A question of physics

[Aeschylus]

You have to put your system of reference either on the center of the EARTH ,make it inmovable and refer the movement of the balls to it, or if you want to take in consideration the acceleration of the earth to these bodies you will have to put the system of reference outside the CENTER of EARTH in wich case the best point to put it will be the CENTER of MASS of the system according to NEWTON.
You can not refer this ball moving to EARTH and then move EARTH to the ball while conserving the CENTER of EARTH as system of reference, IS either one or the other !!!!!!!!!!

Exactly, I can't keep my frame of refernce as the Earth and consider it to be moving BUT this is why I invoked d'Alembert's principle which allows me to treat the Earth as 'static' with the cost of having to consdier inertial forces.

Doing the problem from an inertial from of reference just makes it alot more difficult to solve.

 

[Aeschylus]

JM: It's not just high school textbooks, it's also college and theoretical physics. The fact that in the absence of resistance both accelerate equally and hit and the same time is not really a matter of debate and I can't believe anyone wants to argue otherwise.

Cheers

Joe Meert

Which textbooks and are they specifically taking into consideration both the active mass of the object and the differing accelartions?

If this were the case two body problems would be alot easier to solve.

plus can you show the specifc fault in the proof.

 

[USincognito]

I love all you guys sweating the details. Reminds me of a part of my life long ago and far away there I used to put all this theory into practice.

Disclaimer - I'm not in this photo, it's just representative.

 

[JGMEERT]

Which textbooks and are they specifically taking into consideration both the active mass of the object and the differing accelartions?

If this were the case two body problems would be alot easier to solve.

plus can you show the specifc fault in the proof.

JM: Any freshman physics text should be easy enough to understand. I can't believe how much bandwidth has been spent on such a simple misunderstanding

Cheers

Joe Meert

 

[Aeschylus]

JM: Any freshman physics text should be easy enough to understand. I can't believe how much bandwidth has been spent on such a simple misunderstanding

Cheers

Joe Meert

Consider this situation: what happens when the mass of the ball is so large that the mass of the earth no longer become impornat, is the ball's accelaration still only governed by the mass of the Earth?

 

[The Bellman]

Yes, it does though I have not comphrehensively proved it you should still be able to see that air resistance does mean all things being equal heavier objects fall faster.

No, it doesn't. Air resistance is COMPLETELY unaffected by mass. It is ONLY relevant in regard to different shapes of objects.

 

[The Bellman]

NO my friend,
Both balls hit the ground at the same time and at the same speed , provided they have the same outside geometry (OR should I say topology) and that they fall from the same height.
The force between earth and the heavier ball will be greater but so will be its inertia or opposition of the body to change its state of motion.
This was the basis for EISTEINS Equivalence Principle wich he used to give general form to accelerated movement for its special relativity thus rounding GENERAL RELATIVITY.
PLUS Your experiment has already been carried out by GALILEO GALILEI at the tower of PISA with SPHERES as pendulus thus discovering that all bodies fall at the same rate.
All this thanks to the LORD

For EVERYBODY who keeps saying "try it out, do an experiment", please, stop saying it. The entire point is that if they do fall at different speeds, the difference would be infinitesmal, and could NOT be observed.

 

[Dirac_Notation]

This is the specific fault in your proof--your statement that "C is a constant." The parameter Cd you mention in your formula is not in fact a constant for all objects. Your work obscures any possible mass-dependence of this parameter. (And in fact, drag isn't always proportional to velocity squared either. Fluid dynamics is complicated.) To complete the proof, one of two things must happen

option1: The parameter Cd must be proven to be linearly proportional to mass. Then the acceleration is independent of mass, and the two objects fall at precisely the same rate and strike at precisely the same time.

option2: The parameter Cd must be proven to have any other dependence on mass. Then the acceleration is not independent of the mass, and the two objects can strike at different times.

Enjoy yourselves!

D = Cd * 0.5 * r * v^2 * A

(see the site for an explanation of the formula)

Now we're only inetrested in the effects of mass so we can say:

D = C*v^2

Where C is a constant

 

[The Bellman]

Hmm...so after 50-odd posts, nobody knows...or rather, everybody knows, but they all disagree

 

[Dirac_Notation]

What does this mean? The mass of the earth is always important. It's always part of what determines the gravitational field acting on the ball.

Consider this situation: what happens when the mass of the ball is so large that the mass of the earth no longer become impornat, is the ball's accelaration still only governed by the mass of the Earth?

 

[Aeschylus]

This is the specific fault in your proof--your statement that "C is a constant." The parameter Cd you mention in your formula is not in fact a constant for all objects. Your work obscures any possible mass-dependence of this parameter. (And in fact, drag isn't always proportional to velocity squared either. Fluid dynamics is complicated.) To complete the proof, one of two things must happen

option1: The parameter Cd must be proven to be linearly proportional to mass. Then the acceleration is independent of mass, and the two objects fall at precisely the same rate and strike at precisely the same time.

option2: The parameter Cd must be proven to have any other dependence on mass. Then the acceleration is not independent of the mass, and the two objects can strike at different times.

Enjoy yourselves!

Yes I am aware that that formula is only a simplifcation as any single formula would be that hope to concisely the force due to air resistance as for example at low speeds it is better to model the dependance on v than v^2.

However, it is general result of most models of air resiatnce that when you have simlairly shape objects of different masses the heavier one will reach the ground first.

 

[Aeschylus]

What does this mean? The mass of the earth is always important. It's always part of what determines the gravitational field acting on the ball.

When the mass of the ball is something silly like say a googleplex the times of the Earth we can ignore the active mass of the Earth without much worry.

 

[Aeschylus]

No, it doesn't. Air resistance is COMPLETELY unaffected by mass. It is ONLY relevant in regard to different shapes of objects.

Well brer ket's comment aside, if the force of air resiatnce is independt of mass you should see that this cause heavier objects to fall faster.

 

[The Bellman]

Well brer ket's comment aside, if the force of air resiatnce is independt of mass you should see that this cause heavier objects to fall faster.

Sorry, can you expand on that a bit? I'm not sure I understand.

 

[Aeschylus]

two objects of masses of masses of 50kg and 0.5 kg and an air resisantce 1N, take g as 10N:

resulatnt force on the object of mass 50kg:

= 500 - 1 = 499N

therfore the accelration is:

499/50 = 9.98 m/s^2

the resultant force on the object of 0.5 kg is

5 - 1 = 4 N

therfore it's accelartion is:

4/0.5 = 8 m/s^2

 

[JGMEERT]

Hmm...so after 50-odd posts, nobody knows...or rather, everybody knows, but they all disagre

JM: Some of us know. Others think they know and try to sound sophisticated in their answer. Eventually, those with a willingness to learn and comprehend will quickly come to the same realization as I did when I first waded through all these posts, that most have no clue.

Cheers

Joe Meert

 

[The Bellman]

two objects of masses of masses of 50kg and 0.5 kg and an air resisantce 1N, take g as 10N:

resulatnt force on the object of mass 50kg:

= 500 - 1 = 499N

therfore the accelration is:

499/50 = 9.98 m/s^2

the resultant force on the object of 0.5 kg is

5 - 1 = 4 N

therfore it's accelartion is:

4/0.5 = 8 m/s^2

Okay, so forget air resistance for a moment. Assume a vacuum. Two objects, same size, but one has weighs a ton and the other weighs a billion tons. Drop them both from 1 mile above the earth. Will they hit the ground at PRECISELY the same time? Or different times, but sufficiently close that we wouldn't notice it?

 

[The Bellman]

Hmm...so after 50-odd posts, nobody knows...or rather, everybody knows, but they all disagre

JM: Some of us know. Others think they know and try to sound sophisticated in their answer. Eventually, those with a willingness to learn and comprehend will quickly come to the same realization as I did when I first waded through all these posts, that most have no clue.

Cheers

Joe Meert

Perhaps that's true...but the hard bit is finding out which are the ones that DO have a clue.

 

[Mistermystery]

For EVERYBODY who keeps saying "try it out, do an experiment", please, stop saying it. The entire point is that if they do fall at different speeds, the difference would be infinitesmal, and could NOT be observed.

Well I'm sorry I tried to help you.

 

[JGMEERT]

Okay, so forget air resistance for a moment. Assume a vacuum. Two objects, same size, but one has weighs a ton and the other weighs a billion tons. Drop them both from 1 mile above the earth. Will they hit the ground at PRECISELY the same time? Or different times, but sufficiently close that we wouldn't notice it?

JM: Precisely the same time.

Cheers

Joe Meert