Galileo was technically WRONG

[Trippy]

This is exactly right. Except you're not appreciating that if the Earth falls towards the bowling ball "more quickly" (than, say, towards the golf ball) then the bowling ball has a shorter distance to cover before it impacts the Earth. Not only did I show this in the opening post, Fednis gave a math-based analysis AND we have quotes from actual Physics professors supporting my claim.

The case is closed. Further discussion is futile, but I'm sure that won't stop some folks here.

You're wrong about what I do and do not appreciate, and if you genuinely think that then you haven't understood the post I made in the first place.

How long the earth and the bowling ball take to collide versus how long the earth and the golfball take to collide is a seperate issue to acceleration relative to the center of mass of the system. The time it takes includes the distance of the earth from the center of mass of the system {as well as the relative accelerations} - a factor that you have so far ignored.

While the earth might accelerate towards a bowling ball faster than a golf ball, the earth also has further to fall to reach the center of mass of the earth-golfball system.

 

[Walter L. Wagner]

If I understand you correctly, you're saying you agree with me.

Edit: I don't hide my crazy like Miranda does.

Love that video!

Great video. The thread itself is too cumbersome. Lots of people should go back to Physics 1A to review Newtonian physics.

 

[Trippy]

Great video. The thread itself is too cumbersome. Lots of people should go back to Physics 1A to review Newtonian physics.

Agreed.

 

[Fednis48]

read the post I wrote to you in the other thread. Tidal accelerations can change the free fall path through spacetime not the rate of free fall. That,

Read my post answering yours in the other thread. Tidal accelerationslerations can change the path of the object through spacetime but it doesn't effect the rate of the falling object. The Newtonian analysis you're trying to do isn't something Newton would be doing. If it was then it would show up in the equation of motion derived from his theory.

Huh? Who said anything about tidal accelerations? I just used Hamiltonian dynamics to get the relative acceleration from the potential energy. If you're going to argue, point out a math error, don't try to pick apart the concept. Also, re: the equation of motion derived from Newton's theory; to everyone else reading, brucep repeatedly mentioned the equation dr/dt=[2M/r]^1/2 in the other thread. I finally figured out that it's the equation for the velocity of a small test mass falling into a massive body from infinity. Note that I said "small test mass", because I was able to re-derive it by treating the large mass as fixed and setting kinetic+potential energy equal to zero for the falling mass. Like every other formula that contains M without m, this is an approximation; in reality, the large mass would not be fixed, which is what leads to the effect RJBeery and I are arguing for.

The case is reopened.

If you claim the bowling ball accelerates at the rate of 9.8 m/s^2, then in exactly 1 second the bowling ball traveled 4.9 meters, and has a velocity of 9.8 m/s. Agree?

I agree, and I'm sure RJBeery would as well, assuming we're still just talking about Newtonian mechanics.

You're wrong about what I do and do not appreciate, and if you genuinely think that then you haven't understood the post I made in the first place.

How long the earth and the bowling ball take to collide versus how long the earth and the golfball take to collide is a seperate issue to acceleration relative to the center of mass of the system. The time it takes includes the distance of the earth from the center of mass of the system {as well as the relative accelerations} - a factor that you have so far ignored.

While the earth might accelerate towards a bowling ball faster than a golf ball, the earth also has further to fall to reach the center of mass of the earth-golfball system.

You're quite right here, but I don't know why you think center of mass is the important coordinate. It seems to me that the time it takes a dropped object to fall to Earth could be rigorously defined as the time it takes for that object to collide with the Earth, which will be slightly shorter for a heavier object. The fact that the system's center of mass will be in a different place relative to the point of collision is true, but irrelevant.

 

[danshawen]

The center of mass is an important coordinate in part because Einstein used it for his famous "photon traveling in a tube" analysis which resulted in E=mc^2. There is no more tested theory in all of physics, including and especially those concerning gravitation.

It is exactly because of of classical gravitation arguments like those being discussed here that Einstein eventually formulated GR using the math of fluid dynamics co-opted to express the curvature of space-time, and with much greater success in prediction than any competing classical analysis based on only masses and radii. I don't think many folks would argue that a falling baseball has an effect on Earth's tides, but that effect was neglected along with Coriolis effect (it curves to the right in the Northern Hemisphere on the way down). It just gets into a ridiculously ludicrous and detailed calculation for no good reason very quickly. F = ma is enough already.

Einstein made the wisest choice. Gravitation is evidently not about the property we call mass at all. It is the interaction of vacuum energy between the masses that makes gravity work the way it does. How else could a stone find the exact direction to the center of a large gravitating body? Mathematicians routinely calculate circumferences of circles using geometry, but a stone has no such facility or instrument with which to find or calculate the exact center. Never forget, any theory that ascribes a geometrical center to anything does so only by virtue of human intelligence. Since this has nothing to do with what falling stones can grasp, a theory is needed to explain how they can find centers without intellect. It's not "spooky action at a distance". It's not a "divine hand" that guides things that fall. Anyone who can't figure that something is happening in terms of energy transfer in the space between gravitating bodies is like a stone.

Einstein needed center of mass to justify Special Relativity to an audience of physicists whose knowledge was the equivalent of Sir Isaac Newton. Shorter derivations are possible. I like the one that sets time to unity instead of c; it's less than 1 page and only three steps.

 

[Trippy]

Huh? Who said anything about tidal accelerations? I just used Hamiltonian dynamics to get the relative acceleration from the potential energy. If you're going to argue, point out a math error, don't try to pick apart the concept. Also, re: the equation of motion derived from Newton's theory; to everyone else reading, brucep repeatedly mentioned the equation dr/dt=[2M/r]^1/2 in the other thread. I finally figured out that it's the equation for the velocity of a small test mass falling into a massive body from infinity. Note that I said "small test mass", because I was able to re-derive it by treating the large mass as fixed and setting kinetic+potential energy equal to zero for the falling mass. Like every other formula that contains M without m, this is an approximation; in reality, the large mass would not be fixed, which is what leads to the effect RJBeery and I are arguing for.

I agree, and I'm sure RJBeery would as well, assuming we're still just talking about Newtonian mechanics.

You're quite right here, but I don't know why you think center of mass is the important coordinate. It seems to me that the time it takes a dropped object to fall to Earth could be rigorously defined as the time it takes for that object to collide with the Earth, which will be slightly shorter for a heavier object. The fact that the system's center of mass will be in a different place relative to the point of collision is true, but irrelevant.

You've just reinforced why I think it's important, it also somewhat obviates {in my opinion} this whole 'falling towards a black hole' thing.

There are a whole lot of contexts that are being missed . For example, what Gallileo actually said, as I understand was that all objects fall at the same rate, in opposition to the aristotlean proposition which was popular at the time which was that the rate at which an object falls is dependent on its composition. Nothing that has been offered in this thread so far contradicts the observation that all objects fall at the same rate towards the earth. The only thing the OP has offered is that the earth falls towards more massive objects more quickly - well duh! But this is beside the point. To the best of my ability to discern Gallieo's comments were made, more or less, considering objects (essentially) falling through a bottomless pit. My point being that the rate at which the earth falls towards the objects is irrelevant to anything that gallileo said and the assertion that he is wrong - especially in the experiments described in the thread so far, would require measurements precise enough that quantum effects are going to become important. In fact the level of discrepancy where taling about here is... I don't have the words to describe it.

The average weight of a golfball is 45.93 g
The maximum weight of a Bowling ball is 7260g.
The mass of the earth is (5.97219x10^{27})g
The mean radius of the earth is 637,100,000 cm.

The center of gravity of a golfball and the earth, assuming the earth is a homgenous sphere {it isn't) with the golfball at an altitude of 1km is at a distance of (4.9x10^{-20})m from the center of the earth - this is about (\frac{1}{150}) of the charge radius of a proton.

A bowling ball is 158 times more massive than a bowling ball, so the center of gravity of a bowling ball and the earth, using the same assumptions as above, with the bowling ball at an alttitude of 1km is going to be displaced by approximately the charge radius of a proton.

But this is the maximum distance the earth can fall, the earth isn't even going to move that far - the bowling ball and golf ball can only fall a maximum of 1km before it hits the earth, and the earth and the bowling/golf ball stop moving. The actual distance the earth is going to move in the case of the golf ball is (\frac{1}{150} * \frac{1}{6371}) or on the order of one ten millionth of the charge radius of a proton, for the bowling ball it will be on the order of on ten thousandth of the charge radius of a proton.

Setting aside the fact that for the OP being correct requires taking gallileo's words out of the context they were originally phrased in.
Setting aside the obvious problems regarding accuracy of the numbers to begin with.
We're still left with numbers that are so small that I would go as far as saying that they're practically meaningless anyway.

 

[brucep]

Huh? Who said anything about tidal accelerations? I just used Hamiltonian dynamics to get the relative acceleration from the potential energy. If you're going to argue, point out a math error, don't try to pick apart the concept. Also, re: the equation of motion derived from Newton's theory; to everyone else reading, brucep repeatedly mentioned the equation dr/dt=[2M/r]^1/2 in the other thread. I finally figured out that it's the equation for the velocity of a small test mass falling into a massive body from infinity. Note that I said "small test mass", because I was able to re-derive it by treating the large mass as fixed and setting kinetic+potential energy equal to zero for the falling mass. Like every other formula that contains M without m, this is an approximation; in reality, the large mass would not be fixed, which is what leads to the effect RJBeery and I are arguing for.

I agree, and I'm sure RJBeery would as well, assuming we're still just talking about Newtonian mechanics.

You're quite right here, but I don't know why you think center of mass is the important coordinate. It seems to me that the time it takes a dropped object to fall to Earth could be rigorously defined as the time it takes for that object to collide with the Earth, which will be slightly shorter for a heavier object. The fact that the system's center of mass will be in a different place relative to the point of collision is true, but irrelevant.

You've just reinforced why I think it's important, it also somewhat obviates {in my opinion} this whole 'falling towards a black hole' thing.

There are a whole lot of contexts that are being missed . For example, what Gallileo actually said, as I understand was that all objects fall at the same rate, in opposition to the aristotlean proposition which was popular at the time which was that the rate at which an object falls is dependent on its composition. Nothing that has been offered in this thread so far contradicts the observation that all objects fall at the same rate towards the earth. The only thing the OP has offered is that the earth falls towards more massive objects more quickly - well duh! But this is beside the point. To the best of my ability to discern Gallieo's comments were made, more or less, considering objects (essentially) falling through a bottomless pit. My point being that the rate at which the earth falls towards the objects is irrelevant to anything that gallileo said and the assertion that he is wrong - especially in the experiments described in the thread so far, would require measurements precise enough that quantum effects are going to become important. In fact the level of discrepancy where taling about here is... I don't have the words to describe it.

The average weight of a golfball is 45.93 g
The maximum weight of a Bowling ball is 7260g.
The mass of the earth is (5.97219x10^{27})g
The mean radius of the earth is 637,100,000 cm.

The center of gravity of a golfball and the earth, assuming the earth is a homgenous sphere {it isn't) with the golfball at an altitude of 1km is at a distance of (4.9x10^{-20})m from the center of the earth - this is about (\frac{1}{150}) of the charge radius of a proton.

A bowling ball is 158 times more massive than a bowling ball, so the center of gravity of a bowling ball and the earth, using the same assumptions as above, with the bowling ball at an alttitude of 1km is going to be displaced by approximately the charge radius of a proton.

But this is the maximum distance the earth can fall, the earth isn't even going to move that far - the bowling ball and golf ball can only fall a maximum of 1km before it hits the earth, and the earth and the bowling/golf ball stop moving. The actual distance the earth is going to move in the case of the golf ball is (\frac{1}{150} * \frac{1}{6371}) or on the order of one ten millionth of the charge radius of a proton, for the bowling ball it will be on the order of on ten thousandth of the charge radius of a proton.

Setting aside the fact that for the OP being correct requires taking gallileo's words out of the context they were originally phrased in.
Setting aside the obvious problems regarding accuracy of the numbers to begin with.
We're still left with numbers that are so small that I would go as far as saying that they're practically meaningless anyway.

Huh? Who said anything about tidal accelerations? I just used Hamiltonian dynamics to get the relative acceleration from the potential energy. If you're going to argue, point out a math error, don't try to pick apart the concept. Also, re: the equation of motion derived from Newton's theory; to everyone else reading, brucep repeatedly mentioned the equation dr/dt=[2M/r]^1/2 in the other thread. I finally figured out that it's the equation for the velocity of a small test mass falling into a massive body from infinity. Note that I said "small test mass", because I was able to re-derive it by treating the large mass as fixed and setting kinetic+potential energy equal to zero for the falling mass. Like every other formula that contains M without m, this is an approximation; in reality, the large mass would not be fixed, which is what leads to the effect RJBeery and I are arguing for.

I agree, and I'm sure RJBeery would as well, assuming we're still just talking about Newtonian mechanics.

You're quite right here, but I don't know why you think center of mass is the important coordinate. It seems to me that the time it takes a dropped object to fall to Earth could be rigorously defined as the time it takes for that object to collide with the Earth, which will be slightly shorter for a heavier object. The fact that the system's center of mass will be in a different place relative to the point of collision is true, but irrelevant.

That's the equation of motion for an object falling in the gravitational field [spacetime curvature]. It's also the escape velocity. The only forces present are tidal accelerations. The local spacetime curvature is infinitesimal along the inertial path of the object in free fall. So are the local tidal accelerations along the path. The tidal field is the first derivative of the metric curvature component which happens to be 2M/r also. Your math is irrelevant for understanding this natural phenomena since they're are no forces acting upon the inertial path of the free falling object. Any deviation due to tidal forces are immeasurable until the object reaches the center of the black hole. Your adding irrelevant physics in your derivation. This isn't quantum physics and the equation of motion isn't just for a test particle though the test particle is an easier way to model the events.

 

[Trippy]

That's the equation of motion for an object falling in the gravitational field [spacetime curvature].

Yes, I know. That would be part of the context of Gallileo's original statement that I was referring to.

 

[RajeshTrivedi]

....... Except you're not appreciating that if the Earth falls towards the bowling ball "more quickly" (than, say, towards the golf ball) then the bowling ball has a shorter distance to cover before it impacts the Earth....

This is pure non sense and unadulterated bullshit, if we restrict ourselves to Newtonian Mechanics..

RJbeery, before I answer your question (which is qualitatively answered in my post #36), let me tell you, you have taken all for a long jolly ride and now since you realized the problem in your OP, you first changed the track by chaining etc and then insist on closing the thread...Although I must acknowledge your post has all the ingredients to send many self styled experts to nearby High School for learning ABC of physics, as suggested by some one else also..

Coming to the solution of your Earth + Ball system...

The acceleration of the ball towards the surface of Earth a_o = g - a
The acceleration of the earth towards the ball a_e = a

(where a is the acceleration caused due to ball, your small value.)

(hope you have understood the pseudo force on the ball due to (Earth + ball) being the one body..)

Now dump these values in your formula t = Sqrt(2*d/(a_e+a_0)) then it becomes..

t = Sqrt(2*d/g) irrespective of mass of your ball or balls....no approximation, no negligible value, scope restricted as per your OP.

Where the hell Hamiltonian, curvature, relativity, complex calculus come into picture ??? And please give the authentic details of Ld professor who agreed that the time taken by the heavier ball is less, if it is a fact, he is producing some real good Physicists..I wish to hire them as English Teacher for my school...

 

[Neddy Bate]

I think this means RJBeery is correct.

 

[Dr_Toad]

Hurrah! Rajesh gives a better argument than Neddy. Why am I not surprised?

 

[Neddy Bate]

Hurrah! Rajesh gives a better argument than Neddy. Why am I not surprised?

I guess you are saying the total acceleration is not the sum of the two individual accelerations. So, I guess you think it is one or the other. If so, please tell us which individual acceleration is also the total acceleration. I suppose you think it is the one with the larger mass?

 

[tashja]

Tashja: http://www.sciforums.com/threads/galileo-was-technically-wrong.142700/

Is the author's argument correct?

Prof. Lacy:

Yes. Technically the author is right. Of course, in the time that a 1 kg ball falls 5 m the Earth falls toward the ball only 10^-24 m, or much less than the radius of a proton, so you'd have a hard time measuring the effect. But that is the reason that the orbital period of a planet in an orbit of a given radius depends not just on the mass of the Sun, but on the sum of the planet's mass and the Sun's mass.

John Lacy

 

[RajeshTrivedi]

Prof Lacy is considering this as a two body problem, thats why a reference to orbital motion etc....

To make it very clear, a ball of 100 Kg and another ball of 1 kg, both will take the same time to reach the surface of Earth if dropped from a height of 10 meter (as per OP).

The fundamental thing which is being ignored is that we are considering Ball, away and isolated from Earth while calculating the motion of Earth. The motion or acceleration of Earth, is also of Ball.

To make it further clear, because few guys cannot add up acceleration, downwards force on the Ball is Fb = mg-ma [where ma is the Pseudo force] and the force on the Earth [in isolation with the ball now] is Fe = ma [upwards]. So effective acceleration of the ball is g - a and that of Earth is a. Put this in RJBeery formula and you get the answer. Let me summarize..

1. Time taken is same irrespective of mass of Ball.
2. Earth travel (along with Ball) in upward direction is more if mass is higher.
3. Effective separation between the surface and the ball is not changed due to the motion of the Earth in upward direction.

 

[Fednis48]

Trippy: If you think RJBeery is being pedantic, I don't think even he could disagree with you. I think he pointed out an interesting technicality. Either way, the important part imo is that you're on board with him/me/NeddyBate/tashja regarding the physics. Honestly, I think you and RJBeery have been arguing past each other - rather than against each other - from the start.

That's the equation of motion for an object falling in the gravitational field [spacetime curvature]. It's also the escape velocity. The only forces present are tidal accelerations. The local spacetime curvature is infinitesimal along the inertial path of the object in free fall. So are the local tidal accelerations along the path. The tidal field is the first derivative of the metric curvature component which happens to be 2M/r also. Your math is irrelevant for understanding this natural phenomena since they're are no forces acting upon the inertial path of the free falling object. Any deviation due to tidal forces are immeasurable until the object reaches the center of the black hole. Your adding irrelevant physics in your derivation. This isn't quantum physics and the equation of motion isn't just for a test particle though the test particle is an easier way to model the events.

It's frustrating how you keep trying to argue against my math with just text. As far as I can tell, everything you have said so far relates to the motion of one particle through curved spacetime. But an object falling to Earth comprises a two-body system. If you treat the Earth as static, the whole effect RJBeery is talking about goes away. I guess it's possible that the effect goes away under relativistic corrections; if that's the case, I'd be very interested to see the derivation of it. But unless you can pinpoint what it is about relativity that counteracts the mass-dependence of the falling speed, it's a bit disingenuous to just dismiss my math on the grounds that I'm "adding irrelevant physics."

2. Earth travel [along with Ball] in upward direction is more if mass is higher.

This is completely, irredeemably wrong, and it's the main viewpoint I'm trying to quash. Especially since other posters are agreeing with you, please think long and hard about how you might justify this claim before you make it again.

 

[Aqueous Id]

I think this means RJBeery is correct.

What does total acceleration mean to you?

RJBeery, thanks for bringing my post over to this thread; I didn't mean to contribute to the derailing of a thread on black holes.

tashja's quote of Prof. Wasserman in post #45 is an impeccable qualitative description,

Based on what / how do you know? Did you notice Declan Lunny's remark about how she referred to Sgr A* ?

so I'm going to try and cover the math. One thing I hope everyone can agree on is the expression for the energy of the system, which is a combination of two kinetic energy terms and the Newtonian gravitational potential formula:

E=kinetic+potential=P^2/2M+p^2/2m+GMm/(X-x)

What assumptions have you made here?

Here, positions and momenta are along the earth<->object axis, capital {M, X, P} refer to the Earth's mass, position, and momentum, respectively, lowercase {m, x, p} refer to the same properties of the falling object, and G is the universal gravitation constant. Without loss of generality, I have assumed that Earth's position coordinate is higher than the object's; if the reverse is true, the denominator in the potential energy would be (x-X) instead. The equation for a system's energy is its Hamiltonian, so we can use Hamiltonian mechanics (http://en.wikipedia.org/wiki/Hamiltonian_mechanics) to determine the dynamics. We want to compare accelerations, so we're looking for the rate of change of the relative velocity:

What is your rationale for this? Why are you introducing the Hamiltonian here?

d/dt(v_rel)=d/dt(V-v)==d/dt(P/M-p/m)=dP/dt/M-dp/dt/m

Can you explain this?What is v_rel? What is V? v? What does the notation dP/dt/M mean?

Position and momentum are a canonical coordinate pair, so we can say dp/dt=-dH/dx.

This doesn't follow from the above.

Note that on the Wikipedia page and in textbooks, position is usually denoted as q instead of x. Plugging this into the above expression:
d/dt(v_rel)=-dH/dX/M+dH/dx/m=-Gm/(X-x)^2-GM/(X-x)^2=-G(m+M)/(x_rel)^2

So you end up back at the the law of universal gravitation. I don't understand why you didn't start there and then answer the issue RJ raised.

As RJBeery has been insisting, the result depends on both m and M.

That's an assumption, not a given. Part of the issue here is for RJ to tell us why he is using the 2-body problem at all.

If we assume that M>>m, we can drop m from the equation to good approximation; this is the "Galilean approximation".

But if you do that, RJ will insist that you are ignoring the deflection of the Earth by the bowling ball. So you've gone full circle.

edit: I anyone else experiencing an issue where anything written in parentheses compiles as TeX?

Yep. You can put your tex code inside the parens and it will render nicely.

Imagine one planet is chained to a magical wall which could not be moved.

Do you know the technical name for this? Do physics problems talk about bodies being "chained to a brick wall" or "floating", and what's the terminology for this?

. . . what do you think the outcome of this picture would be? Do you think the fall time of the free planet would be less than, greater than, or equal to the measured fall time if both planets
were free? Because the real answer is that this is what Galileo was describing.

Are you sure about that? What did Galileo know or believe about any constraints placed on the Earth? What assumptions, if any, was he making that you are not making--or vice versa? I'm sure you don't presume to be smarter than Galileo. So what is the central difference in his model from yours?

All objects from feathers to bowling balls to other planets TO BLACK HOLES would all be timed to descend to this chained planet at equal velocities from equal heights.

What has to happen in order for that to happen?

The fact that the Earth is free floating changes the math.

Who said the Earth is free floating? Isn't is constrained to its orbit? Isn't it constrained to its diurnal rotation, and isn't the centroid of the rotation constrained by the proximity and mass of the Moon? What does "free floating" even mean, and what does "chained to a brick wall mean"? And what changes the math that you introduced in the OP? Ignoring the question I asked you earlier about accuracy of the results, what assumptions have you made that are different from Galileo's assumptions?

So, RJBeery, Fednis48, Neddy Bate: was Galileo right or wrong? There seems to be hanging in the air a criticism that experienced people aren't able to solve this problem correctly. In case you're interested, I thought you might want to consider the following results of a study:

Results indicate that (a) the process of making the constraining assumptions needed to convert the real-world problem into a well-defined one was the most difficult step for all, and (b) only the participants who had prior experience making constraining assumptions were able to make the needed assumptions and solve the real-world problem. These findings suggest a need to support physics students develop this important skill.
​

http://onlinelibrary.wiley.com/doi/10.1002/sce.20295/pdf

An open question to all of the readers: Is there a correct response to RJBeery? Alternatively: what is the best answer to the claims made in the OP?

 

[rpenner]

This is not rocket science -- this is word problems.

If you neglect the motion of the Earth, all bodies fall at the same rate.
If you do a two-body solution in either GR or Newton's Universal gravitation, then the rate at which the two bodies close on each other is a function of the total mass. [In GR it's complicated when the masses get very large.] Thus heavier masses fall faster than lighter ones against a reference idealized sphere.
But since it's prohibitively expensive to move stuff into orbit, if you have three bodies, the lighter, the heavier and the Earth, when you drop the weights one at a time, exactly where did you place the other weight, eh?
If you study the limit as the smaller mass goes to zero, there is negligible effect on the falling rate by doubling the mass, so all human-scale bodies effectively fall at the same rate.
If you follow Galileo's reasoning about dropping two point masses at the same time then all quibbles relating to the Earth's motion go away and the two masses hit the ground at the same time.

But if you can't even read the communications of another person and parse those words for the author's intended meaning, asking questions when necessary to clarify, are you really having a conversation?

 

[RajeshTrivedi]

I have a simple open question to all...

1. A person starts at 0600 Hrs in the morning from a location A and reaches location B exactly at 0600 Hrs in the evening. We do not know his speed Vs time profile during this period. On the next day he again starts at 0600 Hrs from location B, follows the same route, and reaches location A exactly at 0600 Hrs in the evening. Again we do not know his speed vs time profile, but we know that his car got stuck and it took him exactly 60 minutes to repair the same, but still he managed to reach at A at 0600 Hrs in the evening. The question is : Would there be any location between A and B, where he was on both the days at the same time ? [time => our wrist watch time, no relativity]

 

[Motor Daddy]

I have a simple open question to all...

1. A person starts at 0600 Hrs in the morning from a location A and reaches location B exactly at 0600 Hrs in the evening. We do not know his speed Vs time profile during this period.

That's crazy talk. Not knowing a uniform acceleration, distance, and time?

Let's put some numbers to the entire deal, with locations in space. We'll use the x,y,z coordinates ((0,0,0)) as the point in space where the x, y, and z axis intersect. We'll use the definition of the meter to imagine ((and measure)) distance along the axis, and furthermore, we can use an expanding light sphere to find the time!! So we have locations, distances, and times, in 3 dimensions. Ready??

 

[RajeshTrivedi]

Fednish48,

this is how you commented on my argument of distance traveled by Earth being dependent on the mass of the Ball.

This is completely, irredeemably wrong, and it's the main viewpoint I'm trying to quash. Especially since other posters are agreeing with you, please think long and hard about how you might justify this claim before you make it again.

I give you mathematics which you are seeking...

A ball of mass m is lying on the surface of the Earth. Now this Ball is lifted up by a height of 10 meter from the surface. …..

The ball is dropped and it starts falling towards the radial line.…

Now the acceleration on the ball is in the downward radial direction and is equal to g…This also cannot be disputed [if we ignore change in g due to height]….So far so good.

The acceleration on the Earth in the radial direction towards ball is g*m/M……I hope you are with me on this too.

The catch…. This acceleration of the Earth [g*m/M] will also apply on the ball [you cannot isolate ball from the Earth]. You agree ?

So effective acceleration of Ball is [g – g*m/M]…..You agree ?

1. The distance travelled by Ball = ½ * [g-g*m/M]*t^2 ..... Can you deny ?

2. The distance travelled by Earth = ½ * [g*m/M]*t^2 ...... Can you deny ?

3. So 10 meter = ½ * g * t^2 [The other components get squared off]

Conclusion..

1. Ball moves less than 10 meter before striking the Earth.
2. Earth moves and distance is dependent on the mass of Ball [Sr# 2]
3. Time taken to strike the Earth is independent of mass of Ball.

Pt#3... Is what Galileo said and he did not know relativity !! He is not wrong !!

Now please tell me what do you want to quash ??

[In Electrical Engineering there is something called Grid Synchronization, 'origin' may throw some light on this, a massive Grid [called Rigid Grid] pulls along a smaller generator even if some hiccups are there in the voltage]. So this small movement of Earth will not cause any disturbance because Earth is otherwise rigidly synchronized with Solar System etc]