Do heavier objects fall faster?

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.

 

Log in or Sign up to hide all adverts.

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.

 

Log in or Sign up to hide all adverts.

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.

 

Log in or Sign up to hide all adverts.

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.

 

I agree.

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

Please Register or Log in to view the hidden image!

[/thread]

 

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.

 

The problem of roughly mixing force and pressure, and density too, maybe...

 

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).

 

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.

 

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.

 

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.

 

Hi Paul,

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.

 

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.

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.

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

 

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.

 

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.

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.

 

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/s², 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.