ItsCalledSoccer
ItsCalledSoccer
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November 16th, 2010 at 6:49:41 AM permalink
OK ... I've given all the responses a fair reading, and they still don't "feel" right to me ... here's why.

If you were to jump off the top of the Stratosphere, you would land at the base of the Stratosphere. You don't "slip" just because you're not rigid with the structure. You're still in the gravitational field. Extending this to being dropped from a helicopter from, oh, 15,000 feet. Neglecting all other effects, you land directly under the helicopter. Extend to the 3,600-mile-high stack of bills and neglecting the effects of air friction, re-entry, etc., and you still land at the base ... I think. You're still in the gravitational field, just like off the Strat and out of the helicopter.

I think the effects that I'm reading are due to some tangential velocity component not due to (angular velocity) * (radius). If that component is zero, I don't think there's any "slip," "slip" = "earth rotating underneath you because you're not being carried along with the gravitational field, which only acts radially."

Admittedly, physics isn't my strong suit, so I'm open to being wrong. I just want to get the "feel" of the right answer.
Doc
Doc
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November 16th, 2010 at 8:13:26 AM permalink
OK, I'll try again. And I apologize for my difficulty in expressing this more clearly -- I really could use some diagrams here.

Compared to jumping off the Stratosphere hotel tower or out of a helicopter, there is a big issue of scale when you are a couple thousand miles up on Bill's money tower. The effects we are discussing might exist, but they would never be noticeable for those low-altitude falls.

The tangential velocity that we are attributing to Bill Gates at the top of that stack of money really is just the angular velocity times radius that you mentioned. Visualize this as if you are looking down on the earth from maybe 50,000 miles above the north pole. You see the earth spinning counter-clockwise with Bill at the tip of the tower. You can also see the base of the tower traveling in a circle -- it would be a smaller circle than the one that Bill is traversing.

Maintaining a circular motion involves continually accelerating toward the center of the circle -- centripetal acceleration. Both Bill and the base of the tower are accelerating toward the axis of the earth (the center of the circle as you view it from above the pole). Their circles, however are not in the same plane -- for the 40 N latitude tower that I suggested, the base is circling in the plane of the 40th parallel, while Bill (due to the height of his money tower) is actually circling the axis beyond the north pole (closer to your vantage point than the ice cap).

Suppose for a moment that the earth were spinning really fast and there were negligible gravity. There would be the tendency for Bill to be flung off into space unless he held tight to the tower. But he would not be flung directly away from the center of the earth -- he would actually be flung along the tangent to his circular path and would remain in that same plane above the pole (parallel to the equatorial plane). Of course, the rotation is not that fast, and there is indeed sufficient gravity to keep him on the tip of the tower.

When Bill steps off the tower, at that instant he will continue moving along the tangent at a high velocity, even before he has started to plummet very fast at all vertically. Gravity will pull him toward the center of mass of the earth. If his tangential velocity were high enough to keep him in orbit like a satellite, his orbit would be around the center of the earth passing alternately over the northern and southern hemispheres. Since his tangential velocity is not great enough to keep him in orbit, he would tend to fall to earth somewhere below that orbit, possibly in the southern hemisphere. During his free fall, he is pulled toward the center of the earth, but there would be no forces attempting to bring him toward that circular path that the base of the tower would continue to follow. He would fail to land near the base of the tower, in some sense as if he had been partially flung into space but finally fell and landed somewhere far away.

Are we making any progress here? It really is difficult to describe this without any graphics.

Perhaps I should admit that I actually tricked myself while writing this description. I started to say that the Coriolis effect would prevent him from falling to earth directly under that hypothetical orbit but would instead cause his path to curve. I believe now that he would indeed fall under that circular (or elliptical) orbit but that his path would appear to be curved when observed from the rotating earth. I could give some description of this Coriolis effect (angular acceleration that results when there is radial motion in a rotating system) in the context of a carousel, but (unless someone is really interested in that) I think that would just add even more unnecessary complications to the discussion of the falling, burning-on-reentry, billionaire. And the discussion has already gone far past the point of being too geeky except for the entertainment of the Wizard.
ItsCalledSoccer
ItsCalledSoccer
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November 16th, 2010 at 9:48:56 AM permalink
Let me ask a different way ... imagine that there's a guy wire running along the stack of bills, and that Bill has an eyebolt on his utility belt, and uses the guy wire to keep him close to his money.

Bill jumps off ... what's causing the force on the guy wire?
Doc
Doc
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November 16th, 2010 at 10:25:33 AM permalink
Quote: ItsCalledSoccer

... what's causing the force on the guy wire?

Bill's momentum in the tangential direction he was originally traveling and the guy wire's attempt to compel him to stay next to the tower as it circles the earth's axis. In two dimensions, it's something like stepping off a merry-go-round while having a leash connecting you to the zebra -- it's not just your radial motion that causes the problem but also the zebra/leash's attempt to keep you circling. I admit, though, there are a few twists in going from 3D to 2D that weaken that analogy.


Edit: added this 2nd paragraph and was typing it while ItsCalledSoccer was posting the next comment.

Suppose for a moment that Bill had no mass and didn't tend to fall. The guy wire would require him to circle the axis of the earth at whatever altitude he happened to be. This would not be circling the center of the earth but circling a point on the earth's axis at the same north/south position. Next, consider that Bill does have mass and is not attached to the guy wire. He would tend to circle the center of the earth rather than its axis. These are two different directions, and he cannot go both ways. If he is attached to the guy wire and has mass, the wire must impose a force to make him stick with the tower.
ItsCalledSoccer
ItsCalledSoccer
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November 16th, 2010 at 10:37:53 AM permalink
Quote: Doc

Bill's momentum in the tangential direction he was originally traveling and the guy wire's attempt to compel him to stay next to the tower as it circles the earth's axis. In two dimensions, it's something like stepping off a merry-go-round while having a leash connecting you to the zebra -- it's not just your radial motion that causes the problem but also the zebra/leash's attempt to keep you circling. I admit, though, there are a few twists in going from 3D to 2D that weaken that analogy.



Wouldn't this cause him to be going faster, since he's jumping towards the center of centripetal acceleration (due to gravity) rather than away from it (like on the merry-go-round)? If his original tangential velocity is (angular velocity) * (radius of earth + stack of bills), wouldn't he be going faster (tangentially) than the bills at, say, 95% of the stack height? He wouldn't be "slipping", he'd be gaining! And that makes even less sense to me ...
Doc
Doc
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November 16th, 2010 at 11:00:28 AM permalink
First, I'm sorry about editing my previous post while you were entering yours. Simultaneous edits are troublesome at times.

Next, consider a variation to my previous description, in this case with the tower of money on the equator. All of Bill's motion during his fall will be over the equator. His initial tangential speed is, as you said, greater than that of any of the bills of currency in the stack. As he descends due to gravity, he does indeed tend to move out ahead of the tower as you described (even though you are not comfortable with that idea). That is what boymimbo was discussing in his calculation back on page 4. This is that Coriolis effect that I was hesitant to go into a few posts back: angular acceleration due to radial motion in a rotating system. I can discuss that more if you like, and it's an interesting topic, but it does cause a bit of stress on the part of new students of dynamics of rotating systems (as well as for some of us geezers), and I'm not sure you really want to get into it. (Let me know on that; if we go there, I'll try to put it in terms of systems you are familiar with.)

When the tower is not on the equator, there is not only this tendency for Bill to move ahead of the tower as he falls but also a tendency to move (initially) closer to the equator, because gravity is forcing him to circle the center of the earth. Both of these effects contribute to keeping him from landing near the base of the tower (in the absence of the guy wire).
ItsCalledSoccer
ItsCalledSoccer
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November 16th, 2010 at 12:58:45 PM permalink
Thanks for your patience, I'm not being contrary, just asking questions to understand better. If this is just going to escape me, so be it.

Maybe I should sleep on this ... sometimes things marinate and I can understand them better. But as best I can describe the difference of opinion (understanding, maybe, since there's a correct answer here), it seems to center on whether or not the rotation even matters.

Yes, there's rotation, but everything's rotating at the same angular velocity so there's no rotation differential between the rigid stack, the falling Bill, and a horrified observer on earth. So, from a point of view on the earth's surface .. say, looking up at the falling Bill ... it would look like he's falling straight down.

I understand that the gravity field weakens as radial distance increases, and this gives the potential for the "slip." Also, there's the "gain" from the initial tangential velocity being greater than all other tangential velocities of the bill stack all the way down. My understanding was, the "slip" was exactly countered by the "gain" (ignoring air resistance, re-entry, etc.), making any relative movemet away from the rigid stack = 0: the falling bear-marble shot experiment. That's what makes intuitive sense to me.

But hey, it doesn't make intuitive sense that you can't turn a spinning bike wheel perpendicularly without a great struggle, but that's true, isn't it?
Ayecarumba
Ayecarumba
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November 16th, 2010 at 7:05:41 PM permalink
Quote: Wizard

I've read different opinions. Some say that over a very long shot, like 1000 yards, it is enough to think about it if accuracy is essential, like a good head shot. I'd be interested to know how many millimeters a 1000-yard shot would be off due to the coriolis effect. It would likely depend on the latitude.



It does depend on your latitude. Think about it as trying to toss a paper airplane at a target while riding a carousel. The closer you are to the center (pole) the more pronounced the required correction.

Here is a formula from the Sniper's Hide website:
Deflection = (Earth's Rate of Rotation*(Range to Target)^2*sin(Latitude))/Projectile's Average Velocity

So, at 45 degrees latitude using a cartridge that sends a projectile 1,000 yards in 1.42 seconds, your required correction is 0.22 feet, or about 6.5 mm.
Simplicity is the ultimate sophistication - Leonardo da Vinci
mkl654321
mkl654321
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November 16th, 2010 at 7:10:37 PM permalink
Quote: Ayecarumba

It does depend on your latitude. Think about it as trying to toss a paper airplane at a target while riding a carousel. The closer you are to the center (pole) the more pronounced the required correction.

Here is a formula from the Sniper's Hide website:
Deflection = (Earth's Rate of Rotation*(Range to Target)^2*sin(Latitude))/Projectile's Average Velocity

So, at 45 degrees latitude using a cartridge that sends a projectile 1,000 yards in 1.42 seconds, your required correction is 0.22 feet, or about 6.5 mm.



So is this effect nonexistent at zero latitude, i.e., the equator, and most pronounced at the poles?
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Ayecarumba
Ayecarumba
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November 16th, 2010 at 7:24:10 PM permalink
According the long range calculators, yes, but only if you are shooting due east or due west at the equator.
Simplicity is the ultimate sophistication - Leonardo da Vinci

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