Galileo's conclusion was that falling bodies of different composition fall at the same rate, he said nothing about the time taken to hit the ground.
That sounds pretty clear, direct and simple to me. However, time is implied by a reference to rate. If that gives you any problem, cite your references and their authority relative to this specific issue. To be clear we are talking about any difference in Galileo's conclusion, Newtonian mechanics and GR, where golf balls and bowling balls represent the masses dropped and the centers of gravity for the two FoR Galileo's {the earth's center of mass} and your two body problem, is less than the diameter of an atom.
Because you're using the wrong equation for motion (as has been explained to you dozens of times now), Galileo's equations are only valid where m1 >>> m2, and the object is falling in a uniform gravitational field. This was the way it was when Galileo first conceived of it, this is the way Galileo stated it. When you consider the two halves of the earth coming together, you are no longer considering a uniform gravity field and you are no longer considering the situation where one mass is much larger than the other and so Galileo's statement IS NOT APPLICABLE
Only if you're stuck in an earth frame of mind where you're dropping things from a fixed height. Not if you're considering a thought experiment where an object can fall indefinitely. Time taken to hit the ground is not the only measure for motion that we can consider in a thought experiment.
No, I understood the part in blue, that this was your construction. As I said, once you have designated X and x as positions in "x-space", you have assigned a reference frame. Therefore x and X are dependent on frame designation. If I come along with a new frame whose origin cuts through the midpoint of (X, x), at and angle θ, then your X and x are going to be equal in magnitude in my system, and opposite in sign. This is why you can't claim frame independence as you say. No I am just letting you know your little scheme is not working. Well, no, it's not going to be valid at all. We can create a special case for the kinematics being independent for any axis parallel to yours, but only for those cases in which the coordinates correspond. However, you are toast trying to map Galileo's experiment onto those parallel axes since the vector r in the Law of Universal Gravitation will vary as you change systems. Therefore you are sunk. Therefore, as I said before, it's incorrect to claim frame independence. You need a coordinate transform (forward and inverse) to get between systems. Let's just agree that you have no idea what a coordinate transformation is and move on. All of the skilled posters here took Linear Algebra and Vector Mechanics, so this is where you get to have conniption fits, go through catharsis, and get the remedial help you need. But seriously: get off the gas. If you want to to convince readers that you've taken the course you're going to have to do better than that. But really, you've shown us your cards. (I say "us" without having read the comments of the people I would call "skilled". So whether or not they are on to your little scheme remains to be seen.) In some higher course, but not down at the level we are talking about. Nice try, though. No, I do not need wikipedia to reinforce what I've posted thus far. That just says you're quote mining wikipedia. It has nothing to do with the structure of college curricula. Not knowing the curriculum is painting the fraud even louder and brighter than your messed up math did. You've left little to the imagination, Fednis. Yeah and I invented the Internet. The problem is that you can't impress a skilled person with your pseudoscience no matter how many credentials you pin on your chest. Isn't accelerating: an accelerometer placed at the origin reads zero. This will not be the case for your proposed coordinate system, since you have stated that the Earth is accelerating relative to your origin. Therefore you are accelerating relative to the Earth, therefore the accelerometer reads nonzero, therefore you are not inertial. There most definitely is a force acting on your system or else it would not be moving at all. Wrong, as follows: Correct. Ergo, QED. What is that supposed to mean? You just admitted your accelerometer is reading nonzero, so no need to try to gloss over it. Gawd you're a nut. Any Earth-centered frame must be inertial, since the accelerometer placed at the origin of any such frame reads zero. Actually, we need a separate thread posed as a contest to the poster who can come up with the most catastrophic consequences of the Earth magically switching to a non-inertial state. I can certainly understand why you would think so. But therein lies the rub.
"In around 1590 Galileo Galilei (1564-1642) climbed up the Leaning Tower of Pisa and dropped some balls to the ground. Two balls of different masses, but of similar shape and density that were released together hit the ground at the same time. Until then it was commonly believed that heavy things fall faster than light things. Many people still believe this, and casual observation of everyday phenomena often does tend to confirm this view." That is one of the story...how far this story is factually correct, I do not know.. Now few observations.. 1. If two balls of different mass are dropped at the same time (released together as stated in the experiment above), then even OP math/Physics of RJBeery will give the same conclusion. Both will hit the ground together. So Galileo was not wrong...his soul can rest in peace. 2. RJBeery puts some dicey figures of acceleration of Earth and comes up with two different times... "For the golf ball: t = sqrt(20/(9.82 + .00008029)) = 1.4271101 seconds For the bowling ball: t = sqrt(20/(9.82 + .00963481)) = 1.4264164 seconds " The difference = 0.0006937 Seconds...... This is a significant difference, and people should and really took notice of this, with a blind faith that RJBeery math was correct. But the fact is RJBeery was "WRONG" on this basic aspect itself, because proper calculations even as per his formula would give a time difference = 0.000000000000000000 keep counting till around 23 zeroes... When exposed about this inaccuracy, RJBeery attempted (and succeeded) in diverting the attention to theoretical aspect, he showed utter intellectual dishonesty by not acknowledging and correcting this misleading figure.. 3. RJBeery in a utter show of intellectual bankruptcy refuses to accept that [Earth + few Kgs Ball at 10 meter Height] is a different ball game, then [Earth + Hypothetical Moon Mass / BH Mass at 10 meter Height] or even chained Earth. Lets conduct some doable experiments..... Two balls of 1 Kg and 2 Kg mass, and let us denote the distance traveled before hitting the surface as d1, d2 and time taken as t1 and t2, and original drop point is 10 meter from the ground.. Let us follow the math/Physics of OP... A. Experiment 1 : Drop together then.. d1 = d2 but both < 10 meter t1 = t2 but both less than Sqrt(20/g) B. Experiment 2 : Drop them individually.. 10 > d1 > d2 Sqrt(20/g) > t1 > t2 C. Experiment 3 : Conduct the test in B inside the flying aircraft, aircraft surface as ground, and assume aircraft big enough for 10 meter height.. 10 =d1 = d2 Sqrt(20/g) = t1 = t2 Separation between aircraft and Earth got reduced as per OP.. (do not bring in Aircraft motion etc..) D. Experiment 4: Drop two balls of 1 Kg (repeat with 6 kg) from either end of Earth (diametrically opposite points) from the same height simultaneously.. (d1' and t1' denotes other side...) 10 = d1 = d1' = d2 = d2' Sqrt(20/g) = t1 = t1' = t2 = t2' Reason : Force on Earth gets squared off due to same mass ball on both sides.. E. Experiment 4: Drop two balls of 1 Kg and 6 Kgs from either end of Earth (diametrically opposite points) from the same height.. d1 > 10 (Earth moved to other direction due to m2 > m1) d2' < 10 t1 > t2' (but t1 is more than what we got in experiment A etc.. E. Just bring in a normal point as well, that is one more drop at a point 90 Degree from opposite point...and denote the parameters by d1" and t1" etc.. Now two balls of 1 Kg from diametrical points and one ball of 1 Kg from normal point simultaneously, so d1 = d1' = 10 > d1" t1 = t1' = Sqrt(20/g) > t1" So, how can we make the absolute and definitive conclusion on OP......
True, but as it applies to the issue I was or may be am having with RJ's OP and the discussion that followed, the earth is the frame of reference, because that was the frame that Galileo arrived at his conclusions from. Really most of the meat of my personal complaint was the implications that the OP some how was proving Galileo TECHNICALLY wrong. Which I don't believe has been done. The rest was a bunch of nit picking the OP and discussion in an attempt to drive the discussion to the flaws, that supports my assertion, that Galileo was correct in his conclusion. Which to be honest, a few others have done a better job at than I have.
I'm impressed. I normally pride myself on being pretty unflappable, but you've actually made me angry. These kind of ad hominems show a high school level of maturity, and your reaction to my bringing up Hamiltonian dynamics shows a high school level of physics education. I'm going to take a stab in the dark and guess you're about 16 years old. Am I close? Regardless, I'm done trying to argue with you, except for one point: Can someone else chime in and confirm for me that this statement is wrong? Because I'm pretty sure gravity is only part of the definition of "inertial" in relativity. I'm pretty sure in Newtonian mechanics, the center of mass is the obvious inertial coordinate origin in this problem. And if I'm wrong, I'd like to know where the error in my reasoning is. Unlike some.
RajeshTrivedi: I tried to respond to your post point-by-point. I really did. But I just don't understand what you're trying to show with all those calculations. Maybe you can answer two questions for me instead? 1. As far as I can tell, you think that we need to add an extra term to the acceleration of the ball because it's "really" just part of the Earth, and that an asteroid (even an asteroid of the same mass, falling from the same height) would experience different acceleration because it's separate from the Earth. Is that accurate? 2. If so, what do you think is the objective, physical standard for determining whether or not two objects are "really" part of the same object, for the purposes of predicting dynamics? If not, please explain to me what I'm misunderstanding about your viewpoint. Thanks!
That's my first point - as I understand it, he didn't. His thought experiment wasn't about dropping things from a tower to the surface of the earth, he was (and the debate was around) effectively objects falling down a bottomless pit. I agree, for the reasons I have already outlined. 1. RJBeery's scenarios ignore the caveats that the original statement comes attached to. 2. The results are physically and practically meaningless. I agree - Galileo was correct in his conclusion, but only if one considers it within its domain of applicability. As soon as you start considering gravity fields that aren't uniform, or situations where one mass isn't much larger than the other you need to use math other than that which RJBeery posted in the OP.
My understanding is that the story isn't factually accurate and was attributed to Galileo at a later date by a biographer.
Did you even indirectly acknowledged that the figures in OP with respect to time are way off ? A time of the order of ms and trillionth of ms makes a difference !! But you switched to supporting RJBeery on technical aspect.. Regarding your asteroid theory, where did you bring the golf ball from (to an height of 10 meter) ?? From moon or from Kuiper Belt ?? Do you understand what it means and what are the dynamics for an asteroid (of whatever mass) to get attracted and come down to earth at 10 meter height in your hand and then you drop it and say lo behold Galileo was WRONG ? Would you assign the same spin and same orbital speed (around Sun) to your pet asteroid at h = 10 meter, at which the golf ball is moving with earth ? May be you are talking about a debris ball of an Asteroid which hit the Arizona few centuries back... you somehow collected it from the desert and took it at h = 10 meters......Suit yourself. Your Q2 becomes infructuous in the light of response to Q1.....
You may be right, but if this story is accurate, then Galileo was right...That's the point I was making.
I don't know what you mean. Do you mean that the OP ignored things like Earth's topography, air resistance, relativistic corrections, etc? If so, RJ and I have both repeatedly admitted that this effect is extremely small, smaller than other sources of noise in the problem and small enough to be of no realistic importance. But it is not zero. If you mean something else, you'll have to clarify for me. The golf ball came from the ground. I do understand the dynamics for a falling asteroid. The asteroid may or may not have the same orbital speed as the golf ball, depending on how it approaches the Earth; a good solution should cover both cases. Now can you please answer my question about where you draw the line between things that are "really" part of the Earth and things that are not? It sounds to me like you're saying "the golf ball is obviously part of the Earth because it came from the Earth, but the asteroid is not because it came from space." Do you really think that the time it takes for an object to fall from a given height depends on whether that object was originally brought up from the ground or brought down from space?
When you are talking about any mass equivalent to what we would associate with a golf or bowling ball, it really would not make any measureable difference, because a mass that much smaller than the earth would not move the two body center of mass, relative to the earth's center of mass, in any significant way. The dropped mass would still need to be large enogh to make the distance between those two centers of mass significant. But for any dropped mass originating from any source other than the earth, it would change the gravitational potential, even if that change were not measureable. So as far as the theoretical disscussion is concerned it does make a difference. We can imagine with a far greater degree of accuracy than we can measure. As long as you are drawing any mass in the two body problem from the initial mass of one of the two resultant bodies, the gravitational potential of the two, as a whole, remains constant. The two problems are like M + (M-m) vs M+m, which again would only become significant when mass m is very much larger than a bowling ball. So, adding mass from an outside source changes the the total mass of the system and the gravitational potential, but only becomes significant to any real world test when the added mass {assumed to be the dropped mass} is large enough to make the distance between the two centers of mass mentioned significant. I added this interpretation earlier to try and focus on the center of mass issue, which is the same as the FoR issue, for the two body problem, and significant to what I interpreted as the OP's intent. I think someone had already mentioned that the change in center of mass would be on the order of the diameter of a proton, which did not seem to be an effective argument. I did not myself verify the proton comment and latter just referenced the change as less than the diameter of an atom.
The following from the OP defines a distance of separation and the drop ending with a collision at the surface of the earth. The disscussion did wander, and for myself cross threads...
Yes, I understand that. My point is that the OP comes to the wrong conclusion because as soon as you start considering the distance the earth falls before colliding with the object you're no longer talking about a uniform gravity field therefore you're no longer talking about Galileo's thought experiment therefore the conclusions of the OP don't actually have anything to say about the thought experiment (and conclusions from it) conducted by Galileo.
I'm not meaning to dodge anything. Perhaps you don't recognize or appreciate my answers. 1) The statement is TECHNICALLY WRONG. If you don't agree then you don't know what TECHNICALLY means. That word has a specific, technical Please Register or Log in to view the hidden image! meaning in this context. 2) The result between feathers and bowling balls may be indiscernible but that's a practical matter. There is obviously a point where the mass of the falling object does matter. 3) Again, you are using the word "realistically" like it falls under the context of something being "technically right or wrong". Two different worlds. This will be the third time I post this. Please read it.
Wait a minute, if the math is wrong I'll correct it no problem. But I'm the one who took a LONG time to derive the analysis in the OP so when "some Tom" comes along and says the numbers are off I'm looking at that Tom to point out specifically where they are off rather than just yelling from the hillside. Additionally, me making an error in the analysis is inconsequential to the underlying point of the OP...unless you can make the fall times equal for objects of varying masses; that's a feat which would be quite impressive indeed!
If the chain is 4.9 meters long what makes you think the brick wall traveled any different distance than 4.9 meters until it hit the earth?
