Galileo was technically WRONG

[RajeshTrivedi]

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

Go on man !! You can use your motor !!

 

[Motor Daddy]

Go on man !! You can use your motor !!

No need to talk about motors for this one, we are using standard definitions of distance and time. We are using a grid coordinate system of locations with exact pinpoint accuracy for the entire infinite volume of space, and infinite continuum of time. I mean, 100% accuracy, so nothing is left to the imagination!! No smoke and mirrors, and no BS!

 

[RJBeery]

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?

rpenner, please don't introduce another variable here. Also, I believe you'll find that the time of descent still differs, being minimized when the masses of the Earth and the falling object are equal.

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.

Again, in my opinion you are clouding the issue a bit. We aren't talking about limits and effective approximations; we're talking about technicalities to the infinite extent, which is why I stated that Galileo was technically wrong.

You are probably considered THE math guy on this forum. Why don't you just explicitly verify that my mathematical analysis in the OP is correct and let the doubters try to refute it?

 

[rpenner]

Because in word problems, the math doesn't matter if you haven't got the right mental model of what is being discussed.

 

[danshawen]

You just can't do this sort of math without an underlying supposition of the absolute space of Euclid's geometry. In the real, actual universe you are modeling, it's like trying to nail jello to a wall, only without the wall or a nail. Tools consisting of bags of hammers will not suffice, yet hammers can easily find their way to what you are seeking.

 

[RajeshTrivedi]

RJBeery OP

Start with variables

a_e.gb = acceleration of the Earth towards the golf ball = .00008029 m/s^2
a_e.bb = acceleration of the Earth towards the bowling ball = .00963481 m/s^2
a_e.bh = acceleration of the Earth towards the black hole = 1.348E13 m/s^2

Third one is not so obvious, but probably he assumed a point mass for BH. How did he get the first two ?? His mathematics only ??

 

[RJBeery]

Because in word problems, the math doesn't matter if you haven't got the right mental model of what is being discussed.

That's a bit of a dodge. If you aren't going to verify the mathematical analysis please supply the wording of the "right mental model" which would make Galileo TECHNICALLY right in regards to the following claim:

FALL TIMES TO EARTH ARE INDEPENDENT OF MASS.

 

[RajeshTrivedi]

have you conveniently dodged my post # 80.

You have really taken every one on a bull ride..

RJBeery dodge...
a_e.gb = acceleration of the Earth towards the golf ball = .00008029 m/s^2
a_e.bb = acceleration of the Earth towards the bowling ball = .00963481 m/s^2


Actual dodge

a_e.gb = acceleration of the Earth towards the golf ball = 0.0000000000000000000000000822 m/s^2

a_e.bb = acceleration of the Earth towards the bowling ball = 0.00000000000000000000000024 m / s^2

Now Beery, please find out the time difference for yourself, by adding above value to 9.82 m / s^2

This is as far as your calculations only, which otherwise is wrong.

And mind you, you are declaring a man wrong, who possibly had no time measurement device at that time, you should extend a thought apology to him.

 

[OnlyMe]

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?

Rpenner, the clarity you often bring to a discussion continues to impress me.

So, I was been wrong when I expanded the discussion to moons and a two earth situation, which would change the total gravitational potential.

My initial disagreement was centered on how RJ presented the issue in another thread... And comparing that to the mountains of experimental evidence, testing the underlying physics. Even then if one accepts the problem as completely theoretical, as it has to be.., from the OP, the clarity of your explanation above is accurate.

The problem in the context of real world experimental tests and any implied real world application {where my issues originated} involving the earth and bowling balls, golf balls or even a car etc., is that none of those objects can be treated as separate and/or added mass, in the field equations. The mass of all of those objects begin and end as part of the earth's total mass. In practice the total mass of the gravitational field, and so the gravitational potential, when a bowling ball or golf ball is dropped, is the same.., and the only thing that is at issue is how the mass {inertial resistance} of the dropped object affects its acceleration toward the earth. Even when any change in the earth's motion is considered it will not change the drop time. It may affect the distance the dropped object travels, if the mass of that object is large enough. The distance the earth moves during the drop being subtracted from the initial distance of separation.

The force being the same in both cases {as long as the dropped object begins as part of the earth's total mass}, the mass of the dropped object has no affect on the time it takes it to hit the ground. Two objects of sufficiently different masses, could travel different distances and have different terminal velocities, and yet experience the same acceleration, as long as their mass began as part of the total mass of the earth. Any difference in the distance traveled and terminal velocity, would be offset by the velocity of and the distance, of the earth's motion during the drop.

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?

Seems you nailed at least a part of my misinterpretation of the discussion.

 

[OnlyMe]

​

That's a bit of a dodge. If you aren't going to verify the mathematical analysis please supply the wording of the "right mental model" which would make Galileo TECHNICALLY right in regards to the following claim:

FALL TIMES TO EARTH ARE INDEPENDENT OF MASS.

RJ, rpenner began with,

If you neglect the motion of the Earth, all bodies fall at the same rate.

Which is what Galieo claimed. Galileo was accepting the earth as the FoR!

But I don't believe it is a technical point. It is accurate based on frame of reference and as I pointed out in my earlier post, for Galileo all masses involved were part of the earth's total mass. The gravitational potential during the drop, is the same for all masses, in that case.

When viewed as a hypotheical, what I understand as the intent of your OP is accurate, but you are treating the mass of each object as independent.., the mass of the bowling ball is in addition to the mass of the earth, which is not a real world situation.

In reading rpenner's post I realized I got carried away with the whole moon drop thing, which would change things considerably. But it still seems to me that you are using hypothetical conditions to make real world conclusions, that are not consistent with experimental results.

Galileo is technically wrong, only if you get your bowling ball from some where other than here on the earth.

 

[Fednis48]

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?

Right as always, rpenner. The bit about the second mass having to be on the Earth while we drop the first was especially clever; I would not have thought of that. [And if my math in post 53 is right, the relative acceleration only depends on M+m, so it won't change even slightly between the two cases.] As far as I can tell, though, there are at least 3 people in this thread who still think that 2-body fall time is independent of the small body's mass, whether because the small body shares the large one's acceleration or because the large body accelerates less in a rest frame. Those are physics errors, not word games, so I will continue arguing against them. Case in point...

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 ?

This is not correct. We are looking at the effects of the Earth's gravity on the ball and vice-versa, so we have to treat them as separate objects. To see one error with your method, multiply the Earth's acceleration by M and the ball's by m to get the forces acting on them. If the forces are not equal in magnitude, you've violated Newton's third law, so something is wrong.

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

I have no idea what you mean here. All I was saying is that the quote tashj posted is a good qualitative explanation of the result I calculated.

What assumptions have you made here?

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

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

This doesn't follow from the above.

I introduced the Hamiltonian because it's a good way to solve very general problems in classical mechanics. Newton's law of gravity tells us that gravitational potential energy between two bodies is (\frac{GMm}{X-x}), and the kinetic energy for a two-body system is just the sum of the individual kinetic energies, (\frac{P^2}{2M}+\frac{p^2}{2m}). The sum of these gives the Hamiltonian in terms of {x,p} and {X,P} which are canonical coordinate pairs. If you look at the Wikipedia page, you'll find formulas relating canonical coordinates via derivatives, which are what I used in my calculation. I used V and v to describe the velocities of the Earth and the ball, and (V_{rel}) to describe the relative velocity between the two. The term dP/dt/M is the time-derivative of P divided by M; I would go back and put it in TeX for clarity, but the edit button seems to have expired.

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.

I ended up at the law of universal gravitation in the relative coordinate, which is not a formula that's easy to find. My goal was to get there from something everyone could agree on [the energy form of Newton's law of gravity] so no one could dispute the result.

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.

Why would you not use the 2-body problem? We're talking about the gravitational attraction between a ball and the Earth; that's two bodies.

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.

RJ would acknowledge, as I did, that the Galilean approximation is an extremely good approximation. But it is still just an approximation, which is the whole point of this thread.

So, RJBeery, Fednis48, Neddy Bate: was Galileo right or wrong?

In response to rpenner, I actually have to change my tune on this one. Galileo was right, if only because any object we could lift up and drop would have to be deducted from the mass of the Earth, so M+m is a constant in the relative acceleration I calculated: (\frac{d}{dt}v_{rel}=\frac{G[M+m]}{x_{rel}^2}). But as RJ formulated the problem in the OP, with a fixed mass of the Earth, the larger ball will indeed fall faster.

 

[Motor Daddy]

Lifting a bowling ball a distance away from the axis is doing work over time, and that is power. Power is not free, there is a cost. The cost of power to lift a bowling ball a specific distance away is much greater than lifting a golf ball the same distance away. End result is that lifting a bowling ball consumes more power than lifting a golf ball. If the power was powered by gasoline from the earth, the total mass of earth and bowling ball after the lift of the bowling ball is less than what is started with. The golf ball and earth total mass is greater than the bowling ball's total after the work is done. Now what?? I hope people aren't trying to lift heavy things for free... (rolls eyes)

 

[OnlyMe]

Lifting a bowling ball a distance away from the axis is doing work over time, and that is power. Power is not free, there is a cost. The cost of power to lift a bowling ball a specific distance away is much greater than lifting a golf ball the same distance away. End result is that lifting a bowling ball consumes more power than lifting a golf ball. If the power was powered by gasoline from the earth, the total mass of earth and bowling ball after the lift of the bowling ball is less than what is started with. The golf ball and earth total mass is greater than the bowling ball's total after the work is done. Now what?? I hope people aren't trying to lift heavy things for free... (rolls eyes)

Hmmm?

Exactly how does burning gasoline to produce kinetic energy convert any of the original mass to energy. Unless you are referring to the photons of light that escape to space, all of the original mass remains part of the system, in a different form.

 

[Motor Daddy]

Hmmm?

Exactly how does burning gasoline to produce kinetic energy convert any of the original mass to energy. Unless you are referring to the photons of light that escape to space, all of the original mass remains part of the system, in a different form.

Force!!

It takes more work to lift a 10 lb object 10 feet than it does to lift a 1 lb object 10 feet. More work lifting the bowling ball means more fuel is used, which means less total mass than the golf ball after the lift. More fuel was used in lifting the bowling ball, so there is less left for the acceleration! Duh?

 

[RajeshTrivedi]

Fednis48 response to my

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 ?

This is not correct. We are looking at the effects of the Earth's gravity on the ball and vice-versa, so we have to treat them as separate objects. To see one error with your method, multiply the Earth's acceleration by M and the ball's by m to get the forces acting on them. If the forces are not equal in magnitude, you've violated Newton's third law, so something is wrong.

This kind of lapse take place when you jump to relativity and Hamiltonian without properly understanding the basic Newtonian Physics...

You did not even think for a moment, that ball is spinning with Earth, ball is revolving with Earth.....but ball is not accelerating with Earth !! Thats quite selective Physics you know....

I see no violation of 3rd law of Newton, It is just that you are messing up a simple problem involving motion of a particle which is a part of a bigger moving object.

 

[OnlyMe]

Force!!

It takes more work to lift a 10 lb object 10 feet than it does to lift a 1 lb object 10 feet. More work lifting the bowling ball means more fuel is used, which means less total mass than the golf ball after the lift. More fuel was used in lifting the bowling ball, so there is less left for the acceleration! Duh?

Other than the light produced, which in a gasoline engine is contained, all of the energy is chemical. No atoms are converted to energy just a rearrangement of atoms in molecules. The mass remains constant.

You seem to be confusing force, energy and mass, as if they were all interchangeable. If that were the case you could lose weight just by having someone push you down! No wait, you might wind up loosing weight that way, depending on the severity of your injuries... But that would be a different process and would not change the mass of the earth.

While all forms of mass can be thought of as an expression of energy, not all forms of energy can be thought of as mass.

And no I still cannot pull out the reference for that paraphrased quote, but it was from a discussion/commentary on the equation E=mc^2.

 

[Motor Daddy]

Other than the light produced, which in a gasoline engine is contained, all of the energy is chemical. No atoms are converted to energy just a rearrangement of atoms in molecules. The mass remains constant.

I didn't say the word energy, I said the word power, which is work/time. You seem to want to eat the cake and have it too. Not!

 

[RJBeery]

And mind you, you are declaring a man wrong, who possibly had no time measurement device at that time, you should extend a thought apology to him.

Rajesh, frankly, you ramble a lot. I'm not dodging you; you simply haven't given anything substantive to respond to. If you believe I made an error in my math, and your correction technically makes Galileo RIGHT, then I will grant Galileo a "thought apology". The numbers that you provided still do not do that, because the accelerations are not EQUAL.

 

[OnlyMe]

I didn't say the word energy, I said the word power, which is work/time. You seem to want to eat the cake and have it too. Not!

Below is the portion of your earlier post I was responding to,

If the power was powered by gasoline from the earth, the total mass of earth and bowling ball after the lift of the bowling ball is less than what is started with. The golf ball and earth total mass is greater than the bowling ball's total after the work is done.

You are in the above asserting that burning gasoline in some kind of engine, converts mass to energy {in the form of a force} which is lost to the system. Burning gasoline does not change the mass of the system it just rearranges the molecular structure.

Lift the object with an atomic bomb and you would then change the mass of the system by a very small amount.

Whether work done changes the mass of a system is not solely dependent on the force exerted. It depends on where the force comes from or how it is produced. Burning gasoline does not change the mass of the earth, even when it does change the mass of a car.., as that car spews exhaust out its tail pipe. BTW the mass of the exhaust is the same as the mass of the gasoline and the volume of air, consumed to produce it.

Granted there is probably a very small amount of mass lost to space in infrared heat loss, but really!

 

[Motor Daddy]

Below is the portion of your earlier post I was responding to,



You are in the above asserting that burning gasoline in some kind of engine, converts mass to energy {in the form of a force} which is lost to the system. Burning gasoline does not change the mass of the system it just rearranges the molecular structure.

Lift the object with an atomic bomb and you would then change the mass of the system by a very small amount.

Whether work done changes the mass of a system is not solely dependent on the force exerted. It depends on where the force comes from or how it is produced. Burning gasoline does not change the mass of the earth, even when it does change the mass of a car.., as that car spews exhaust out its tail pipe. BTW the mass of the exhaust is the same as the mass of the gasoline and the volume of air, consumed to produce it.

Granted there is probably a very small amount of mass lost to space in infrared heat loss, but really!

Does the earth ever lose any heat to its environment? Does entropy increase the more work that is being done? You need to step back and think about what the heck energy is, because from my observations, you're clueless, or you have an agenda to pursue.