Why don't heavy objects fall more SLOWLY?

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Gravity accelerates all objects at the same rate regardless of their mass due to the balance between mass and gravitational force. While a heavier object resists acceleration more than a lighter one, it is also acted upon by a greater gravitational force, resulting in equal acceleration for both. The principle of equivalence states that the initial accelerations of different masses in a gravitational field will be the same, independent of their properties. This is demonstrated in experiments where objects of varying masses fall at the same rate in a vacuum, highlighting that air resistance is the primary factor affecting falling speeds in non-vacuum conditions. Thus, the acceleration due to gravity is consistent across all masses when other forces, like air resistance, are negligible.
  • #31
TurtleMeister said:
You're not alone. I've been trying to make sense of this for a long time. The more I learn, the more confused I am. If I'm understanding you correctly, then I agree with you that the distinction between passive gravitational mass (mp) and inertial mass (mi) seems pointless.
The distinction becomes clearer if you were to represent one of the object's gravitational field in the form of a potential.

The best accuracy to support this equivalency is the Kreuzer experiment of 1966 (1 part in 5x105).
Direct test, yes. However, lunar laser ranging provides an indirect but much more precise test of the equivalence between passive and active gravitational mass. The centers of mass of the Moon's crust and core are not collocated. The Moon would self-accelerate if passive and active gravitational mass were not equivalent (at least for the distinctly different material that form the Moon's crust and core). That self-acceleration was not observed, at least to within the incredible accuracy of the range measurements.
Consider the following thought experiment: ... We will always meet at our COM.
This is not true if passive and active mass are not equivalent. The non-equivalence would make gravity will disobey Newton's third law, or alternatively, the conservation laws.
 
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  • #32
D H said:
Turtle said:
You're not alone. I've been trying to make sense of this for a long time. The more I learn, the more confused I am. If I'm understanding you correctly, then I agree with you that the distinction between passive gravitational mass (mp) and inertial mass (mi) seems pointless.
The distinction becomes clearer if you were to represent one of the object's gravitational field in the form of a potential.
A gravitational field is the product of active gravitational mass (ma). We were talking about mp and mi. Are you saying that mp has gravitational potential?
D H said:
Turtle said:
The best accuracy to support this equivalency is the Kreuzer experiment of 1966 (1 part in 5x105).
Direct test, yes. However, lunar laser ranging provides an indirect but much more precise test of the equivalence between passive and active gravitational mass. The centers of mass of the Moon's crust and core are not collocated. The Moon would self-accelerate if passive and active gravitational mass were not equivalent (at least for the distinctly different material that form the Moon's crust and core). That self-acceleration was not observed, at least to within the incredible accuracy of the range measurements.
Yes, I am aware of this test. It is based on Newtons laws, which assume WEP to be true. Why should we test WEP based on an equation that already assumes it to be true?
D H said:
Turtle said:
Consider the following thought experiment: ... We will always meet at our COM.
This is not true if passive and active mass are not equivalent. The non-equivalence would make gravity will disobey Newton's third law, or alternatively, the conservation laws.
Again, this can also be used to show that Newton's laws only work when WEP is true. One can easily look at the equations and see that they were not designed to work with mass that has an unequal ratio of ma to mp / mi.

Newton did not write the laws and then have the universe obey. He built his laws based on observation and experimentation. And modern experimentation has already proven them wrong (relativity).
 
  • #33
TurtleMeister said:
A gravitational field is the product of active gravitational mass (ma). We were talking about mp and mi. Are you saying that mp has gravitational potential?
Oops. My response was off-topic to that particular issue. Describing an object's gravitational potential applies to active gravitational mass.

Yes, I am aware of this test. It is based on Newtons laws, which assume WEP to be true.
That test did not assume Newton's laws. It was (is) a test of the strong equivalence principle: general relativity. The measurements are measurements of the distance between the Earth and Moon; about the only thing assumed in those measurements is the constancy of the speed of light and the correctness of the clock. There is nothing in those measurements that assumes any form for gravity. Gravitation, and particularly the equivalence principle, predicts values for those measurements. In particular, a particular wobble would have been observed if the equivalence principle was not true. No such wobble was observed. What this means is that the the equivalence principle is consistent with the observed data. That is the best science can do. While scientific theories can be disproved (all it takes is one stupid experiment), they cannot proven true.

If you're interested, you want to read up on falsification and statistical hypothesis testing.
 
  • #34
D H said:
That test did not assume Newton's laws. It was (is) a test of the strong equivalence principle: general relativity. The measurements are measurements of the distance between the Earth and Moon; about the only thing assumed in those measurements is the constancy of the speed of light and the correctness of the clock. There is nothing in those measurements that assumes any form for gravity. Gravitation, and particularly the equivalence principle, predicts values for those measurements. In particular, a particular wobble would have been observed if the equivalence principle was not true. No such wobble was observed. What this means is that the the equivalence principle is consistent with the observed data.

Is this the test you are referring to: http://prola.aps.org/abstract/PRL/v57/i1/p21_1?
A limit is established for the violation of the equality of passive and active gravitational mass. Our test is based on an asymmetry in the composition of the moon. We hypothesize that the 2-km offset between the moon’s center of figure and center of mass indicates an asymmetry in the distribution of Fe and Al. Unless the Fe on one side and the Al on the other attract each other with the same force, the moon will not be in the orbit predicted by classical mechanics. Using the results from laser ranging and a model for the moon’s interior we find that the ratios of active to passive mass for Fe and Al are the same to a precision of 4×10-12.
If so, then I think you are wrong. The highlighted portions pertain to Newton's laws. In effect, what they are saying is that if Fe and Al do not attract with the same force (ma) then there will be a self acceleration of the moon, causing it's orbit to be different from the one predicted by classical mechanics.

This what I am talking about in my post #30. Please look at the equations and the three sentences that precede and follow it.
 
  • #35
Ah, yes, I see now the issue with active gravitational mass. Gravity is caused by the presence of matter, but this matter may have different compositions (different internal arrangements) and the question is whether different arrangements of the same quantity of matter (or at least the same inertial mass) might have different gravitational active capacities...

I have not thought enough about the experiment with the Moon that DH mentioned and about which you argue that it does not exclude the possibility of the above discrepancy.

Just a semantic comment for the time being: strictly speaking, what the WEP and the experiments that back it up require is that a given source accelerates all bodies in free fall at the same rate, that is to say, that the passive mass plays no role. This is not threatened by the above mentioned issue: if the passive mass is irrelevant, it’s so in all respects, quantity and composition.

As to the need that the two bodies interacting gravitationally meet at the COM of the system… Well, if active gravitational mass were different from inertial mass, then the bodies would meet at their “gravitational” COM, eventually different from their “inertial” COM.

The big problem, as you point out, would be for Newton’s Third Law. Choosing units so that G/r^2 = 1, what the Law requires is that:

m0 (inertial) • m1 (gravitational) = - m1 (inertial) • m0 (gravitational)

So if the ratio inertial (m0/m1) <> gravitational (m0/m1), the equation breaks down…

But, that is my point, this and only this would be what is at stake. “Equivalence” can refer to two different things: (i) when faced with an attractive given mass, all objects are attracted with “equal” acceleration and (ii) the ratio of gravitational accelerations is “equal” to the ratio of inertial accelerations in the context of a collision. Only (ii) would be discussed here.
 
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  • #36
Saw said:
Ah, yes, I see now the issue with active gravitational mass. Gravity is caused by the presence of matter, but this matter may have different compositions (different internal arrangements) and the question is whether different arrangements of the same quantity of matter (or at least the same inertial mass) might have different gravitational active capacities...
Thanks for taking the time to interpret my ramblings Saw. I am a little lacking in math and communication skills, so you've made my day in being able to see what I have been puzzled over for so long. One of the questions involved in all of this is whether or not leptons have active gravitational mass. If you find this discussion interesting then you may also find this http://www.gravityresearchfoundation.org/pdf/awarded/2001/unnikrishnan_gillies.pdf" interesting.

Saw said:
I have not thought enough about the experiment with the Moon that DH mentioned and about which you argue that it does not exclude the possibility of the above discrepancy.
I believe that this test is invalid, because it is based on Newton's laws, which as you have already discovered, breaks down when there is an inequality between ma and mi.

Saw said:
Just a semantic comment for the time being: strictly speaking, what the WEP and the experiments that back it up require is that a given source accelerates all bodies in free fall at the same rate, that is to say, that the passive mass plays no role. This is not threatened by the above mentioned issue: if the passive mass is irrelevant, it’s so in all respects, quantity and composition.
Yes, this has also been an area of confusion for me. I have stated in my previous posts that a test of the equivalence of ma is also a test of WEP. But I've never been sure of that. Since, as I've already stated, I have not studied GR, I'm not sure as to how this issue affects or threatens it.

Saw said:
As to the need that the two bodies interacting gravitationally meet at the COM of the system… Well, if active gravitational mass were different from inertial mass, then the bodies would meet at their “gravitational” COM, eventually different from their “inertial” COM.
I tend to disagree here. What is gravitational COM? The barycenter of two bodies is determined by their inertial mass. If that were not the case, then it would lead to the same result as the moon experiment, where the third law of motion is violated and the moon self accelerates. And that would open up a Pandora's box of nonsense, such as perpetual motion and unlimited free energy.

Saw said:
The big problem, as you point out, would be for Newton’s Third Law. Choosing units so that G/r^2 = 1, what the Law requires is that:

m0 (inertial) • m1 (gravitational) = - m1 (inertial) • m0 (gravitational)

So if the ratio inertial (m0/m1) <> gravitational (m0/m1), the equation breaks down…

But, that is my point, this and only this would be what is at stake. “Equivalence” can refer to two different things: (i) when faced with an attractive given mass, all objects are attracted with “equal” acceleration and (ii) the ratio of gravitational accelerations is “equal” to the ratio of inertial accelerations in the context of a collision. Only (ii) would be discussed here.
I agree that (i) is not the issue here. I think the issue is whether or not Newton's laws can be used to describe a non equivalence between ma and mp / mi. Either the non equivalence cannot exist, or we must have another equation, or theory, to describe it.
 
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  • #37
TurtleMeister said:
Is this the test you are referring to: http://prola.aps.org/abstract/PRL/v57/i1/p21_1?

If so, then I think you are wrong. The highlighted portions pertain to Newton's laws. In effect, what they are saying is that if Fe and Al do not attract with the same force (ma) then there will be a self acceleration of the moon, causing it's orbit to be different from the one predicted by classical mechanics.
That is not what they are saying.

Suppose, for the sake of argument, that ma=mp=mi for iron, while for aluminum, mp=mi but ma=0. The iron core will exert a gravitational force on the aluminum crust, but the crust will exert zero force on the core. (Note well: Newton's third law fails here.) Because the core and crust are structurally attached to one another, the Moon will self-accelerate along the directed line from the crust's center of mass to that of the core. This self-acceleration would make the orbit of this equivalence principle-violating Moon about the Earth differ from that of a Moon that obeys the equivalence principle. This in turn would manifest as an discrepancy between the measured and predicted values of the distance between the Earth and the Moon.

That is just an extreme example. A much smaller deviation between active and passive gravitational mass would still be observable. An extremely small (but still non-zero) deviation would not be observable because there are some uncertainties in both the measurements and predictions. What this experiment did was to place an upper limit (a very, very small upper limit) on what the deviation could be.

A similar situation occurs in experiments to verify that photons have zero rest mass. Physicists cannot absolutely prove this experimentally. What they can do experimentally is to (a) confirm that the theoretical zero rest mass is consistent with experimental results, and (b) establish an upper limit on the photon's rest mass.

Going back to the case at hand, physicists similarly found the strong equivalence principle to be consistent with experimental results and they were able to place an upper limit on |1-m_p/m_a|.
 
  • #38
But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?

Is it so? Could you be more specific? If you did, I don’t think you’d be breaking forum rules. We are just trying to better understand Newton’s Laws by considering an unlikely hypothesis that would make them fail and discussing whether any alternative Law would be theoretically compatible with logic and experiment. That’s a very orthodox intellectual tool, which makes study amusing…
 
  • #39
Saw said:
But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?
Perfectly stated.

Saw said:
Is it so? Could you be more specific? If you did, I don’t think you’d be breaking forum rules. We are just trying to better understand Newton’s Laws by considering an unlikely hypothesis that would make them fail and discussing whether any alternative Law would be theoretically compatible with logic and experiment. That’s a very orthodox intellectual tool, which makes study amusing…

F = \frac{G(m^\text{g}_1+m^\text{g}_2)}{r^2(\frac{1}{m^\text{i}_1}+\frac{1}{m^\text{i}_2})}

A_1 = \frac{F}{m^\text{i}_1}

A_2 = -\frac{F}{m^\text{i}_2}
 
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  • #40
Today I watched the first two lectures in this series on General Relativity:

http://www.youtube.com/view_play_list?p=6C8BDEEBA6BDC78D

which I found referenced on another thread. Near the end of the second lecture there is a discussion of the equivalence principle which I understand initially as follows. There may be more implications in further lectures which I'll view soon.

Using the flat Earth assumption gravity field g is constant. A person accelerating at rate g in a remote region of space would observe events within the accelerated frame of reference as equivalent to a field of gravity. Notice in the flat Earth assumption there is no divergence of the gravitational field at a distance from the earth.

The equivalence principle holds if there is no way to know whether one is accelerating through space or experiencing a gravitational field. But gravitational fields around a sphere have a divergence of the field lines at a large radius, while an accelerating frame of reference has no such divergence in the observer's experience. What can be detected in a divergent gravity field is "tidal forces." So, for example, the way the moon imposes tidal forces on the oceans would not be reproduced by accelerating the Earth at a rate in one direction equal to the moon's gravitational acceleration.

The tidal forces are a barrier to the application of the equivalence principle is how the lecture concludes the analysis. I haven't read the Wiki (it is not necessarily written by the most informed authorities but often includes accurate information) so I can't compare to what is discussed there. Hope this at least brings something informative to the thread.
 
  • #41
Saw said:
But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?
It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass. Violations of Newton's third law also arise if gravitational and inertial mass are not equal (or better stated, proportional to one another). These violations would also show up in the lunar laser ranging experiments. What is measured here is the time it takes for a laser pulse to travel to the Moon, hit the targeted retroreflector, and return to the Earth. Multiplying those timing measurements by the speed of light yields the distance between the Earth and the Moon. While those inferred distances do depend on the assumption of the constancy of the speed of light, there is no dependency on the equivalence principle. Because those measurements are so precise and because they are independent of the equivalence principle, they serve as an excellent foil for testing the validity of the equivalence principle.
 
  • #42
TurtleMeister said:
F = \frac{G(m^\text{g}_1+m^\text{g}_2)}{r^2(\frac{1}{m^\text{i}_1}+\frac{1}{m^\text{i}_2})}

A_1 = \frac{F}{m^\text{i}_1}

A_2 = -\frac{F}{m^\text{i}_2}

Ah, you don't play with Newton's 3 Laws of Dynamics, only with the Law of Gravitation. It's hard for me to imagine what this formula might lead to, but I have one early doubt: which reference frame are you considering? Looking at the A formulas, it seems it should be an inertial frame, like the COM frame, just like in Newton’s formula, is it?

On the other hand, regarding:

D H said:
Suppose, for the sake of argument, that ma=mp=mi for iron, while for aluminum, mp=mi but ma=0. The iron core will exert a gravitational force on the aluminum crust, but the crust will exert zero force on the core. (Note well: Newton's third law fails here.) Because the core and crust are structurally attached to one another, the Moon will self-accelerate along the directed line from the crust's center of mass to that of the core. This self-acceleration would make the orbit of this equivalence principle-violating Moon about the Earth differ from that of a Moon that obeys the equivalence principle. This in turn would manifest as an discrepancy between the measured and predicted values of the distance between the Earth and the Moon.

I have a stupid question: I suppose it’s not possible that a hypothetical difference between the ma of core and crust originated rotation (or at least served to maintain historical rotation alive)…
 
  • #43
D H said:
It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass.

I understand that what would to lead to a self acceleration of the Moon would be a hypothetical difference between the active gravitational masses of different parts of the Moon, assuming the “X” masses of those two portions were equal. What is “X”? You call “X” passive gravitational mass… Well, I called it simply “inertial” mass because of what I said in post 29, which I reproduce here for your convenience:

Saw said:
As to “passive gravitational mass” (PGM)… what is that? The equivalence principle (WEP, let’s leave aside the subtleties of GR) means that the acceleration caused by the Earth (towards an inertial reference) on anybody is equivalent, regardless its mass. That entails that whether there or many or few atoms or particles on that side is irrelevant. No matter how many they are, they do not manage to oppose any concerted action to gravity. Yes, they also cause the acceleration of the other side, but that’s an independent Action, rather than a Reaction. I see here no resistance, no opposition, but mere surrender. So the concept looks void to me: yes, mass, when acted upon by gravity, is passive…, so passive that it does nothing at all, other than fully yielding to an effect that will be stronger or weaker depending on the active mass, but not on the passive one. Can you measure that passivity? The idea looks meaningless: zero is always zero…

A dynamometer or a spring scale measure “weight” (force of gravity on a mass m = GmM/r^2). It’s sometimes said that it can also be used to measure “PGM” (m) by keeping everything else constant: you observe the stretching or compression of the spring (l) with one mass m1; you repeat the experiment with mass m2, also in the Earth (M is constant) and at the same altitude and latitude (r is also constant). If the stretching or compression is 2l, that must be because the PGM of m2 = 2 * m1…

I don’t think this way we have made the concept of PGM more meaningful. It seems that in the end what we are measuring is simply inertial mass. Both m1 and m2 have accelerated due to gravity with identical accelerations. That’s their “reaction” to the Earth’s gravity: nothing. And that’s the end of gravity’s role in the story. Now we have a new episode that is the contest between m and the spring: m pulls the spring and the spring reacts by pulling m. Here we do have resistance or inertia, from both sides. If the spring is more rigid, it will resist more and stretch less. If m is greater, it will resist more the pull of the string and descend further. But this new fight is a pure collision, a match of contact forces, nothing to do with gravity. Hence it seems to me that the instrument measures inertial mass, not PGM, which still remains a void concept…

Probably this dispensing with the concept of passive gravitational mass is wrong, since the more I search, the more I see it’s a well established concept. But really I find it hard to assimilate it, for those reasons.
 
  • #44
Saw said:
Probably this dispensing with the concept of passive gravitational mass is wrong, since the more I search, the more I see it’s a well established concept. But really I find it hard to assimilate it, for those reasons.

I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.
 
  • #45
This is false. No matter what the mass, all objects accelerate at the rate of 32 ft. per second squared in a vacumn.

Note. If there are only two objects in the entire universe, our sun a me both objects move towards each other. With no other reference there is no reference point. They just get closer together.
 
  • #46
Big O said:
This is false. No matter what the mass, all objects accelerate at the rate of 32 ft. per second squared in a vacumn.

SI Units man! They make life so much easier.
 
  • #47
D H said:
It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass.
I am assuming that mp = mi. p can be substituted for i in the equation for F. The g represents "active gravitational mass".

Saw said:
Ah, you don't play with Newton's 3 Laws of Dynamics, only with the Law of Gravitation. It's hard for me to imagine what this formula might lead to, but I have one early doubt: which reference frame are you considering? Looking at the A formulas, it seems it should be an inertial frame, like the COM frame, just like in Newton’s formula, is it?
Yes, it is the COM (center of inertial mass). And that is the reason it works1. Newton's law only preserves the COM when it is used in the manner in which it was meant to be used, where ma = mi = mp.

atyy said:
I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.
Yes, I tend to agree with that. Newton's law was not designed to work with masses that have unequal active/passive mass. The equality was assumed.

The mutual gravitational force acting between two bodies should be equal in magnitude and opposite in direction. And indeed, that is what the universal law of gravity shows, except when you try and force it to work with a situation where mi = mp <> ma.

F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}

F_2 = -\frac{Gm^\text{g}_1m^\text{p}_2}{r^2}

IF m^\text{g}_1 &lt;&gt; m^\text{p}_1 THEN ABS(F1) <> ABS(F2).

In the example above, the forces are opposite, but not equal in magnitude. Another way to look at it is that the frame of reference is not at the COM. And that is what leads to the violation of the third law.

1 This equation works in all cases where ma is less than, greater than, or equal to, mi and mp. mi and mp are assumed to be equal. The COM is preserved as the frame of reference when mp = mi <=> ma. You may want to go back and review my thought experiments in post #30. This will give a visualization of how the COM is preserved. My equation shows gravity to work the same way, eliminating the nonsensical results when Newton's universal law of gravitation is used in cases where ma <> mi = mp.
 
  • #48
atyy said:
I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.

Thanks for the pointer, Atty.

I googled following your hint and found this quote:

“Because of the symmetry of eqn (1.8) (due to Newton’s third law), no essential difference between active and passive gravitational mass exists in Newton’s theory. In GR, on the other hand, the concept of passive mass does not arise, only that of active mass –the source of the field.”

(Essential relativity: special, general, and cosmological, Wolfgang Rindler, page 16)

So, with regard to GR, he says that passive gravitational mass, as I was arguing, “does not arise”, does not play any role in the film, it’s a superfluous concept.
 
  • #49
TurtleMeister, I submitted your equation to a little obvious test (namely, whether it reduces to Newton's equation if active gravitational = inertial mass) and of course it does:

F = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{1}{{m_1^{\rm{i}} }} + \frac{1}{{m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{{m_2^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }} + \frac{{m_1^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{{m_1^{\rm{i}} + m_2^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{m_1^{\rm{i}} m_2^{\rm{i}} (m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{G}{{r^2 }}<br />

By the way, I've just tried MathType for this purpose and, like your equation, it's a nice toy. I never thought I could be able to do algebra so quickly in a computer...
 
  • #50
Thanks for testing the equation Saw. I've been typing my LateX in manually, but I can see how that tool would be very handy if I had a lot of it to do.

Here's a few relavent quotes from a book I've been reading (Concepts Of Mass In Contemporary Physics - by Max Jammer)

Just as a violation of the mi = mp, equality would be fatal to Einstein's general relativity, a violation of the mp = ma equality would be fatal to Newtonian physics, for it would invalidate Newton's third law of motion.

...It is therefore not surprising that little attention has been paid to an experimental confirmation of the mp = ma equality...

In 1992 William B. Bonnor argued that nevertheless the general theory of relativity allows a violation of the equality between mp and ma, or rather requires such a violation for massive bodies.

He then goes on to write that other gravitational theories allow for, or predict, a non equivalence of mp and ma.

So it appears that the apparent violation of the third law of motion has been a major reason for the lack of laboratory tests for the equivalency of mp and ma, even though other theories predict it. And I can agree that it does seem strange to test for something for which a positive result would mean the acceptance of perpetual motion machines and unlimited free energy. So, to me it seems logical to conclude that either an inequality between mp and ma is impossible, or something is wrong with the methodology (which includes my argument: that Newton's universal law of gravitation is incompatible in cases where ma <> mp).
 
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  • #51
Turtle, I've been looking at your equation under this arrangement:

\begin{array}{l}<br /> F = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_1^{\rm{i}} m_2^{\rm{i}} }}{{r^2 }} \\ <br /> A_{m1} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_2^{\rm{i}} }}{{r^2 }} \\ <br /> A_{m2} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_1^{\rm{i}} }}{{r^2 }} \\ <br /> \end{array}<br /> \]

It has an advantage: the part on the right side looks like Newton's Law of Gravitation; you just add a ratio on the left, the ratio between the gravitational and pasive/inertial masses of the system, which would be precisely the correction proposed for the hypothesis under consideration (that one day an experiment proved that gravitational mass may occasionally differ from passive/inertial mass).

The only problem with this is that, like this, it appears as if the cause of the accceleration were passive/inertial mass, which is obviously not the case...
 
  • #52
You are correct Saw. The equation is Newton's law of gravitation with a correction for the ratio between active gravitational and passive/inertial mass. In fact, the arrangement that you show for F is my original arrangement of the equation. My rearrangement for the final version was made to eliminate the duplication of the two mi variables.
The only problem with this is that, like this, it appears as if the cause of the accceleration were passive/inertial mass, which is obviously not the case...
There is no provision for m1g and m2g in the Newtonian part of the equation. The normal method is to split into two different forces, one for A1 and another for A2. But that would lead us right back to the 3rd law violation (when ma <> mi). So in effect, what I have done is combine the two equations into one by leaving out m1g and m2g for the Newtonian part. This could very well be a flaw for the equation when ma <> mi. But I'm not sure. For example, setting a value for m1i and m2i and then adjusting the values of m1g and m2g for the desired ratio, seems to give the correct results.
 
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  • #53
TurtleMeister said:
what I have done is combine the two equations into one by leaving out m1g and m2g for the Newtonian part. This could very well be a flaw for the equation when ma <> mi. But I'm not sure. For example, setting a value for m1i and m2i and then adjusting the values of m1g and m2g for the desired ratio, seems to give the correct results.

Why do you need to adjust the values of m1g and m2g for the desired ratio, in order to get the correct results? If by "correct results" you mean (i) to comply with your requirement that the two masses must meet at their inertial centre of mass (CM) and (ii) that there's no violation of Newton's 3rd Law (the two forces are equal and of opposite sign), it seems you satisfy those conditions with this equation.

My only problem is to find a physical meaning for the math. The formula is telling us that each mass will get accelerated in proportion to the other's passive/inertial mass (that's why they meet at the inertial CM, as you had ordered) and that, however, if total inertial mass > total gravitational mass, the meeeting will take place later (or earlier in the opposite case). Well, this sounds odd, because what is really causing the acceleration (the gravitational field) does not determine where the objects meet, it simply contributes to determine when, while the resistance of the objects plays the fundamental role.

We noted that small disagreement before. Of course, I do not want to end up with absurd, nonsensical resuls. Of course, the probable end of all this may be that we admit that the assumption that passive = inertial mass is correct. But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.

Take this example. Two identical masses, as measured in a balance scale. In between them, two springs of equal mass and equal length, attached to one another. The CM of the two masses is in the mid-point between them. But the two springs have different elasticity coefficients. Then the two balls are pushed towards each other by identical forces, until they fully compress the springs and meet. Will they meet (i) at the mid-point (CM) or (ii) somewhere closer to the more rigid spring?

It's not a rhetorical question. I really don't know the answer. But my guess is that the ball by the more elastic spring reaches the CM earlier and starts compresing the more rigid one, so the answer would be (ii).

In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?
 
  • #54
These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.
 
  • #55
D H said:
These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.

Sorry, DH, but I respectfully disagree. I have little physics, but I am a good lawyer:smile:. Nobody is posting personal theories here. We assume and accept mainstream physics and we are just trying to grasp the logic of its principles by noticing how alternative routes lead to a dead-end. That's called reductio ad absurdum and for the benefit of the standard approach. If in the course of this discussion, I made incorrect statements, I would most welcome correction. In particular, I really appreciate your interventions, each of which contains a little jewel of information...
 
  • #56
Those equations are a personal theory.
 
  • #57
DH said:
These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.
Thanks for the warning. I will be more careful with my posts. Also, please feel free to add to the discussion or correct my errors, as you have done in the past. Your input is greatly appreciated.

Saw said:
what is really causing the acceleration (the gravitational field) does not determine where the objects meet, it simply contributes to determine when, while the resistance of the objects plays the fundamental role.

We noted that small disagreement before. Of course, I do not want to end up with absurd, nonsensical resuls. Of course, the probable end of all this may be that we admit that the assumption that passive = inertial mass is correct. But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.

Take this example. Two identical masses, as measured in a balance scale. In between them, two springs of equal mass and equal length, attached to one another. The CM of the two masses is in the mid-point between them. But the two springs have different elasticity coefficients. Then the two balls are pushed towards each other by identical forces, until they fully compress the springs and meet. Will they meet (i) at the mid-point (CM) or (ii) somewhere closer to the more rigid spring?

It's not a rhetorical question. I really don't know the answer. But my guess is that the ball by the more elastic spring reaches the CM earlier and starts compresing the more rigid one, so the answer would be (ii).
If I'm understanding this thought experiment correctly, the answer is (i). You partially answered it yourself in the first highlighted sentence that I quoted from you. The springs only supply the force, they do not determine the meeting point (considering the springs are massless and have no substance). This is true even when using multiple springs with different elasticities. The meeting point is the CM, which is determined by the inertial mass of the two objects. The simple equation for this is:

m_1d_1 = m_2d_2

where d1 = the distance from m1 to the CM and d2 = the distance from m2 to the CM.

Saw said:
But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.
Because if they don't meet at the CM then the third law is violated (considering only the two masses are involved). But I guess your question is, how would not meeting at the CM be nonsensical? Well, I'm having trouble thinking of an example for that because you really need to have a good understanding of the third law to get it. And since you're questioning this, I'm assuming you do not have a good understanding of it. But I can tell you that if I had that magic spring from your thought experiment, I could build a device that could propel itself (in the vacuum of space) simply by moving weights around and applying the spring. And that would be nonsensical. You can find lots of stuff on the web about Newton's third law of motion.

Saw said:
In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?
Sorry, I do not understand that.
 
  • #58
Why all the fuss.

In a vacumn all objects fall at the same rate. If there is an atmosphere it will offer resistance to the object falling just as the planets surface will continously halt the object from falling. If the planet changes in size the object adjust with it.
 
  • #59
Big O said:
Why all the fuss.

In a vacumn all objects fall at the same rate. If there is an atmosphere it will offer resistance to the object falling just as the planets surface will continously halt the object from falling. If the planet changes in size the object adjust with it.
That's not what we're talking about. We have gotten a little off topic to the OP, but still related.
Saw said:
In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?
Ok, I think I understand this now. You are saying, by analogy, that the spring can have the same effect as changing the inertial mass of the object it's connected to? No, that is not true. As you have already stated yourself, the springs (source of the force) only has an effect on when they meet, not where. Even though you have two springs with different elasticities connected together, the force applied to each object will be equal in magnitude and opposite in polarity. Where they meet is determined only by their inertial mass.

And by analogy, if we have two objects attracted to each other by gravity, the force on each object will be equal in magnitude and opposite in polarity, even if one has a stronger gravitational field than the other. The source of the force ma has no effect on where they meet.

And that is the reason the following method of applying Newton's laws fails when m1g <> m1p or m2g <> m2p

F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}

F_2 =- \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}

A_1 = \frac{F_1}{m^\text{i}_1}

A_2 = \frac{F_2}{m^\text{i}_2}

Just like in your springs thought experiment, where you have two springs with different elasticities connected together, F1 <> F2, when m1g <> m1p or m2g <> m2p. This causes a problem if we apply F1 to the acceleration of m1, and F2 to the acceleration of m2. If we do that then it would be analogous to your thought experiment where the spring affects the point where they meet. The third law will be violated.

Without inventing new equations for the universal law of gravitation, and for demonstration purposes only, I will attempt to show how this can be done without violating the third law.

F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}

F_2 = \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}

F = \frac{F_1+F_2}{2}

A_1 = \frac{F}{m^\text{i}_1}

A_2 = -\frac{F}{m^\text{i}_2}

The force is now equal in magnitude for each object but opposite in polarity. The objects will meet at their COM.

edit: The averaging of the forces has the effect of restoring the frame of reference to the COM.
 
Last edited:
  • #60
TurtleMeister said:
F_1 = \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}

F_2 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}

F = \frac{F_1+F_2}{2}

A_1 = \frac{F}{m^\text{i}_1}

A_2 = -\frac{F}{m^\text{i}_2}

The force is now equal in magnetude for each object but opposite in polarity. The objects will meet at their COM.
What is this F=F1+F2 stuff? What justifies that step, and then using that force as the force on each of the two objects? That the result is consistent with Newton's third law is not surprising: You implicitly assumed Newton's third law here.
 

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