Author Topic: "All objects fall at the same rate." Really?  (Read 3667 times)

Offline poxy

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"All objects fall at the same rate." Really?
« on: September 04, 2010, 12:36:08 AM »
Something along those lines I've read from Stephen Hawking's book. Seems like common knowledge.

I think it was Plato that hypothesized heavier objects fall faster, and Galileo proved otherwise by rolling balls of different weights down a slope. There was the famous dropping of the hammer and feather on the surface of the moon to demonstrate this.

But I can't wrap my head around it. It can't be wrong if a genius says so. Though the math says otherwise:

Newton's law of universal gravitation:
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Therefore:
Ffeather < Fhammer
And it seems like they fall at the same rate because the difference is so small.
Hypothetically, if one dropped two objects at the opposite poles of the moon, of the same size say, one with a mass 80% of the moon and the other a feather, which would fall faster (hit the moon first)? What am I missing?
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Offline Aqua

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Re: "All objects fall at the same rate." Really?
« Reply #1 on: September 04, 2010, 12:53:59 AM »
F = ma
The forces might be different, but a is constant (at ~9.8 m/s on the Earth).
The m is what changes the F

But objects don't fall at the same speed; they only do so in a vacuum and only if they're not so massive that they heavily distort a gravitational field.
Air resistance, terminal velocity, and other things can make things fall at different speeds.

About your hypothetical example...
The object with 80% of the moon would hit the moon first because that object has enough mass to actually drag the moon toward it (though the moon would drag more)
The feather would be there along for the ride XD
« Last Edit: September 04, 2010, 12:57:32 AM by Aqua »

Offline poxy

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Re: "All objects fall at the same rate." Really?
« Reply #2 on: September 05, 2010, 02:56:56 AM »
Yea, I applied the wrong formula...
But I suppose the problem is that in F=ma the a is the acceleration, assuming the objects falling have near nil gravity of their own.

Well, I looked up the definition for "fall" on Wikipedia (though not truly an authoritative source).
"Free fall is any motion of a body where gravity is the only or dominant force acting upon it, at least initially."
And the problem is that a single "body" is described, and the force or acceleration produced by that body is not taken into account.

I guess the point I was trying to make, is that that definition is not balanced, in that it considers gravity for one object where two are involved.

That is why in my example I placed the falling objects at opposite poles, to demonstrate the idea I had in mind. The feather would be along for the ride, but it would it's landing would be the last to take place. Granted, it's not a common situation that would usually occur... :p
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Offline Aqua

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Re: "All objects fall at the same rate." Really?
« Reply #3 on: September 05, 2010, 03:12:25 AM »
But for something to enter "free fall," the other object's gravitational pull is so small that it's negligible.
It needs to be small or else it wouldn't be "free fall."
If the second object actually affected the first object, then it'd be more like an electron and a proton reacting with each other XD

Edit to help clarify:
The second object DOES have a gravitational pull on the first object.
But since F = ma and the first object has a HUGEEEEE m, then the a has to be small (like close to 0).
« Last Edit: September 05, 2010, 03:20:02 AM by Aqua »

Offline Ryex

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Re: "All objects fall at the same rate." Really?
« Reply #4 on: September 05, 2010, 03:51:30 AM »
G((m1*m2)/d^2)
G = 6.67300 × 10^-11 (m^3 kg^-1 s^-2)
mass Earth = 5.9742 × 10^24 kg
mass object = 10 kg
distance from surface = 10,000 m
radius of earth = 6,378,100 m
remember that when using this law you must use the distance between their center points as it assumes point masses
total distence: 10 000 + 6 378 100 = 6 388 100
f = (6.67300 * (10^(-11))) * (((5.9742 * (10^24)) * 10) / (6 388 100^2)) = 97.6916557  kg*(m/s^2)
f = ma
f/m = a
97.6916557 / 10 = 9.76916557
oh look at that, the acceleration due to gravity. also, you can see why it is rounded to 9.8 most of the time (or 10 for really lazy calculation)

all objects fall at the same rate. the acceleration however isn't constant is gets greater the closer the objects are together but for two different objects the same distance form a third much larger mass those two will reach the third mass at the same time assuming that there are no other forces that are acting differently on the objects (friction due to air for example).

f = (6.67300 * (10^(-11))) * (((5.9742 * (10^24)) * 10 000) / (6 388 100^2)) = 97 691.6557  kg*(m/s^2)
f = ma
f/m = a
97691.6557 / 10000 = 9.76916557
as you can see a difference of 100* more weight made no difference to the result down to 1 100 millionth m/s^2
i got Google to make calculation with the object being 10^154 kg and it still didn't change...
thats really weird
also it should be noted that the force is applied to BOTH masses but wen such a small force is applied to a mass like the earth it results is an insignificant acceleration

also you see that the force is relative to distance? that is why the force of gravity in the upper atmosphere isn't the same as the surface or the same as outer orbit.

also I feel obligated to point this out. this law is Newtonian. Newtonian physics doesn't work with modern scientific knowledge, however it is still accurate in day to day situations. but, it should be noted thinks like gravity don't work when you get into the celestial scale thats where General relativity or the general theory of relativity comes in.
« Last Edit: September 05, 2010, 04:12:43 AM by Ryexander »
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Offline winkio

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Re: "All objects fall at the same rate." Really?
« Reply #5 on: September 05, 2010, 04:46:42 AM »
I'm pretty sure you get the equation explanation by now, but I just thought I'd add in the verbal one:

Gravity is defined as an attractive force between two objects.  On planet Earth, there is an interesting case that the mass of one object, the earth, accounts for almost 100% of the total mass of the system.  Additionally, all objects are on or very close to the Earth's surface.  This creates a situation where the massive object is held in a relatively stable position, while all the other objects experience more or less the same gravitational acceleration.  Nevertheless, the Earth still feels an attractive force to each object.  But when an object of 5.9742 × 10^24 kilograms feels a force of 1000 newtons, it's acceleration is practically zero.  And when it feels many of these forces pulling in all directions, the net acceleration is even less.

@ryex: Newtonian equations can be derived from quantum mechanics, and are the most accurate approximation for large scale phenomena.

Offline WhiteRose

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Re: "All objects fall at the same rate." Really?
« Reply #6 on: September 05, 2010, 05:56:56 AM »
I may be wrong, but I was under the impression that the Newtonian equations stop applying when objects near the speed of light, and that it was the speed of the object - not the mass - that made the difference. Of course, when things are moving that quickly, the scale has to be massive, so there is a correlation.

Offline Aqua

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Re: "All objects fall at the same rate." Really?
« Reply #7 on: September 05, 2010, 06:19:06 AM »
It is much easier for something of small mass to move very quickly due to F = ma.

Atomic particles can move very fast... They just have a ton of stuff that get in their way usually.

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Re: "All objects fall at the same rate." Really?
« Reply #8 on: September 05, 2010, 07:06:43 AM »
I never said the Newtonian physics was wrong, just that it stops working when you get up to the celestial scale or when objects move near the speed of light. I believe it also stops working when you get to the sub-atomic level too
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Offline winkio

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Re: "All objects fall at the same rate." Really?
« Reply #9 on: September 05, 2010, 07:12:26 AM »
I may be wrong, but I was under the impression that the Newtonian equations stop applying when objects near the speed of light, and that it was the speed of the object - not the mass - that made the difference. Of course, when things are moving that quickly, the scale has to be massive, so there is a correlation.

You are right that Newtonian physics stop working near the speed of light (whenever anything moves at a speed > 0.9c).  If a large mass moves near the speed of light (>0.9c), it will break down into very small particles.  Thus, generally, relativistic effects are observed in small particles, not large masses.

I never said the Newtonian physics was wrong, just that it stops working when you get up to the celestial scale or when objects move near the speed of light. I believe it also stops working when you get to the sub-atomic level too
Okay, whatever.

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Re: "All objects fall at the same rate." Really?
« Reply #10 on: September 05, 2010, 05:41:58 PM »
The force is different, but the acceleration is not:

F=G m1*m2 / r^2

a2 = F / m2
a2 = (G m1*m2) / (r^2 * m2)
a2 = G m1 / r^2

The acceleration of Earth is different, but for both the hammer and feather it is so insignificant that you can practically consider it not to be there. I don't think we have precise enough instruments to measure the difference yet. At least not without some other noise overshadowing the effect.

@ryex:
I think the reason 9.8 m/s^2 is typically used has more to do with the standard acceleration of gravity being 9.80665 m/s^2. (At least according to Precise Measurement of Mass)

Also try looking at the acceleration of Earth as you increase/decrease the weight of the other object ;)


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Offline poxy

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Re: "All objects fall at the same rate." Really?
« Reply #11 on: September 06, 2010, 07:43:40 AM »
I stand corrected. All objects fall at the same rate if they fall from the same height and towards the same body of mass.

Ryex pointed out the acceleration of different masses towards the Earth. However, if one were to calculate the rate of the Earth falling toward that object the accelerations would probably have to be added to yield the actual time of impact.

Taking my above example, if an object that is falling towards the Earth and is 80% of the Earth's mass, it will accelerate toward it at the same speed as a feather would (assuming no drag). However, the Earth would also be "falling" towards the heavy mass at slightly lower rate, and the time of impact would actually be less than that of the Earth and feather.
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