A Flaw In Using Fall?

[xyzt]

If the masses are the same size and at the same location they will hit at the same time.

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

[Mordred]

With 3 bodies in general the bodies don't accelerate toward the center of mass. The nearer of two bodies will have a greater pull per unit mass. I don't follow your line of reasoning here.

 

 

It is an old descriptive of how Newton reconciled, gravity with orbiting bodies, I got the descriptive from my 1923 1st year University physics textbook. (lol coincidentaly how they modeled the atom didn't include the neutron, just a side note) All bodies are falling due to gravity. Their angular momentum keep them in orbit. Your right on the nearer of two bodies however if you look at say for example 17.5.1 page 15 the center of mass of a system is not at an actual body. Its between body A and body B. so both A and B will orbit the effective center of mass.. In the case of our solar system though or center of mass coincides near or at the sun, However local objects to a stronger mass source will orbit the local strong mass source over the solar systems center of mass. example planets and its moons

 

"Thus each body undergoes a motion about the center of mass in the same manner that the reduced body moves about the central point" The reduced body has a similar geometry application figure 17.2 page 3

 

http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf

Edited June 13, 2014 by Mordred

[MigL]

Let me get this straight.

If we have an object composed of two different parts of very different densities such that they have a coincident centre of mass, and we drop said object from great height, xyzt is claiming that there will be a measurable shearing force between the parts of differing density ????

This is absurd !

 

Keep in mind that mathematical models are just that, they can be applied wrong, and the ultimate test is weather they conform with experimental reality.

[xyzt]

 

It is an old descriptive of how Newton reconciled, gravity with orbiting bodies, I got the descriptive from my 1923 1st year University physics textbook. (lol coincidentaly how they modeled the atom didn't include the neutron, just a side note) All bodies are falling due to gravity. Their angular momentum keep them in orbit. Your right on the nearer of two bodies however if you look at say for example 17.5.1 page 15 the center of mass of a system is not at an actual body. Its between body A and body B. so both A and B will orbit the effective center of mass.. In the case of our solar system though or center of mass coincides near or at the sun, However local objects to a stronger mass source will orbit the local strong mass source over the solar systems center of mass. example planets and its moons

 

"Thus each body undergoes a motion about the center of mass in the same manner that the reduced body moves about the central point" The reduced body has a similar geometry application figure 17.2 page 3

 

http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf

It is an excellent book. Unfortunately, none of the stuff applies directly to the radial fall discussed in this thread. You would need to do your own calculations, as I did.

[Mordred]

It is an excellent book. Unfortunately, none of the stuff applies directly to the radial fall discussed in this thread. You would need to do your own calculations, as I did.

quite honestly the calculation is already in the references I pointed out in the article. I do not need to merely copy or paste them. The related formulas are already derived.

17.2 clearly shows how a 2 body problem can be reduced to a one body problem.

 

"We shall begin our solution of the two-body problem by showing how the motion of two bodies interacting via a gravitational force (two-body problem) is mathematically equivalent to the motion of a single body with a reduced mass given by"

[latex] \mu=\frac{m_1m_2}{m1+m2} [/latex]

the steps on how to derive from this formula to arrive at the formula at 17.3.12. which all conform to Newtons 3 laws.

17.3.12 is the force between two bodies

When I have such a clear reference I will post that reference and point out the related sections. In this case there is nothing that needs to be further derived or reduced.this article clearly shows all the textbook related formulas in how to derive both the force between gravitational bodies, the centrifugal force between those bodies, including the elliptical case. It clearly takes you from the force related calculations to arrive at Kepler's laws of rotation.

by the method commonly found in textbooks. textbooks such as,

 

Extragalactic Astronomy and Cosmology by Peter Schneider though this one doesn't go into detail in this area

Introduction to Astronomy and Cosmology by Ian Morison chapter 1.9.1 Derivation of Kepler’s third law

 

I am aware that the Kepler solutions has limitations,(particularly in 3 body problems) however its covered extensively in textbooks.

 

I do have 2 other Astronomy textbooks. my main interest is in the field of Cosmology though, in that arena I own 12 textbooks, In the related metrics such as relativity, I own Ronald M Wald's General relativity book (excellent highly recommend it) and lecture notes on General Relativity by Mathius Blau. (handy 925 page article. (its not a textbook as per se but close enough) http://www.blau.itp.unibe.ch/newlecturesGR.pdf "Lecture Notes on General Relativity" Matthias Blau

 

I don't believe there is a completely reliable solution to the 3 body problem, I have some papers covering the circular restricted 3 body problem, and the Poincare solutions, as well as how N-body can be used to treat the 3 body problem. (including one textbook on N-body Gravitational treatments, can't recall the authors name though)

 

However Astronomy is not particularly my primary interest. (except in its relation to cosmology applications lol)

 

by the way I do follow your formula's my problem was in misinterpreting how I thought you were describing it in terms of the equivalence principle but I'm clear on that now

Edited June 14, 2014 by Mordred

[swansont]

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]z_1-z_2=0[/math] is a boundary condition. Your equation has an acceleration perpendicular to the separation. It's wrong. Newton's law is one of attraction along the direction connecting the centers of mass.

 

There's no point in continuing if you are going to make such a fundamental error.

[xyzt]

 

[math]z_1-z_2=0[/math] is a boundary condition.

No, according to you [math]z_1(t)=z_2(t)[/math] (the two bodies fall as one), so [math]z_1-z_2=0[/math] is NOT (only) a boundary condition, it is the permanent condition. It is actually your ongoing, not supported mathematically, claim. Once again, I would love to see a mathematical proof of your claim , all you have been doing is offering unsupported claims. Back up your claims with math, would you?

Edited June 14, 2014 by xyzt

[swansont]

No, according to you [math]z_1(t)=z_2(t)[/math] (the two bodies fall as one), so [math]z_1-z_2=0[/math] is NOT (only) a boundary condition, it is the permanent condition. It is actually your ongoing, not supported mathematically, claim.

 

 

 

I agree that there is no point in continuing debating with you. As you can see, the errors are all yours.

 

So you keep insisting, but as your equation doesn't reduce to the basic form of Newton's law when it should, you've not presented a valid rebuttal. If you apply the boundary condition, you get incorrect behavior: acceleration in the wrong direction. That means it's wrong. I can't see how you can possibly claim that it's right.

[xyzt]

 

So you keep insisting,

[math]z_1=z_2[/math] is your claim, not mine. I pointed out that I am skeptical of it, you presented nothing in terms of supporting your claim.

 

 

 

but as your equation doesn't reduce to the basic form of Newton's law when it should, you've not presented a valid rebuttal.

 

You have not presented any proof, so there is very little to rebut.

Edited June 14, 2014 by xyzt

[swansont]

[math]z_1=z_2[/math] is your claim, not mine. I pointed out that I am skeptical of it, you presented nothing in terms of supporting your claim.

 

 

 

 

You have not presented any proof, so there is very little to rebut.

 

 

My initial claim is that IF z1=z2 (a condition, not a conclusion) then there is no force in the z direction. Since F = GMm/r^2 is in the r12 direction (and attractive), there is no vector component in the z direction for the co-located masses.

 

Your equation has an acceleration in the z direction for separation in an orthogonal direction, which is not consistent with Newton's law of gravitation, or compliant with his third law of motion.

[xyzt]

 

 

My initial claim is that IF z1=z2 (a condition, not a conclusion) then there is no force in the z direction. Since F = GMm/r^2 is in the r12 direction (and attractive), there is no vector component in the z direction for the co-located masses.

 

Your equation has an acceleration in the z direction for separation in an orthogonal direction, which is not consistent with Newton's law of gravitation, or compliant with his third law of motion.

The equations of motion along the Z (radial) axis are:

 

[math]M\frac{d^2Z}{dt^2}=+(\frac{GMm_1}{(Z-z_1)^2}+\frac{GMm_2}{(Z-z_2)^2})[/math]

[math]m_1\frac{d^2z_1}{dt^2}=-(\frac{Gm_1M}{(Z-z_1)^2}+\frac{Gm_1m_2cos \alpha}{(z_1-z_2)^2+(r_1+r_2)^2})[/math]

[math]m_2\frac{d^2z_2}{dt^2}=-(\frac{Gm_2M}{(Z-z_2)^2}+\frac{Gm_1m_2cos \alpha}{(z_1-z_2)^2+(r_1+r_2)^2})[/math]

[math]cos \alpha=\frac{z_1-z_2}{\sqrt{(z_1-z_2)^2+(r_1+r_2)^2}}[/math]

 

with the initial conditions:

 

[math]z_1(0)=z_2(0)=D[/math]

[math]Z(0)=0[/math]

[math]\frac{dZ}{dt}=\frac{dz_1}{dt}=\frac{dz_2}{dt}=0[/math] at [math]t=0[/math]

The "times to collision" are obtained from the conditions [math]Z(t)-z_i(t)=R+r_i[/math], [math]i=1,2[/math]

 

Please solve the above and prove your claim that [math]z_1(t)=z_2(t)[/math]. You have all the information. Personally, I do not think that the system of equations is solvable, at least symbolically.

Edited June 14, 2014 by xyzt

[swansont]

Personally, I do not think that the system of equations is solvable, at least symbolically.

 

alpha is 90º. Those terms containing cos(alpha) vanish.

[xyzt]

 

alpha is 90º. Those terms containing cos(alpha) vanish.

Correction, [math]\alpha=\alpha(t)[/math]. All you know is that [math]\alpha(0)=\frac{\pi}{2}[/math].

What you are doing is just another way of your asserting with no proof that [math]z_1(t)-z_2(t)=0[/math]. You need to prove that. Proof by assertion is not a valid form of proof.

Edited June 14, 2014 by xyzt

[swansont]

Correction, [math]\alpha=\alpha(t)[/math]. All you know is that [math]\alpha(0)=\frac{\pi}{2}[/math].

What you are doing is just another way of your claiming that [math]z_1(t)-z_2(t)=0[/math]. You need to prove that. Proof by assertion is not a valid form of proof.

 

 

I was under the impression that kinematics was understood.

 

If the relative acceleration is zero, there will be no relative velocity and thus no acceleration. The integral of zero is zero. Alternately, the equations that apply at t=0 are identical for each mass. The uniqueness theorem tells me that there is only one solution to the problem.

[xyzt]

 

 

I was under the impression that kinematics was understood.

 

If the relative acceleration is zero,

You do not know that. The two test probes have the same acceleration only when they freefall by themselves, not when there are two probes present, you need to prove your claim You keep trying to extrapolate assertions without any basis.

As a matter of fact , if you wanted to do things in a rigorous way, you shouldn't even operate off acceleration, you need to write down the complete equations of motion.

 

 

 

Alternately, the equations that apply at t=0 are identical for each mass. The uniqueness theorem tells me that there is only one solution to the problem.

 

Actually, the equations of motion for the test probes have identical form and have identical boundary conditions. Therefore, the two equations have solutions that can only differ by a constant. Since the initial conditions are the same, the constant is zero. This might be the beginning of a valid proof.

Edited June 14, 2014 by xyzt

[swansont]

You do not know that. The two test probes have the same acceleration only when they freefall by themselves, not when there are two probes present, you need to prove your claim You keep trying to extrapolate assertions without any basis.

As a matter of fact , if you wanted to do things in a rigorous way, you shouldn't even operate off acceleration, you need to write down the complete equations of motion.

 

Answer this: what is the relative acceleration of the test masses at t=0?

[xyzt]

 

Answer this: what is the relative acceleration of the test masses at t=0?

Zero. Totally irrelevant, you keep insisting on working off acceleration , when the rigorous way is to work with the complete equations of motion. This is exactly like the people that try to solve SR problems using length contraction instead of using the full Lorentz transforms.

Edited June 14, 2014 by xyzt

[swansont]

 

Actually, the equations of motion for the test probes have identical form and have identical boundary conditions. Therefore, the two equations have solutions that can only differ by a constant. Since the initial conditions are the same, the constant is zero. This is a valid proof.

 

Are you saying that you accept that they will hit at the same time?

[xyzt]

 

Are you saying that you accept that they will hit at the same time?

No, but the approach through studying the properties of the equations of motion is a better, more rigorous approach to solving the problem, one that I have tried to push you all along, away from the hacks based on looking at relative acceleration.

I just realized though that there is no theorem of existence and uniqueness for the type of non-linear equations we are examining, so you cannot conclude that the two solutions are identical. For starters, there is nothing to prove that the solution even exists.So, this approach needs more work (but it is more promising).

Edited June 14, 2014 by xyzt

[Mordred]

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})[/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.

 

 

can you clarify statement one.? As far as I know EP always applies for falling bodies. (this is rigorously tested and is in every textbook and related article I have ever read in over 20 years (non controversial that is) m_i =m_g I don't need to prove this as it is well known and established.

 

now statement 2 I agree with, take 2 objects at rest. object A exerts a force upon object b , however that force is also applied to object a (forces are applied in equal and opposite directions) however f=ma so the the math will show that mass A's rate of acceleration will be different than mass B

 

now statement 3 and 4 is where I have a problem,

 

[math]m_1+m_2[/math] is combined to determine

[latex]\mu=G(m1+m2)[/latex] where [latex]\mu [/latex] is the standard gravitational parameter

 

[latex]\mu =GM[/latex]

 

[latex]\mu=G m_1+G m_2 =\mu_1+\mu_2 [/latex]

 

now here is my problem the right hand of your equations uses the law of universal gravitation

 

[latex] f_1=f_2=G\frac{m1m2}{r^2}[/latex]

 

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

 

so therefore your equation should work for gravitational bodies..

 

[latex]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/latex]

 

[latex]\frac{d^2 x}{dt^2} =\frac{dv}{dt} =\frac{dx}{dt}\frac{dv}{dx} =v\frac{dv}{dx} =\frac{GM}{x^2}[/latex]

 

if [latex]\frac{d^2 x}{dt^2} =\frac{GM}{x^2}[/latex]

 

why do you have

 

[latex]M\frac{d^2 x}{dt^2}[/latex] ?

 

in the above equation?

Edited June 15, 2014 by Mordred

[xyzt]

 

 

can you clarify statement one.? As far as I know EP always applies for falling bodies.

 

You are confused, the EP establishes the equivalence between bodies stationary in a uniform gravitational field and uniformly accelerated bodies. This is the EEP. There is another form, the WEP , that establishes the equivalence between the inertial and gravitational mass.

 

 

(this is rigorously tested and is in every textbook and related article I have ever read in over 20 years (non controversial that is) m_i =m_g I don't need to prove this as it is well known and established.

 

now statement 2 I agree with, take 2 objects at rest. object A exerts a force upon object b , however that force is also applied to object a (forces are applied in equal and opposite directions) however f=ma so the the math will show that mass A's rate of acceleration will be different than mass B

 

now statement 3 and 4 is where I have a problem, first off the paper I posted where your arguing against its validity is based on Kepler's laws

 

[math]m_1+m_2[/math] is combined to determine

 

by Kepler's laws [latex]\mu=G(m1+m2)[/latex] where [latex]\mu [/latex] is the standard gravitational parameter

 

[latex]\mu =GM[/latex]

 

[latex]\mu=G m_1+G m_2 =\mu_1+\mu_2 [/latex]

 

now here is my problem the right hand of your equations uses the universal force of gravity

 

[latex] f_1=f_2=G\frac{m1m2}{r^2}[/latex]

 

so therefore your equation should work for gravitational bodies.

 

and yet you seemingly chose to ignore the articles I posted on Kepler's laws can I ask why?

 

Because the problem being discussed has nothing to do with "Kepler's laws". I explained to you that your attempt at doing physics via quote-mining doesn't work, this is not how physics is done.

 

 

The orbit of a planet is an ellipse with the Sun at one of the two foci.

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

 

[latex]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/latex]

 

[latex]\frac{d^2 x}{dt^2} =\frac{dv}{dt} =\frac{dx}{dt}\frac{dv}{dx} =v\frac{dv}{dx} =\frac{GM}{x^2}[/latex]

 

if [latex]\frac{d^2 x}{dt^2} =\frac{GM}{x^2}[/latex]

 

why do you have

 

[latex]M\frac{d^2 x}{dt^2}[/latex] ?

 

in the above equation?

 

 

 

Simple, this is how the equations of motion get formed, the LHS contains the mass X acceleration, the RHS contains the expression of the force, be it gravitational (in the case of this thread), Coulomb, Lorentz, etc.

 

 

 

 

 

 

 

that would be like saying

 

[latex]\frac{G(2M)}{x^2}[/latex]

 

Nope, you are much mistaken, about most everything you posted.

Edited June 15, 2014 by xyzt

[Mordred]

I'm asking for clarity about how you derived the above equation

 

I can't figure out where you got the extra M term

Edited June 15, 2014 by Mordred

[xyzt]

I'm asking for clarity about how you derived the above equation

 

I can't figure out where you got the extra M term

There is no "extra M" term. M is the mass of the Earth, [math]M\frac{d^2X}{dt^2}[/math] is the force exerted by the two test probes onto the Earth. That is obviously equal to

 

[latex]\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2}[/latex]

Edited June 15, 2014 by xyzt

[Mordred]

how is re-posting the same equation count in showing how you derived it?

Newtons gravitational law

 

Newton's law of universal gravitation states that any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

 

 

so I have

force between particle A and Earth

 

[latex]f=\frac {GMm}{r^2}[/latex] the mass of the Earth is already combined with the mass of the test particle to determine the force between them

 

same formula for particle B same relation

 

now if I drop these two at the same time from the same location

 

the two masses are essentially combined in its influence on Earth

 

which is the same as saying

 

[latex]f=\frac{G(m1+m2)}{r^2}[/latex]

 

so why is the M term on the left hand side needed? in terms of acceleration? please show me how you went from this to your equation

Edited June 15, 2014 by Mordred

[xyzt]

how is re-posting the same equation count in showing how you derived it?

This is basic introductory physics. It simply says [math]m \frac{d^2x}{dt^2}=NetForce[/math]

 

 

now if I drop these two at the same time from the same location

 

the two masses are essentially combined in its influence on Earth

 

which is the same as saying

 

[latex]f=\frac{G(m1+m2)}{r^2}[/latex]

 

so why is the M term on the left hand side needed? in terms of acceleration?

 

You are rolling multiple basic errors in your post. To start with, [latex]f=\frac{G(m1+m2)}{r^2}[/latex] doesn't have the dimension of a force. Would you please let me continue the discussion with swansont undisturbed?

Edited June 15, 2014 by xyzt