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You can't have same size if one is INSIDE the other one as you have set it.

No, you have just added a bogus case, where the two test probes are GUARANTEED not to hit at the same time. By design.

Rabinowitz does not even set the equations of motions correctly, let alone solve them. I can see that now, that I pointed out your errors, you changed your tune. So, you accept my correction. Yes. As I pointed out repeatedly, "same acceleration" doesn't mean "same time to collision". You NEED to form the equations of motion and you need to SOLVE them. This gives you the correct answer. You keep repeating that but, the only way to have this is to have one tst probe INSIDE the other one. when you do that, they are GUARANTEED NOT TO HIT the Earth at the same time, a totally uninteresting scenario.

This is if you make your assumption that [math]z_1-z_2=0[/math]. I demonstrated your error in your assumption by reduction to absurd, so , it simply proves my point.

[math]t_i=\frac{D^{3/2}}{\sqrt{2G(M+m_i)}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math] tells you that the time to collision is always dependent on the mass (and the radius) of the probe, so the statement is always correct. Whether the influence is measurable , that is a function of the sensitivity of measuring devices. The formula IS Newton's law. No "discrepancy". There is no "dependence on the size of the object", [math](z_1-z_2)^2+(r_1+r_2)^2[/math] is the distance between the centers of the test probes, this is the second time you fail to recognize it. Two touching spheres of radii [math]r_1,r_2[/math] will attract each other with the force [math]\frac{Gm_1m_2}{(r_1+r_2)^2}[/math],.

Yes, the two objects have the same acceleration. This doesn't mean they hit the Earth in the same time. The larger object hits first because it forces the Earth to close the distance between them at a faster rate. This is the third time I am explaining this to you.

A very good reference. Unfortunately, it doesn't support your views. You can't do physics by quite mining, you need to be able to calculate for yourself, something that you have not done. Nope, I didn't. The equations of motion I posted , look at the LHS, see that [math]\frac{d^2 z}{dt^2}[/math]? What do you think it represents? Good, you realized that. I was very clear about the setup. Now, you introduced a DIFFERENT setup, with the smaller test probe INSIDE the larger test probe. It is obvious i this case that the larger test probe hits first, so, contrary to your claim, they don't hit at the same time.

Well, the language of physics is math, since, by your own admission , you cannot do any math (and you haven't posted any) you have no valid argument, physics is not a collection of assertions. Even if you repeat them, it doesn't make them right.

The EP says the bodies have the same acceleration, INDEPENDENT of the body mass. What I am saying is that basic physics teaches you that the two objects hit the Earth at DIFFERENT time, the heavier object the faster. Do you understand why there is no contradiction between what I am saying and mainstream physics? Nope, basic physics (see the equations of motion I posted) also teaches you that you are simply repeating a false claim with no basis.

The shell theorem has nothing to do with two separate bodies. The bodies cannot be one inside the other, they are side by side. I did, this is how I know is riddled with mistakes.

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?

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

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.

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

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. 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.

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. 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).

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]

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. 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).

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

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.

OK, you will need to solve the system of equations, if I understand correctly, you claim that [math]x_1(t)-x_2(t)=0[/math]. I am skeptical. So , solve the system and prove your point. Here, subtract the two equations: [math]\frac{d^2x_1}{dt^2}=-(\frac{GM}{(X-x_1)^2}+\frac{Gm_2}{(x_1-x_2)^2})[/math] [math]\frac{d^2x_2}{dt^2}=-(\frac{GM}{(X-x_2)^2}+\frac{Gm_1}{(x_1-x_2)^2})[/math] You get: [math]\frac{d^2(x_1-x_2)}{dt^2}=-(\frac{GM}{(X-x_1)^2}-\frac{GM}{(X-x_2)^2}+\frac{G(m_2-m_1)}{(x_1-x_2)^2})[/math] Actually, the problem is even more complicated because we must take into consideration that the centers of the test probes are laterally separated as well: [math]\frac{d^2x_1}{dt^2}=-(\frac{GM}{(X-x_1)^2}+\frac{Gm_2}{(x_1-x_2)^2+(r_1+r_2)^2})[/math] [math]\frac{d^2x_2}{dt^2}=-(\frac{GM}{(X-x_2)^2}+\frac{Gm_1}{(x_1-x_2)^2+(r_1+r_2)^2})[/math] So, the correct equation is: [math]\frac{d^2(x_1-x_2)}{dt^2}=-(\frac{GM}{(X-x_1)^2}-\frac{GM}{(X-x_2)^2}+\frac{G(m_2-m_1)}{(x_1-x_2)^2+(r_1+r_2)^2})[/math] 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).

I think that it is a LOT more complicated. You need to solve for [math]X(t), x_1(t),x_2(t)[/math] the system of coupled differential equations: [math]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/math] [math]m_1\frac{d^2x_1}{dt^2}=-(\frac{Gm_1M}{(X-x_1)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2})[/math] [math]m_2\frac{d^2x_2}{dt^2}=-(\frac{Gm_2M}{(X-x_2)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2})[/math] with the initial conditions: [math]x_1(0)=x_2(0)=D[/math] [math]X(0)=0[/math] [math]\frac{dX}{dt}=\frac{dx_1}{dt}=\frac{dx_2}{dt}=0[/math] at [math]t=0[/math] The "times to collision" are obtained from the conditions [math]X(t)-x_i(t)=R+r_i[/math], [math]i=1,2[/math]

But the forces exerted by the Earth on the two test probes are NOT identical, as a matter of fact they are different so the test probes are highly likely to SEPARATE in flight, such that they are NO LONGER subjected to the same acceleration. You need to solve a very complicated "three body problem" in order to get the equations of motion. I can post the solution for the two body case, if you wish, you can try to generalize to the case of the three bodies, the generalization is not trivial, you are oversimplifying things.

Sorry, swansont's claim.

I am talking about radial separation, meaning that they do NOT hit the Earth at the same time. Contrary to swansont's claim above that started it all.