Why don't heavy objects fall more SLOWLY?

[Saw]

Well, then, I tend to think that what needs a physical explanation is not your math but the standard math: why is gravity so different from any other interaction?

As commented, the standard math assumes that gravitational interaction is not really so but a sort of “sum of two actions”, which are unconnected to each other: due to the WEP, “M” causes on any “m” an acceleration that is independent from any characteristic of “m”, either inertial or gravitational mass, whether passive or active; the overall effect is affected by “m”, because the latter accelerates in turn “M”, but again at a rate that has nothing to do with M’s features. In view of this, one could perfectly apply the usual motto: “this is the way the universe is, as per experiments, whether you like it or not”. But I do not think it is intellectually unsound to highlight how much that approach is inconsistent with other well-established features of nature. One could even say that such approach does not truly satisfy Newton’s Third Law: on both sides we have an action without any REACTION of the passive object, without any opposition or softening of the effect due to resistance, due to inertia of the affected body.

In line with your own thought experiments (now that I better understand them), I would like to recall the example of collisions. Let’s see if I put it without many mistakes (I should say from time to time I am no expert):

In any collision:

$$\frac{{m_1 }}{{m_2 }} = \frac{{a_2 }}{{a_1 }} = \frac{{\frac{{\Delta v_2 }}{t}}}{{\frac{{\Delta v_1 }}{t}}} = \frac{{\Delta v_2 }}{{\Delta v_1 }}$$

If the system formed by the two bodies is closed (no external force) and the collision is perfectly elastic (no internal dissipation of energy), then the relative speed of the two bodies does not change. For example, in the frame of m1, m2 was approaching at a certain v and after the collision it recedes in the opposite direction. This means that:

$$\Delta v_1 + \Delta v_2 = 2v_{rel}$$

After some algebra:

$$\begin{array}{l} \frac{{m_1 }}{{m_2 }} + 1 = \frac{{\Delta v_2 }}{{\Delta v_1 }} + 1 = \frac{{m_1 + m_2 }}{{m_2 }} = \frac{{\Delta v_2 + \Delta v_1 }}{{\Delta v_1 }} = \frac{{2v_{rel} }}{{\Delta v_1 }} \to \Delta v_1 = 2v_{rel} \frac{{m_2 }}{{m_1 + m_2 }} \\ \frac{{m_2 }}{{m_1 }} + 1 = \frac{{\Delta v_1 }}{{\Delta v_2 }} + 1 = \frac{{m_2 + m_1 }}{{m_1 }} = \frac{{\Delta v_1 + \Delta v_2 }}{{\Delta v_2 }} = \frac{{2v_{rel} }}{{\Delta v_2 }} \to \Delta v_2 = 2v_{rel} \frac{{m_1 }}{{m_1 + m_2 }} \\ \end{array}$$

So we see that that the proportion between the two velocity increases is determined by the inertial masses (m1 accelerates in proportion to m2’s share in the total mass and vice versa), while their magnitude depends on the nature of the cause of the interaction (in this case, relative motion).

Thus, shifting back to gravity, one is inclined to expect that something similar happens: that the ratio between the accelerations is given by the (inverse) ratio of the masses, but also that the magnitude is affected by the nature of the interaction (the sum of the “gravitational masses”?). Ok, it’s not so, but it’s a pity, isn’t it? For us, students, it’d make our life easier… I would see a coherent system where now I find a juxtaposition of (in some case, mysterious) rules...

 

[TurtleMeister]

Thanks Saw. I never considered a comparison between gravity and elastic collisions. Interesting observation.

I think the duality of gravitational forces goes back to the time of Newton, but I'm not sure. I'll try to find a reference for this when I return home in a few days.

 

[Saw]

Thanks Saw. I never considered a comparison between gravity and elastic collisions. Interesting observation.

Well, that was something that had also occupied me for some time and I had been thinking of ways to reconcile the two things, to no avail. But your equation has given me the right orientation: it is not only interesting that it paves the way for an investigation about a hypothetical difference between gravitational and inertial mass (by eliminating the fear that such difference would violate Newton’s Laws of Dynamics and hence the principle of Conservation of Momentum or Energy); leaving aside that issue, its first advantage is that it’s a good explanation about many obscure things in gravity, it removes many concerns I had because of the way that the subject is usually taught. In this second sense, far from being a “personal theory”, it’s an elementary truth. It’s a truism, albeit a very revealing one. And far from being a daring modification of Newton’s Law of Gravitation, it’s just its expression in full form.

I’ll explain myself. Consider this arrangement of Newton’s formulas for gravitation (= yours but without distinction between gravitational and inertial mass):

$$\begin{array}{l} F_{on{\rm{ m}}} = m\frac{M}{{(m + M)}}\frac{{G(m + M)}}{{R^2 }} = - F_{on{\rm{ M}}} = - M\frac{m}{{(m + M)}}\frac{{G(m + M)}}{{R^2 }} \\ a_m = \frac{M}{{(m + M)}}\frac{{G(m + M)}}{{R^2 }} \\ a_M = \frac{m}{{(m + M)}}\frac{{G(m + M)}}{{R^2 }} \\ \end{array}$$


Obviously, the ratio (m+M)/(m+M) is 1 and can be left out. But keeping it there for a moment reveals what you’ve been pointing out: the fraction on the right is the full effect, the fraction on the left is the share that each body will take in it. Here it simply happens that (m+M) is on both sides, up and down, so it cancels out. There’s a compensation of factors. In the light of this, now I do understand the answer to a number of traditional questions. For example:

- The principle that “all bodies m, regardless their masses (m1,m2…), fall towards a given mass M at the same rate” looks less mysterious. Ok, it’s true, if you replace m1 with a more massive m2, you get the same acceleration rate, but all the ingredients of this rate have changed. The meeting time with M, as commented in the thread, is reduced. This is due to the (m+M) of the right, which increases the relative acceleration. But the meeting path for m2 is also reduced. This is due to the (m+M) of the left. If and to the extent that the right-hand term = the left-hand term, one thing compensates the other and it “appears as if” nothing had changed, although in fact it’s just the opposite: everything changed, albeit proportionally! Note: in elastic collisions it is the accelerated frame of any of the bodies participating in the interaction the one that suffers this sort of quirk (the magnitude of the relative velocity doesn’t change), here it is the CM frame, but the phenomenon is similar.

- In the formula for the acceleration of m, why is R (the full path between the centres of mass of the two bodies) in the equation, instead of the distance between m and the CM of the system (r_m)? Actually, r_m is there disguised as RM/(m+M), as follows:

$$a_m = R\frac{M}{{(m + M)}}\frac{{G(m + M)}}{{R^3 }} = R\frac{M}{{(m + M)}}\frac{1}{{t^2 }} = \frac{{r_m }}{{t^2 }}$$


which fits with Newton’s modification of Kepler’s Third Law for the case of orbital motion.

- Last but not least, this leads us back to your subject: what if the left-hand (m+M) is different from the right-hand (m+M) term? This is an issue to be settled by experiment but the only thing I can say is that I agree there is no logical objection for exploring such possibility. For someone who proclaims that the acceleration of m is GM/R^2 and full stop (m is nowhere in the equation), it seems logical to infer that m’s features are irrelevant in all respects, quantity and composition. But if one looks at Newton’s formula in full-fledged form (which is the really explanatory one), then one realizes that m IS relevant: it reduces the common meeting time, as well as m’s meeting path. And if this doesn’t show up as a difference in acceleration rate, it’s only because there is a compensation of effects. But that is an accident, not a necessity. It could perfectly happen that the composition of the bodies ruined the offset, so that there is an increase or decrease in meeting time (affected by the gravitational masses of the right), but the meeting point (which only depends on the inertial masses of the left) remains unaffected… I wonder if there are any experiments going on looking for THIS particular effect.

Enjoy your holidays and take your time to comment!

 

[TurtleMeister]

Very nice post Saw. I am pleased that my equation has helped you reconcile the laws of gravity with Newton's other laws of dynamics. Your post was very informative for me. My original intent was to find a way to show that a non equivalence between active gravitational mass and inertial/passive mass would not necessarily result in a violation of the third law of motion. But as your arrangements have so eloquently demonstrated, the equation shows more than that.

You say that the equation is a truism and it is just an expression of the universal law of gravitation in it's full form. If it's a truism then it's correctness should be self evident. It certainly appears that way to me, but the opinions of others would be helpful.

- Last but not least, this leads us back to your subject: what if the left-hand (m+M) is different from the right-hand (m+M) term? This is an issue to be settled by experiment but the only thing I can say is that I agree there is no logical objection for exploring such possibility. For someone who proclaims that the acceleration of m is GM/R^2 and full stop (m is nowhere in the equation), it seems logical to infer that m’s features are irrelevant in all respects, quantity and composition. But if one looks at Newton’s formula in full-fledged form (which is the really explanatory one), then one realizes that m IS relevant: it reduces the common meeting time, as well as m’s meeting path. And if this doesn’t show up as a difference in acceleration rate, it’s only because there is a compensation of effects. But that is an accident, not a necessity. It could perfectly happen that the composition of the bodies ruined the offset, so that there is an increase or decrease in meeting time (affected by the gravitational masses of the right), but the meeting point (which only depends on the inertial masses of the left) remains unaffected… I wonder if there are any experiments going on looking for THIS particular effect.

No. As far as I know there are currently no experiments being conducted to test for a non equivalence of active gravitational mass. If we discount the Bartlett Buren experiment (because it is based on a violation of the third law of motion) then the best we have to date is the Kreuzer laboratory experiment of 1966. And as I have already stated, current torsion balance technology can exceed the level of sensitivity of that experiment by many orders of magnitude. The lack of experimental work in this area has baffled me for a long time. My first post at PF was concerning this. And as I have already stated, and quoted from another source, the apparent conflict with Newton's laws may have been a factor for the lack of experimentation in this area. I believe there is potential for discovery and a better understanding of gravity in experiments of this nature.

And again, nice work with the equations Saw. Your math and communications skills have helped me also. Thanks.

 

[TurtleMeister]

Just wanted to add that by including the distinction between the gravitational and inertial mass you could avoid identifying it as "the fraction on the left" and "the fraction on the right". After all, it is this distinction that sets this form of the universal law of gravitation apart from it's traditional form. Even when the two are assumed to be proportionally equivalent, it avoids confusion when the equation takes on different arrangements. For example.

$$F = G \frac{M_g+m_g}{R^2} \frac{M_i m_i}{M_i + m_i}$$

This form clearly shows how the gravitational mass and the inertial mass affect the force. The contribution of the gravitational mass is determined by the sum of the gravitational mass of the two objects divided by the square of the distance between them. And the contribution of the inertial mass is determined by the product over the sum formula for the inertial masses of the two objects.

 

[Big O]

Question from an uneducated inquisitive soul, your post seem to acknowledge a difference between acceleration and gravity. Doesn't Einstein prove that acceleration and gravity to be the same thing?

 

[TurtleMeister]

Question from an uneducated inquisitive soul, your post seem to acknowledge a difference between acceleration and gravity. Doesn't Einstein prove that acceleration and gravity to be the same thing?

Are you referring to Einstein's "equivalence principle proper"? I do not know the relevance to my posts in this thread. Can you be more specific about the post(s) you are questioning?

 

[Big O]

"TurtleMeister wondered what would happen if the composition of the masses had some influence in how objects accelerate due to gravity"

Why use "accelerate" and "gravity" as if they are different concepts?

 

[TurtleMeister]

TurtleMeister wondered what would happen if the composition of the masses had some influence in how objects accelerate due to gravity.

This statement is a little misleading and can be viewed in a way that is incorrect. The statement seems to imply that I am questioning whether the composition can affect the "Universality of Free Fall", which is not necessarily the case. This effect would not be detectable for ordinary sized objects in Earth free-fall. The statement should read: "TurtleMeister wondered what would happen if the composition of the masses had some influence on the magnitude of the gravitational field produced by the masses (active gravitational mass), or the ratio of "active gravitational mass" to "inertial / passive gravitational mass".

The g in the equations represents active gravitational mass.

edit:
Notice that in the equation, if small m represents the object in free fall and the large M represents the earth, then changing m_g would have undetectable effect on the value of F. In fact, the object in free fall could have no gravitational mass at all and the change would be undetectable. Now if we could somehow change the ratio of M_a to M_i for the earth, then that would be a different story.

I think Saw understands this, and the sentence is correct. It's just phrased in a way that can be misleading.

 

[Saw]

Turtle, I agree with your posts. The formula with the gravitational potential masses has an obvious potential added value over the truism. I just wanted to highlight the basic principle underlying it, which you have yourself identified: in an interaction, the source of the effect (the common source, comprised of its two elements, since we talk about an interaction) determines its magnitude, while the distribution of this effect between each participant depends on its share in the total inertial mass [eg: m/(m+M)].

You mentioned these two examples:

I am sitting in a chair on a frictionless surface. You are sitting in a chair across from me on the same frictionless surface. I reach out and pull you toward me. The result is that we both meet at our COM, or barycenter. Now we repeat the experiment, except this time you reach out and pull me toward you. The result is the same. We both meet at our COM. We repeat the experiment again, except this time we both pull simultaneously, me pulling with a slightly greater force than you. The result is exactly the same. We will always meet at our COM.

Two magnets, M1 and M2, are sitting on a frictionless surface with their opposite poles facing each other. Both magnets have the same inertial mass m_i, but M1 produces a stronger magnetic field than M2. Now, will M2 have a higher acceleration than M1? Where will they meet? Of course they will have the same acceleration and they will meet at their COM (which is determined by their m_i).

Following, I think, your trend of thought, I proposed an analogy with elastic collisions. Taking the acceleration formulas as reference: in elastic collisions, the a of m is a quota [M/(m+M)] in the source of the acceleration (relative motion = 2 times the relative velocity); more relative velocity means that the source is stronger and hence the “separation time” (time needed to create a certain distance) decreases, although the “separation spatial reference” is still the COM (the distribution of the total effect is done according to the share of each body in the sum of inertial masses).

Can you express your above mentioned thought experiments in similar formulas? I mean, can it be mathematically shown that in those examples, while the “meeting point” is the COM, the “meeting time” (I presume) is decreased if you increase, for example, the strength of any of the magnetic fields?

It’d also be interesting to mathematically express the experiment I proposed above about the springs. Its advantage is that it’s a close analogy with gravity. Think of this clearer re-formulation: the springs are first in their natural length, stuck to one another; then the balls are moved apart from each other by equal forces (thus stretching the springs) and then released (so that the springs rapidly recover their original states and, I think, are further compressed and bounce back and so on, to the extent there are no external forces). Here the springs’ restoration forces play the part of a factor that is added to the inertial masses but depends on their specific compositions... I’ll try to think of the math, but really I stretch my math ability to its elasticity limit, although I am quite rigid… Thinking aloud: if k augments, you need more force (kx) to separate the balls a given distance; so the force on return (ma) will include a higher acceleration and hence less meeting time. Or through conservation of energy: if k augments, the spring stores more potential energy, which is afterwards given back in the form of more kinetic energy. But could we put this in a common formula, analogous to that of gravity or collisions, something including (m+M) as well as (k+K)?

 

[Saw]

I studied a little and a formula for the case of springs attached in series would be (with some simplification):


$$F = \frac{{k_1 k_2 }}{{k_2 + k_1 }}\left( {\Delta x_1 {\rm{ }} + {\rm{ }}\Delta x_2 } \right)$$

It bears a resemblance with your formula:

$$F = \frac{{m_1^{\rm{i}} m_2^{\rm{i}} }}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{G(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{r^2 }}$$

although the roles are exchanged (I wanted k1 and k2 to play the part of gravitational masses, not of inertial masses). Still I think the analogy can be saved through some rearrangement. I'll think about it.

Edit: Well, this is also true:

$$F = \frac{{\Delta x_1 \Delta x_2 }}{{\Delta x_1 + \Delta x_2 }}(k_1 + k_2 )$$

But still not what I looked for, since the masses are missing. Sometimes analogies are useful. Sometimes it’s too troublesome to construct them...

 

[TurtleMeister]

Sorry Saw, I am unable to formulate my thought experiment mathematically. The equations for the magnetic forces are too complex for me, and in my opinion unnecessary to illustrate the point I was trying to make. I chose the magnet thought experiment because the magnetic field acting at a distance is somewhat analogous to a gravitational field acting at a distance. The magnets having the same inertial mass but different magnetic field strengths is analogous to two objects with the same inertial mass in mutual gravitational attraction, where one has a hypothetical different active gravitational mass than the other one. In the case of the magnets, it is obvious (without mathematics) that they will meet at their COM (the third law of motion will not be violated). Otherwise, we would all have free energy generators in the basements of our homes. :) So my question then was, if the magnets obey the third law of motion then why shouldn't gravity? My logic was that any time you get results where the third law of motion is being violated then it should throw a red flag that something is wrong with your methodology. And in the case of the current methodology using Newton's classical law of universal gravitation, it means that either a non equivalence of active gravitational mass and inertial mass is impossible, or something is wrong with the methodology. I chose the latter and tried to show another way. I am poor with math, but I knew it was the most effective way to communicate my thoughts to others. So I put in considerable effort to come up with the equation we are now discussing.

Your analogies with elastic collisions and the springs have been very interesting and have given more validity to the equation by showing it's similarities to other laws of dynamics. However, there is a limit to how far you can take the analogies.

I think your most interesting contribution is your arrangement of the equation in post #73. It was very illustrative of how much more effective it is over the traditional form. For example, my response to the OP in post #2 is my traditional answer to the question:

I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?

It is true that the more massive object will resist acceleration more than the less massive object. However, the more massive object has a greater force acting on it. The two are perfectly balanced so that the effect is that all objects will accelerate at the same rate regardless of their mass.

With the traditional form of the acceleration equation it was difficult to see the balancing of inertia and gravitational forces. But with your arrangement, the effect is clearly visible in the equation.

$$a_m=\frac{M_i}{(m_i+M_i)}\frac{G(m_g+M_g)}{R^2}$$

 

[Saw]

Thank you for the comments, Turtle. You had the thought, I am just exploring its implications, for fun, because really the conversation is enjoyable, both in terms of contents and tone. If you also like the idea, I’d like to share some thoughts on the subject from time to time.

there is a limit to how far you can take the analogies.

Yes, the trick for making analogies (this is an idea borrowed from the law field) is identifying the relevant elements of the two terms being compared. “Relevant” is what matters for a certain purpose. There may be 99% differences between two situations but if the 1% similarities are the ones that matter for the purpose at hand, then the analogy works. You take it “too far” if you apply it to the non-relevant aspects, but if you use it only for the relevant ones, it’s OK.

In this line, I still think that the analogy of the springs may be useful.

Taking for example this form of the equations that you mentioned:

$$a_m=\frac{M_i}{(m_i+M_i)}\frac{G(m_g+M_g)}{R^2}$$

If you were to assume that experiments (those sensitive torsion balances that you talk about) ever showed a difference between the active gravitational masses (m_g ≠ M_g) of two equal inertial masses (m_i=M_i), then the question would be: how do you express that? With a new unit of gravitational mass? Or with a coefficient of “gravitational activity” of materials, stipulating that k = 1 for a certain reference material and a portion of 1 for the others? If the latter idea makes any sense, then we do have a similarity between the two situations (gravity and springs), in the relevant aspect: in both cases, both the quantity of matter and its composition, in both sides, “matter”.

As commented, I am thinking of a system with two masses, attached to springs of different constants, attached to one another. The masses are placed apart (the springs stretch out) and then released. I dare to think that (with a number of simplifications and idealizations), we would create here “simple harmonic motion”, where these equations would apply (T = period of oscillation, X = total change of length of the springs and K = effective coefficient):

$$\begin{gathered} a_{rel} = \frac{{v_{rel} ^2 }} {X} = \frac{{(2\pi X/T)^2 }} {X} = \frac{{4\pi ^2 X}} {{T^2 }} \hfill \\ a_{rel} = \frac{{k_1 x_1 }} {{m_1 }} + \frac{{k_2 x_2 }} {{m_2 }} = \frac{{m_2 k_1 x_1 + m_1 k_2 x_2 }} {{m_1 m_2 }} = \frac{{KX(m_1 + m_2 )}} {{m_1 m_2 }} \hfill \\ \frac{{4\pi ^2 X}} {{T^2 }} = \frac{{KX(m_1 + m_2 )}} {{m_1 m_2 }} \hfill \\ T^2 = \frac{{4\pi ^2 X(m_1 m_2 )}} {{KX(m_1 + m_2 )}} = \frac{{4\pi ^2 (m_1 m_2 )}} {{K(m_1 + m_2 )}} = \frac{{4\pi ^2 (m_1 m_2 )}} {{\frac{{k_1 k_2 }} {{k_1 + k_2 }}(m_1 + m_2 )}} = \frac{{4\pi ^2 (m_1 m_2 )(k_1 + k_2 )}} {{(k_1 k_2 )(m_1 + m_2 )}} \hfill \\ a_{rel} = \frac{{4\pi ^2 X}} {{T^2 }} = \frac{{4\pi ^2 X}} {{\frac{{4\pi ^2 (m_1 m_2 )(k_1 + k_2 )}} {{(k_1 k_2 )(m_1 + m_2 )}}}} = \frac{{X(k_1 k_2 )(m_1 + m_2 )}} {{(m_1 m_2 )(k_1 + k_2 )}} = X\frac{{k_1 k_2 }} {{k_1 + k_2 }}\frac{{m_1 + m_2 }} {{m_1 m_2 }} \hfill \\ a_{m_1 } = \frac{{m_2 }} {{m_1 + m_2 }}X\frac{{k_1 k_2 }} {{k_1 + k_2 }}\frac{{m_1 + m_2 }} {{m_1 m_2 }} = \frac{{KX}} {{m_1 }} \hfill \\ a_{m_2 } = \frac{{m_1 }} {{m_1 + m_2 }}X\frac{{k_1 k_2 }} {{k_1 + k_2 }}\frac{{m_1 + m_2 }} {{m_1 m_2 }} = \frac{{KX}} {{m_2 }} \hfill \\ \end{gathered}$$

I am quite unsure as to whether this is correct. It looks nice, however: more stiffness, more acceleration; more inertial masses, less acceleration; a possibility for a compensation of effects...

 

[Flatland]

It is true that the more massive object will resist acceleration more than the less massive object. However, the more massive object has a greater force acting on it. The two are perfectly balanced so that the effect is that all objects will accelerate at the same rate regardless of their mass.

I'm sure this question has been asked many times at PF, so you may be able to use the search function to learn more. Or, please feel free to ask more questions here if you like.

But mass will only resist acceleration if there is inertia. During free-fall there is no inertia.

 

[cragar]

there are physicists now that are trying to prove that more massive objects fall fasters because the gravitational attraction between the Earth and the object would be greater.

 

[Saw]

But mass will only resist acceleration if there is inertia. During free-fall there is no inertia.

Not really. You seem to be identifying inertia with the sensation you feel when standing on the ground of the Earth: the Earth pulls you towards its centre of mass, you are pressed against the ground and the ground reacts with a normal force of equal magnitude and opposite direction and you seem to resist that, in a sort of loop, which leads to a quite stable equilibrium (net force = 0). Instead, when in free-fall, like when you go down in a rapidly falling elevator or down a hill in a roller coaster (that is to say, when there is no opposition of the counter-force of the mass you are attracted to), you don’t have the sensations mentioned at the beginning. But that doesn’t mean that either the weight or the inertia is missing in free-fall:

- First, you are not weightless. It’s only that the weight is not apparent in that way, it’s not detectable through that particular method. But it’s there because you are accelerating towards the Earth. And to the extent that the body we consider is big enough, it becomes apparent due to the non-homogeneity across distance of the gravitational field: particles closer to the Earth are attracted more than those farther away. That’s why you have tides (tidal effects, in general).

- Second, if there is weight, there is inertia. Otherwise, you would accelerate at a much higher rate.

In the standard explanation, the justification goes as follows:

The force of gravity is:

$$F = \frac{{GM_g m_g }}{{R^2 }}$$

Your acceleration is:

$$\begin{gathered} F = m_i a_m = \frac{{GM_g m_g }} {{R^2 }} \to \hfill \\ a_m = \frac{{\frac{{GM_g m_g }} {{R^2 }}}} {{m_i }} = \frac{{m_g }} {{m_i }}\frac{{GM_g }} {{R^2 }} = \frac{{GM_g }} {{R^2 }} \hfill \\ \end{gathered}$$

As you can see, in the standard explanation, your gravitational mass is balanced out by your inertial mass and so you are attracted to the Earth at a rate only related to the (gravitational) mass of the Earth.

What we are saying here (not objected so far) is that the explanation should be slightly different, for example as follows:

$$\begin{gathered} F = m_i \frac{{M_i }} {{(m_i + M_i )}}\frac{G} {{R^2 }}(m_g + M_g ) \to \hfill \\ a_m = \frac{F} {{m_i }} = \frac{{m_i \frac{{M_i }} {{(m_i + M_i )}}\frac{G} {{R^2 }}}} {{m_i }} = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}} \hfill \\ \end{gathered}$$

The difference is that here your inertial mass is canceled out by your inertial mass. So that’s unavoidable. And then there is the possibility that the sum of the two intervening gravitational masses is canceled out by the sum of the two inertial masses… or not, if they were not really equivalent. The advantage of this formulation is that it opens the latter possibility without threatening Newton’s Third Law or conservation principles. Additionally, it’s not some ad hoc invention, but it links coherently with the logic of the whole system. In the last posts we were simply considering the possibility of replacing the idea of gravitational mass with a sort of “coefficient of gravitational charge” specific to each material, in the understanding of course that this is just an intellectual exercise, since no experiments have ever detected a difference between inertial and gravitational masses.

Maybe you are pointing at Einstein’s idea that free-fall motion is (not exclusive of but) equivalent to inertial motion, in a sufficiently small region of space and time (hence no tidal effects) and so you can apply the same laws of physics in both types of frames. That’s another issue. Maybe connected to this, but we were having a purely Newtonian discussion.

there are physicists now that are trying to prove that more massive objects fall fasters because the gravitational attraction between the Earth and the object would be greater.

Can you be more specific? The fact that, given two mutually attracting masses M and m1, if you replace m1 with a more massive m2, the meeting time is reduced… is not discussed here. That’s a non-controversial phenomenon. In that case, the relative acceleration (wrt M, for example) increases. It only happens that the centre of mass of the system slides a little towards m2. Due to this, the individual accelerations (wrt this centre of mass) turn out to be the same, no matter whether the falling object is m1 or m2. That’s mainstream physics, although Turtle and I were arguing that it’s better seen, more didactically shown with the longer formulation of Newton’s Law of Gravitation written above.

 

[ZapperZ]

there are physicists now that are trying to prove that more massive objects fall fasters because the gravitational attraction between the Earth and the object would be greater.

Please cite who these "physicists" are, because they're trying to "prove" what is known already. As saw has stated, this isn't new and certainly isn't puzzling at all, since it comes from the fact that one no longer makes the simplification of another mass m being significantly smaller than the Earth's mass M (see the last entry in the FAQ in the General Physics forum).

Zz.

 

[Saw]

Thank you, ZapperZ, for the comment. It’d be useful to know if you find any fault with the arrangements of Newton’s formula that Turtle produced and I have been commenting (I mean, with its logic = the acceleration of a body is a share in the relative acceleration or sum of the two accelerations, based on gravitational masses; in particular, this share = the other body’s share in the sum of inertial masses).

It’s interesting that this formula casts doubts on the validity of one of Newton’s experiments. To check the equivalence (or at least the proportionality) between inertial and gravitational mass, he constructed two pendulums, whose bobs were cylindrical containers. He filled the cylinders with different substances but “the same weights” and set the pendulums in oscillation. Despite many tries with various substances, he found that the periods of the two pendulums were always the same. Apparently he stated (I haven’t found the original text but a citation): “in these experiments, in bodies of the same weight, a difference of matter that would be even less than a thousandth part of the whole could have been clearly noticed”.

The modern mathematical explanation is:

$$\begin{gathered} F_{Gravity} = \frac{{GM_g m_g }} {{R^2 }} \hfill \\ F_{Gravity} {\text{ }} - {\text{ }}Tension = F_{net} = \frac{{GM_g m_g }} {{R^2 }}sin\theta \hfill \\ If{\text{ }}\theta {\text{ is very small }} \to sen\theta \simeq \theta \to \frac{s} {l}{\text{ }} \simeq \frac{x} {l} \hfill \\ F_{net} = m_g \frac{{GM_g }} {{R^2 }}\frac{x} {l} = m_g g\frac{x} {l} = m_i a_m \to \hfill \\ a_m = \frac{{m_g }} {{m_i }}g\frac{x} {l} \hfill \\ a_m = \frac{{m_g }} {{m_i }}g\frac{x} {l} = \omega ^2 x \to \hfill \\ \omega ^2 = \frac{{m_g }} {{m_i }}\frac{g} {l} \hfill \\ \omega = \sqrt {\frac{{m_g }} {{m_i }}\frac{g} {l}} \hfill \\ T = \frac{{2\pi }} {{\sqrt {\frac{{m_g }} {{m_i }}\frac{g} {l}} }} = 2\pi \sqrt {\frac{{m_i }} {{m_g }}\frac{l} {g}} \hfill \\ \end{gathered}$$

So the observed equality of the periods of different bobs relies on the condition that m_i = m_g or at least the ratio between any m_i and any m_g is always the same.

However, there is something odd here. If inertial mass has to be canceled out with gravitational mass, let it be so. But why the cancellation only for m? What about M_g versus M_i?

In other (to a certain extent) similar examples, the way that T is affected by mass varies, but you always find the two masses:

- In the case of “free fall”, as commented, increasing any of the intervening masses augments the relative acceleration and reduces the meeting time.

- In the case of “orbital motion” of the planets, it was Newton himself who corrected Kepler’s Third Law and produced this new formula:

$$T = 2\pi \sqrt {\frac{{R^3 }} {{G(M + m)}}}$$


This means that increasing any of the masses reduces the orbital period. It is true that this also changes the centre of mass of the system and that may affect the orbit, unless one adjusts some other value, like distance or tangential velocity. But the conceptual principle is there: the period is affected by the two masses and it’s only when one mass << the other that the formula reduces to Kepler’s, where the masses are left out.

- Instead, in “simple harmonic motion”, like in a mass-spring system, the period increases if you increase the mass:

$$T{\text{ }} = {\text{ }}2{\text{ }}\pi \sqrt {\frac{m} {K}}$$


For the case of two springs, attached to their respective masses, I derived above this formula (correct?), where it’s seen that increasing any of the masses increases the period:

$$T = 2\pi \sqrt {\frac{{(m_1 m_2 )(k_1 + k_2 )}} {{(m_1 + m_2 )(k_1 k_2 )}}}$$


- And in the “simple pendulum” example, like in Newton’s experiment, why should only the mass of the bob be relevant? The simple pendulum is said to follow the pattern of simple harmonic motion, except for the fact that it’s a “small angle approximation”. But for the purpose considered here (is the period affected?), that doesn’t seem to be relevant. If it weren’t, in order to bring the two masses back, the longer formula for Gravitation commented here would turn out to be helpful, since it would enable this other derivation:

$$\begin{gathered} F_{Gravity} = m_i \frac{{M_i }} {{(m_i + M_i )}}\frac{G} {{R^2 }}(m_g + M_g ) \to \hfill \\ a_m = \frac{F} {{m_i }} = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}} = g\frac{{(m_g + M_g )}} {{(m_i + M_i )}} \hfill \\ F_{net} = m_i g\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{x} {l} = m_i a_m \to \hfill \\ a_m = g\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{x} {l} = \omega ^2 x \to \hfill \\ \omega ^2 = \frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{g} {l} \hfill \\ \omega = \sqrt {\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{g} {l}} \hfill \\ T = \frac{{2\pi }} {{\sqrt {\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{g} {l}} }} = 2\pi \sqrt {\frac{{(m_i + M_i )}} {{(m_g + M_g )}}\frac{l} {g}} \hfill \\ \end{gathered}$$


Should this be true, the teaching would be that Newton could have played with his bobs as much as he wanted and he would not have found any difference in the relevant ratio, even if it actually existed, because both m_g and m_i are negligible wrt M_g and M_i. He should have played with the composition of the Earth for that purpose…! A curious implication of the formula, isn´t it?

 

[TurtleMeister]

I'm still away from home, but I just wanted to make a few comments. Saw, to my delight, it appears that you now have the same view of gravity that I do. It's amazing what a simple equation can do.

Should this be true, the teaching would be that Newton could have played with his bobs as much as he wanted and he would not have found any difference in the relevant ratio, even if it actually existed, because both mg and mi are negligible wrt Mg and Mi. He should have played with the composition of the Earth for that purpose…! A curious implication of the formula, isn´t it?

It has been confusing to me as to what past and current experimenters are really looking for. If Newton was looking for a difference between "passive gravitational mass" and "inertial mass" then his experiments were correct. But it is "my opinion" that this distinction (and the associated experiments) is pointless. It seems to me that "passive gravitational mass" is simply another name for "inertial mass". And I'm not alone in this belief (I will find references when I return home).

Most all of the current and more recent experiments (the Eotvos experiments, the experiments at Washington University - The Eot-Wash Group, and STEP) are actually modern versions of Newton's experiment. As you've already discovered, if one is to detect a difference between "active gravitational mass" and "inertial mass" for different compositions of matter, then you must be able to control the composition of the "attractor" mass. And in the case of Newton's experiment (and STEP), that would have been the earth. Fortunately, we do not have to use the earth. Torsion balance experiments could be easily designed for this purpose. But no one is doing it! If anything is to come out of this equation, it is my hope that it will stimulate interest in this type of experiment by eliminating the belief that a positive outcome would result in a violation of the laws of motion.

 

[Saw]

Saw, to my delight, it appears that you now have the same view of gravity that I do. It's amazing what a simple equation can do.

Well, yes, as the thread shows, my views evolve very quickly. At the beginning I didn’t really understand the principle of equivalence. When I did, I didn’t see the need for the centre of (inertial) mass being always the meeting point of an attraction. When I got that, I saw that it matched with an idea I had about collisions. And now I like your point very much. Just want to remind from time to time that we don’t want to break forum rules. The subject seems to be a relatively open one and so it seems there is room for discussion, accepting of course any guidance we may receive from experts...

If Newton was looking for a difference between "passive gravitational mass" and "inertial mass" then his experiments were correct. But it is "my opinion" that this distinction (and the associated experiments) is pointless. It seems to me that "passive gravitational mass" is simply another name for "inertial mass". And I'm not alone in this belief (I will find references when I return home).

Precisely that’s one point I would rescue from my earlier comments.

People may wonder why we only talk about gravitational versus inertial mass, without distinction between active and passive gravitational mass.

Referring to Newton’s experiment of the pendulum, I suppose that a more precise form of the standard view (more precise than the one I used above) would be:

$$\begin{gathered} F_{{\text{on m}}} = m_{grav}^{passive} M_{grav}^{active} \frac{G} {{R^2 }} \to \hfill \\ a_m = \frac{F} {{m_{inertial} }} = \frac{{m_{grav}^{passive} M_{grav}^{active} \frac{G} {{R^2 }}}} {{m_{inertial} }} = \frac{{m_{grav}^{passive} }} {{m_{inertial} }}M_{grav}^{active} \frac{G} {{R^2 }} = \frac{{m_{grav}^{passive} }} {{m_{inertial} }}g \hfill \\ \end{gathered}$$But no matter whether one calls it active or passive gravitational mass, or simply gravitational mass, the question is still: We are talking about an interaction and its effects. And in the result of an interaction, both sides play their parts. I see in the above formula how the inertial mass of m is canceled by its passive gravitational mass. But what about the ratio between the inertial and the passive gravitational mass of M? Instead, if m’s inertial mass is canceled against its own inertial mass, that leaves no room for discussion. And then there is the ratio between *the sum of* inertial and gravitational masses, which may respond to a fixed proportion or not… That sounds more consistent with the nature of an interaction. And in that case the experiment might be useless.

I’ll quote Newton’s words with regard to inertia, to which sometimes he referred as vis inertiae (the force of inactivity):

But a body only exerts this force when another force, impressed upon it, endeavours to change its condition; and the exercise of this force may be considered as both resistance and impulse; it is resistance so far as the body for maintaining its present state, opposes the force impressed; it is impulse so far as the body, by not easily giving way to the impressed force of another endeavours to change the state of that other.

I think he was implying that the distinction between “active” and “passive” is artificial. It’s just the two sides of the same coin. Our mind tends to analyze phenomena by decomposing them in little pieces (concepts), which usually comes very helpful, but that doesn’t mean that those pieces really exist… We don’t talk about an active and a passive inertial (or electromagnetic, after all) mass, so why distinguishing between a passive and an active gravitational mass?

 

[Saw]

I’ve been thinking about the simple pendulum and doubt about the validity of the last formula I posted in #88. But really it’s not a question of loading the forum with more tries without knowing whether they are true or not.

Maybe the experts can help.

The question is: The simple pendulum is, after all, a case of “attenuated gravity”, just like in Galileo’s inclined plane example, isn’t it? If so, it seems we should observe in it the usual pattern observed in gravity, only with lesser magnitude. This pattern is that increasing *any* of the involved masses does affect time. If analyzing the situation from the CM frame, acceleration is not affected, because the system's CM changes if you change any of the masses, but the change of time is there. If analyzing the situation from the frame of any of the intervening masses, that becomes more apparent, because the relative acceleration increases. But the mathematical derivation could be done from the perspective of any frame and the impact on time should always shine up. For the case of orbital motion, I have seen those derivations and they actually converge on the conclusion that the orbital period varies as a function of any of the masses. Shouldn’t this apply as well to the simple pendulum experiment? If a bob is set to oscillate on the Moon, its period augments with regard to the same bob oscillating on the Earth, because the Moon is less massive than the Earth. And if you used the Moon as the bob of a pendulum on the Earth, shouldn’t that make the Earth oscillate as well around the CM of the system and shouldn’t both planets complete their cycles more frequently (lower period) than if the bob were a 1-kg mass? Of course, if one mass is the Earth and the other is a much smaller object, the latter is negligible for practical purposes, but it’d be still interesting to know the full formula (accounting for the *two* masses), because that may relevant, for example, for the purpose discussed here (to what extent does this kind of experiments serve to detect lack of proportionality between inertial and gravitational mass?).

 

[TurtleMeister]

Of course, if one mass is the Earth and the other is a much smaller object, the latter is negligible for practical purposes, but it’d be still interesting to know the full formula (accounting for the *two* masses), because that may relevant, for example, for the purpose discussed here (to what extent does this kind of experiments serve to detect lack of proportionality between inertial and gravitational mass?).

I do not know how to do the math you are asking for (nor do I have the time right now), but I can tell you that in order to use this method to detect a lack of proportionality between inertial and gravitational mass you would need to be able to control the composition of the "attractor" (large) mass. A torsion balance experiment can use this technique, where the period, or frequency, of the pendulum is measured. Of course the torsion balance experiment would not use the earth, but a controlled composition attractor mass. The larger this mass is in relation to the pendulum mass the better.

 

[Saw]

I can tell you that in order to use this method to detect a lack of proportionality between inertial and gravitational mass you would need to be able to control the composition of the "attractor" (large) mass. A torsion balance experiment can use this technique, where the period, or frequency, of the pendulum is measured. Of course the torsion balance experiment would not use the earth, but a controlled composition attractor mass. The larger this mass is in relation to the pendulum mass the better.

I agree with all that. It's just that I was not happy with the derivation of Period (T) for the "simple pendulum" that I wrote at the end of post #88, because it suffers from the same problem as the the standard derivation, namely:

- the derivation is done on the basis of the individual acceleration of the mass m of the bob;

- that means the considered reference frame is the centre of mass of the system bob-Earth;

- given so, it's not strange that such individual acceleration turns out to be independent of the bob's mass;

- however, period is a different thing: period SHOULD be affected by the bob's mass and that should shine up, regardless whether you make the derivation in the CM's (inertial) frame or in the (accelerated) frame of the bob or the Earth;

- ok, in practical terms, if m (bob's mass) is negligible wrt to M (Earth's mass), the contribution of the bob's mass to the period is virtually inexistent, but conceptually it's there and should appear in the formula and

- last but not least, the concept that the bob's mass does affect the period becomes fully relevant when you try to use the experiment to discuss on the proportionality between inertial and gravitational mass.

The use of the "modified" Law of Gravitation initially seemed to solve the issue, but as I said above it's not sufficient. Another correction is needed. I think I know now how to do it. To be consistent, if one plays with individual acceleration, one also has to play accordingly with the "individual radius" of the bob's mass. Alternatively, one can reason from the accelerated frame but then play with relative acceleration and the whole radius. Both routes should converge on the same period of oscillation. I'll try to post the derivations later (I am also short of time now) but the results should look as follows:

$$T = \frac{{2\pi }} {{\sqrt {\frac{{G(m_g + M_g )}} {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}} {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + \frac{{GM_g }} {{R^2 }}}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + g}}}$$

 

[TurtleMeister]

Ok, I see what you're trying to do. And I think you're on the right track. You need to consider the relative acceleration instead of the COM based one.

 

[Saw]

Ok, I see what you're trying to do. And I think you're on the right track. You need to consider the relative acceleration instead of the COM based one.

Yes, judging from the accelerated frame (of the Earth, for example) and hence considering the relative acceleration, everything looks easier:

$$\begin{gathered} a_{relative} = \frac{{G(m_g + M_g )}} {{R^2 }}\frac{x} {l} = \omega ^2 x \hfill \\ \omega ^2 = \frac{{G(m_g + M_g )}} {{R^2 l}} \hfill \\ \omega = \sqrt {\frac{{G(m_g + M_g )}} {{R^2 l}}} \hfill \\ T = \frac{{2\pi }} {{\sqrt {\frac{{G(m_g + M_g )}} {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}} {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + \frac{{GM_g }} {{R^2 }}}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + g}}} \hfill \\ \end{gathered}$$

The second form is similar to the accepted formula for the period of orbital motion (with just R^2•l instead of R^3) and the last one shows how little different the outcome is from what you would obtain with the standard pendulum formula.

But what worried me is that judging from the inertial reference of the COM you should also obtain the same outcome, since in Newtonian mechanics, period (time), unlike acceleration, is frame-independent. Finally, I looked carefully at the derivation, from the COM frame, of Kepler’s 3nd Law (Newton’s version) that I found in http://www.vikdhillon.staff.shef.ac.uk/teaching/phy105/celsphere/phy105_derivation.html. The key is that in the COM frame you must consider both individual acceleration and individual radius.

Starting with your “modified” Law of Gravitation (with the standard one the result is similar, although with the difference we can comment at the end), the expected result is obtained as follows:

$$\begin{gathered} F_m = m_i \frac{{M_i }} {{(m_i + M_i )}}\frac{G} {{R^2 }}(m_g + M_g ) \to \hfill \\ a_m = \frac{F} {{m_i }} = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}} \hfill \\ F_{net} = m_i \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{{x_m }} {{l_m }} = m_i a_m \to \hfill \\ a_m = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{{x_m }} {{l_m }} = \omega ^2 x_m \to \hfill \\ \omega ^2 = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{1} {{l_m }} = \frac{{GM_i }} {{R^2 }}\frac{{(m_g + M_g )}} {{(m_i + M_i )}}\frac{1} {{l\frac{{M_i }} {{(m_i + M_i )}}}} = \frac{{G(m_g + M_g )}} {{R^2 l}} \to \hfill \\ \omega = \sqrt {\frac{{G(m_g + M_g )}} {{R^2 l}}} \hfill \\ T = \frac{{2\pi }} {{\sqrt {\frac{{G(m_g + M_g )}} {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}} {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + \frac{{GM_g }} {{R^2 }}}}} = 2\pi \sqrt {\frac{l} {{\frac{{Gm_g }} {{R^2 }} + g}}} \hfill \\ \end{gathered}$$

If we started instead with the official version of Newton’s formula…

$$\begin{gathered} F_m = \frac{{GM_g^a m_g^p }} {{R^2 }} \to \hfill \\ F_{net} = \frac{{GM_g^a m_g^p }} {{R^2 }}\frac{{x_m }} {{l_m }} = m_i a_m \to \hfill \\ a_m = \frac{{GM_g^a }} {{R^2 }}\frac{{m_g^p }} {{m_i }}\frac{{x_m }} {{l_m }} = \omega ^2 x_m \to \hfill \\ \omega ^2 = \frac{{GM_g^a }} {{R^2 }}\frac{{m_g^p }} {{m_i }}\frac{1} {{l_m }} = \frac{{GM_g^a }} {{R^2 }}\frac{{m_g^p }} {{m_i }}\frac{1} {{l\frac{{M_i }} {{(m_i + M_i )}}}} = \frac{{G(m_i + M_i )}} {{R^2 l}}\frac{{M_g^a }} {{M_i }}\frac{{m_g^p }} {{m_i }} \to \hfill \\ \omega = \sqrt {\frac{{G(m_i + M_i )}} {{R^2 l}}\frac{{M_g^a }} {{M_i }}\frac{{m_g^p }} {{m_i }}} \hfill \\ T = \frac{{2\pi }} {{\sqrt {\frac{{G(m_i + M_i )}} {{R^2 l}}\frac{{M_g^a }} {{M_i }}\frac{{m_g^p }} {{m_i }}} }} \hfill \\ \end{gathered}$$

… then the result is awkward, because –apart from some funny need to cancel M active grav. with M inertial but m passive grav. with m inertial- the period is determined by inertial masses, while it seems it should depend on the gravitational masses. The same problem would arise in the derivation of Kepler’s Third law. So, if true, this would be good feedback for the “modified” version of Newton’s Law of Gravitation.

It'd also confirm, as you hold, that the pendulum experiment is not apt for validating the WEP, because any change in the composition of the bob would have an undetectable impact on the period... even without any modification of Newton's Law of Gravitation.

 

[TurtleMeister]

… then the result is awkward, because –apart from some funny need to cancel M active grav. with M inertial but m passive grav. with m inertial- the period is determined by inertial masses, while it seems it should depend on the gravitational masses. The same problem would arise in the derivation of Kepler’s Third law. So, if true, this would be good feedback for the “modified” version of Newton’s Law of Gravitation.

Yes, the modified version handles the separation of gravitational mass and inertial mass much better.

It'd also confirm, as you hold, that the pendulum experiment is not apt for validating the WEP, because any change in the composition of the bob would have an undetectable impact on the period... even without any modification of Newton's Law of Gravitation.

I assume you meant to say: any change in the active gravitational mass of the bob as a result of a change in it's composition would have an undetectable impact on the period. Yes, I have realized this for a long time. But not until this equation was I able to relate it to others. The bob could have no active gravitational mass at all and it would still be undetectable. That is why I am so puzzled by experiments such as STEP.

I appreciate all the mathematical experimenting you've done with the equation Saw. It all looks good to me. But as I have already stated, I am poor with math. So you should not rely solely on my opinion.

 

[Saw]

I assume you meant to say: any change in the active gravitational mass of the bob as a result of a change in it's composition would have an undetectable impact on the period. Yes, I have realized this for a long time. But not until this equation was I able to relate it to others. The bob could have no active gravitational mass at all and it would still be undetectable. That is why I am so puzzled by experiments such as STEP.

Right. Maybe you could explain STEP more in detail.

I appreciate all the mathematical experimenting you've done with the equation Saw. It all looks good to me. But as I have already stated, I am poor with math. So you should not rely solely on my opinion.

Well, the math itself, the algebra may be correct. What may be wrong is the assumptions. The first part (taking the standard formula but applying to it the modified Law of Gravitation) looks less objectionable. About the second part (deriving on the basis of relative acceleration or individual acceleration with an individual radius), I am far less sure now. Maybe it's wrong because it relies on a bad understanding of the physical situation. (I am thinking that the mass of the bob affects the tension, perhaps that's the key...). It's a pity that the experts do not pop into comment, but probably that would require a specific question, in a more concise manner. I may post it at some time.

 

[TurtleMeister]

Right. Maybe you could explain STEP more in detail.

http://www.sstd.rl.ac.uk/fundphys/step/"
http://www.npl.washington.edu/eotwash/intro/intro.html"
Both of the above are modern tests of the equivalence principle, similar to Newton's pendulum experiment that you mentioned earlier. And both have/will produce the same null result for the same reason you mentioned earlier (even if the gravitational mass of one object is <> the other). The Eot-Wash experiment could work if the controlled composition mass were the attractor instead of the pendulum. And the satellite experiment can never work because we cannot control the composition of the earth. Can anyone tell me if I am missing something here?

 

[Saw]

Before making any further progress… I’ve realized that in post #73 I wrote a formula for free-fall time that is an over-simplification. I think the underlying idea posted there is still valid, but anyhow the good formula, as written in Wikipedia (http://en.wikipedia.org/wiki/Free-fall_time), which assumes mass m is negligible, would be:

$$t_{orbit} = \frac{{2\pi }} {{\sqrt {GM} }}\left( {\frac{R} {2}} \right)^{3/2} = \frac{{\pi R^{3/2} }} {{\sqrt {2GM} }}$$

$$t_{ff} = t_{orbit} /2 = \frac{{\pi R^{3/2} }} {{2\sqrt {2GM} }}$$

 

[TurtleMeister]

I recently noticed an interesting link between the modified universal law of gravitation equation and Newton's second law.

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

The equation above can also be expressed as:

$$F = \mu a$$

where,

$$\mu = \frac{m^{i}_1m^{i}_2}{m^{i}_1+m^{i}_2} = \text{reduced mass}$$

$$a = \frac{G(m^{g}_1+m^{g}_2)}{r^2} = \text{relative acceleration}$$

 

[Saw]

That’s a good point, bringing in the concept of reduced mass! It converges with the progress I was making in the domain of collisions (see the thread “Acceleration in an elastic collision”), where the conclusion seems to be that the force intervening in the collision can also be defined as:

$$F = \mu (1 + \varepsilon )v_{rel}^{initial}$$

where:

$$\begin{gathered} \mu = \frac{{m_1 m_2 }} {{m_1 + m_2 }} = {\text{reduced mass}} \hfill \\ \varepsilon = \frac{{v_{rel}^{final} }} {{v_{rel}^{initial} }} = {\text{coefficient of restitution}} \hfill \\ a = (1 + \varepsilon )v_{rel}^{initial} = {\text{relative acceleration}} \hfill \\ \end{gathered}$$

On the other hand, if you want to consider the possibility that gravitational masses (included in the relative acceleration term) differ from inertial masses (those included in the reduced mass term), then you have to set an upper limit to the former.

In the collision formula, that limit, derived from the conservation of kinetic energy, is that relative acceleration cannot exceed 2 times relative velocity, since the coefficient of restitution can range from 0 to 1. Certainly, if you know that coefficient based on the ratio final/initial relative velocity, the formula is not very useful. But the idea is that you could also obtain it empirically as a coefficient inherent to the material that each body is composed of. In particular, one would have to use, together with “reduced or effective mass”, the “effective coefficient” of the two materials, as if they were one = k₁k₂/(k₁+k₂). I still haven’t thought, however, how to translate this to the gravity formula.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

Oops. I correct again: where I said momentum, I should have said change of velocity and hence of momentum in a given time (collision time) = impulse, isn't it?

 

[Saw]

I have been thinking that the concept of “gravitational mass” may be somehow misleading. Really the underlying issue is: does the composition of the bodies involved in a gravitational interaction have any bearing in the outcome of a gravitational interaction? And, well, there is no need to mess that up with the concept of mass. One could perfectly let the sum of the masses of the “relative acceleration” term be divided out by the sum of the masses of the “reduced mass” term and still ask that question.

Of course, then the issue is, how could you make that factor (composition of the bodies) relevant, if (speculating about the idea) you wished to? In the context of collisions (what we could call contact force), such factor is no doubt relevant and it shines up mathematically in the coefficient of restitution, which is a dimensionless number. In the context of springs (elastic force), it appears as the spring constant, which is measured in N/m. In the context of gravity, we have the constant G, measured in Nm^2/kg^2. Thus… if we considered the possibility that G were multiplied by a certain (dimensionless) coefficient, would that be simply illogical, would it clash against physics principles?

It seems it would have to be a “reductive coefficient”, ranging between 0 and 1, like the coefficient of restitution (let’s call it “e”). Thus we could have less but not more relative acceleration than in the standard formula. On top of that, it would be empirically constructed, like I gather that it happens with “e”, on the basis of the properties (composition) of the two involved bodies. In fact, it would be “e”.

For example, if we have two bodies “dropped” from a relative distance R and acquiring a certain relative velocity at collision time and if the collision is partially elastic and partially plastic, the bodies would bounce off each other with outgoing relative velocity = (1+e) times their incoming relative velocity… Then they separate until running out of outgoing kinetic energy due to the gravitational force between them. Since that outgoing KE is now lower (part of it has been dedicated and is kept busy heating or deforming the bodies), gravity will stop them and revert their direction sooner, before they reach the initial distance R. Thus they’ll return with lower velocity, which will diminish even more after collision. And so on until the bodies are brought to rest with each other.

But does all that mean that the gravitational interaction is fully “elastic” (it only transfers to the bodies energy that is useful for motion, that is either kinetic or potentially convertible into kinetic energy)? In principle, I do not se why it should be necessarily so. Note this: when the bodies are separating from each other, they are being stretched, due to the non-uniformity of the gravitational field. That is the well-known phenomenon of tides. And when they descend or get closer to each other, I think that they progressively lose that condition, because (as the force increases due to the inverse square law) the difference of strength of the field between the two extremes becomes narrower. Given this, should the bodies perfectly recover their original form? Why should they? They will not when they are later compressed due to the collision. Should the gravitational stretching be different? Shouldn’t it also generate, due to the composition of the bodies, some conversion of the energy transferred by gravity into other non-mechanical forms? I wonder if there is any logical, not purely experimental reason banning that.

 

[boit]

If I lived in a planet where more massive objects were falling much slower, I will create a perpetual motion energy generator in no time. How? Since by implication more massive object will have less potential energy, I'll rig the contraption to haul stones in a single bag and then throw them down sepatately. On Earth surely 1/2(1+1+1)10 m^2/s^2 can not be less than 1/2(3)10 m^2/s^2 [from E=1/2mv^2]. This is because whatever you do the centre of mass will be a point source. The trick could be to make your more massive object so tall that the COG is in a zone where accelaration is say half 'g'. But unfortunately you can't make them rigid enough.