Galileo was technically WRONG

[RJBeery]

The heavier item may have a greater attraction but it also has great resistance to change...Intertia:
The lighter item has less attraction, and also less Inertia and less resistance to change/movement/falling.

So we have one item with great gravitational attraction and more intertia, fighting that attraction...and the other have less gravitational attraction but also less Inertia fighting the gravitational pull.
Guess what? The two effects balance each other out, so that they hit the ground at the same time.

Galileo was right!

Dammit paddoboy, you are right if the Earth is NOT AFFECTED BY THE ACCELERATION. That is the critical assumption by Galileo, et al, because it's such a small factor when we're discussing small objects. Use the following equation

t = sqrt(2d/(a_o.e+a_e.o))

but consider a_e.o (the acceleration of Earth towards the object) t0 be ZERO. That equation will give you the same answer regardless of what mass the object is INCLUDING A BLACK HOLE!

You are blindly hoping that Earth moving towards the object somehow cancels out but you're not exactly sure how. I'm explicitly showing you the math to do so, and the math shows that you are mistaken. Period.

 

[paddoboy]

Except that while the earth is pulled up toward the bowling ball shortening the distance.., in the case of the golf ball, IT (the golf ball) falls toward the earth faster than the bowling ball does.., also shortening the distance...

This is never an issue, if you always use the earth as the frame of reference for the fall in both cases.., because all of the closing velocity is then attributed to the smaller object. When you try to look at it from the rest frame before the drop you have to take into account that the golf ball falls faster toward the earth than the bowling ball does.., which is exactly balanced by the difference in how fast the earth falls toward the smaller object in each case.

You are wrong! And as been pointed out Newton and Einstein are both right. Take the time to look over, The force of gravity is the same for atoms and baseballs. That should be a larger difference in mass than your bowling and base ball fantasy. And I am sure than even though I don't want to strain myself with the math, the researchers involved had no issue with the math... Without a reference I think I also remember a proposed test with neutrons, but I am unsure and a little skeptical given the short half life of a free neutron.

Makes reasonable sense OnlyMe.
The inferences of the thread, are as invalid as the inferences in the "BH'S can't exist " thread.
Mainstream cosmology are rather lucky they don't have to listen to such dribble.
We have it all to ourselves!

 

[RJBeery]

Motordaddy, your equations only include a single acceleration term. How do you justify that when there are presumably two masses involved in gravitational attraction?

 

[Aqueous Id]

The 9.82 m/s^2 is unchanged in my analysis. The numbers I use in my analysis in the OP are DERIVED via F=Gm1*m2/r^2. They are NOT simply given by convention! The difference in g(r(t)) is indeed ignored, but that is inconsequential because the field gradation between 10 meters and the ground of the Earth would be the same for all objects, so it is considered to be a constant at the value derived at 10 meters. The result (or more specifically, the conclusion) would be the same either way. I'm holding out hope that you, Aqueous Id, are clever enough to grasp this proof because it appears the others who are (i.e. rpenner) don't have the intellectual fortitude to publicly admit it.

You begin by telling us that you have a better way to model Galileo's experiment: namely, the two-body model. But my concern was that, as the Earth rises to meet the ball, the distance closes. As it does, the g-field immersing the ball is going to increase. If you say "let's ignore this -- it's negligible" then that looks to me like you're falling back to the position of the people you are trying to correct. How does ignoring this negligible amount make your model better than anyone who constructs this as a simple one-body problem? That's what I'm asking. The effect of the ball on the Earth is either negligible or it isn't. A related question: in solving a problem of this kind, how does a person looking for the exact answer know what to leave in and what to ignore?

 

[RJBeery]

Paddoboy, I've given you ample explanation. At this point you are being willfully ignorant of the facts.

 

[RJBeery]

You begin by telling us that you have a better way to model Galileo's experiment: namely, the two-body model. But my concern was that, as the Earth rises to meet the ball, the distance closes. As it does, the g-field immersing the ball is going to increase. If you say "let's ignore this -- it's negligible" then that looks to me like you're falling back to the position of the people you are trying to correct. How does ignoring this negligible amount make your model better than anyone who constructs this as a simple one-body problem? That's what I'm asking. The effect of the ball on the Earth is either negligible or it isn't. A related question: in solving a problem of this kind, how does a person looking for the exact answer know what to leave in and what to ignore?

OK! Fair enough. At the very least your answer is not dismissive. Let me consider a decent response and get back to you tomorrow, it's rather late here in Nebraska, US.

 

[Motor Daddy]

Motordaddy, your equations only include a single acceleration term. How do you justify that when there are presumably two masses involved in gravitational attraction?

Because each object accelerates at a different rate according to you, no? And what do you suppose that acceleration is relative to for each of them? Do you understand what a closing speed is?

 

[paddoboy]

Paddoboy, I've given you ample explanation. At this point you are being willfully ignorant of the facts.

You've given me nothing, except observational evidence to how far some will go to try and deride accepted cosmology and physics.

You begin by telling us that you have a better way to model Galileo's experiment: namely, the two-body model. But my concern was that, as the Earth rises to meet the ball, the distance closes. As it does, the g-field immersing the ball is going to increase. If you say "let's ignore this -- it's negligible" then that looks to me like you're falling back to the position of the people you are trying to correct. How does ignoring this negligible amount make your model better than anyone who constructs this as a simple one-body problem? That's what I'm asking. The effect of the ball on the Earth is either negligible or it isn't. A related question: in solving a problem of this kind, how does a person looking for the exact answer know what to leave in and what to ignore?

 

[RajeshTrivedi]

First of all I think Earth cannot rise to meet the ball (or Balls). For the simple reason that any suicide prone nerd may take the entire earthlings with him. If by throwing a ball from x meter height, moves up the Earth by dx, then keep repeating the exercise till you are off the orbit.... That does not happen so, Galileo must be right.

The point probably we are missing is that, The Earth is actually = Earth of RjBeery + those balls, for a larger system under which our actual Earth is moving. Whatever RjBeery is saying would have been right, if we were talking about two objects completely isolated, then both the objects would have moved towards each other depending on the other objects mass. [A crude analogy would be movement of two charges towards each other, gap closing time depends on the magnitude of charges.]

But once we bring in a massive object like BH (his third calculation), then situation changes drastically, it kind of becomes exclusively 2-body solution in some kind of isolation, so his calculation is still correct, but has got nothing to do with Galileo.


Edit : I presume his maths about 2-body motion is otherwise correct.

 

[Trippy]

Dammit paddoboy, you are right if the Earth is NOT AFFECTED BY THE ACCELERATION. That is the critical assumption by Galileo, et al, because it's such a small factor when we're discussing small objects. Use the following equation

t = sqrt(2d/(a_o.e+a_e.o))

but consider a_e.o (the acceleration of Earth towards the object) t0 be ZERO. That equation will give you the same answer regardless of what mass the object is INCLUDING A BLACK HOLE!

You are blindly hoping that Earth moving towards the object somehow cancels out but you're not exactly sure how. I'm explicitly showing you the math to do so, and the math shows that you are mistaken. Period.

Only if the blackholes mass was negligible compared to the mass of the earth, for that is the caveat applied to the assertion that you are ignoring.

 

[RajeshTrivedi]

The effect of the ball on the Earth is either negligible or it isn't.

It is nil, big Zero.

Its a trick question: Force on Earth = GMm/(r^2) + GM(-m)/(r^2) = 0.

(Only Galileo aspect)

 

[Aqueous Id]

It is nil, big Zero.

Its a trick question: Force on Earth = GMm/(r^2) + GM(-m)/(r^2) = 0.

(Only Galileo aspect)

Yeah, I'm beginning to understand the trickery here. For every bowling ball of mass m, there is a bowling ball of equal and opposite mass -m. Newton's 4th Law. That, and that other thing Newton said about ΣF = 0. Yeah, all those people who took kinematics and differential calculus really don't have a clue, do they. And they sure don't know how to smell a trick question either. So the real answer is to throw away the books and let the dropouts teach class. Yeah, just promote Homer Simpson from the control room to the Lucasian Chair and we can also plug up all those pesky black holes, and do away with the invariance of c, and pretty soon every one will be back to singing A Mighty Fortress is Our God from hymnals made of recycled Calculus and Biology texts.

Good ole Homer. Who would ever want to take him out to the woodshed. Kinda makes me misty thinking I ever complained about these Sock Puppet Army of Zealots [SPAZ] attacks.

 

[RajeshTrivedi]

Aqueous ID,

Before you go misty eye...choose the one for bowling ball drop from the height 10 m...

1. Force on Earth is Zero.
2. Force on Earth is non Zero.

 

[Aqueous Id]

Aqueous ID,

Before you go misty eye...choose the one for bowling ball drop from the height 10 m...

1. Force on Earth is Zero.
2. Force on Earth is non Zero.

How would anyone know? You introduced a negative mass. You mean you're not going to explain where it comes from?

 

[danshawen]

Galileo said that heavy objects and light objects, neglecting air resistance, accelerate at the same rate under the influence of the Earth's gravity near the surface of the Earth.

Are you saying that the Earth accelerates at different rate toward a dropped baseball than a pea?

Galileo was right in an even deeper sense than he intended. There is no absolute space or time, including origins or centers of mass that terrestrial mathematicians would like to idealize in exactly the same manner they tend to idealize straight lines and circles in Euclid's geometry.

The Earth has inertia. A great deal more of it than either a baseball or a pea, but exactly the same in terms of inertia imparted to electrons, bosons, W and Z and their antiparticles of each mass. According to the latest physics, this inertia derives of the Higgs mechanism. According to the Standard Model, this inertia derives of the Higgs boson slowing down these particle, and gravity has nothing to do with it. They are right, after a fashion, but conservation of energy-wise, they seem determined to pass up this opportunity to endow virtual particles with a means of sustaining inertia the way that a photon trapped between two mirrors evidently does (and the same experiment that led to E=mc^2 by the way, which particle physicists use on a daily basis.

So, how exactly was Galileo wrong again?

 

[RajeshTrivedi]

RJBeery offers some calculations and states that heavier mass will travel less distance before hitting the ground as Earth also moves towards it and thus will take lesser time. This way he offers that Galileo was wrong !!

But the Galileo was / is not wrong !!

My negative mass did not go well down and it choked some one, but the fact is as stated in my earlier post that the falling ball is a part of Earth, if we look at the problem from this perspective, then there is no problem to Galileo's legacy.

Our Earth is spinning, revolving, kind of moving in the space....ditto with its parts, that is ditto with that falling ball. So any motion of Earth is the motion of ball also. This way if the Earth moves dx in the direction of ball, then it moves along with its parts, that is along with the ball, so the distance between the ball and earth remained unchanged due to the motion of Earth..

 

[billvon]

1. Force on Earth is Zero.
2. Force on Earth is non Zero.

Force on the Earth is nonzero for any real object with mass.

 

[RJBeery]

There is an absurdity in the idea that the "greater mass" is the mass which solely dictates falling times. Galileo's proclamation that falling times are independent of mass were only approximations due to the relative insignificance of the objects he was studying. They DID and DO fall to the Earth at different rates (when dropped separately) because they exert a reactive force (i.e. "every action has an equal and opposite reaction") upon the Earth. If this reactive force did not exist then Observer B in the following picture could arbitrarily decide which planet of equal mass "fell towards" the other one by simply jumping between the two, thereby tilting the scales.

 

[RJBeery]

A while back I went from posting here to just lurking for time management reasons, but I feel like RJBeery really deserves a friendly voice here. Newton's law of gravity says that in the Earth's gravitational field at a given height oranyfixedgravitationalfieldforthatmatter, any object will experience the same gravitational acceleration. On the other hand, the Earth's acceleration toward the object will depend on the gravitational field produced by that object, which in turn depends on its mass. The closing acceleration between the Earth and the object is the sum of these two terms, so it also depends on the object's mass (very slightly).
↑

RJ, that is taking only the effect of the different masses of the falling objects, on the earth's mass and inertial resistance, into account. You are correct in stating that the larger mass will "pull" the earth toward it more than the smaller mass would, but what is being neglected is that the earth "pulls" the smaller mass toward it more than it does the larger mass. You have to account for the inertial resistance of all objects.

It bothers me that this quote was not only posted, but quoted and liked multiple times. The whole point of Galileo's experiment was that the acceleration of an object under gravity is completely independent of its mass; in other words, gravitational mass is the same as inertial mass. The Earth pulls on any two masses equally, and the object's pull on the Earth must be added to that constant value.

As soon as you step up to the sun or a black hole the result changes.., because the larger mass becomes the origin of the dominate gravitational field. The gravitational potential 1 meter above the surface of a star is far greater than it would be 1 meter above the earth. That greater potential changes the way the inertial resistance plays out. You start introducing greater terminal velocities. They would only fall toward each other , in the same manner as a smaller object falling toward the earth, until they reach a point where the gravitational potential is equivalent to that at the surface of the earth.

Unless you're dealing with thermodynamics in the large-N limit, just about every function in nature is smooth if you look closely enough. It makes no sense to say that any objects less massive than the Earth fall at exactlythe same rate, but as soon as they get more massive than the Earth they start to fall faster. Instead, you should be looking for a function that correctly predicts the small- and large-mass limits but interpolates smoothly between them. RJBeery's description meets this criterion.

And brucep: I would also like a citation for dr/dt = 2M/r^1/2. For one thing, it doesn't have the right units; I assume M here is standing in for some combination of the Earth's mass and the gravitational constant? Also, the equation as written gives velocity as a function of r independent of the height from which the object was dropped, which can't be correct.

Thanks for the attempt Fednis. Perhaps you can speak their language!​

 

[OnlyMe]

Now you start with Farsight's tactic, cherry picking quotes.., as in cross posting comment from another thread without including the response, or a functional link.