A Flaw In Using Fall?

[xyzt]

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. .

The shell theorem has nothing to do with two separate bodies. The bodies cannot be one inside the other, they are side by side.

 

 

 

If you have more questions, have a go at trying to read the paper

 

I did, this is how I know is riddled with mistakes.

Edited June 13, 2014 by xyzt

[Mordred]

equivalence principle

 

http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf

 

the problem is your trying to state the equivalence paper is wrong, the equivalence principle clearly states 2 objects of differing mass removing all friction will fall to the center of mass at the same rate.

 

your statement that the heavier mass will fall faster is plain wrong.

 

but instead of listening to the truth you misinterpret your formula and state every textbook and test in GR is wrong. You even go so far as to call a well established and rewarded physicist a crack pot.

 

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

 

but its form is more correctly

 

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

 

 

this formula may work great where the 3 bodies are of similar mass to each other as in 3 bodies orbiting each other, however IT does not mean the equivalence principle does not apply.

 

if you took particle a and place it near to particle b so they drop at the same time from the same location their mass is effectively combined. If the center of mass is still the center of the Earth due to the size of mass being so insignificant to the mass of the Earth. the two will land at exactly the same time.

 

if mass a is 4kg and mass b is 10 kg and they are dropped from opposite ends of the Earth whose mass is far greater the center of mass is the center of the Earth effectively. the two object will land at the same time. hence equivalence principle

 

here is your 2 body problem with the center of mass correlations

 

http://en.wikipedia.org/wiki/Gravitational_two-body_problem

 

notice where your center of mass is?

Edited June 13, 2014 by Mordred

[xyzt]

equivalence principle

 

http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf

 

the problem is your trying to state the equivalence paper is wrong, the equivalence principle clearly states 2 objects of differing mass removing all friction will fall to the center of mass at the same rate.

 

your statement that the heavier mass will fall faster is plain wrong.

 

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?

 

 

 

if you took particle a and place it near to particle b so they drop at the same time from the same location their mass is effectively combined.

 

Nope, basic physics (see the equations of motion I posted) also teaches you that you are simply repeating a false claim with no basis.

Edited June 13, 2014 by xyzt

[Mordred]

your interpretation of that equation is wrong period

[MigL]

I think it was waaaay back in post #11 that xyzt stated that two test probes of different masses, if dropped simultaneously, would separate during their descent. He stated that swansont must be wrong in his assertions.

 

It seems to be very easy to prove. Put two differing test weights in a bag and drop them from altitude. Does the bag rip and allow the heavier weight to land first ?

If you go over a bridge in your car are you pinned to the roofliner during your descent because of the differing weight of you and your car ? Is that why we wear seatbelts, to hold us down ?

 

Someone was definitely wrong, but I don't think it was swansont.

[Mordred]

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.

 

basic physics also teaches you that inertial mass is equivalent to gravitational mass

the article I posted above shows that test of this are accurate to 1 part in 10¹³ (this just allows for possible systematic errors)

http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf

 

in other words how objects fall depend on its gravitational acceleration not upon its mass

 

the problem is your not considering the effective center of mass. yes the mass of the 3 objects is important but in the sense its needed to calculated the effective center of mass, not in the sense of heavier objects falling faster than lighter ones

 

here is the proper 2 body problem equations

 

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

 

all objects in a 2,3,x number of objects always fall towards an effective center of mass.

 

the reaction of [latex]M\frac{d^2X}{dt^2}[/latex] is a consequence of an offset center of mass relative to the Earths center.

 

it does not mean that the heavier object fall faster, it means the Earth also has to fall towards the center of mass

 

just an amusing side note here is a brief NASA coverage for the feather and hammer test on the moon.

http://science1.nasa.gov/science-news/science-at-nasa/2007/18may_equivalenceprinciple/

 

 

also [latex]f=\frac{GMm_2}{r^2}[/latex] is a newtonian law

 

in terms of instantaneous acceleration your usage doesn't consider all the factors.

(this is more for others reading the post informational purpose)

[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]

 

Newtonian law works great at short distances not so great at longer ones such as spacecraft. (granted you probably wasn't trying to include the varying gravity due to distance from the gravitational source, and for the purpose of your equation assumed constant gravity)(this is more for others reading the post informational purpose)

 

http://en.wikipedia.org/wiki/Equations_for_a_falling_body

"A set of dynamical equations describe the resultant trajectories when objects move owing to a constant gravitational force under normal Earth-bound conditions. For example, Newton's law of universal gravitation simplifies to F = mg, where m is the mass of the body. This assumption is reasonable for objects falling to earth over the relatively short vertical distances of our everyday experience, but is very much untrue over larger distances, such as spacecraft trajectories. Please note that in this article any resistance from air (drag) is neglected."

 

please note none of these equations at the various gravity levels include a mass term for the object falling on that wiki link

 

wonder why that is??

 

http://www.npl.washington.edu/eotwash/sites/www.npl.washington.edu.eotwash/files/webfiles/publications/pdfs/schlamminger_AAPT07.pdf

http://www.umich.edu/~mctp/SciPrgPgs/events/2008/SS08/EP_Mich.pdf

 

 

"On a Theoretical Proof of the Weak Equivalence Principle from within the Confines of Newtonian Gravitation"

http://vixra.org/pdf/1111.0082v1.pdf

 

"the weak equivalence principle is the statement that bodies of different mass and composition will – as first demonstrated by Galileo Galilee; fall at the same rate in a gravitational field. Current measurements on the equality of gravitational and inertial mass indicate that this equality holds on a level of one part to 10¹³ "

Edited June 13, 2014 by Mordred

[md65536]

It seems to be very easy to prove. Put two differing test weights in a bag and drop them from altitude. Does the bag rip and allow the heavier weight to land first ?

If you go over a bridge in your car are you pinned to the roofliner during your descent because of the differing weight of you and your car ? Is that why we wear seatbelts, to hold us down ?

You can't go over a bridge while in your car. You have to get out and fall side-by-side (see post #51).

 

IF they are initially separated, "they would separate (radially)" does not refer to separation from each other, but an increase in the difference of their respective distances to Earth. The initial "radial separation" is 0.

 

The increasing "radial separation" comes from the fact that the Earth accelerates slightly more toward the heavier of the falling masses (and then it might also give the heavier mass a faster acceleration because it's closer).

 

Note that if the falling objects are initially separated and equidistant from Earth, the lighter one is initially farther from the center of mass of the system.

 

I think that xyzt's right about what would happen in that set up, but wrong about it being the only possibility (I don't see why you can't fall inside a car while being separate from the car). I think swansont's right when the separation of the falling objects is negligible.

 

To be fair, the original question didn't say negligible separation or side-by-side, but with impossibly great separation, "far away from each other so that the marble sized object with greater mass did not have a gravitational effect on the lesser mass marble sized object". However that doesn't apply to every case discussed in this thread.

 

 

Edit: Actually, I think that the only way to satisfy the conditions of the original question would be if the objects could occupy the same space, and were exactly co-located.

Edited June 13, 2014 by md65536

[xyzt]

your interpretation of that equation is wrong period

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.

[Mordred]

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.

fine where in any of your terms or math did you apply acceleration? or did you forget that step

Edited June 13, 2014 by Mordred

[xyzt]

 

basic physics also teaches you that inertial mass is equivalent to gravitational mass

the article I posted above shows that test of this are accurate to 1 part in 10¹³ (this just allows for possible systematic errors)

http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf

 

in other words how objects fall depend on its gravitational acceleration not upon its mass

 

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.

fine where in any of your terms or math did you apply acceleration? or did you forget that step

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?

 

I think that xyzt's right about what would happen in that set up,

 

 

Good, you realized that.

 

 

 

but wrong about it being the only possibility (I don't see why you can't fall inside a car while being separate from the car). I think swansont's right when the separation of the falling objects is negligible.

 

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.

Edited June 13, 2014 by xyzt

[Mordred]

try it in straight newton physics

 

mass a=1,000 kg

mass b=10 kg

 

force of gravity at Earths surface= 9.8 N/KG lets just round it off at 10 newtons/kg

 

f=ma

 

therefore object a=10,000 netwons

object b= 100 newtons

 

f=ma therefore a=f/m

 

object a acceleration=10,000newtons/1000kg =10 m/s²

object b acceleration=100N/10kg=10 m/s²

 

^{wow they have the same acceleration therefore they land at the same time}

Edited June 13, 2014 by Mordred

[xyzt]

try it in straight newton physics

 

mass a=1,000 kg

mass b=10 kg

 

force of gravity at Earths surface= 9.8 N/KG lets just round it off at 10 newtons/kg

 

f=ma

 

therefore object a=10,000 netwons

object b= 100 newtons

 

f=ma therefore a=f/m

 

object a acceleration=10,000newtons/1000kg =10 m/s²

object b acceleration=100N/10kg=10 m/s²

 

^{wow they have the same acceleration therefore they land at the same time}

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.

[Mordred]

the newton mass is cancelled by the weight can't you see that?? are you that blind????

 

=equivalence principle

 

now I am not about to go through all these equations for you in how to PROPERLY use the 2 body problem in regards to gravitational bodies as the 2 body problem applies to 2 bodies with similar mass ie they each have their own gravity, for object falling within Earths atmosphere the Euclidean forms are more than adequate where the mass of the falling obect do not significantly offset the effective common gravity.

 

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

 

its up to you to read them

 

Kepler’s Laws:

Each planet moves in an ellipse with the sun at one focus.

•The radius vector from the sun to a planet sweeps out equal areas in equal time.
•The period of revolution T of a planet about the sun is related to the major axis of the ellipse by T² =kA³
=where k is the same for all planets

 

"We shall begin our solution of the two-body problem by showing how the motion of twobodies 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 shell theorem has nothing to do with two separate bodies. The bodies cannot be one inside the other, they are side by side.

this statement is also wrong

 

"These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with calculus. (Alternatively, Gauss's law for gravity offers a much simpler way to prove the same results.)"

 

http://en.wikipedia.org/wiki/Shell_theorem

Edited June 13, 2014 by Mordred

[swansont]

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.

 

Yes, and if one compares your result with Newton's law, there is an obvious discrepancy. There should be no dependence on the size of the object. So whatever follows is wrong, since the equation is wrong.

[Mordred]

. 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

 

 

this is correct in certain circumstances,

 

1) the mass of the falling object must be large enough to have a measurable influence upon the effective center of gravity, (must be sufficient to move the Earth appreciably)

[xyzt]

 

this is correct in certain circumstances,

 

1) the mass of the falling object must be large enough to have a measurable influence upon the effective center of gravity, (must be sufficient to move the Earth appreciably)

[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.

 

Yes, and if one compares your result with Newton's law, there is an obvious discrepancy. There should be no dependence on the size of the object. So whatever follows is wrong, since the equation is wrong.

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],.

Edited June 13, 2014 by xyzt

[Mordred]

The question boils down to a matter if practicality.
For every day circumstances the Earth reaction is not a factor.
With Kepler's law all bodies are understood in GR to be considered falling towards the effective center of mass. Their angular momentum offsetting their rate of fall.

However this is more a statement that mass affects the location of the effective center of mass. Not that different mass objects fall at different rates.
So in this aspect I agree with you

Edited June 13, 2014 by Mordred

[swansont]

 

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],.

 

No, because we are looking at acceleration in the z direction. Your equation implies that two test masses at z1=z2 will accelerate perpendicular to the line drawn between their centers owing to their mutual attraction. That is most decidedly NOT consistent with Newton's law.

[xyzt]

 

No, because we are looking at acceleration in the z direction. Your equation implies that two test masses at z1=z2 will accelerate perpendicular to the line drawn between their centers. That is most decidedly NOT consistent with Newton's law.

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.

[md65536]

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.

Yes, exactly. As swansont said, as the Rabinowitz paper points out, the answer depends on how the question is stated and how the experiment is set up. I hope we can all agree on that.

 

Yes, in this case the larger object hits first, not the more massive object. The point is that they have the same acceleration with respect to Earth, and that it is physically reasonable to allow the test masses to be close together without interacting gravitationally.

[xyzt]

Yes, exactly. As swansont said, as the Rabinowitz paper points out, the answer depends on how the question is stated and how the experiment is set up. I hope we can all agree on that.

 

 

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.

 

 

 

Yes, in this case the larger object hits first, not the more massive object.

 

So, you accept my correction.

 

 

 

 

The point is that they have the same acceleration with respect to Earth,

 

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.

 

 

 

 

and that it is physically reasonable to allow the test masses to be close together without interacting gravitationally.

 

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.

Edited June 13, 2014 by xyzt

[md65536]

With Kepler's law all bodies are understood in GR to be considered falling towards the effective center of mass.

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.

So, you accept my correction.

[...]

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.

I never said that the "mass in a shell" masses would hit the Earth at the same time. It was only mentioned to correct your statement:

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.

So I'm not sure what you're correcting, but sure, I'll accept your correction anyway. Thanks. Do you accept my correction too?

[DimaMazin]

Yes, exactly. As swansont said, as the Rabinowitz paper points out, the answer depends on how the question is stated and how the experiment is set up. I hope we can all agree on that.

 

Yes, in this case the larger object hits first, not the more massive object. The point is that they have the same acceleration with respect to Earth, and that it is physically reasonable to allow the test masses to be close together without interacting gravitationally.

xyzt is right even if the falling objects have identical sizes but different masses.We can connect the objects by pipe to eliminate action of mutual attraction of the falling objects. Distinction of masses and length of the pipe define various times of collisions.Because center of falling masses is nearer to big mass and the center attracts Earth to itself.

[xyzt]

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.

I never said that the "mass in a shell" masses would hit the Earth at the same time. It was only mentioned to correct your statement:

So I'm not sure what you're correcting, but sure, I'll accept your correction anyway. Thanks. Do you accept my correction too?

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

Edited June 13, 2014 by xyzt

[md65536]

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

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