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Nebula
09-07-1999, 07:19 AM
I was reading the old mailbag Arceive and came across the typical question of feather and hammer. If they are dropped togather which will hit the earth before?
http://www.straightdope.com/mailbag/mgravity.html

SDSTAFF member Ian promised to set someone straight but now I would like to set him straight 8-)

He gave this function which gives the force on the hammer and the feather.

F = G m1 m2
-------
r^2

Which is quite correct.
He then canceled out all the factors and found the quite correct fact that the acceleration of the hammer and the feather are exacly the same. To do this he used this equation. a = F/m

Now for the correction. Ian forgot that the equation also tells us the force that acts upon the earth do the the feather or the hammer. The force that acts on the earth form the Hammer is surly larger than the force from the feather so the earth will accelerate faster toward the hammer than the feather and there for the hammer should hit first.

Don't try to tell me that because the feather and the hammer are droppet togather that the earth accelertion should be canceled out since the feather and the hammer are not at the same place and therefor the earth accelertion matters.

By the way. The earth acceleration is small, tiny, smaller then tiny, but it is there!!!


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Gunnar Valur Gunnarsson

Pleonast
09-07-1999, 11:00 AM
Strictly speaking, Nebula, you are correct, but Ian was very well-justified in his response. Let's work through the math.

Suppose we have a 2.00kg hammer. Near the earth's surface (say h=1.00m), the hammer's acceleration is about a=9.81m/s^2. The force of the earth on the hammer is F=19.6N (F=ma). It will take the hammer t=0.452s to fall the distance h=1.00m (t^2=2h/a).

From Newton's Third Law, we know the force of the hammer on the earth is also 19.6N (but in the opposite direction). The mass of the earth is about 5.98e24kg, giving an acceleration of a=3.28e-24m/s^2. During that same time t=0.452s, the earth will move 3.34e-25m (2x=at^2).

This is very small. Atomic sizes are on the order of 1e-10m. Nuclear sizes are on the order of 1e-15m. I'm not certain how it would be even theoretically possible to measure a distance as small as 1e-24m. For any intent or purpose, the earth does not move.

John W. Kennedy
09-07-1999, 11:32 AM
Not to mention that, since we are talking about two simultaneous drops, that the earth will most definitely be moving at the same speed as itself.

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John W. Kennedy
"Compact is becoming contract; man only earns and pays."
-- Charles Williams

moriah
09-07-1999, 07:31 PM
Well, yeah the earth will be traveling at exactly the same speed as the earth if the hammer and feather are dropped simultaneously at the same location. But how about if they are dropped simultaneously at opposite sides of the earth? Then you'd have the earth shifting a quark or two towards the hammer... but then the gravitational force of the hammer will be pulling the feather faster than if they were side by side... and, hey, what are you doing with that hammer? Get away from me you freak. Stop it! HELP!! HELP!! OWW!! ugh...

Polycarp
09-09-1999, 09:55 AM
Well, Moriah, he warned you in the original post that the hammer was "surly."

HeadlessCow
09-09-1999, 03:51 PM
Fg= M1*M2*G / R^2 - Universal Law of Gravitation

F= M1*a - Newton's Law

F=Fg b/c the only force on a falling object is gravity

thus M1*a = M1*M2*G / R^2

Which simplifies to a=M2*G / R^2

In other words the mass of the falling object is irrelevant to it's acceleration from gravity. The falling objects gravitaional attraction to the planet is irrelevant.

Fg=Force of gravity on an object
F=Force acting on an object
M=Mass ;M1=object M2=planet
a=acceleration
G=Gravitational constant
R=Radius

Hammers fall faster on earth because feathers have more drag in an atmosphere. On the moon though or in an evacuated container you can see them fall at the same rate.

AuraSeer
09-10-1999, 02:28 AM
HeadlessCow, you missed the point. The OP had nothing to do with air resistance; it's just a misapplication of the law of gravitation.

Nebula wrote:
He then canceled out all the factors and found the quite correct fact that the acceleration the hammer and the feather are exacly the same.

This is not quite exactly true. The acceleration of the hammer and the earth toward each other, is the same as that of the feather and the earth toward each other. The gravitational force always acts equally on Earth and the dropped object, so they move together and meet in the middle someplace.

We usually simplify this, and assume that Earth remains stationary while the dropped object accelerates at 9.8/m/s/s. But in reality, the dropped object accelerates toward Earth at (9.8 minus epsilon) m/s/s, and Earth accelerates toward the object at (epsilon) m/s/s, where epsilon is a teeny tiny puny little number. Epsilon is a little bit bigger in the case of the hammer, but the total acceleration is the same.

So, Nebula, you are incorrect. If you drop the feather and the hammer from the same height at the same time, they both hit the ground at precisely the same instant.

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Of course I don't fit in; I'm part of a better puzzle.

C K Dexter Haven
09-10-1999, 07:32 AM
I don't think Headless Cow missed the point at all. I think Cow's formulas indicate that the acceleration of the earth towards the object is irrelevant, because the gravitational constant is... well... constant.

Cow's comment about air resistance is the prelude to the comment about on the moon (replace "mass of earth" in the problem with "mass of moon").

Nebula
09-10-1999, 07:41 AM
AuraSeer Wrote:

The gravitational force always acts equally on Earth and the dropped object

This is true but the gravitational force is not constant.
Lets make a closer look on how this will affect the accelerations.

F = G m_earth m_object / r^2

We can easily see that the force is in direct relation to the mass of the falling object. So the heavier the object the more force will act upon both the earth and the object. Now since

a_object = F/m_object
a_object = G m_earth / r^2

We can see that the acceleration on the object is constant and in no relation to it's mass. ( it is in direct relation to the earths mass )

How ever the acceleration of earth will be

a_earth = G m_object / r^2

What we can learn from this.

1. The acceleration of the object is constant since it is depended on the earths mass
2. The acceleration of the earth is not constant since it is depended on the objects mass.
3. Therefore the total acceleration of the earth and the object toward each other is not constant

As Pleonast noted the acceleration of the earth is very small do to our normal live objects and could therefore be ignored since it is within atomic scales.

But hey this is physics and we can make use of the perfect world where everything is possible so lets say that the hammers mass is the same as the earths mass. Now both the earth and the hammer will have an acceleration of around 9.81 toward each other and meet much quicker than if we dropped a 2Kg hammer.

I wouldn't want to live on the earth if this happened so please don't try this at home. 8-)

By the way IF YOU DROP 2 OBJECTS THEY CAN'T BE AT THE SAME LOCATION.


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Gunnar Valur Gunnarsson

TheIncredibleHolg
09-10-1999, 08:51 AM
"If a feather and a hammer are dropped together, ..." -- this reminds me of the old saying: "Birds of a feather drop together."

Momotaro
09-11-1999, 05:32 AM
Gravity is proportional to mass. Naturally the force acting on the hammer is greater because of the greater mass, but there's more mass to move so it cancels out. That's why the acceleration is the same, in vacuum that is. Of course, if you do that experiment in your living room, air resistance will interfere significantly with the fall of the feather.

That reminds me of this trick question: what's heavier, a ton of bricks or a ton of feathers?

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Only humans do inhuman things.

moriah
09-11-1999, 11:03 PM
"The hammer and feather can't be in the same place at the same time..."

Blah, blah, blah.

Hey, if you're going to parse the two inches between the hammer and feather, what are you going to do with the distance between the mass at the top of the hammer with the mass at the bottom of the hammer, which are surely farther than two inches apart?

What if all the atomic bonds between the molecules in the hammer suddenly came apart in midfall? Would that change the acceleration of the 'hammer' over all? Would each seperate molecule have a unique acceleration?

Whether the parts are linked together or not, they're all not in the same place at the same time, so, if you're going to split hairs in this experiment, you might as well split quarks.

Peace.

TheIncredibleHolg
09-13-1999, 03:55 AM
That reminds me of this trick question: what's heavier, a ton of bricks or a ton of feathers?Oh, what a giveaway! Why did you say it was a trick question? :)

AuraSeer
09-13-1999, 09:41 AM
Nebula, scroll back up a bit, and this time read my post before trying to talk about it.

Assuming you can measure accurately enough, the acceleration of any dropped object is slightly less than 9.8 m/s/s. The Earth also accelerates slightly "upward", toward the object. Together, these two acclerations will always add up to 9.8 m/s/s.

I'll use your own example. If you drop a hammer with the same mass as the Earth, it will accelerate toward the planet at 4.9 m/s/s. The Earth will accelerate toward the hammer at 4.9 m/s/s. The total acceleration is 9.8 m/s/s.

To reiterate: the acceleration due to gravity depends solely on the mass of the Earth. It doesn't matter whether you drop a hammer, a feather, a marble or the Moon; they will all experience the exact same acceleration.

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Of course I don't fit in; I'm part of a better puzzle.

DSYoungEsq
09-13-1999, 11:01 AM
You know, if y'all were a little more open to listening, a little less open to being pissed off at people who don't accept what you say as definitive, and a lot more willing to be specific and comprehensive in your posts, this place might actually solve some questions ;)

I propose for consideration the following problem which should solve the issue. In one case, we have two point masses of equal mass at rest a distance x apart. In the other case, we have two point masses of equal mass at rest the same x apart, but in this case, the masses are 100 times as heavy as the first case.

Now, if I read what Nebula is saying correctly, he feels the smaller masses will take less time to meet as a result of the attraction of gravity than the larger masses. Nebula then carries this over to the feather and the hammer, noting that the combined mass of the Earth-hammer system is greater than the combined mass of the Earth-feather system, if only infinitismally so. More mass = faster acceleration.

Now Aura is saying that the rate of accelleration is not affected by the difference in mass between the hammer and the feather, being totally dependent on the mass of the Earth. In practicality, this is correct. But for Nebula to be 'wrong,' then the rate of accelleration for the Earth-hammer system has to be the same as for the Earth-feather system. If this is true, then the time for the two masses to meet in each case in my problem would be the same.

NOW, please, someone solve the problem posed and relate it back to the original post with something OTHER than shrill rhetoric?

HeadlessCow
09-13-1999, 11:36 PM
The Universal Law of Gravitation doesn't tell you the attraction of the earth on the hammer or feather...It tells the force of the attraction between the two objects. It already takes into account the attraction of the earth on the hammer and the attraction of the hammer on the earth. It doesn't matter how large or small the object being attracted is there is ALWAYS the same acceleration in a perfect case. As the mass increases the force of gravitation increases so does the inertia of the object and they increase at the same rate. This is why the acceleration is constant.

HeadlessCow
09-13-1999, 11:38 PM
The falling mass is the one that is canceled out and since it is the smaller mass that is percieved as falling in the example of the earth and the sun the earth would be the falling object and the acceleration would then depend on the mass of the sun.

Nebula
09-14-1999, 01:03 AM
AuraSeer:


I'll use your own example. If you drop a hammer with the same mass as the Earth, it will accelerate toward the planet at 4.9 m/s/s. The Earth will accelerate toward the hammer at 4.9 m/s/s. The total acceleration is 9.8 m/s/s.


This is not true and If you carry out calculation you will find that out.
Now think about this. What if the hammer was the mass of the sun? will the total acceleration still be 9.8 m/s/s even though the gravity of the sun is much more? Is the gravity of the earth some universal constant in your physics? Why don't 2 soap bars attract each other at the same rate?

Thanks DSYoungEsq for your posting. The two 100times heavier objects will meet faster and I can do calculations for you if you want but I think you already know.


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Gunnar Valur Gunnarsson

C K Dexter Haven
09-14-1999, 01:12 AM
So would the grammar hit the hound first?*

* Dept of subtle spoonerisms

Nebula
09-14-1999, 06:15 AM
Object with mass of earth falls on earth!!!
Object has very small radius. ( very denced )

F = G * m_earth * m_object / r^2

This force acts upon the earth and acts upon the object

F = 6.67e-11 * 5.98e24 * 5.98e24 / 6.37e6^2
= 5.88e17N

Then

a_earth = F/m_earth = 5.88e25 / 5.98e24 = 9.83m/s^2
a_object = F/m_object = 5.88e25 / 5.98e24 = 9.83m/s^2

Now show me where I am wrong. Show me numbers!



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Gunnar Valur Gunnarsson

Nebula
09-14-1999, 06:17 AM
Sorry a typo F = 5.88e25 not 5.88e17

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Gunnar Valur Gunnarsson

Pleonast
09-14-1999, 04:04 PM
There's some poor physics above, and some misunderstandings. The acceleration of the hammer and the acceleration of the feather will be the same. The acceleration of either depends does not depend on their masses. What the original question ( http://www.straightdope.com/mailbag/mgravity.html ) asks is whether or not the time is takes for each object to hit the earth is the same. It is not exactly identical, but is close enough for anyone but a stubborn philosopher (perhaps I'm one of them).

If you really want to know how long it takes two freely moving objects to collide, here's the answer. Not for the math impaired. Setup: two objects at rest with respect to each other and only experiencing their mutual gravitational attraction. Without loss of generality we can assume they are spheres and consider motion in only one dimension. We'll give the masses m1 and m2, radii r1 and r2, positions x1 and x2. Initial conditions are initial distance is d=x2(0)-x1(0) and initial velocities v2(0)=v1(0)=0.

The Langrangian for this system is L=(1/2)*(m1*v1^2+m2*v2^2)+G*m1*m2/(x2-x1). From here we can derive the equation of motion (and I have done so correctly): (d2/dt2)x=-G*M/x^2, where x=x2-x1 (the distance between the objects) and M=m1+m2 (the total mass). If x does not change much and one mass is much larger than the other, this will reduce to the constant acceleration approximation we've been using until now. The solution to this equation with the given initial conditions is t=sqrt[d^3/(8*G*M)]*(2*sqrt[u*(1-u)]+arctan[2*sqrt[u*(1-u)/(u-(1-u))]). The total time until collision is t, and u=(r1+r2)/d. Be careful to pick the correct quadrant when you calculate the arctangent.

Now anyone(?) can calculate how long it takes something to fall. For the same situation I gave in my earlier post, the difference in time between a 2kg hammer and 1g feather is 7.54e-26 seconds. Of course this is ignoring that the earth and the objects do not start at rest with respect to each (e.g., the earth's rotation), but the corrections will not greatly change the answer.

Sorry for the math overkill, but there are too many misunderstandings arising from the approximations many of the posters are making implicitly.

Strainger
09-14-1999, 04:34 PM
Yeah, but what if you drop the feather and fire the hammer out of a gun at the same time?

Pleonast
09-15-1999, 10:40 AM
Yeah, but what if you drop the feather and fire the hammer out of a gun at the same time?

That would be orbital mechanics--you're no longer guaranteed the hammer will hit the earth.

Nebula
09-15-1999, 12:49 PM
Thanks Pleonast
If someone didn't understand what I said in my original post, ( I see that you did ) they can be quite assured now that what I was trying to say is true and you have proven it. You have also shown that the difference is very small so we can easily ignore it but we shouldn't since this is theory not practis.


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Gunnar Valur Gunnarsson

Geezer
09-15-1999, 04:56 PM
I think that much of the difficulty we are having here arises from a failure to specify the coordinate system in which we are measuring acceleration. Keep in mind that Newtonian mechanics only applies in an inertial reference frame. Given that the question allows for acceleration of the Earth, the conventional choice of an Earth-fixed coordinate system is inappropriate and will yield bizarre results.

I would suggest that the best coordinate system to choose is one with origin fixed at the center of mass of the combined system. In this coordinate system, the acceleration on each particle due to the others is independent of its own mass, depending only on the masses of and distances to the other objects. This point has been demonstrated several times. The acceleration computed is not the acceleration with which two particles move toward each other; it is the acceleration towards the system CM.

Now, there is another point of confusion. Are we talking about simultaneous arrival at the CM, or simultaneous arrival at the surface of the earth? If we're talking about simultaneous arrival at the CM, then (in a system with zero initial kinetic energy and angular momentum...) all particles arrive at the CM simultaneously. If you don't want to do the detailed calculation, think of it this way: if one particle got to the CM first, it wouldn't be the CM anymore. If you do want the full calculation, keep in mind that, although all particles accelerate at different rates, they have different distances to traverse.

But the original question is, which hits the ground[/] first? Leaving out issues like the fact that the Earth isn't spherical and doesn't have uniform density and making the hammer and feather point particles and so forth, in this case, I [i]think the hammer always has the edge, but I'm hard-pressed to prove it. We're talking three-body problem, here, folks, and it's famous for not having a nice general solution like the two-body problem. If you admit of the rotation of the earth, then it's really complicated since the three particles are in orbit around each other after release of the hammer and feather. (This is why they launch rockets east from Cape Canaveral.)

To me, though, the really interesting question is this: why is it that everyone is perfectly willing to cancel the mass that appears in Newton's Second Law (F=ma) with the mass in Newton's Universal Law of Gravitation (F=GmM/r^2)? Why do you think that inertial mass and gravitational charge are mathematically identical?

--Y

Geezer
09-16-1999, 09:49 PM
[quote]Simply put, there's no reason to think inertial and gravitational masses are different.[quote/]

Well, the experimental evidence strongly indicates that they are identical, anyway. The question I was raising, however, is that of [i]why[i/] they should be identical when they arise from two [i]different[i/] phenomena--inertia and gravitation. The fact that gravitational charge and mechanical inertia appear to be exactly equal indicates that gravitation and inertia are at least related, or even the same phenomenon. Welcome to General Relativity! And to think we started with feathers and guineas (errr.. hammers). Ain't science great?

Pleonast
09-17-1999, 01:36 AM
``Is inertial mass identical to gravitational mass?'', that is the fundamental question everyone is asking (even if they don't realize it). Newton's Laws of Motion and Gravity and both of Einstein's Theories of Relativity assume they are mathematical identical. Experimentally, the answer is that our best measurements cannot find any difference to within error bonuds (about one part in 10^10). These experiments carefully compare accelerations of differenct objects under gravity. Simply put, there's no reason to think inertial and gravitational masses are different.

BTW, there is a typo in the time formula of my previous post. Both sqrt's have the same argument: [u*(1-u)]. I left out the closing bracket of the second root. And, I should mention that my Lagrangian dynamics is wholly classical: no special or general relavity and no quantum mechanics.

John W. Kennedy
09-17-1999, 12:42 PM
Newton did not merely assume that inertial mass and gravitational mass were identical; he observed that both the two-weight experiment and Keplerian orbital mechanics were explained thereby.

General Relativity is partly a result of Einstein asking, "OK, so they are identical, but why?"

------------------
John W. Kennedy
"Compact is becoming contract; man only earns and pays."
-- Charles Williams

Geezer
09-17-1999, 07:22 PM
And Einstein's conclusion was that the "force" of gravity is as fictitious as centrifugal "force" or the Coriolis "force." What we perceive (and measure) as the force of Gravity is just inertia--the simplest possible physical law: things don't change if you don't do anything to them. Orbiting objects coast along straight lines in curved space. What a concept!

The Ryan
03-17-2000, 09:12 PM
I found this thread from this newer thread (http://boards.straightdope.com/ubb/Forum6/HTML/000280.html), and I was surprised at the ignorance of physics.

AuraSeer posted 09-13-1999 08:41 AM

Assuming you can measure accurately enough, the acceleration of any dropped object is slightly less than 9.8 m/s/s. The Earth also accelerates slightly "upward", toward the object. Together, these two acclerations will always add up to 9.8 m/s/s.
First of all, 9.8 + 9.8 =19.6 != 9.8 Second of all, in the object frame, the Earth accelerates at 9.8, and the object is stationary. In the Earth frame, the Earth is stationary, and the object is accelerating at 9.8. In no frame are both accelerating towards each other at 9.8. Pick a frame and stick with it!

I'll use your own example. If you drop a hammer with the same mass as the Earth, it will accelerate toward the planet at 4.9 m/s/s. The Earth will accelerate toward the hammer at 4.9 m/s/s. The total acceleration is 9.8 m/s/s.
Are you saying that a hammer with the mass of the Earth would have less acceleration than a normal hammer? Why?

To reiterate: the acceleration due to gravity depends solely on the mass of the Earth. It doesn't matter whether you drop a hammer, a feather, a marble or the Moon; they will all experience the exact same acceleration.
Why is the Earth so special? Why does the acceleration depend on the mass of the Earth and not on the mass of the object? Is the Earth made of some special substance?

HeadlessCow posted 09-13-1999 10:36 PM

The Universal Law of Gravitation doesn't tell you the attraction of the earth on the hammer or feather...It tells the force of the attraction between the two objects.
It gives both the force on Earth and the hammer. To find the acceleration, you have to divide by the appropiate mass.
It doesn't matter how large or small the object being attracted is there is
ALWAYS the same acceleration in a perfect case. As the mass increases the force of gravitation increases so does the inertia of the object and they increase at the same rate.
Wat? The more massive two objects are, the more they attract. Increasing the mass of the Earth increases the acceleration of the object (but not the acceleration of the Earth, at least not directly). Increasing the mass of the object increases the the acceleration of the Earth.

As for the original question: since the term "hitting the Earth" refers to the time at which the valence shells of the Earth and the object would start to have significant interference, and the difference would be about a dozen orders of magnitute smaller than the valence shell diameter, there would be no detectable difference.
If you really want the exact difference, you would have to solve a three-body problem. The frist order answer, I suppose, would be that the mass of the two objects has absolutely no effect. The second order answer would be that the more massive the objects are, the faster they fall, but since they're dropped at the same time, they'll hit at the same time. I suppose as a third order answer we can take the mass of the feather to be negblible. Then Pleonast's formula would apply to the motion of the hammer and earth, but not the feather. For the feather, set the inititial condition to be velocity=(0,0,0), position=(2 meters, arm's length, 0). Then use the hammer and Earth's positions to calculate the force at each intsnat, and integrate. I don't think can be done in any manner other than numerically, and I don't have access to a program that can keep track of all the significant digits necessary to get any answer other zero (I suppose I could write my own program, but it doesn't seem all that important). My guess is that seeing as how the horizontal separation of the hammer and feather is about four order of magnitudes smaller than the radius of Earth, this will lead to an answer four orders of magnitude smaller than Pleonast's answer (assuming, of course that it's correct).