Galileo, a Hammer, & a Feather

Cecil's Columns/Staff Reports
1

I have a comment that is surely a waste of somebody’s time, about an answer IAN gave to a mailbag question…

He stated that a hammer & feather will always fall at the same rate in a perfect vacuum. Not exactly true, since the earth accelerates toward the hammer slightly more than towards the feather, giving it an imperceptibly slanted vector that I’m too tired to figure out.

If they’re dropped separately, though…

Grav. constant= 6.673e-11 m^3 Kg^-1 s^-2
Mass(earth)= 5.976e24 Kg
Mass(hammer)= 2.000 Kg
Mass(feather)= 0.001 Kg (1 g)
Radius(earth)= 6.378e6 m (equator)
Distance dropped= 2.000 m (height of some random guy who’s doing the dropping)

I’m not ignoring the significant digits rule, but since I can’t find the mass of the earth to 27 digits anywhere, and we need to carry it out to at least 25 decimal places to see the difference, let’s use pretend objects: a pretend earth that is precisely 5.976e24 Kg, a pretend hammer that is precicely 2 Kg, etc. all to arbitrary precision…let’s say the values are good to at least 30 decimal places. And we’ll pretend the hammer and feather have no volumes, to avoid the hassle of which one has farther to fall.

Gravitational acceleration towards any body:

G*m

= ------ (mass 2 cancel out since we’re
d^2 figuring acceleration not force)

Acceleration of any body towards our pretend earth at 2m above the equator=
(6.67310^-11)(5.97610^24)

     6378002^2

= 9.803076945371995882725286955486 m/s^2
(yes, for the real earth it’s 9.81something, but dropping 24 significant digits makes a significant difference. this is theory here. and not the real earth)

Acceleration towards the hammer=
2*(6.673*10^-11)

 6378002^2

= 0.000000000000000000000003280815 m/s^2

Acceleration towards the feather= 0.001*(6.673*10^-11)

  6378002^2

= 0.000000000000000000000000001640 m/s^2

So the total acceleration between the earth and the hammer (both bodies accelerate towards each other) is:

9.803076945371995882725290236301 m/s^2,

and the total between the earth and the feather is:

9.803076945371995882725286957126 m/s^2.

Plugging these numbers into the equation d=(1/2)(at^2) and solving for time, the feather lands after
0.638776293290439842764991676717 seconds,

and the hammer after only
0.638776293290439842764991569880 seconds:

a full 1.06837e-25 seconds faster! if I could just save that much time every day, I’d…well I guess I’d never know the difference.

Of course there’s the integral you have to mess with, since the acceleration increases as the objects get closer together, the interference caused by the gravitational attraction of the guy dropping the hammer & feather, magnetic effects, etc, but we’ll ignore those because they’re petty.

Ok, so is this, but so what? :slight_smile:

2

OOPS. Forgot the link. here it is:
http://www.straightdope.com/mailbag/mgravity.html

3

Ahh, but in the classic test, the hammer and feather are dropped at the same time. Given that, the earth will accelerate towards the gravitational center of both objects combined. (Of course, most of the earth’s acceleration will be because of the hammer, but that’s just the cause, not the effect.)

4

You might want to check out this earlier topic on the same subject… http://boards.straightdope.com/ubb/Forum6/HTML/000137.html

5

Wow. What are the odds? Are they greater than 10^-25?

rocks

6

Joe_Cool, I’m sure your math is impeccable (i.e. it sounds about right, and no way am I cranking through those numbers :rolleyes: ), but there is a fatal flaw in your calculations. When you drop the feather, where is the hammer? It’s laying on the ground, at your feet! In other words, you should add the hammer mass to the Earth mass when calculating how fast the feather falls. Similarly, the Hammer is falling to the Earth, feather combination.

As an admitted precision fiend (come on, you can’t deny it after your OP), I’ll leave it to you to work through the details. Assume the hammer’s mass is evenly distributed throughout the Earth when the feather is dropped.

Speculation: you’ll either get a difference in time about 2000 times smaller (ratio of the hammer and feather masses), 10^25 smaller (I’ll put my money here), or exact cancellation.

Further speculation: if you really had the hammer laying on the ground, the proximity of its mass to the feather will make the feather hit the ground sooner than the hammer falling with the feather on the ground.

It is too clear, and so it is hard to see.

7

ZenBeam

An even more important factor is the shape of the objects. Since gravity acts as if the mass were concentrated at the center only in approximations, and we’re dealing with effects the size of 10^-25, I’m pretty sure that those effects are swamped by other considerations.

That was the point you were making in the shape of the earth topic, wasn’t it? (Which, is still unresolved I guess. I thought the mailbag column was going to be updated.)

rocks

8

I was going to reply that JoeCool specified a spherical Earth, but looking through his OP, I see that he didn’t.

Regardless, I don’t think that shape is important here. Whatever the Earth’s field is, both the hammer and feather see the same one. He specified the feather and hammer had no volume, which I took to mean point objects. He was looking at the difference between the two cases, not so much what either actual result was. Certainly the effect of the shape of the Earth swamps the difference he came up with, but it should effect both cases the same, so the difference shouldn’t change.

Not specifying where the hammer is when the feather is dropped, and vice-versa, means the problem itself is ambiguous at the precision he is working at. That’s why I was careful to specify that he distribute the hammer mass evenly throughout the Earth when droppng the feather.

I’ve been waiting for the “Which way is down” update also.

It is too clear, and so it is hard to see.

9

True, I forgot to specify in so many words that my example earth is a perfect sphere. but I thought it was implied clearly enough, since “radius” is the only dimension I specified. the only shape I know whose only dimension is radius would be a sphere.

Ok, ok, I didn’t specify three-dimensional space (four, since we’re figuring time as well), but that’s pretty well implied, since in two-dimensional space there is no such thing as mass, only area, so gravity would not be applicable.

As for ZenBeam’s comment, I meant for the objects to be dropped separately. So separately, in fact, that neither one affects the other’s drop. The objects experience a quantum state fluctuation, instantaneously bringing one (being dropped) into existence 2 meters above the surface of the earth and changing the other’s (not being dropped) position to a point so far away that its effects are negligible. For argument’s sake, let’s define negligible this way: gravitational acceleration caused by the object at this distance is < 10^-1,000,000,000 m/s^2.

Good enough? Why not? If we can imagine a perfectly sized and shaped sphere and arbitrarily precise masses, why not a gross misapplication of quantum mechanics? hehe :slight_smile:

10

By the way, ZenBeam, I didn’t actually crank through the numbers. I wrote a bc script (under Linux). :slight_smile: <shameless plug>bc is a great little program. You can do calculations to any precision in theory.</shameless plug>

All told it took about 5 minutes.

11

I missed that! I wasn’t referring to the shape of the earth (although I mentioned it)–I was referring to the different shapes of the hammer and feather. The OP converted them to points to avoid the “distance dropped” ambiguity–but that makes them awful hard to nail down.

What about the general relativity corrections? At this level of accuracy, GMm/r² doesn’t work anymore, does it?

rocks

12

Yes, but I still want to know what happens with the case I described.

Excellent. It shouldn’t be too hard to rerun it with the hammer mass added to the Earth for the feather dropping, and the feather mass added to the Earth for the hammer. Just to answer my “speculation” question.

I had been wondering that also, but I wasn’t going to bring it up. You are volunteering to work it out, aren’t you? :wink:

It is too clear, and so it is hard to see.

13

Speculation? Yeah, right. (sotto voce: ZenBeam doesn’t speculate, ZenBeam calculates). In JoeCool’s calculation, he moved the hammer away to nearly infinity.

Since the mass of the earth is only 3x10^24 times the mass of the hammer, the added mass of the hammer would change the answer by approx. 1/3x10^24, which is twice the difference that Joe_Cool found in the OP. Y’all both are extremely sneaky: ZenBeam for calculating, and Joe_Cool for throwing away “vital” mass.

rocks

14

Ok, ZenBeam, here ya go.

Dropping the feather with the hammer’s mass distributed evenly throughout the earth, or vice versa, you were right, of course. either dropped object falls in
.638776293290439842764991569826 seconds. That’s a difference on the order of 10^-29 for the hammer, and 10^-25 for the feather.

With the thing not being dropped at the dropper’s feet (or, since we’re ignoring the mass of the dropper, collinear with the dropped object and the center of mass of the earth), the difference for the hammer is on the order of 10^-11 for both. But the feather does fall faster, obviously, since all that matters is the attractive mass, not the falling mass.

By the way, I cheated a bit. For the last part (object at feet), I didn’t calculate the acceleration towards the center of mass of the entire system, I figured the attractions for the two systems separately and added them.

15

Joe_Cool wrote:

Actually, that’s exactly correct. Cheating would be to just figure on the center of mass, because, as RM Mentock pointed out, the
“all mass concentrated at the center of mass” approximation is only valid for spherically symmetric mass distributions, i.e., not a (theoretical) spherically symmetric Earth + (theoretical) point mass on its surface. And the GR effects do come in way sooner than 10^-24; I would be tempted to actually calculate them, but at that level of precision, you’d need a theory of quantum gravity, which, alas, has not yet been developed.

“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein

16

I find this hard to believe. Any errors due to using Newtonian physics ought to cancel out*. What GR effects are going to affect the hammer and feather differently?

  • Unless you’re considering both the feather and hammer as point masses, and you’re dropping one micro-black-hole into another. But this hardly seems like a valid approximation for the original case. :slight_smile:

It is too clear, and so it is hard to see.

17

Joe Cool bites again. The OP said the hammer and feather have no volumes. I guess that’s the same as point-mass.

When do we actually get to do a real experiment?

rocks

18

Doh!

19

You’re right, ZenBeam, the relativistic effects should affect the hammer and feather (almost) identically… I was just thinking about the effect on either. The black holes raise their own problems… If I recall my hawking correctly, a 2-kg or 1-g black hole would probably explosively evaporate before it hit the ground.

20

Further information on the Hawking evaporation: According to some back-of-the-envelope calculations I was just doing, a hole of mass 2 kg would radiate power at about 10^34 watts, which would result in total evaporation within 10^-17 seconds (less, in fact, since the power increases with decreasing size), much less than the time it would take to hit the Earth, so I think that we have to say that they’re not QUITE point masses. If anyone really wants, I can post the calculations.

“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein