A question of physics

[JGMEERT]

You are absolutely correct that they will hit at the same moment when you drop both at the same time and in the same place. However, this is not true when you drop them at different times.

JM: Well DUH! I was never arguing about whether or not you dropped them at different times. No one needs math to figure out the answer to that question. The original question was as follows:

One is hollow, so it weighs 1/5 the amount the other one weighs. Drop them both from a height. They hit the ground at precisely the same time. Or do they? Does not the heavier one fall faster because its greater mass exerts a greater gravitational pull on the earth?

JM: I was assuming that the question was about dropping them from the same height at the same time. Otherwise, the question is no fun at all to answer.

Cheers

Joe Meert

 

[Mekkala]

JM: Well DUH! I was never arguing about whether or not you dropped them at different times. No one needs math to figure out the answer to that question.

Cheers

Joe Meert

Yes, Joe, you do need math if you understand what is being said here, by Aeschylus and myself. I think that you have the impression that we're saying that they will hit the ground at different times if you drop them at different times, because they started falling at different times. While that's true, that's not what's being said.

If you drop a heavy cannonball, and measure the time it takes to hit the ground, and then you drop a light cannonball from the exact same height later, and measure the time it takes to hit the ground, you will find that the two times are very slightly different (if you had instruments sensitive enough, that is). This is due not to the acceleration of the cannonballs towards the earth, which is the same for both, but due to the acceleration of the earth towards the cannonballs, which is not the same for both.

 

[JGMEERT]

If you drop a heavy cannonball, and measure the time it takes to hit the ground, and then you drop a light cannonball from the exact same height later, and measure the time it takes to hit the ground, you will find that the two times are very slightly different (if you had instruments sensitive enough, that is). This is due not to the acceleration of the cannonballs towards the earth, which is the same for both, but due to the acceleration of the earth towards the cannonballs, which is not the same for both.

JM: That is wrong and no different from calculating whether they would hit at the same time if dropped them at the same time. You can do all the math you want, but if you start out with an incorrect concept, it's not going to help you. You should both ask for a refund for your physics classes. Gentleman's bet is still on.


Cheers

Joe Meert

 

[The Son of Him]

Yes, Joe, you do need math if you understand what is being said here, by Aeschylus and myself.
If you drop a heavy cannonball, and measure the time it takes to hit the ground, and then you drop a light cannonball from the exact same height later, and measure the time it takes to hit the ground, you will find that the two times are very slightly different (if you had instruments sensitive enough, that is). This is due not to the acceleration of the cannonballs towards the earth, which is the same for both, but due to the acceleration of the earth towards the cannonballs, which is not the same for both.

WRONG !! YOUR ARE MOVING THE POINT OF ORIGIN OF YOUR SYSTEM OF REFERENCE WHEN TRYING TO CONSIDER THE PULL EXERTED ON EARTH!!
JOE IS RIGHT !!!

 

[Mekkala]

WRONG !! YOUR ARE MOVING THE POINT OF ORIGIN OF YOUR SYSTEM OF REFERENCE WHEN TRYING TO CONSIDER THE PULL EXERTED ON EARTH!!
JOE IS RIGHT !!!

No, I am not. The point of origin when using the gravitational formula is the center of mass of the system, NOT the center of mass of the earth, or of the cannonball.

 

[Mekkala]

JM: That is wrong and no different from calculating whether they would hit at the same time if dropped them at the same time. You can do all the math you want, but if you start out with an incorrect concept, it's not going to help you. You should both ask for a refund for your physics classes. Gentleman's bet is still on.


Cheers

Joe Meert

Gentlemen's bet? Joe, I've already asked a physics professor to confirm my work. You are refusing to believe me when I tell you what he said -- so again, I think you should go ask one yourself, because there is no way for you to determine that I'm telling the truth on the Internet -- nor do I have any way to prove to you that I'm telling the truth.

If you like, I can try to have a professor email you, but again, how will you know that it's really a physics professor emailing you? Go to a local university and ask someone, ok? That's the only way you'll be convinced, I promise you.

 

[Tomk80]

WRONG !! YOUR ARE MOVING THE POINT OF ORIGIN OF YOUR SYSTEM OF REFERENCE WHEN TRYING TO CONSIDER THE PULL EXERTED ON EARTH!!
JOE IS RIGHT !!!

Son of him, could you please stop typing with capslock on all the time. I think you point out some usefull things now and then, but I just get immensily annoyed every time someone does that. It makes it harder for me (and probably some other too) to take your posts seriously this way.

 

[The Son of Him]

No, I am not. The point of reference when using the gravitational formula is always the center of mass of the system, NOT the center of mass of the earth, or of the cannonball.

THEN DRAW GRAPHICS AND YOU SHOULD SEE THAT EARTH INERTIA OPOSES THE FORCE EXERTED BY EITHER BALL SO EARTH AND EITHER BALL COME IN CONTACT AT THE SAME EXACT TIME !!!!!!!!!!!!!!!!!
OTHER WISE YOU ARE PROPOSING THAT THESE OBJECTS TRAVEL THE SAME DISTANCE IN DIFFERENT AMOUNTS OF TIME, WICH MEANS DIFFERENT ACCELERATION FOR EACH BALL, AND THAT MEANS DIFFERENT VALUES OF g(GRAVITATIONAL ACCELERATION) WICH IS NOT TRUE !!!!

 

[Onesiphorus]

11 pages of this is all well and good, but.....

Does anyone know the average wing velocity of a coconut-laden swallow?



-- I seek the Truth... nothing more.

 

[The Son of Him]

Son of him, could you please stop typing with capslock on all the time. I think you point out some usefull things now and then, but I just get immensily annoyed every time someone does that. It makes it harder for me (and probably some other too) to take your posts seriously this way.

I apologize for that I got so into the problem that I did not see that.

 

[Tomk80]

I apologize for that I got so into the problem that I did not see that.

's okay. I just get icky if I see things like these (or rants, man I hate rants (uneless they are my own of course )).

 

[The Son of Him]

's okay. I just get icky if I see things like these (or rants, man I hate rants (uneless they are my own of course )).

I did not mean to have caps-lock ,it was an accident.Sorry

Tom, what is your take on the little problem ?

 

[Mekkala]

THEN DRAW GRAPHICS AND YOU SHOULD SEE THAT EARTH INERTIA OPOSES THE FORCE EXERTED BY EITHER BALL SO EARTH AND EITHER BALL COME IN CONTACT AT THE SAME EXACT TIME !!!!!!!!!!!!!!!!!
OTHER WISE YOU ARE PROPOSING THAT THESE OBJECTS TRAVEL THE SAME DISTANCE IN DIFFERENT AMOUNTS OF TIME, WICH MEANS DIFFERENT ACCELERATION FOR EACH BALL, AND THAT MEANS DIFFERENT VALUES OF g(GRAVITATIONAL ACCELERATION) WICH IS NOT TRUE !!!!

Are you proposing that the earth's inertia is so great that it does not move at all?

 

[Tomk80]

I did not mean to have caps-lock ,it was an accident.Sorry

Tom, what is your take on the little problem ?

I do not know. I'm only following the debate from the side, without thinking to much about it. Got other debates here to focus on (and I'm expected to produce some work also ). The debates are a welcome change from the IPCC-reports I'm now going through Thanks for asking though.

 

[Aeschylus]

Again, this is the proof, if you wish to show it is incorrect you must address it directly:

Okay I'll clear this up this is why a heavier object always hits our idealized Earth regradless of air resitsnace (i.e. (1 in my last post):

Newtons universal law of graviationa states that (note M can be thought of as the mass of the Earth):

F = -GMm/r^2

this means that both the Earth and our object will be subject to this same force but in opposite directions, i.e. the object will feel force F push it towards the Earth and the Earth will feel force F push it towards the object.

Now we consider things from the non-inertial refernce frame of the Earth. The Earth is being accelarted with force F, so we will have to use something called D'Alembert's principle which allows us to treat the Earth's frame as static by introducing a 'fictious' inertial force acting on the object.

From D'Alembert's principle we can now write the accelartion of the object in our non-inertial reference frame:

a = -G(M + m)/r^2

So we now have the insatneous accelartion of the object with respect to the Earth.

If we now call R the radius of the Earth and we call h the height of the object above the Earth's surface and R the radius of the Earth we can find the average accelartion by inergrating the last equation with respect to r with limits h + R and R and dividing by h to find the average accelartion, which is:


1/h * G(M + m)[1/h - 1/(h + R)]

From this equation it should be obvious that the average accelartion is directly proportional to M + m, so the higher the value of m the higher the average accelartion of the object.

It should also be obviosu that the higher the average accelartion the sooner that the object will hit the Earth, therfore a heavier object will hit the Earth befpre a lighter object (again excepting the case when tey are dropped at the same time and the same place).

 

[The Son of Him]

Are you proposing that the earth's inertia is so great that it does not move at all?

My friend: In both cases with either ball the EARTH moves towards the CENTER of mass of the system.(If you want to take that point as origin of system of reference).
But be consistent where you put your point of origin of your system of reference.
As a matter of fact you could put the system of reference in the center of the balls and consider that only the earth moves !!!!!!!!!!!!!!!!!

 

[Aeschylus]

My friend: In both cases with either ball the EARTH moves towards the CENTER of mass of the system.(If you want to take that point as origin of system of reference).
But be consistent where you put your point of origin of your system of reference.
As a matter of fact you could put the system of reference in the center of the balls and consider that only the earth moves !!!!!!!!!!!!!!!!!

But the Eartjh accelartes more when the object is heavier which is the whole point. As you can see I've created a suitable frame of reference above in which to solve the equations, as time is never transformed unless you're doing relativistic physics the time measured for the objects to hit the ground will be the same in all refernce frames, with heavier object hitting the ground in the shortest time period.

 

[JGMEERT]

Ok, before asking a physicist, let me make sure I have the question right.

(1) if two balls of different mass are dropped from the same height at the same time, they will hit the earth at the same time (ignoring air). There will be absolutely no difference between the two.

if however

(2) One ball is dropped 5 minutes before the other and we measure the time (assuming different masses and ignoring air) then the more massive one will hit the ground slightly faster than the less massive one.

Is this the claim?

Cheers

Joe Meert

 

[Aeschylus]

You have to be careful when posing the question, something along the lines of:

"If an object is dropped from a certain height and we measure the time for it to hit the ground and then repeat exactly the same experiment, but this time with an object of greater mass, measuring the time again; will the two times be equal (assuming the system is isolated, no air resistance, modelling the object and the Earth as point masses, etc.), taking the accelartion of the earth due to objects' active gravitational mass into consideration?"

I assure you though my proof is sufficent as it considers all factors involved in the idealized set-up.

 

[JGMEERT]

You have to be careful when posing the question, something along the lines of:

"If an object is dropped from a ceratin height and we measure the time for it to hit the ground and then repeat exactly the same experiment twice measuring the time will the two times be equal (assuming the system is isolated, no air resiatnce, modelling the object and the Earth as point masses, etc.), taking the accelartion of the earth due to objects active graviational mass into consideration?"

I assure you though my proof is sufficent as it considers all factors involved in the idealized set-up.

JM: your question says nothing about different masses.