Why don't heavy objects fall more SLOWLY?

TurtleMeister

You are correct Saw. The equation is Newton's law of gravitation with a correction for the ratio between active gravitational and passive/inertial mass. In fact, the arrangement that you show for F is my original arrangement of the equation. My rearrangement for the final version was made to eliminate the duplication of the two mⁱ variables.
There is no provision for m1^g and m2^g in the Newtonian part of the equation. The normal method is to split into two different forces, one for A₁ and another for A₂. But that would lead us right back to the 3rd law violation (when m_a <> m_i). So in effect, what I have done is combine the two equations into one by leaving out m1^g and m2^g for the Newtonian part. This could very well be a flaw for the equation when m_a <> m_i. But I'm not sure. For example, setting a value for m1ⁱ and m2ⁱ and then adjusting the values of m1^g and m2^g for the desired ratio, seems to give the correct results.

Saw

Why do you need to adjust the values of m1^g and m2^g for the desired ratio, in order to get the correct results? If by "correct results" you mean (i) to comply with your requirement that the two masses must meet at their inertial centre of mass (CM) and (ii) that there's no violation of Newton's 3rd Law (the two forces are equal and of opposite sign), it seems you satisfy those conditions with this equation.

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

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

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

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

In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?

D H

These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.

Saw

Sorry, DH, but I respectfully disagree. I have little physics, but I am a good lawyer. Nobody is posting personal theories here. We assume and accept mainstream physics and we are just trying to grasp the logic of its principles by noticing how alternative routes lead to a dead-end. That's called reductio ad absurdum and for the benefit of the standard approach. If in the course of this discussion, I made incorrect statements, I would most welcome correction. In particular, I really appreciate your interventions, each of which contains a little jewel of information...

D H

Those equations are a personal theory.

TurtleMeister

Thanks for the warning. I will be more careful with my posts. Also, please feel free to add to the discussion or correct my errors, as you have done in the past. Your input is greatly appreciated.

If I'm understanding this thought experiment correctly, the answer is (i). You partially answered it yourself in the first highlighted sentence that I quoted from you. The springs only supply the force, they do not determine the meeting point (considering the springs are massless and have no substance). This is true even when using multiple springs with different elasticities. The meeting point is the CM, which is determined by the inertial mass of the two objects. The simple equation for this is:

[tex]m_1d_1 = m_2d_2[/tex]

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

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

Sorry, I do not understand that.

Big O

Why all the fuss.

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

TurtleMeister

That's not what we're talking about. We have gotten a little off topic to the OP, but still related.
Ok, I think I understand this now. You are saying, by analogy, that the spring can have the same effect as changing the inertial mass of the object it's connected to? No, that is not true. As you have already stated yourself, the springs (source of the force) only has an effect on when they meet, not where. Even though you have two springs with different elasticities connected together, the force applied to each object will be equal in magnitude and opposite in polarity. Where they meet is determined only by their inertial mass.

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

And that is the reason the following method of applying Newton's laws fails when m1^g <> m1^p or m2^g <> m2^p

[tex]F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}[/tex]

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

[tex]A_1 = \frac{F_1}{m^\text{i}_1}[/tex]

[tex] A_2 = \frac{F_2}{m^\text{i}_2}[/tex]

Just like in your springs thought experiment, where you have two springs with different elasticities connected together, F_{1 <> F}2, when m1^g <> m1^p or m2^g <> m2^p. This causes a problem if we apply F₁ to the acceleration of m1, and F₂ to the acceleration of m2. If we do that then it would be analogous to your thought experiment where the spring affects the point where they meet. The third law will be violated.

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

[tex]F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}[/tex]

[tex]F_2 = \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}[/tex]

[tex]F = \frac{F_1+F_2}{2}[/tex]

[tex]A_1 = \frac{F}{m^\text{i}_1}[/tex]

[tex] A_2 = -\frac{F}{m^\text{i}_2}[/tex]

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

edit: The averaging of the forces has the effect of restoring the frame of reference to the COM.

D H

What is this F=F₁+F₂ stuff? What justifies that step, and then using that force as the force on each of the two objects? That the result is consistent with Newton's third law is not surprising: You implicitly assumed Newton's third law here.

TurtleMeister

It's just a demonstration to help us understand the problem. Yes, I am assuming Newton's third law.

D H

Why didn't you use the obvious [tex]a_1 = F_2/m_1[/tex] and [tex]a_2 = F_1/m_2[/tex] here? Why that force sum? You did not justify it, and it most certainly does not jibe with reality. If the equivalence principle is true, this summation makes the acceleration be off by a factor of two.

TurtleMeister

First, I noticed an error in my equations in post #59. I had m1 and m2 reversed for Newton's universal law of gravitation. I have fixed it now, so you may want to change the [tex]a_1=F_2/m_1[/tex] and [tex]a_2=F_1/m_2[/tex] in your post #62 to [tex]a_1=F_1/m_1[/tex] and [tex]a_2=F_2/m_2[/tex]. I don't think this affects the questions you are asking.

Referring to the first four equations in post #59, and assuming that m1^p = m2^p, do you agree that:

[tex]A_1 = A_2[/tex] when [tex]m^\text{g}_1 = m^\text{p}_1[/tex] and [tex]m^\text{g}_2 = m^\text{p}_2[/tex]

[tex]A_1 <> A_2[/tex] when [tex]m^\text{g}_1 <> m^\text{p}_1[/tex] or [tex]m^\text{g}_2 <> m^\text{p}_2[/tex]

Referring to the second set of five equations in post #59, and assuming that m1^p = m2^p, do you agree that:

[tex]A_1 = A_2[/tex] when [tex]m^\text{g}_1 = m^\text{p}_1[/tex] and [tex]m^\text{g}_2 = m^\text{p}_2[/tex]

[tex]A_1 = A_2[/tex] when [tex]m^\text{g}_1 <> m^\text{p}_1[/tex] or [tex]m^\text{g}_2 <> m^\text{p}_2[/tex]

If you do not agree with any of the questions above, then please explain where you think I am wrong. Note: edited to correct errors in above questions.

If you do agree with all of the questions, then you should understand the reason I posted them. The first four equations is the normal way of showing the accelerations of A₁ and A₂ with a variable m^g. It shows that this method causes a violation of the third law of motion (self accelerating moon) when m1^g <> m1^p or m2^g <> m2^p. The second set of five equations show that if the frame of reference is restored to the COM [tex]F = \frac{F_1+F_2}{2}[/tex] then the third law of motion is not violated.

I do not know how you arrived at that conclusion. If the equivalence principle is true then F = F1 = F2.

D H

What motivates that force averaging? It is far to ad hoc.

TurtleMeister

It's just a demonstration to show that the cause of the third law violation is the frame of reference not being at the COM. The averaging puts the frame of reference at the COM and eliminates the third law violation.

Also, for you information. I fixed some errors in the questions in post #63. So you may want to review it again.

Saw

Turtle, I am ready to give up with any arguing based on my thought experiment about the springs. But I would need some further explanation.

When a force is applied to a spring, such force is defined as:

[tex]F = k\Delta l[/tex]

k being the spring's constant of rigidity and ∆l being its stretching or, in this case, compression

Another way to put it is that the streching / compression is directly proportional to the force and inversely proportional to k

[tex]\Delta l = \frac{F}{k}[/tex]

Here ∆l seems to play the part of acceleration and it appears that k plays the part of m.

Hence in the experiment where two springs are compressed against each other, I expected that Newton's Third Law would be fulfilled as follows:

[tex]k'\Delta l' = - k\Delta l[/tex]

What did I miss?

TurtleMeister

To make sure I'm not misunderstanding you, I've made a little graphic of my interpretation of your thought experiment.



Let's say that A and B have the same inertial mass and are attracted to each other, either gravitationally or magnetically. There are no outside forces involved. Also, we will ignore the mass of the springs. My understanding of your position is that because the spring on the left has a higher k than the spring on the right that the two masses will meet at a point closer to "point1" on the left side of the COM.

Newton's third law of motion states that for every action there is an equal and opposite reaction. So if A moves to the right 1cm, then to satisfy this law, B must move to the left 1cm. Since they both move equally in distance and opposite in direction then they must meet at the COM point. The springs have nothing to do with where they meet. The 1k spring will compress more than the 2k spring and the point where they are attached will move to the right.

Does this help to clear things up, or did I misinterpret your thought experiment?

danielatha4

I didn't read all the replies to this thread but this picture always helps me understand the difference between acceleration due to gravity and force due to gravity.



now just break F1 up into M1 * A1 and F2 = M2 * A2