A Flaw In Using Fall?

[md65536]

I have put my thoughts in separate sentences to allow delineation of where the fault arises.

[...]

From outside perspective; Mouse travels a greater distance in a greater time (cos the earth isnt rushing up to greet it ) and thus has a higher speed on impact

[...]

The elephant hits first if viewed from an outside perspective - therefore it must hit first from all perspectives.

Sounds right, I see no fault. It's easier if you consider two arbitrary masses.

 

Yes, from the outside perspective the mouse has a higher speed on impact but only relative to the outside observer. The closing speed of the two masses is greater for the elephant.

 

By analogy, imagine accelerating a car into a parked truck, vs. letting the truck accelerate toward the car from the same spot, starting at the same time. The speed of the car relative to the ground will be greater if the truck is parked, but the relative impact speed will be greater if the truck also accelerates.

 

 

 

Also... we're talking "point mouse" and "point elephant" here, right? Earth's gravitational gradient over the height of an elephant would surely be more significant than the gravitational pull of the elephant.

 

Also also... just to avoid confusion this is the case where the elephant and the mouse are dropped separately, not simultaneously as swansont and xyzt are discussing...

Edited June 12, 2014 by md65536

[swansont]

Correct. See the rigorous analysis in this post.

Now, if one drops both the elephant and the mouse simultaneously, contrary to Swansont's claim , they do not hit the Earth simultaneously.

 

 

Which one hits first and why?

 

Looking at the above, it becomes immediately apparent that it is impossible to have [math]x_1-x_2=0[/math] because it would result into:

 

[math]0=-\frac{G(m_2-m_1)}{(r_1+r_2)^2}[/math]

 

So, contrary to your claim, the test probes hit the ground simultaneously if and only if [math]m_1=m_2[/math], a trivial conclusion.

While it has become clear that it is difficult, if not impossible to find the equations of motion [math]X(t),x_1(t),x_2(t)[/math] we can safely conclude that the two probes hit the ground simultaneously if and only if they are identical, they must not only have the same mass (see above) but the also must have the same radius (otherwise, the one with larger radius touches down first).

 

That goes away if you separate the terms into a radial and a tangential component. That tangential term is irrelevant, as it is orthogonal to the radial term. It can't cause a radial acceleration.

 

If the two masses are to accelerate at different rates, there must be some interaction between them radially, which can only exist if they have a different r, but that's eliminated by the boundary conditions. Because the acceleration imparted by the earth has to be the same.

[xyzt]

 

 

Which one hits first and why?

I simply pointed out the error in your statement. Care to acknowledge it? Or do you still maintain that they hit simultaneously?

Edited June 12, 2014 by xyzt

[swansont]

I simply pointed out the error in your statement. Care to acknowledge it?

 

I don't think I've made an error. When you apply all of the boundary conditions to the problem, the solution for the two test masses are going to be identical. How can you get a different answer for one?

 

And to the other note: yes, I had already assumed they are identical other than their mass, and essentially co-located

 

edit: co-located stated here

[Delta1212]

 

I don't think I've made an error. When you apply all of the boundary conditions to the problem, the solution for the two test masses are going to be identical. How can you get a different answer for one?

 

And to the other note: yes, I had already assumed they are identical other than their mass, and essentially co-located

 

edit: co-located stated here

I guess my question would be: if we're already dealing with an effect that is so small that negligible would be a generous description, is it really reasonable to treat the two as being co-located. Obviously there's not much difference between being in the same spot and being a few feet apart, but "not much difference" applies to most parts of the problem.

 

I understand and agree they'd both hit at the same time if co-located. I'm just wondering if that's a reasonable simplification to make in this case?

[xyzt]

 

I don't think I've made an error. When you apply all of the boundary conditions to the problem, the solution for the two test masses are going to be identical. How can you get a different answer for one?

The math says that the equation does not admit a zero solution. If you think otherwise, I would like to see your mathematical proof. Restating the same claim over and over is not a valid proof.

 

 

That goes away if you separate the terms into a radial and a tangential component. That tangential term is irrelevant, as it is orthogonal to the radial term. It can't cause a radial acceleration.

 

If the two masses are to accelerate at different rates, there must be some interaction between them radially, which can only exist if they have a different r, but that's eliminated by the boundary conditions. Because the acceleration imparted by the earth has to be the same.

There is no "tangential" component. In the equation [math]r_1,r_2[/math] are constants, the radii of the two test probes. The two test probes move along one axis (I chose "x" to represent their axis of motion).

Edited June 12, 2014 by xyzt

[swansont]

The term that has x₁ - x₂ goes to zero. It has no radial component, and can be ignored. The mutual attraction of the test masses can only change the speed at which they fall to earth of the radial components are unequal, which they are not. A shortcoming of your equations is that they are set up as scalars, and are ignoring that we're dealing with vectors. The tangential acceleration is irrelevant, and x₁ - x₂ is tangential. The term should not be in the system of equations that we are solving.

[md65536]

I understand and agree they'd both hit at the same time if co-located. I'm just wondering if that's a reasonable simplification to make in this case?

It's not just a simplification but a specification. It depends on what you're asking about. If you only want to know about the gravitational acceleration then it's reasonable. If you want to know how a system will behave in general then you need to use the more complicated setup.

 

However, more complicated won't necessarily answer the question you're asking. If you have a general 3-body system, then you will have lateral acceleration, which means you have to include inertial pseudo forces, right? So xyzt's complicated setup is still not complicated enough. For example, if we have 3 point masses, Tiny, Large, and Huge, with Tiny and Large very near so that tiny accelerates mainly toward Large (instead of to Huge), then... doesn't the acceleration direction change over time due to the changing location of the other body, so Tiny would end up orbiting Large? And then couldn't you adjust things, eg. place Huge closer or farther away so that the time of minimal distance between the bodies (ie. "impact") can happen with Tiny at a different point of its orbit? In other words, in general the question isn't answerable and depends on the specifics. So the complicated answer is not useful unless you're considering a specific case, and you set up the specific case by choice to avoid such complications.

 

I might be wrong.

[xyzt]

The term that has x₁ - x₂ goes to zero.

You keep saying that. Saying does not constitute a valid proof. Actually, the equation of motion clearly does not admit [math]x_1(t)-x_2(t)=0[/math] as a solution as I have proven it mathematically.

The tangential acceleration is irrelevant, and x₁ - x₂ is tangential. The term should not be in the system of equations that we are solving.

I think that you misunderstand, [math]x[/math] is the radial coordinate, not the tangential. Once again, I labelled the axis of motion as "x".

 

If it makes it easier, here is the problem statement with the axis of radial motion chosen to be "z":

 

 

 

[math]\frac{d^2(z_1-z_2)}{dt^2}=-(\frac{GM}{(Z-z_1)^2}-\frac{GM}{(Z-z_2)^2}+\frac{G(m_2-m_1)}{(z_1-z_2)^2+(r_1+r_2)^2})[/math]

 

Looking at the above, it becomes immediately apparent that it is impossible to have [math]z_1-z_2=0[/math] because it would result into:

 

[math]0=-\frac{G(m_2-m_1)}{(r_1+r_2)^2}[/math]

Edited June 12, 2014 by xyzt

[swansont]

You keep saying that. Saying does not constitute a valid proof. Actually, the equation of motion clearly does not admit [math]x_1(t)-x_2(t)=0[/math] as a solution as I have proven it mathematically.

I think that you misunderstand, [math]x[/math] is the radial coordinate, not the tangential. Once again, I labelled the axis of motion as "x".

 

If it makes it easier, here is the problem statement with the axis of radial motion chosen to be "z":

 

 

 

[math]\frac{d^2(z_1-z_2)}{dt^2}=-(\frac{GM}{(Z-z_1)^2}-\frac{GM}{(Z-z_2)^2}+\frac{G(m_2-m_1)}{(z_1-z_2)^2+(r_1+r_2)^2})[/math]

 

Looking at the above, it becomes immediately apparent that it is impossible to have [math]z_1-z_2=0[/math] because it would result into:

 

[math]0=-\frac{G(m_2-m_1)}{(r_1+r_2)^2}[/math]

 

 

Why does that acceleration term depend on r₁ and r₂?

[Mordred]

interesting discussion, I happen to have an article that describes the variations for everyone's review

Falling Bodies: the Obvious, the Subtle, and the Wrong

http://arxiv.org/ftp/physics/papers/0702/0702155.pdf

 

statement with supportive mathematics included

 

"So, it shouldn't be too perplexing if we find that all three possibilities: heavy faster than light, heavy and light equally fast, and even light
bodies faster than heavy bodies can all occur in the falling body problem."

 

copy and paste from the conclusion section of the article

 

I recall once having other related articles in my archives, I'll see if I can locate them

Edited June 12, 2014 by Mordred

[xyzt]

 

 

Why does that acceleration term depend on r₁ and r₂?

Earlier in the derivation I showed that the attraction force between the two test bodies is [math]\frac{Gm_1m_2}{(z_1-z_2)^2+(r_1+r_2)^2}[/math] since the squared distance between the test bodies is [math](z_1-z_2)^2+(r_1+r_2)^2[/math]. Now, according to you [math]z_1-z_2=0[/math], so the rest follows.

interesting discussion, I happen to have an article that describes the variations for everyone's review

Falling Bodies: the Obvious, the Subtle, and the Wrong

http://arxiv.org/ftp/physics/papers/0702/0702155.pdf

 

The above reference falls in the "not even wrong, totally crackpot category". Beware when engineers masquerade as physicists, extreme crankiness , in the vein of "violations of Einstein's Equivalence Principle" emerges. The author managed the unenviable feat of getting ALL his equations (and conclusions) wrong. To wit:

 

1. There is no violation of EP

2. Bodies with larger mass still hit the Earth faster than bodies with lighter mass (The explanation is simple, the larger mass forces the Earth to close the distance to the falling body).

Edited June 12, 2014 by xyzt

[Mordred]

how is this paper in disagreement with the EP? when it specifically states

 

VII. EQUIVALENCE PRINCIPLE (EP)
"In the previous 3-body analyses, it appears as if the principle of the equivalence of inertial and gravitational mass in Einstein's General Relativity is being violated
because the lightest mass has the highest acceleration.As shown this is not a violation of the EP since the inertial and gravitational masses are equal. The lightest mass m_l goes the fastest in the center of mass 3-body system because it is acted by both M and m2 ,where as m₂ is acted on by both M and m_l "

 

"2. Bodies with larger mass still hit the Earth faster than bodies with lighter mass (The explanation is simple, the larger mass forces the Earth to close the distance to the falling body)."

 

huh, how does this even make sense, ? if two objects are falling to Earth at the same time their combined masses would influence the motion of Earth in other words their combined mass.

 

secondly does it make sense that the Earth would move faster to the heavier body or does it not make more sense that the lighter body would move faster to the larger mass? a per the 2 body problem

Edited June 13, 2014 by Mordred

[xyzt]

how is this paper in disagreement with the EP? when it specifically states

 

VII. EQUIVALENCE PRINCIPLE (EP)

"In the previous 3-body analyses, it appears as if the principle of the equivalence of inertial and gravitational mass in Einstein's General Relativity is being violated

because the lightest mass has the highest acceleration.As shown this is not a violation of the EP since the inertial and gravitational masses are equal. The lightest mass m_l goes the fastest in the center of mass 3-body system because it is acted by both M and m2 ,where as m₂ is acted on by both M and m_l "

 

Errr, looky here:

 

PHYSICS OVERVIEW

Despite the outstanding success of Einstein's General Relativity (EGR), there can be

exceptions to the Equivalence Principle (EP) that is the very cornerstone upon which EGR is

based. Due to increasing acceleration because of increased gravitational force as bodies fall

toward each other, gravitational radiation with a concomitant retarding gravitational radiation

reaction force can be expected. The EP is violated because the gravitational acceleration

is independent of the accelerated mass, whereas the gravitational radiation reaction acceleration

depends on the accelerated mass. This violation of the EP may only be perceptible for very

high gravitational fields. There may also be quantum mechanical(QM) deviations from the

EP. Particularly so since QM violates the EP.

 

"2. Bodies with larger mass still hit the Earth faster than bodies with lighter mass (The explanation is simple, the larger mass forces the Earth to close the distance to the falling body)."

 

huh, how does this even make sense, ? if two objects are falling to Earth at the same time their combined masses would influence the motion of Earth in other words their combined mass.

 

secondly does it make sense that the Earth would move faster to the heavier body or does it not make more sense that the lighter body would move faster to the larger mass? a per the 2 body problem

 

But, the crankish author says the opposite, that the lighter body falls faster. BTW, you are wrong as well, if two objects are dropped onto the Earth simultaneously, the math says that they DON'T hit the Earth simultaneously IF they have different masses or different radii.

 

One last thing, you glossed over the fact that not ONE of the author's calculations is correct, he managed to make mistakes even in the most elementary calculations. No wonder no physics journal accepted his "masterpiece". So, he published in IEEE.

Edited June 13, 2014 by xyzt

[Mordred]

yes the paper I posted mentions that. in the 3 body section.

 

however he did reference that statement

 

A Theory of Quantum Gravity may not be possible because Quantum Mechanics violates the Equivalence Principle

http://arxiv.org/abs/physics/0601218

 

 

Deterrents to a Theory of Quantum Gravity

http://arxiv.org/abs/physics/0608193

 

by the way here is numerous papers written by the author the above is also his papers.

 

http://arxiv.org/find/all/1/all:+AND+Mario+Rabinowitz/0/1/0/all/0/1

 

http://www.amazon.com/Mario-Rabinowitz/e/B005S81GJS

Mario Rabinowitz has a Ph.D. in physics, with over 170 publications on a wide variety of subjects, including articles in the McGraw-Hill Encyclopedia of Science and Technology as well as a number of its Yearbooks. Google Books lists about 55 results for his book contributions, such as book chapters and published reports. He has also written invited book reviews. He has been an Adjunct Professor at five universities. He received the 1992 Washington State University Alumni Achievement Award.

[xyzt]

yes the paper I posted mentions that. in the 3 body section.

 

however he did reference that statement

 

A Theory of Quantum Gravity may not be possible because Quantum Mechanics violates the Equivalence Principle

http://arxiv.org/abs/physics/0601218

 

 

Deterrents to a Theory of Quantum Gravity

http://arxiv.org/abs/physics/0608193

 

by the way here is numerous papers written by the author the above is also his papers.

 

http://arxiv.org/find/all/1/all:+AND+Mario+Rabinowitz/0/1/0/all/0/1

 

http://www.amazon.com/Mario-Rabinowitz/e/B005S81GJS

Mario Rabinowitz has a Ph.D. in physics, with over 170 publications on a wide variety of subjects, including articles in the McGraw-Hill Encyclopedia of Science and Technology as well as a number of its Yearbooks. Google Books lists about 55 results for his book contributions, such as book chapters and published reports. He has also written invited book reviews. He has been an Adjunct Professor at five universities. He received the 1992 Washington State University Alumni Achievement Award.

Yes, when it comes to relativity and the theory of gravitation he's a total crank.

[Mordred]

oh really and what are your qualifications to assert that?

 

 

2. Bodies with larger mass still hit the Earth faster than bodies with lighter mass (The explanation is simple, the larger mass forces the Earth to close the distance to the falling body).

 

tell you what find me a peer reviewed paper that supports this claim in regards to two bodies of different mass falling to a larger gravitational mass body at the same time.

Edited June 13, 2014 by Mordred

[xyzt]

oh really and what are your qualifications to assert that?

 

tell you what find me a peer reviewed paper that supports this claim in regards to two bodies of different mass falling to a larger gravitational mass body at the same time.

You got it backwards, for the third time, the body of larger mass hits FIRST, they do NOT hit "at the same time". This is basic physics that neither you nor the author of the paper you linked seem to know. I explained the derivation in an early post in this thread.

Edited June 13, 2014 by xyzt

[Mordred]

were talking 3 bodies not 2 bodies, I know what the 2 body problem states.

 

2 bodies falling towards Earth according to the principle of equivalence of acceleration will fall to Earth at the SAME rate. When you do the calculations based on two bodies simultaneously dropping towards a center of mass such as the Earth. The Earth will react to their combined mass.

 

now if you drop them separately and measure the time of fall then your scenario is correct.

Edited June 13, 2014 by Mordred

[xyzt]

2 bodies falling towards Earth according to the principle of equivalence of acceleration will fall to Earth at the SAME rate. When you do the calculations based on two bodies simultaneously dropping towards a center of mass such as the Earth. The Earth will react to their combined mass.

Several mistakes:

 

1. The bodies falling towards the Earth have nothing to do with EP.

2. While the two bodies have acceleration independent of their mass (therefore IDENTICAL) when dropped INDIVIDUALLY, correct application of elementary mechanics teaches us that the two bodies no longer have identical accelerations when dropped together, simultaneously. This is due to the fact that the bodies interact with each other in ADDITION to interacting with the Earth.

3. In fact, the correct application of math shows (see my posts to swansont) that the two bodies will NOT hit the Earth at the same time UNLESS they have SAME mass (and same radius).

4. It is true that "the Earth will react to the combined mass" of the two dropped bodies, this is also demonstrated rigorously in my posts. Yet, the "Earth reaction" is not purely additive, i.e. we cannot simply add [math]m_1+m_2[/math] (as Rabinowitz does in his "paper"), the dependency is much more complicated. So, this is also a mistake in your claims. You cannot simply make claims, you must be able to back them up with math and you have done none of this.

 

[math]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/math] describes the rather complex reaction of the Earth (denoted by its acceleration [math]\frac{d^2X}{dt^2}[/math]) and the two masses [math]m_1,m_2[/math]. As you can see, the masses do NOT add.

 

 

 

now if you drop them separately and measure the time of fall then your scenario is correct.

 

My treatments of the two cases are both correct. Rabinowitz , on the other hand, managed to bungle both of them.

Edited June 13, 2014 by xyzt

[Mordred]

Quite frankly I'm going to pay more attention to what the textbooks teach, and what someone as distinquished as Mario Rabinowitz is over your opinion

 

no offense, but quite frankly you started insulting his paper just from the letterhead initially. In case you didn't notice he stated in the paper that all 3 views in a certain sense is correct. Depending on the treatment.

 

In case you haven't figured it out I don't need the mathematics as I'm not claiming either of the 3 views are more correct than the other. I'm not the one making claims you are. I merely posted the paper showing that depending on its treatment all 3 views can be accurate

[md65536]

2. While the two bodies have acceleration independent of their mass (therefore IDENTICAL) when dropped INDIVIDUALLY, correct application of elementary mechanics teaches us that the two bodies no longer have identical accelerations when dropped together, simultaneously. This is due to the fact that the bodies interact with each other in ADDITION to interacting with the Earth.

I realize you don't need to read the paper to know that everyone else is wrong, but try anyway:

 

VI. SIMULTANEOUS FREE FALL OF COINCIDENT BODIES

When the concurrently falling bodies were separated, the lighter mass fell faster than the

heavier mass relative to the CM, and the same rate relative to the earth. For our next thought

experiment, we want to see what happens when we drop m2 and m1 from the same point

simultaneously. This 0o case has to be done carefully to avoid the criticism, "of course they fell

at the same rate since they were stuck together gravitationally and acted as one body." Let m2

be a large hollow transparent sphere with ml a tiny sphere inside it at its center. (The inner and

outer spheres have no net gravitational attraction between them for all points inside the outer

sphere.) The center of mass of the system is in line with m2 and ml, and equally distant from

both, as is M. Therefore they will fall at the same rate with respect to both the earth and the

center of mass of the system. (m2 and ml could be reversed for further verification.)

http://arxiv.org/ftp/physics/papers/0702/0702155.pdf

 

 

Your assertion that they must interact gravitationally is incorrect.

 

 

Edit: Fixed link? (Yes, Mordred it's the Rabinowitz paper, I'd copied the link from your post too, haha)

Edited June 13, 2014 by md65536

[Mordred]

link isn't working for me can you repost it? or is that the Mario Rabinowitz article?

 

yeah works now

Edited June 13, 2014 by Mordred

[xyzt]

I realize you don't need to read the paper to know that everyone else is wrong, but try anyway:

 

http://arxiv.org/ftp/physics/papers/0702/0702155.pdf

 

 

Your assertion that they must interact gravitationally is incorrect.

I see, you think that there is no attractive force between the two bodies. So, Newton works only between the Earth and each gravitational body. Why is that?

Edited June 13, 2014 by xyzt

[md65536]

I see, you think that there is no attractive force between the two bodies. So, Newton works only between the Earth and each gravitational body. Why is that?

No net attractive force. It's called the shell theorem. http://en.wikipedia.org/wiki/Shell_theorem

 

It's true that you can create a different setup that gives a different answer than other setups. You can also set it up to avoid those extra complications, and have two masses centered at the same place without a net attractive force between them. As mentioned, the paper explains this. If you have more questions, have a go at trying to read the paper.