View Full Version : Do heavier objects fall faster?


mountainhare
08-02-04, 06:07 AM
I know that objects with the same surface area but different densities fall at the same speed in a vacuum.

However, there is actually a bit of an argument between my science teachers.

One science teacher thinks that heavier objects do fall faster (he provided calculations). The other one (appears) to think that this is nonsense.

I decided to try it for myself. I dropped a piece of paper, and a piece of cardboard with similiar SA. The cardboard fell much faster.

I could explain this by stating that heavier objects have more force to 'push' air molecules out of the way.
In otherwords, heavier objects are not as affected by air resistance.

Whaddya think? :confused:

Blindman
08-02-04, 06:12 AM
Yes.. The acceleration due to gravity is a result of the total mass of the two objects.

dmcm01
08-02-04, 06:29 AM
is this in air or in a vacuume?

i know that the speed it falls at is affected by weght, gravity and air resistance,
.....(put very simply)....air reistance is created by surface area. Ther similar SA effectivly counts for nothing as it is the same on both if u get me?
so the heavier object should fall faster as that is the only 'difference' between the two objects.
also paper and card is not a good way to test it as the paper tends to float about and go wierd places, try two boxes with one empty and one full of stuff.....

and yes, put in school terms N=Kg X M/s/s
force = wieght x acceleration
so the heavier one would have more force or 'pushing power'

Blue_UK
08-02-04, 06:44 AM
I hate this one, it's so obvious.


Force = G * (m1.m2 / d^2)
G = constant
m1 = mass of earth
m2 = mass of object
d = distance between them
Force does not = acceleration (F = ma)

Heavy objects will have a force applied by grav proportional to their masses. So objects will fall towards the earth at the same speed.

However,

The earth will also be drawn to the objects in question with accel proportional to mass of object. So assuming you drop one after the other and have a very accurate timer to hand you will notice the heavier object fall faster if you reference the earth as a fixed point.

Galileo eat your heart out.

Rick
08-02-04, 07:17 AM
Was leaning Tower Of P. the actual place where Galileo performed his experiment?...I wonder...

bye!

James R
08-02-04, 08:17 AM
The force of air resistance on an object is usually given as:

F = apAv<sup>2</sup>

where a is a constant which depends on the shape of the object, A is the cross-sectional area of the object, p is the density and v is the speed.

The rate of fall in air is affected both by gravity and the air resistance on an object. Since the air resistance is somewhat complex, it is not surprising that paper and cardboard fall at different rates.

Astronauts on one of the Apollo missions actually did the experiment on the Moon in a vacuum, using a feather and (I think) a rock. Both fell at the same rate and hit the ground at the same time.

AD1
08-02-04, 08:32 AM
Astronauts on one of the Apollo missions actually did the experiment on the Moon in a vacuum, using a feather and (I think) a rock.

It was Dave Scott, commander of Apollo 15, and it was a falcon feather and a geology hammer.

Blue_UK
08-02-04, 09:17 AM
It was Dave Scott, commander of Apollo 15, and it was a falcon feather and a geology hammer.

Had he conducted the experiment by dropping the hammer and the feather seperatly there would have been a (immeasurably small, but present) difference.

MacM
08-02-04, 09:39 AM
Had he conducted the experiment by dropping the hammer and the feather seperatly there would have been a (immeasurably small, but present) difference.

I have seen this arguement before and have posted it here about a year or so ago. It was done by a guy named Eric. I had his e-mail and used to be in contact with him but due to having to have my hard drive replaced after a crash, I have lost contact.

He made a strong arguement for the different rate of fall but I still am not convienced. The issue of the earth moving toward the free-falling object is a good one but I suspect that the formula is correct and that it represents the "Closure Rate" between the earth and the object, which means they actually close at the same rate and if the imperceptable motion of the earth could be measured (which it is in a timing free-fall to contact) the times would be the same.

The (what I believe is a falacy) idea that an object falls at a given (equal rate) but that the earth responds to different mass with different counter motion creates the illusion that the rate should be different. But if you treat the problem symmetrically and view it as a formula for "Closure Rate" between the object and Earth then the result changes and is consistant with the idea that the free-fall time to contact is the same.

Take a bowling ball (bb) and a soccer ball (sb) of the same size and shape. Compared to the earths mass the difference in the earth's response to such objects is negligable but whatever it is, it is the closure rate that determines the apparent free-fall rate.

That is expressed as F = G*m1*m2/r^2 and F=ma. The truth of the matter can better be visualized if you forget earth and consider cases of bb's and sb's in deep space.

Two bb's will close in a given amount of time. Two soccer balls will close in the same time given the same seperation. Likewise you can see that if you release one of each, the bb and a sb will close at the same rate regardless of which one is to the left or right on your graph and assumed to free-fall. Granted the amount of motion of the objects varies with its mass but the closure time is the same.

mountainhare
08-03-04, 06:39 AM
Oh, I tried a different experiment. I dropped a cube of metal and wood, and they both hit the ground at the same time.

Contradicts my paper and cardboard experiment (pretty crap, I admit. But it's all I had on hand.)

Pete
08-03-04, 07:28 AM
I've seen an extremely rigorous treatment of this problem somewhere on the web. Let me see if I can find it.

I recall that it considered the case of dropping the two items as the same time on different parts of the globe!

Pete
08-03-04, 07:40 AM
Nope... can't find it quickly.

But! I did find this gem published in the European Journal of Physics in 1987:

Aristotle was right: heavier objects fall faster (http://www.iop.org/EJ/abstract/0143-0807/8/2/006)
Donoghue, JF & Holstein, BR, European Journal of Physics, Vol 8 Issue 2, April 1987

Blue_UK
08-03-04, 07:58 AM
MacM, I have to admit I don't really understand the last paragraph of your post.

I will use the two forumlas you have given to show that 'closure time' will be shorter for objects of greater mass.

For simplicities sake, lets make up the constants so they are easier to work with:
Earth = 100kg
Bowling ball = 10kg
Soccer ball = 1 kg
Distance (r) = 1m
G = 1 (instead of 6.6742 ◊ 10^−11)

So we have our made up world, Nonceworld.

Force = G. m1.m2 / r^2

Bowling Ball

Force = 1 . 100 . 10 / 1 = 1000 Noncicles of force

F = ma, so a = 1000/10 = 100 m/s^2

Soccer Ball

Force = 1 . 100 . 1 / 1 = 100 Noncicles of force

F = ma, so a = 100/1 = 100 m/s^2

So they accelerate at the same rate towards earth

THE EARTH

OK, this bad boy when placed a metre away from the BB
F = ma, so a = 1000/100 = 10 m/s^2

and from the SB
F = ma, so a = 100/100 = 1 m/s^2

So the earth accelerates towards the BB faster than towards the SB.

And so the heavier object lands first.

This is Nonceworld, so the figures will be different in the real world but only in scale. And if the objects were dropped at the same time, they would both land at the same time, it's only if dropped seperately that this happens.

MacM
08-03-04, 09:05 AM
MacM, I have to admit I don't really understand the last paragraph of your post.

I will use the two forumlas you have given to show that 'closure time' will be shorter for objects of greater mass.

For simplicities sake, lets make up the constants so they are easier to work with:
Earth = 100kg
Bowling ball = 10kg
Soccer ball = 1 kg
Distance (r) = 1m
G = 1 (instead of 6.6742 ◊ 10^−11)

So we have our made up world, Nonceworld.

Force = G. m1.m2 / r^2

Bowling Ball

Force = 1 . 100 . 10 / 1 = 1000 Noncicles of force

F = ma, so a = 1000/10 = 100 m/s^2

Soccer Ball

Force = 1 . 100 . 1 / 1 = 100 Noncicles of force

F = ma, so a = 100/1 = 100 m/s^2

So they accelerate at the same rate towards earth

THE EARTH

OK, this bad boy when placed a metre away from the BB
F = ma, so a = 1000/100 = 10 m/s^2

and from the SB
F = ma, so a = 100/100 = 1 m/s^2

So the earth accelerates towards the BB faster than towards the SB.

And so the heavier object lands first.

This is Nonceworld, so the figures will be different in the real world but only in scale. And if the objects were dropped at the same time, they would both land at the same time, it's only if dropped seperately that this happens.

Ok. I surrender. :D I agree it appears to be so. I remember that I also agreed with Eric at the time but had forgotten the details of his presentation which were simular to yours and was swayed to think it was in error.

My last paragraph really wasn't that important in that it merely showed the closure rate between two masses were always idendtical but the inclusion of the third mass restores the arguement.

Blue_UK
08-03-04, 09:28 AM
:)

One thing I don't understand though, is Pete's link. I have no knowledge of quantum mechanics what so ever. I've heard of spin and understand what it means to say 1/2 spin, but it obviously contradicts what our brains would like us to think of the world. I'll wait until uni before pretending to know anything about that.

MacM
08-03-04, 09:48 AM
:)

One thing I don't understand though, is Pete's link. I have no knowledge of quantum mechanics what so ever. I've heard of spin and understand what it means to say 1/2 spin, but it obviously contradicts what our brains would like us to think of the world. I'll wait until uni before pretending to know anything about that.

Yea, I'm not into the mathematical solutions either but I couldn't get to the paper. It requires subscription.

Pete
08-03-04, 08:05 PM
I could neither understand the abstract nor access the paper...

Alpha
08-03-04, 09:40 PM
All their papers are unavailable ATM:
"Brief interruption to IOP Electronic Journals due to essential maintenance
Access to Electronic Journals will be temporarily interrupted to enable vital maintenance work to be undertaken. This work is part of a programme of updates designed to maximise the reliability and performance of IOP's Web services."

Don't see any copies elsewhere either.

Facial
08-03-04, 09:47 PM
Hmmm... I wonder what the terminal velocity of a 30lb bowling ball is...

shmoe
08-03-04, 10:11 PM
I could access the paper just fine. The paper has the abstract in a language that I don't know (looks like Spanish?) in addition to the English version. I got as much out of the Spanish side as I did the english. :bugeye:

Brandon9000
08-04-04, 08:48 AM
Aside from air resistance, all objects near the surface of the earth fall at an acceleration of:

GM/r^2

Where G is the gravitational constant, M is the mass of the Earth, and r is the distance between the falling object and the center of the Earth.

This is approximately 9.8 meters per second squared, and it is the same for all objects, regardless of their mass.

J_Wilson
08-04-04, 01:52 PM
That is not right Brandon. There is an equation that looks something like that to give you force. f=ma so the acceleration is f/m. We've already established this.

Brandon9000
08-04-04, 01:57 PM
That is not right Brandon. There is an equation that looks something like that to give you force. f=ma so the acceleration is f/m. We've already established this.
A = F/m where F, m, and A refer to the falling object.

F = (GMm)/r^2 where M = mass of Earth and m = mass of the falling object.

Plugging the 2nd equation into the first, which is equivalent to dividing the 2nd by m, gives, A = GM/r^2, as in my original post, regardless of the mass of the falling object. This particular acceleration is called "the acceleration due to gravity." Note that the only mass which appears in this equation is the mass of the Earth.

Incidentally, you can find this material in virtually every high school Physics book in the world.

Blue_UK
08-05-04, 08:40 AM
Brandon, check one of my above posts (lots of colour) - we all agree that objects accel towards earth at the same rate, it's the earth's accel towards the object that changes!

Brandon9000
08-05-04, 10:22 AM
Brandon, check one of my above posts (lots of colour) - we all agree that objects accel towards earth at the same rate, it's the earth's accel towards the object that changes!
This is so, but the acceleration of the Earth toward the object is infinitessimal. Someone may find this interesting theoretically, but the Earth's acceleration is so tiny as to be of no practical significance. The force exerted on the Earth by the object is identical to the force exerted on the object by the Earth, but the Earth has incomparably greater mass.

Blue_UK
08-05-04, 01:13 PM
I agree.

Needless to say I am more interested in the actual physics than the practical observation. :)

[/thread]

newbie56k
08-11-04, 11:14 PM
What r u guys doing, the ratio of force to mass is the same for large and small objects.
Force = mass x acceleration (force is equal to acceleration, w=ma [same thing])

98N Force
--------- = 9.8m/s/s
10kg Mass

4900N Force
----------- =9.8m/s/s
500Kg Mass

See the acceleration is the same, for large and small. I got most of this from my midterm exam notes for physics and i got a 103% in it.
People think that heavier objects fall faster because it hurts much more when it lands on their foot.

curioucity
08-11-04, 11:45 PM
The problem of roughly mixing force and pressure, and density too, maybe...

MacM
08-11-04, 11:55 PM
What r u guys doing, the ratio of force to mass is the same for large and small objects.
Force = mass x acceleration (force is equal to acceleration, w=ma [same thing])

98N Force
--------- = 9.8m/s/s
10kg Mass

4900N Force
----------- =9.8m/s/s
500Kg Mass

See the acceleration is the same, for large and small. I got most of this from my midterm exam notes for physics and i got a 103% in it.
People think that heavier objects fall faster because it hurts much more when it lands on their foot.

You need to be careful in what you stipulate. Falling faster meaning F=ma of a free falling object, it is correct to say they all fall at the same rate.

However, even getting that 103% on your mid-term fails to acknowledge that the earth indeed also moves toward the object and does so to a greater extent for a more massive object such that the closure rate and contact with the earth from an initial height (in theory) actually contacts earth sooner because the earth moved more toward the heavier object.

I refer you to "Blue UK's" post on 8/3/04 at 6:58 AM above. He shows you mathematically why this is the reality (even though it is immeasureable).

vslayer
08-13-04, 07:09 AM
the objects are both being pulled by the same amount of force(gravity) so the only factor to slow them down is wind resistance, if you dropped a ten ton weight and a parachute in a vacuum they would land at the same time.
tho only extro factor is the force on imparct, in which case you would multiply the mass by the amount of gravity.

in coclusion:

during a freefall with no resistance, the force of all objects are equal, it is only when they land that their mass or surface area becomes a factor

MacM
08-13-04, 08:36 AM
the objects are both being pulled by the same amount of force(gravity) so the only factor to slow them down is wind resistance, if you dropped a ten ton weight and a parachute in a vacuum they would land at the same time.

tho only extro factor is the force on imparct, in which case you would multiply the mass by the amount of gravity.

in coclusion:

during a freefall with no resistance, the force of all objects are equal, it is only when they land that their mass or surface area becomes a factor

I normally tend to agree with you on things but here you have made some errors.

1 - Free falling objects do NOT all have the same force. i.e. - a 1 gram ball falls under a 1 gram force. A 10 kg ball falls under a 10 kg force. That is the meaning of the equivelence principle between inertia and gravity. Since objects acclerate by amounts of force which are always equal to the mass the rate of acceleration is the same but not the accelerating force.

2 - However it can be seen that objects only contact the earth at the same time in such tests if the test is conducted simultaneously. And in that case both objects contact the earth sooner than if either had been tested independantly.

3 - Since the force on the object is 1 gram or 10 kg causing their acceleration, the complimentary force on earth must be equal. It is obvious that if you apply a series of forces to earth with a 10,000/1 ratio of magnitude you are going to get an acceleration of the earth toward the objects which have a 10,000/1 magnitude. That is to say the earth moves 10,000 times as far toward the free-falling object which has the 10 kg mass.

Since the earths motion in response to free-falling objects is different in each case the closure rate and contact of the free falling object with earth MUST change and the time is not the same.

vslayer
08-14-04, 08:53 AM
but lets say the earths gravity has a power of 10
the earth can only pull things at a speed of 10 because that is all its power allows
no matter how much an object weighs it can only ever go as fast as 10
the only thing to make a light object fall slower is the fact that it cannot displace as much air as a heavier object

in a vacuum both objects would be pulled at a speed of 10, and since they do not need to displace any other particles there is nothing to give the heavier one an advantage

MacM
08-14-04, 09:58 AM
but lets say the earths gravity has a power of 10
the earth can only pull things at a speed of 10 because that is all its power allows
no matter how much an object weighs it can only ever go as fast as 10
the only thing to make a light object fall slower is the fact that it cannot displace as much air as a heavier object

in a vacuum both objects would be pulled at a speed of 10, and since they do not need to displace any other particles there is nothing to give the heavier one an advantage

What you are missing is this: (Air resistance ignored).

If filmed against a scaled backdrop (i.e. - a tape measure and a clock) where the tape hangs from a point in space above earth (not attached to earth) and a light and heavy object are dropped individually, you could then see that each crossed markings of the tape at identical times.

But what you miss is that while these objects are in free-fall the earth is either moving (F=ma) toward theses objects with a force that is either heavy or light. Clearly the earth during such free-fall time will move more under the influence of the force from the heavy object. Hence the heavy object ultimately travels less distance before contacting the earth because the earth move closure to it that it did when the lighter oject was free-falling.

Because the havy object must travel less distance (at the same speeds and acceleration) as the lighter object it requires less time for free-fall to contact with earth.

Now if we make the scale a solid ruler mounted to earth one would see the collective closure rate between earth and the free-falling objects and low and behold we would see that heavy objects were falling faster in such a test.

Paul T
08-14-04, 06:26 PM
What you are missing is this: (Air resistance ignored).

If filmed against a scaled backdrop (i.e. - a tape measure and a clock) where the tape hangs from a point in space above earth (not attached to earth) and a light and heavy object are dropped individually, you could then see that each crossed markings of the tape at identical times.

But what you miss is that while these objects are in free-fall the earth is either moving (F=ma) toward theses objects with a force that is either heavy or light. Clearly the earth during such free-fall time will move more under the influence of the force from the heavy object. Hence the heavy object ultimately travels less distance before contacting the earth because the earth move closure to it that it did when the lighter oject was free-falling.

Because the havy object must travel less distance (at the same speeds and acceleration) as the lighter object it requires less time for free-fall to contact with earth.

Now if we make the scale a solid ruler mounted to earth one would see the collective closure rate between earth and the free-falling objects and low and behold we would see that heavy objects were falling faster in such a test.

MacM, I would say those are BS. Say those two objects were let to fall at the same time. Let ignore other falling objects, consider just those two objects and also, of course, assume that earth movement toward those objects is signaficant at all. Shouldn't the earth accelerate toward the falling object at one certain rate (not difference rate for each falling object)? Force resultant to earth is due to both of the falling object. You can't expect earth to move at one accelleration rate to object A and another to object B, while A and B move side by side at the same acceleration.

Pete
08-14-04, 09:23 PM
Hi Paul,

Shouldn't the earth accelerate toward the falling object at one certain rate (not difference rate for each falling object)? Force resultant to earth is due to both of the falling object. You can't expect earth to move at one accelleration rate to object A and another to object B, while A and B move side by side at the same acceleration.

A larger mass requires greater force than a smaller mass to accelerate at the same rate.
If A and B are of different masses, but experience the same acceleration, then they must be subject to different forces.
This means that the Earth also experiences greater force when the larger mass object is falling.

A few notes:
Objects falling side-by-side is a different problem. Since they are falling together, the Earth is moving toward both. You also need to consider the attraction between the objects...
The problem changes again depending on exactly how far away from each other are the two objects falling at the same time...
The problem changes again if you consider the direction of separation between the objects, as any North-South separation will involve Coriolis forces, and attraction between the objects will have curious effects on East-West separation...
When considering the objects' effect on the Earth, you should also consider tidal effects. The near part of the Earth is attracted much more strongly than the far part, so the Earth will be stretched. This means that the Earth's material properties (bulk modulus etc) will need to be considered...
Further complicated by differences in material properties throughout the Earth...

Of course, all these problems are insignificant beside local variations in gravity. For reasonably sized objects, such things as random air mass movements (ie weather) will have more effect on the fall rate than any of the above mentioned issues.

Paul T
08-14-04, 10:34 PM
Hi Pete



A larger mass requires greater force than a smaller mass to accelerate at the same rate.
If A and B are of different masses, but experience the same acceleration, then they must be subject to different forces.
This means that the Earth also experiences greater force when the larger mass object is falling.


This is certainly okay to me.



Objects falling side-by-side is a different problem. Since they are falling together, the Earth is moving toward both.


This was exactly the case mentioned by MacM. Therefore, earth should accelerate to both object at one rate only.

Even if the film was taken one at a time for object A and B, which has different mass, there is one thing remain the same...the system center of mass. If we don't consider the variation of gravitational acceleration due to the change of distance, the acceleration of object A and B relative to the system center of mass is not affected by the rate of earth acceleration toward the object (or the system center of mass). Therefore, I didn't see MacM's idea indicating that somehow heavier object fall faster. He just picked the wrong reference point.



You also need to consider the attraction between the objects...
The problem changes again depending on exactly how far away from each other are the two objects falling at the same time...
The problem changes again if you consider the direction of separation between the objects, as any North-South separation will involve Coriolis forces, and attraction between the objects will have curious effects on East-West separation...
When considering the objects' effect on the Earth, you should also consider tidal effects. The near part of the Earth is attracted much more strongly than the far part, so the Earth will be stretched. This means that the Earth's material properties (bulk modulus etc) will need to be considered...
Further complicated by differences in material properties throughout the Earth...

Of course, all these problems are insignificant beside local variations in gravity. For reasonably sized objects, such things as random air mass movements (ie weather) will have more effect on the fall rate than any of the above mentioned issues.

I think we can put a side all those issues for the time being.

MacM
08-14-04, 11:31 PM
MacM, I would say those are BS. Say those two objects were let to fall at the same time. Let ignore other falling objects, consider just those two objects and also, of course, assume that earth movement toward those objects is signaficant at all. Shouldn't the earth accelerate toward the falling object at one certain rate (not difference rate for each falling object)? Force resultant to earth is due to both of the falling object. You can't expect earth to move at one accelleration rate to object A and another to object B, while A and B move side by side at the same acceleration.

1 - First you must understand that the affect only occurs if the objects are dropped independantly at different times.

2 - The mass of Earth is 5.9742E24Kg

3 - If dropped independantly a 1 gram mass places a 1 gram force on earth during free-fall. F = ma, a = F/m:

a = 0.001kg/5.9742kg = 1.6738E-28

4 - If you now drop a 10kg mass during free fall it generates a force on Earth of 10kg:

a = 10kg/5.9742E24kg = 1.6738E-24

The acceleration by the earth (however miniscule) during a free-fall period of a mass is 10,000 times greater for a mass 10,000 times as large. The acceleration of the free-falling masses is always the same because the accelerating force equals the inertial mass.

5 - Dropped simultaneously the collective mass in free-fall is 10.001kg and induces a force of 10.001kg on the Earth :

a = F/m = 10.001kg/5.9742E24kg = 1.6740E-24

In this case both objects contact the earth at the same time but the total time is less than if either are dropped independantly.

The point that people seem to be missing is that this is "Closure Rate" from point of free-fall to contact. The velocity of free-falling objects relative to the point of initial release is always the same but the velocity relative to earth varies because earth also moves (at differnt rates) towards the free-falling objects. The magnitude of motion during free-fall is inversly proportional to the masses involved. That is the total distance moved is divided between the Earth and the object in proportion to their masses.

If a 10kg object is placed 16 feet (using rounded generic numbers here) above earth, due to the formula used we expect it to travel 16 feet in one second and make contact but the reality is that it only free-falls for 1 second minus the 1.6738E-24 acceleration time reduced because Earth moved in response to the 10kg force.

For the 1 gm object it free-falls for 1 second minus 1.6738E-28 acceleration time. You can see that the reality is that the closure rate is less for heavier objects, even though it is true that the acceleration of the object and its terminal velocity relative to its free-fall origin are the same for either mass. The impact velocity with Earth however is greater and the time period of the free-fall is less because Earth acclerates, hence moves greater distance towqrd heavier objects and attains a greater velocity during the free-fall.

MacM
08-14-04, 11:37 PM
Hi Pete

This is certainly okay to me.

This was exactly the case mentioned by MacM. Therefore, earth should accelerate to both object at one rate only.

You have clearly mis-read my post. I stated objects falling simultaneously will contact Earth at the same time but that that time will be less than if they were dropped indenpendantly at different times.


Even if the film was taken one at a time for object A and B, which has different mass, there is one thing remain the same...the system center of mass. If we don't consider the variation of gravitational acceleration due to the change of distance, the acceleration of object A and B relative to the system center of mass is not affected by the rate of earth acceleration toward the object (or the system center of mass). Therefore, I didn't see MacM's idea indicating that somehow heavier object fall faster. He just picked the wrong reference point.

Not at all. You are picking a different reference point than is used in the F = G * m1 * m2 / r^2 and F = ma formulas. See my previous response to your post to me above. The "Closure Rate" (time of free-fall) is less for heavier objects.

Paul T
08-15-04, 12:11 AM
1 - First you must understand that the affect only occurs if the objects are dropped independantly at different times.


Okay. Let's go with that situation.



2 - The mass of Earth is 5.9742E24Kg

3 - If dropped independantly a 1 gram mass places a 1 gram force on earth during free-fall. F = ma, a = F/m:

a = 0.001kg/5.9742kg = 1.6738E-28

4 - If you now drop a 10kg mass during free fall it generates a force on Earth of 10kg:

a = 10kg/5.9742E24kg = 1.6738E-24

The acceleration by the earth (however miniscule) during a free-fall period of a mass is 10,000 times greater for a mass 10,000 times as large. The acceleration of the free-falling masses is always the same because the accelerating force equals the inertial mass.

5 - Dropped simultaneously the collective mass in free-fall is 10.001kg and induces a force of 10.001kg on the Earth :

a = F/m = 10.001kg/5.9742E24kg = 1.6740E-24

In this case both objects contact the earth at the same time but the total time is less than if either are dropped independantly.

The point that people seem to be missing is that this is "Closure Rate" from point of free-fall to contact. The velocity of free-falling objects relative to the point of initial release is always the same but the velocity relative to earth varies because earth also moves (at differnt rates) towards the free-falling objects. The magnitude of motion during free-fall is inversly proportional to the masses involved. That is the total distance moved is divided between the Earth and the object in proportion to their masses.


This is a meaningless argument with respect to "Do heavier object fall faster?" Your argument doesn't prove anything. It's like:

At one time, you walk slowly toward a car accelerating toward you at rate 9.81m/s<sup>2</sup>, then at other time you run toward that car and you argued that when you run, the car closure rate is higher -- equivalent to having higher acceleration. This argument has not much value as it prove nothing about whether the car acceleration increase because you are running (the car acceleration, of course, not affected by your running!). You just picked an incorrect reference point to assess the problem.



If a 10kg object is placed 16 feet (using rounded generic numbers here) above earth, due to the formula used we expect it to travel 16 feet in one second and make contact but the reality is that it only free-falls for 1 second minus the 1.6738E-24 acceleration time reduced because Earth moved in response to the 10kg force.

For the 1 gm object it free-falls for 1 second minus 1.6738E-28 acceleration time. You can see that the reality is that the closure rate is less for heavier objects, even though it is true that the acceleration of the object and its terminal velocity relative to its free-fall origin are the same for either mass. The impact velocity with Earth however is greater and the time period of the free-fall is less because Earth acclerates, hence moves greater distance towqrd heavier objects and attains a greater velocity during the free-fall.

Your argument, as I said tell us nothing about whether heavier object fall faster or not...hence, it's a pointless argument.

MacM
08-15-04, 01:55 AM
Okay. Let's go with that situation.

This is a meaningless argument with respect to "Do heavier object fall faster?" Your argument doesn't prove anything. It's like:

At one time, you walk slowly toward a car accelerating toward you at rate 9.81m/s<sup>2</sup>, then at other time you run toward that car and you argued that when you run, the car closure rate is higher -- equivalent to having higher acceleration. This argument has not much value as it prove nothing about whether the car acceleration increase because you are running (the car acceleration, of course, not affected by your running!). You just picked an incorrect reference point to assess the problem.

Your argument, as I said tell us nothing about whether heavier object fall faster or not...hence, it's a pointless argument.

Not at all. I clearly pointed out the arguement; including the fact that from the reference point of the start of free fall the acceleration and velocity of all objects are the same.

I also pointed out the FACT that the time to impact with the earth IS NOT THE SAME. Should you actually set up a test capable of measuring this insignifigant difference you will find that due to the variable closure rate your test will yield results which show that the heavier objects impact the earth sooner.

I didn't pick an incorrect reference point I am simply showing that it is only one view and that a timed free-fall over an equal distance is less for the heavier object.

Your test being referenced from earth will result in a velocity calculation which is higher. The momentum of impact if you measure it at the earth's surface will be greater.

Remember, all velocity is relative. If earth is the rest reference for the test (which it generally and normally is), then you stand corrected.

Paul T
08-15-04, 08:51 AM
Not at all. I clearly pointed out the arguement; including the fact that from the reference point of the start of free fall the acceleration and velocity of all objects are the same.

I also pointed out the FACT that the time to impact with the earth IS NOT THE SAME. Should you actually set up a test capable of measuring this insignifigant difference you will find that due to the variable closure rate your test will yield results which show that the heavier objects impact the earth sooner.

I didn't pick an incorrect reference point I am simply showing that it is only one view and that a timed free-fall over an equal distance is less for the heavier object.


I agree on this. But, see below:



Your test being referenced from earth will result in a velocity calculation which is higher. The momentum of impact if you measure it at the earth's surface will be greater.

Remember, all velocity is relative. If earth is the rest reference for the test (which it generally and normally is), then you stand corrected.

The same wrong way of thought as what you did before (when you talked about accelerating rocket). A reference point for velocity must be an inertial reference frame. Earth, in this respect, become a non-inertial reference frame and therefore should not be used as a reference point. Picking a fixed point such as the origin of the objects (or the system center of mass) would not yield any differential of velocities or acceleration. That's why I said, you had picked the wrong reference point in your assessement.

MacM
08-15-04, 09:51 AM
I agree on this. But, see below:

The same wrong way of thought as what you did before (when you talked about accelerating rocket). A reference point for velocity must be an inertial reference frame. Earth, in this respect, become a non-inertial reference frame and therefore should not be used as a reference point. Picking a fixed point such as the origin of the objects (or the system center of mass) would not yield any differential of velocities or acceleration. That's why I said, you had picked the wrong reference point in your assessement.

Well we clearly disagree on this .

1 - I have yet to see anyone use a hypothetical point in space as a reference point for calculating the relative speed of a free falling object.

If they did however then and only then would you see the acceleration and terminal velocity, time to impact, momentum, etc., always come out correctly because you would see from that vantage point not only the free-falling mass accelerate but also the earth.

2 - All prior reference to the free-fall acceleration, etc., have been Earth.

If you use Earth as your reference and could actually measure the minute difference doing so causes then you find that heavier objects accelerate faster and contact earth sooner.

To try and put this issue into proper perspective.

1 - We all agree that the mathematics are generally valid.

2 - It is the general claim being made for the experiments and data that is false.

The following link correctly notes the flaw regarding the fact that the earth also moves and points out that conclusions based only on the acceleration of the mass are only pseudo correct.

3 - Since Galileo, Newton virtually all tests are stipulated for the simualtaneous free-fall of two objects side by side, in which case the conclusion is correct. They contact earth at the same time.

4 - However, there is ample proof and evidence that tests based on earth as the reference point where it is claimed that light and heavy objects attain the same velocity in free fall to the earth are equal is invalid.

All motion is relative. All velocity is relative and by using the earth as their reference it becomes a "0" rest reference and its motion toward a free-falling object compounds the net velocity, distance traveled/time or basis for acceleration, do not support the arguement that light and heavy objects hit the earth at the same time.

So the math is correct but the general statement or conclusion made from that are mis-stated.

http://encarta.msn.com/encnet/refpages/RefArticle.aspx?refid=761556362#endads

Paul T
08-15-04, 02:07 PM
Based on my earlier encounter with you on similar matter, where you hopelessly unable to cope with the reference frame issue, I can now understand your failure to accept my comment that you shouldn't use earth as the reference frame for case under discussion here.

When we deal with gravitational interaction between two objects (earth and another object that we thought as free falling toward the earth), the correct -- possibly the easiest -- reference point is the center of mass of the system. Since earth is very much more massive than the falling object (say with mass in the excess of a few kgs to a few tonnes), this system center of mass is located very close to the earth center of mass. In practice, people ussualy choose earth as the reference point. But, with respect to your argument where you want to take into consideration the very tiny acceleration of earth toward the falling object, earth cannot be used as the reference point (as it move and it accelerate). The only point that we know remain unmoved is the system center of mass. Your statement that nobody use such "hypothetical" point as reference point indicate that you either do not understand the issue or you just want to defend your pointless argument.

System center of mass is not a hypothetical point. If you understand classical mechanics well enough, you would understand that such point preserve system momentum conservation. It goes like this. There is no external force act on the two objects under gravitational interaction. Viewed by someone fixed to star far far away, those two objects gain velocity, but the whole system is stay at the same position....otherwise, momentum conservation is violated (we are not talking about such violation here). The point where the system unmoved is the system center of mass.

MacM
08-15-04, 08:44 PM
Paul T,

You can stop your preaching. We all already know what you have just said. I have also correctly pointed out the view from earth's perspective and from the free falling mass's initial drop point perspective.

What I have said in fact is that the current trend of teaching and usage of the concept of gravity is being misrepresented. It is claimed that a light object and a heavy object falling from an equal height (excluding air, etc) will strike earth in the same amount of time. THAT IS SIMPLY INCORRECT. The heavier object will strike earth sooner.

It is not incorrect to say the acceleration of gravity is the same for it is. The issue and the problem is that people DO use the incorrect reference when refering to the acceleration of gravity. Using the COM would make it correct.

But re-read this thread my statements are qualified and correct. I do not have a problem with reference frames. That is your favorite distractor for a discussion to claim somebody doesn't understand and you then like to pretend to be teaching.

QUIZ: (exclude extraneous interferences such as air resistance, etc).

1 - So does a 1 gram object and a 10 kg object dropped from equal heights hit the ground in the same amount of time?

2 - Does an object released from a height of 9.8 meters contact the earth in 1 second?

Paul T
08-15-04, 11:32 PM
What I have said in fact is that the current trend of teaching and usage of the concept of gravity is being misrepresented. It is claimed that a light object and a heavy object falling from an equal height (excluding air, etc) will strike earth in the same amount of time. THAT IS SIMPLE INCORRECT. The heavier object will strike earth sooner.


From the practical stand point, statement that "a light object and a heavy object falling from an equal height will strike earth in the same amount of time" is perfectly acceptable. Based on any imaginable experiment on earth, the displacement of earth is so so miniscule, much smaller than even the size of a proton. This, I would say, next to imposible for any practical measurement and therefore indicating that for any practical purposes, taking earth as the reference point and saying that those two objects would strike the earth at the same time are fairly acceptable. We know that earth also falls toward the falling object and theoretically a more massive object strike the earth surface in lesser time due to the earth moving closer to the object (although not measurable).



It is not incorrect to say the acceleration of gravity is the same for it is. The issue and the problem is that people DO use the incorrect reference when refering to the acceleration of gravity. Using the COM would make it correct.


Whatever you say the original problem "Do heavier objects fall faster?" remain unsolved. There is nothing fundamental about more massive object would hit earth in shorter time. We don't need any fancy theory to explain such phenomenon. Standard classical mechanics is just enough for that.



So your claim is that a 1 gram object and a 10 kg object dropped from equal heights will hit the ground in the same amount of time?

You should know by now what my view on such problem is. My point from the start remain the same: The theoretical fact that more massive object takes lesser time to hit the ground does not mean that more massive object fall faster and therefore such argument of yours is pointless with regard to answering "Do heavier objects fall faster?"

In case you want to know, my answer to "Do heavier objects fall faster?" is "NO". Now, I ask you: Say we dropt objects A and B consequitively (object A heavier). Will A take lesser time to fall a distance of 1m than B? Ignore variation of gravitational acceleration and any other disturbance.

Okay, my answer is: "both A and B will take the same amount of time to fall that 1 m distance" and therefore "HEAVIER OBJECTS DO NOT FALL FASTER!"

Nasor
08-16-04, 12:10 AM
What I have said in fact is that the current trend of teaching and usage of the concept of gravity is being misrepresented. It is claimed that a light object and a heavy object falling from an equal height (excluding air, etc) will strike earth in the same amount of time. THAT IS SIMPLY INCORRECT. The heavier object will strike earth sooner.

It is not incorrect to say the acceleration of gravity is the same for it is. The issue and the problem is that people DO use the incorrect reference when refering to the acceleration of gravity. Using the COM would make it correct. Well, youíre right about this. But if you want to get really picky about it you could say that itís incorrect to teach kids that F(g)=m1*m2*G/d^2, since that doesnít incorporate what we know about relativity. You always have to decide how much you want to dumb things down in the interest of making time for other things.

MacM
08-16-04, 12:27 AM
From the practical stand point, statement that "a light object and a heavy object falling from an equal height will strike earth in the same amount of time" is perfectly acceptable.

We of course agree. But that hasn't been the issue.


Based on any imaginable experiment on earth, the displacement of earth is so so miniscule, much smaller than even the size of a proton. This, I would say, next to imposible for any practical measurement and therefore indicating that for any practical purposes, taking earth as the reference point and saying that those two objects would strike the earth at the same time are fairly acceptable. We know that earth also falls toward the falling object and theoretically a more massive object strike the earth surface in lesser time due to the earth moving closer to the object (although not measurable).

Again we agree.


Whatever you say the original problem "Do heavier objects fall faster?" remain unsolved.

Here we only partially agree. I suspect you are correct in that it has not been proven by testing but I disagree that it is untestable. I personally have participated in tests which measured the gravity affects of 60 trillionths of a pound force.


There is nothing fundamental about more massive object would hit earth in shorter time.

This is incorrect. It is indeed fundamental and inherent in current theory and mathematics if applied correctly using relativity - i.e. All motion is relative, all velocity is relative and you are measuring from earth as a reference.


We don't need any fancy theory to explain such phenomenon. Standard classical mechanics is just enough for that.

We agree properly applying current classical mechanics is all that is required to show the reality vs the false standard claim.


You should know by now what my view on such problem is. My point from the start remain the same: The theoretical fact that more massive object takes lesser time to hit the ground does not mean that more massive object fall faster and therefore such argument of yours is pointless with regard to answering "Do heavier objects fall faster?"

Now that is interesting. You agree that heavier object hit the ground in lesser time but refuse to admit that from that perspective they fell faster. :bugeye:

You are a self proclaimed relativists. Why do you now choose to reject relative velocity as a basis for stating a relative speed (i.e. - faster)?


In case you want to know, my answer to "Do heavier objects fall faster?" is "NO". Now, I ask you: Say we dropt objects A and B consequitively (object A heavier). Will A take lesser time to fall a distance of 1m than B?

I set the trap, don't expect to lure me into it. It all depends on your definition of "fall". Its spatial distance traveled is not the same as the closure distance between the earth reference and the object.

1 - From the perspective of earth as a reference the answer would be "Yes".

2 - From the perspective of the COM or origin of the free-fall the answer is "No".


Okay, my answer is: "both A and B will take the same amount of time to fall that 1 m distance" and therefore "HEAVIER OBJECTS DO NOT FALL FASTER!"

I agree that the time to travel 1 m relative to the origin of the free-fall or any other fixed point in space is the same. However, that avoids the issue.

The time to free-fall 1 m from earth to contact with earth requires less time. With earth being the standard reference, free-fall rates are now variable based on mass.

You can only get by using earth as a reference because you are generally free-falling substantially smaller objects such that the response of earth is negligable.

However, the point is (and has been) that this is bad science. The actual process should be taught and simply point out that for all practical purposes it may be considered equal if the mass differential is substantial and stop preaching a false concept.

To clarify one need only consider a test of a bowling ball dropped 4.9 m in free-fall to earth and an object the diameter and mass equal to the earth being dropped in free-fall 4.9 m from the earth's surface.

Do you still claim these two items will reach the earth consuming the same amount of time? Of course not, The heavier object will consume approximately 0.5 second not 1 second. So why make the claim light and heavy objects will free-fall to earth in the same amount of time?. It simply is an invalid statement. Teach it as it is.

I'm glad to see you qualify your answers however, that is the correct thing to do.

PS: I recognize that the acceleration of earth during any free-fall experiment means it is no longer a valid inertial reference since it is accelerating however, that is not taught. What is taught is that light and heavy objects hit the ground in the same amount of time from equal heights and that is false.

The most correct thing to teach would be the COM reference and then stipulate earths motion may be disregarded in most cases.

MacM
08-16-04, 12:48 AM
Well, youíre right about this. But if you want to get really picky about it you could say that itís incorrect to teach kids that F(g)=m1*m2*G/d^2, since that doesnít incorporate what we know about relativity. You always have to decide how much you want to dumb things down in the interest of making time for other things.

No the formula is fine for that level of teaching but they teach it incorrectly. I don't see that it takes that great of an effort to point out the truth of the function of the formula.

As it is kids grow up making false claims and as you have seen here will still argue against the truth because they were mis-taught.

Nasor
08-16-04, 01:17 AM
No the formula is fine for that level of teaching but they teach it incorrectly. I don't see that it takes that great of an effort to point out the truth of the function of the formula.Well, if you understand the formula it's self-evident how it works. That's the beauty of formulas. Usually telling kids that light objects fall at the same rate as heavy objects is something you do at a very young age in order to correct naive misconceptions in kids who donít know algebra yet.

MacM
08-16-04, 01:11 PM
Well, if you understand the formula it's self-evident how it works. That's the beauty of formulas. Usually telling kids that light objects fall at the same rate as heavy objects is something you do at a very young age in order to correct naive misconceptions in kids who donít know algebra yet.

Funny how that translates into the many adults; including scientists, that will still argue that reference earth they fall to earth in the same amount of time however.

soccerdvy
08-21-04, 11:57 PM
This is very simple... It just occured to me today, while I was sitting around the house, and I found this post as I was looking for more information on the subject. I'm 15 years old, and have never been taught anything other than "any 2 objects will take the same time to fall to Earth." It occured to me today that this is rediculous.. While both objects will accelerate towards Earth at the same time, Earth will also be accelerating towards them. Using Earth as a point of reference is completly valid for all "practical" (a word that has been used many times to defeat this argument) purposes becuase Earth is the point of impact, and the entire question is how long will it take for the object to reach the point of impact, not how long it will take to reach come in contact with a body which is no longer the center of matter...

soccerdvy
11-13-04, 12:18 PM
Actually, you wouldn't even have to drop the two said objects at different times. You could drop them both at the same time and the acceleration of the two objects would be proportional to the sine of their masses. Consider two obejcts m1 and m2 of two distinct masses. If these two objects were to be dropped from said height H, the mean of their distance would be a point at which a line (the adjacent), running through the center of the Earth would be perpindicular to a line running through the two objects. Therefore we would have two congruent angles formed by this line, which would also be an angle bisector.
Because it's a bisector and the adjacent side of both angles would be equal (becuase its the same segment for both angles), and the height, H, was already defined to be equal. Thanks to the side angle side theorum, we can now determine that these two triangles formed would be congruent.

The force determined by the two masses would be this:
F1 = ((20/3 x 10 ^ -11)(m1)(mass of earth))/h^2
and
F2 = ((20/3 x 10 ^ -11)(m2)(mass of earth))/h^2

It can be seen clearly here that these two objects will have distinct Forces, based on the function of their masses. The objects will exert their distinct forces at an angle (a) determined by the equation:

F1 sin a = distance from adjacent
and
F2 sin a = distance from adjacent

It can be infered that for the two objects forces to attract the Earth to their midpoint, they would have to be at a distance from the adjacent proportional to their masses. However, because we have already determined that the adjacent would be an angle bisector, we can infer that the rate at which the Earth was attracted to each object would be proportional to the masses of the angles, and only if their masses were equal would they both hit the Earth at the same time. Since it has been determined earlier that the masses are distinct, we can gather that the two objects would fall at different rates even if dropped at the same time.

RawThinkTank
11-16-04, 04:46 AM
I ill b back

RawThinkTank
11-20-04, 07:27 AM
I am back.

Which will fall faster :-

1) A tiny blackhole (with 10G) droped on earth ?

OR

2) Moon ?

soccerdvy
11-23-04, 02:44 PM
trick question, there's no such thing as a black hole with a mass of 10 g becuase a 10g mass wouldn't have enough gravitational force to create a black hole in the first place

TruthSeeker
11-23-04, 03:48 PM
I think his "G" doesn't mean "grams"....

Besides, the mass doesn't matter. A single gram can create a black hole if it is small enough. It is not the mass that determines the black hole, it is the density - that is relationship between the mass and the space occupied by it, which is what create the gravity necessary to form the black hole.

Nasor
11-23-04, 07:35 PM
trick question, there's no such thing as a black hole with a mass of 10 g becuase a 10g mass wouldn't have enough gravitational force to create a black hole in the first placeWhether or not something qualifies as a black hole is more a question of density then mass. In theory you could have a black hole of any mass, if you had a way to compress it enough.

shoffsta
11-26-04, 10:50 PM
falling acceleration only depends on air-density, airodynamics of the object and the gravity of the planet, no more.

FreeMason
12-28-04, 03:49 AM
I would just like to point out that the argument that had they dropped the hammer and feather seperately there would be a difference in their rates of acceleration (by an unnoticeable and thus unimportant amount) is just a dumb argument.

The conditions of the experiment are thus:

If two objects are dropped at the same time, they will hit the ground at the same time.

This statement is true.

MacM
12-28-04, 04:57 PM
I would just like to point out that the argument that had they dropped the hammer and feather seperately there would be a difference in their rates of acceleration (by an unnoticeable and thus unimportant amount) is just a dumb argument.

The acceleration would apear to be the same but the issue is free fall time from a given height.


The conditions of the experiment are thus:

If two objects are dropped at the same time, they will hit the ground at the same time.

This statement is true.

That is correct. However, that statement doesn't properly translate to the claim that all masses have the same free fall time from a given height, which was the root of the arguement.

In fact dropping the hammer and feather at the same time results in both having an equally shorter free fall time than either independantly.

The other issue was then how meticulus science wants to be in some areas using atomic clocks to a second in 300 years etc., but then be cavilier about gravity using overly generalized (and actually false) statements.

Ace
12-28-05, 09:57 PM
Thank you MacM,

I had this argument with my physics teacher in High School, and the moron could never get in into his puny pea sized brain.

He compared me to my classmates, asking them why they weren't "more like Mr. Hall"? What a jerk, he ought to be strung up by the toes and beaten to death with a pillowcase filled with feathers and pennies.

His name was Arthur Schang.

Ace

URI
12-29-05, 04:56 AM
see Allias effect

More dense objects in orbit decay faster than less dense objects.

CANGAS
12-31-05, 03:35 AM
F = ( G x M1 x M2 ) / r.

F = M x a: a = F/M

CANGAS
12-31-05, 03:36 AM
URI:

They fall apart into constituent particles quicker?

MacM
12-31-05, 08:56 AM
F = ( G x M1 x M2 ) / r.

F = M x a: a = F/M

F = G * m1 * m2 /. r<sup>2</sup> :)

CANGAS
12-31-05, 10:41 AM
Thank you, MacM.

Sooner or later I will learn to avoid writing important things when I am too tired to be alert and accurate. :o

I am glad you have helped me get my r's squared away. ;)

alain
12-31-05, 11:23 AM
someone may already have said this
gravitational force on an object on earth = universal gravitational constant x mass of the object x mass of the earth / distance squared

and acceleration = force / mass

so if you make the mass twice as big, the force will be twice as big (from the first equation) however, acceleration will be the same (due to the second equation)

the reason that the cardboard fell faster, is that the upwards force on both paper and cardboard of the same size is equal, and so, by the second equation, the upwards acceleration on the card board is much less, as its mass is much greater

leopold
12-31-05, 12:54 PM
there is some video shot on the moon that will answer your question.
a hammer and a feather were dropped from the same height
they both hit the ground at the same time

MacM
12-31-05, 06:13 PM
there is some video shot on the moon that will answer your question.
a hammer and a feather were dropped from the same height
they both hit the ground at the same time

Correct, assuming precise same height over the surface and dropped simultaneously.

However, the actual time between being dropped and contact with the ground is slightly less than had the hammer or feather been dropped from the same height at seperate times.

Also, dropped seperately the hammer will hit the ground in less time than the feather when dropped seperately.

Blue_UK
01-01-06, 09:23 AM
Guys, if you look on page 1 or 2, you'll see I've already provided a nice proof. I even used colours to mark the different numbers.

Here (http://sciforums.com/showthread.php?p=649184#post649184)