71 · #**102** (**permalink**) 16-July-2005, 12:33 AM

Quote:

Originally Posted by Tassel

No.

W = 20kg * 23m/s^2 = 460N which is NOT what we would observe, if we were wrong about G at Jupiter.

What happened to the 5 G? You're missing it.

(1G*m) = (5G*0.2m)

Your correct example should have read:

W = 5(20kg) * 23m/s^2 = 2300 N.

Quote:

Originally Posted by Tassel

Quote:

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earth’s N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N.

You are saying that W(Jupiter) is W(Earth) * 5 * g(Jupiter). This is just horrendously wrong. In the real, constant G universe, you don't multiply W(earth) by g(Jupiter) to get W(Jupiter). You multiply the mass of the object by the "g" of the planet you are weighing it on. You further mangle the issue by multiplying by the increase in G, which also makes no sense, since g(Jupiter) already includes G (old or "new").

You are correct, my mistake. I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

The way you see it, the 5 G doesn't exist anywhere, and if I did it your way, if I understand what you're saying, factoring in the 5 G, it would look like this:

W'" = 5(100 kg * 23.m^2) = 11,500 N , which is wrong.

Metricyard · #**103** (**permalink**) 16-July-2005, 12:55 AM

Quote:

Originally Posted by nutant gene 71

You are correct, I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 kg * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

The way you see it, the 5 G doesn't exist anywhere, and if I did it your way, if I understand what you're saying, factoring in the 5 G, it would look like this:

W'" = 5(100 kg * 23.m^2) = 11,500 N , which is wrong.

But where is the mass going? In the above example, you've lost 80kgs of mass. Why and where??