A question of physics

[Mekkala]

You must understand though that I'm not really convinced. I can clearly remember my physics things from school, and if I recall it correctly the outcome was that if the objects have the same shape, but a diffrent weight, the objects will fall to the ground in exactly the same speed, and will hit the ground in exactly the same time. Do you have any other source that might confirm your side of it?

I don't believe he's disagreeing with you on that. If he is, he's wrong, but my understanding has been that he's saying that as long as both objects are falling at the same time and from the same starting point, they will hit the ground at precisely the same moment -- correct, disregarding air resistance, which is not a factor that we can predict with the mathematical precision of gravity.

 

[Aeschylus]

There's two seperate issues at work here:

1) we have the issue relating to the OP and the accelartion due to gravity of the Earth this can be proved mathematically as long as you accept Newtonian physics

2) Air resistance which can again be proved mathematically, but as long as you are willing to accept certain formulas, for example this Nasa site has a formula for drag:
http://www.grc.nasa.gov/WWW/K-12/airplane/falling.html

So do you disagree with what I've said about the first issue, the second issue or both?

 

[Mistermystery]

I disagree that I'm utterly confused now... no wait.. I mean I agree with that. To be honest I need some sleep first, I'll try to sort out my mind tomorrow...

 

[JGMEERT]

There's two seperate issues at work here:

1) we have the issue relating to the OP and the accelartion due to gravity of the Earth this can be proved mathematically as long as you accept Newtonian physics

2) Air resistance which can again be proved mathematically, but as long as you are willing to accept certain formulas, for example this Nasa site has a formula for drag:
http://www.grc.nasa.gov/WWW/K-12/airplane/falling.html

So do you disagree with what I've said about the first issue, the second issue or both?

JM: In my answer I ignore air resistance since I don't believe that was what was confusing anyone. The part that seems to have people confused is the similar shape (different mass). This part should also not be confusing since both would accelerate at the same rate and hit the ground at the same time (ignoring air).

Cheers

Joe Meert

 

[ForeRunner]

I think everyone is confused.

He was referring to pull of the cannon balls on the earth. Basically, the earth and the Cannon balls move towards eachother. The earth accelerates the cannon balls equally towards it, the difference lies in the way the balls accelerate the earth towards themselves.

If you dropped both balls at the same time, there wouldn't likely be any difference, as it seems to me that the gravity of the two balls combine would pull on the earth. However, if you dropped both balls from the same height one after the other, the heavier one would accelerate the earth towards itself slightly faster due to its heavier mass. Essentially, the more massive object would touch the earth more quickly than the less massive object.

This effect is beyond our ability to measure though, so we cannot actually test it.

 

[Aeschylus]

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).

QED

 

[The Bellman]

I don't think so.. And I think this is a question that you can find out for yourself. You do have 2 coke bottles, right? fill one up with something heavy (like sand), and throw them both out of an elevated platform. Perhaps you can throw it from a really high building with someone else monitoring which bottle hits the ground first... If I recall my tests correctly they both hit the ground at the same time.

No, they won't. See everyone else's posts above. Testing won't help, the difference is so small as to be undetectable.

 

[vajradhara]

Namaste all,

ok... so let me see if i understand this properly..

if i drop Marlon Brando and Gary Coleman out a window, Brando will impact the ground first.

this is due to the mass of both Brando and Coleman having their own gravitational attractions to the earth and the earths gravational attractions to Brando and Coleman.

each object exerts its own gravitational attractions, omni-directionally, and thus, Brando is pulling Coleman towards him as they are falling!

is that close?

 

[The Bellman]

Yep this is true:

a) due to the reaction force

b) due to the fact that air resiatnce means in general that heavier objects fall faster.

(b) is completely wrong. Air resistance does NOT mean that heavier objects fall faster.

 

[The Bellman]

uh... Hold on.. let's take this one step at the time:

what do we have:

- 2 objects simmilar in size, diffrent from mass. (one empty, and one full coke bottle for instance)
- A certain height. (let's say for instance 3 meters. but this can also be 50.)
- Normal air (that means no vaccuum).

right?

Now if you drop both bottles at the exact same time, under the exact same conditions (meaning that there is not external windforce on the bottles) what will happen to those bottles? I think that they hit the ground at the same time because of the wind resistance. I fail to see what action/reaction on the earth has to do with it...

Wind resistance has nothing to do with it.

 

[Aeschylus]

Now to prove (2 - that air resistance means heavier objects (with all else being equal) fall faster.

The formula for drag from this site is given as:

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

this means that the instanous resultant force acting on an object (we're going to ignore my last post as relatsically the effects outlined in it are negligible) is :

mg - Cv^2

The instaneous acceleration is then:

a = g - Cv^2/m

unfortunately this gives us a nonlinear differential equation, but from this we notice that when our two objects have the same velocity the heavier one is subject to more acceleration as Cv^2/m term is less so we can reasonably conclude that the heavier object will reach the ground before the lighter object.

 

[JGMEERT]

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):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).

QED

JM: And that...is completely incorrect. They will (in the absence of air resistance) hit at exactly the same time. This is becoming an almost comical misunderstanding of gravity. Oops, sorry I see the 'same time same place exception'. Isn't that the case in question?

Cheers

Joe Meert

 

[Aeschylus]

(b) is completely wrong. Air resistance does NOT mean that heavier objects fall faster.

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.

 

[Aeschylus]

JM: And that...is completely incorrect. They will (in the absence of air resistance) hit at exactly the same time.

Cheers

Joe Meert

Okay which step was incorrect?

Yes I know that in all high school textbooks it says that with differnet mass dropped from the same height will, in the absence of air resitance hit the ground in the same time. This is not quite correct as the objects are considered as test particles and the active masses of the objects are effectively ignored. This gives a very, very good approximation (certainly for all parctical purposes) when M >> m, but it's still not quite the whole story.

 

[The Son of Him]

Sorry to bore you, but this question has nothing to do with creation or evolution - it's purely about science. I'm posting it here because this at least used to be a science forum, and I know there'll be several here who can answer it.

Two bodies of unequal mass fall at exactly the same rate (ignoring wind resistance). Virtually everyone knows this. In a vacuum, a cannon ball and a feather fall at the same speed.

But...imagine this. Two cannon balls, exactly the same size (so the same amount of wind resistance). 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? Obviously, the difference would be imperceptible (if it's true) - so it's purely a theoretical question, something I've wondered about for a while. Bodies fall toward the earth based on the earth's gravitational attraction...but doesn't the bodies' gravitational attraction (infinitesmal though it is in comparison to that of the earth) play a part, too, making heavier objects fall very slightly faster?

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

 

[Aeschylus]

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 acrried out by GALILEO GALILEI at the tower of ****A with SPHERES as pendulus thus discovering that all bodies fall at the same rate.
All this thanks to the LORD

I refer you back to post #46, if you treat the problem as a 2 body problem, you'll see this is not always the case. for example imagine situation if one of the balls had the same mass as the Earth.

 

[The Son of Him]

I refer you back to post #46, if you treat the problem as a 2 body problem, you'll see this is not always the case. for example imagine situation if one of the balls had the same mass as the Earth.

If you want to treat the problem with a ball of that dimension it does not change the outcome.
In any case if you feel confused , you should treat the problem like NEWTON would and refer everything to the CENTER of MASS of the system composed by both bodies and the symetries should be clear to understand

 

[Aeschylus]

If you want to treat the problem with a ball of that dimension it does not change the outcome.
In any case if you feel confused , you should treat the problem like NEWTON would and refer everything to the CENTER of MASS of the system composed by both bodies and the symetries should be clear to understand

What I am saying is you're essientially ignoring the accelration due to the gravity of the ball of the Earth, I don't see the relevance of the centre of mass, you're going to have to be more precise, remeber I've also taken the gradient of the accelartion into account.

 

[The Son of Him]

What I am saying is you're essientially ignoring the accelration due to the gravity of the ball of the Earth, I don't see the relevance of the centre of mass, you're going to have to be more precise, remeber I've also taken the gradient of the accelartion into account.

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 !!!!!!!!!!

 

[JGMEERT]

Okay which step was incorrect?

Yes I know that in all high school textbooks it says that with differnet mass dropped from the same height will, in the absence of air resitance hit the ground in the same time. This is not quite correct as the objects are considered as test particles and the active masses of the objects are effectively ignored. This gives a very, very good approximation (certainly for all parctical purposes) when M >> m, but it's still not quite the whole story.

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 deabte and I can't believe anyone wants to argue otherwise. In simple terms, the more massive object will speed up more slowly than the less massive object resulting in an equivlanecy of the time it takes to 'hit' the ground. I've not really encountered quite a misunderstanding as this.

Cheers

Joe Meert