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Why don't heavy objects fall more SLOWLY?

danielatha4, the drawing is very illustrating, although we were now discussing a further issue:

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

D H pointed out that certain experiments prove that such thing would lead to nonsensical results. If I interpret well, portion A having passive but no active gravitational mass would be attracted to portion B without attracting B in turn; that would mean that A is accelerated towards B and collides with it; thus the body would self-accelerate away from its centre of mass, in the direction of B, breaking Newton’s Laws of Dynamics.

TurtleMeister commented that such nonsensical result could be avoided if the bodies were gravitationally accelerated according to a Law of Gravitation that’d be slightly different from Newton’s: so as to meet at the centre of mass of the system, but at a rate which would depend on the ratio between the total of gravitational masses and the total of inertial masses.

I said: but why is it so important that the bodies meet at the inertial centre of mass? And proposed a thought experiment where it might not be so.

I was wrong due to faulty understanding of Newton’s Laws and TurtleMeister has just taken the trouble to explain why. My thought experiment does not prove what I thought but the opposite: between two balls affected by equal attracting forces there are two massless springs of equal lengths but different spring constants; in spite of that, the balls tend to meet at their centre of mass.

I attach an image with a try at a numerical example and the outcome of the experiment, for correction. (For simplicity, and lack of a better software, I drew the springs as rectangles.)
Attached Thumbnails

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 Blog Entries: 8 Recognitions: Gold Member 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...
 Recognitions: Gold Member 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.

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 Quote by TurtleMeister 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 (rm)? Actually, rm 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.

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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.

 Quote by Saw - 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.

 Recognitions: Gold Member 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.
 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?

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 Quote by 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?
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?

 "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?

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 Quote by Saw 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 mg 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 Ma to Mi 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.

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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:

 Quote by TurtleMeister 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.
 Quote by TurtleMeister 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 mi, 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 mi).
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)?

 Blog Entries: 8 Recognitions: Gold Member 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...

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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.

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

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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.

 Quote by TurtleMeister 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 (mg ≠ Mg) of two equal inertial masses (mi=Mi), 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...

 Quote by TurtleMeister 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.

 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.