Bill Gates math problem....

A question I was asked......

If Bill Gates was stood atop a pile of $1 bills equivalent to his net wealth (approx $42Billion) and jumped off, but the ground moved away from the bottom of the pile at the standard rate of interest, would he ever hit the floor? If so how long would it take?

Assume the following.
The thickness of a $1 dollar bill is 0.1mm so the pile would be 4.2 million metres high.
Bill will fall at a rate of 55 metres per second (ignore the fact that he would need to reach this speed first)
Bill will earn 5% of his money every year, but he will be paid the going rate 'pro rata' as he falls.
I am sure he will not earn the $550,000 a second he needs to survive hitting the floor, but how long will he survive?
Bare in the mind that he is earning money all the time he is falling.
I think he makes it just over 21 hours without interest.

Yes, he would reach the bottom. Let's ignore the fact that his 4200 kilometer pile of money would put him well into space, for the sake of the problem. The terminal velocity of a falling human body facing the earth is 195 kilometers an hour. Bill's money would only grow at a rate of 24 meters per hour, or about $240,000/hr. So he would fall much faster than his money grew. By the way, it would take him about 21.5 hours to hit the bottom.

Quote: Wizard

Yes, he would reach the bottom. Let's ignore the fact that his 4200 kilometer pile of money would put him well into space, for the sake of the problem. The terminal velocity of a falling human body facing the earth is 195 kilometers an hour. Bill's money would only grow at a rate of 24 meters per hour, or about $240,000/hr. So he would fall much faster than his money grew. By the way, it would take him about 21.5 hours to hit the bottom.

This one creeps me out. The earth would complete 90% of a rotation under Bill while he fell. Assuming he fell on a straight line toward the center of the earth (I think, if he's in the atmosphere, this is NOT the case), he could jump from Grand Island, NE and land in Scranton, PA.

Quote: Wizard

Yes, he would reach the bottom. Let's ignore the fact that his 4200 kilometer pile of money would put him well into space, for the sake of the problem. The terminal velocity of a falling human body facing the earth is 195 kilometers an hour. Bill's money would only grow at a rate of 24 meters per hour, or about $240,000/hr. So he would fall much faster than his money grew. By the way, it would take him about 21.5 hours to hit the bottom.

I don't think so.

At 55 m/s it'll take him 76363 seconds to get to where the bottom of the pile originally was. By then the pile will be about 10^1617 meters high, and grow at a rate of about 5*10^1615 meters per second - about 2*10^1607 faster than speed of light. He'll never catch up at his measly 55 m/s rate (even if you consider the gravitational acceleration, it does not help, because at the constant acceleration rate, the relative percentage of the speed gain will diminish, while the speed of the ground getting away will keep increasing by 5% every second).

A similar problem (but with a very different solution) goes like this:
At the end of a 1 kilometer long rubber strip there is a little worm, moving towards the other end at the speed of 1 millimeter per second. At the end of every second, the band is stretched to become 1 kilometer longer. Will the worm ever reach the other end?
s m

Quote: weaselman

By then the pile will be about 10^1617 meters high, and grow at a rate of about 5*10^1615 meters per second - about 2*10^1607 faster than speed of light.

Can you show me your math on that?

4,200,000 (m) /55 (m/sec) = 76,363.(63) (seconds)
1.05^76,000 * 4,200,000 (m) = 1.023 * 10^1617 (m)
1.023 * 10^1617 (m) * 0.05 (1/sec) = 5.115 * 10^1615 (m/sec)
Did I screw something up again? :)

Quote: rdw4potus

This one creeps me out. The earth would complete 90% of a rotation under Bill while he fell. Assuming he fell on a straight line toward the center of the earth (I think, if he's in the atmosphere, this is NOT the case), he could jump from Grand Island, NE and land in Scranton, PA.

Orbital mechanics confuses me, but, I THINK that inertia will cause him to rotate with the stack of money as he falls right next to it.

Of course, this assumes that the stack is structually solid, which, it probably would not be. Except that the basic premise of the problem would have to force rigidity onto the stack, so he can stand up there prior to jumping.

And he falles in a straight line only until he hits the atmosphere.


Yeah, it creeps me out too, in a goofy, fun kinda way.

Quote: weaselman

4,200,000 (m) /55 (m/sec) = 76,363.(63) (seconds)
1.05^76,000 * 4,200,000 (m) = 1.023 * 10^1617 (m)
1.023 * 10^1617 (m) * 0.05 (1/sec) = 5.115 * 10^1615 (m/sec)
Did I screw something up again? :)

I think the interest rate is 5% per year, not per second. I wish I could get 5% per second on my money.

Quote: Wizard

I think the interest rate is 5% per year, not per second. I wish I could get 5% per second on my money.

It's the mention of $550,000 per second in the original post that made me think the condition was to receive a yearly rate every second.
If it's only 5% per year, that makes the problem trivial, because the speed of receding ground becomes negligible.
What do you think of my worm problem?

Quote: weaselman

What do you think of my worm problem?

That is a good problem. The way it is usually told is that the rubber band stretches at a certain rate all the time, which would require partial differential equations to solve. I've seen it before, but the math was a bit too hard for me to understand the solution. I should revisit it.

Quote: DJTeddyBear

Orbital mechanics confuses me, but, I THINK that inertia will cause him to rotate with the stack of money as he falls right next to it.

Nope. If you stepped off a 4,200 km tower you'd fall straight down. You'd have to be higher to attain orbital speed.

Quote:

Of course, this assumes that the stack is structually solid, which, it probably would not be. Except that the basic premise of the problem would have to force rigidity onto the stack, so he can stand up there prior to jumping.

Side problem: how much shorter would Bill's bill stack be if he had to use carbon nano-tubes to keep it rigid? :P

Quote:

And he falles in a straight line only until he hits the atmosphere.

Sort of. The Earth would rotate under him. Depending of what side of the stack he steps off from, he'd either collide with it shortly afterward or it would move away from him.

Without an atmosphere to slow him down, he's accelerate at 9.81 m/s every second. I've no idea how fast he'd go by the time he ran into the upper layers of the ionosphere, but he'd surely burn up on re-entry; unless he shortens his stack some more by devising a re-entry suit or capsule that could take the heat (not impossible, I suppose, with enough ablative layers).

Quote: Wizard

That is a good problem. The way it is usually told is that the rubber band stretches at a certain rate all the time, which would require partial differential equations to solve. I've seen it before, but the math was a bit too hard for me to understand the solution. I should revisit it.

I may be wrong, but I don't see how stretching 1 km/sec vs. 1 km every second changes the (qualitative - yes/no) answer.

Quote: weaselman

If it's only 5% per year, that makes the problem trivial, because the speed of receding ground becomes negligible.

5% per year is $2.1billion/year. It depresses me that you're right and the effect on the problem is negligible...

Quote: weaselman

I may be wrong, but I don't see how stretching 1 km/sec vs. 1 km every second changes the (qualitative - yes/no) answer.

I think the difference is in whether the growth is constant or instantaneous.

Quote: rdw4potus

I think the difference is in whether the growth is constant or instantaneous.

Yes, I understand that that's the point, but I can't see what difference it makes for the solution. It seems to me that the answer to the problem is the same in either case.

Quote: Nareed

...Without an atmosphere to slow him down, he's accelerate at 9.81 m/s every second. I've no idea how fast he'd go by the time he ran into the upper layers of the ionosphere, but he'd surely burn up on re-entry; unless he shortens his stack some more by devising a re-entry suit or capsule that could take the heat (not impossible, I suppose, with enough ablative layers).

Now, if we're going to take the original problem and start twisting it away with the realities of stepping off a 4,200 km tower, there are a lot of other factors to consider. Here are a few that relate to Nareed's analysis of falling toward the atmosphere:

(1) At 4,200 km above the surface, the initial acceleration of gravity is less than 9.81 m/s^2 because of his greater distance from the center of mass of the earth.

(2) Unless the base of the vertical tower is on the equator, while he is standing on top of the tower the circle he is moving in is not an orbit around the center of the earth but at some fixed latitude slightly different from the latitude of the base, possibly even beyond the pole. Thus, maintaining his orbital momentum after jumping would not keep him vertically above the area near the base of the tower, because he would then go in a different direction, tending to orbit the earth's center of mass while falling; i.e., it becomes a problem of decaying orbital motion. (I think this is an issue with ICBM trajectories; perhaps we could get the Wizard's dad as an expert witness here?)

(3) Even if he were on the equator or we otherwise ignored #2 so that his fall is straight down along side the tower, the "centrifugal force" effects (I know, it's not really a force) would decrease his initial rate of radial acceleration even further.

I hope this added geekiness can be tolerated considering the original subject matter. Anyone want to add more obsurdities, or should we go back to discussing 5%/second interest rates?

Wait ... is this a trick question?

Once he stepped off of the pile and started downward, however much higher the pile gets doesn't matter. So, if answers.com is right and a billion $1 bills stacked would be 67.7 miles high, then Gates's $54 billion would be 54 * 67.7 = 3,656 miles high, which I think is about the same altitude as the earliest transatlantic relay satellites.

Quote: weaselman

I may be wrong, but I don't see how stretching 1 km/sec vs. 1 km every second changes the (qualitative - yes/no) answer.

If the question is whether the ant ever reaches the end, it probably wouldn't matter. However, the question is usually phrased as how long will it take. Here is how it is usually told:

"An ant starts to crawl along a taut rubber rope 1km long at a speed of 1cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1km per second (so that after 1 second it is 2km long, after 2 seconds it is 3km long, etc). Will the ant ever reach the end of the rope?" -- source: en.wikipedia.org/wiki/Ant_on_a_rubber_rope

Quote: Doc

(1) At 4,200 km above the surface, the initial acceleration of gravity is less than 9.81 m/s^2 because of his greater distance from the center of mass of the earth.

Right, I overlooked that fact. In any case Bill would accelerate to a very high speed and would burn up on re-entry, etc, etc.

Quote:

(2) Unless the base of the vertical tower is on the equator,

I assumed it would be built on the Equator, because no other location makes sense for such a structure due to the reasons you mentioned.

Quote: Wizard

If the question is whether the ant ever reaches the end, it probably wouldn't matter. However, the question is usually phrased as how long will it take. Here is how it is usually told:

"An ant starts to crawl along a taut rubber rope 1km long at a speed of 1cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1km per second (so that after 1 second it is 2km long, after 2 seconds it is 3km long, etc). Will the ant ever reach the end of the rope?" -- source: en.wikipedia.org/wiki/Ant_on_a_rubber_rope

The solution I had in mind was to use the harmonic series as described on the page you quoted, and then approximate the partial sum to N as log(N), which gives an answer, that differs from the exact analytical solution by exactly 1. A pretty good accuracy at the scale of e^100,000 :)

Quote: Nareed

I assumed it would be built on the Equator, because no other location makes sense for such a structure due to the reasons you mentioned.

Ooops! It just occurred to me -- would building the tower at one of the poles (almost) eliminate the problem of not dropping vertically? If Bill were wearing his re-entry suit, perhaps his descent path would just spiral around the tower.

I think that his downward drop would not be affected no matter where he was unless he had some sort of tangential velocity. If he dropped straight down, the earth would turn under him but he would still be affected by the downward pull of gravity ... so he, the stack, and the earth would turn at the same rate, right? You don't have to be connected to the earth to turn with it.

In any event, there would be some altitude where this would be the case. If you base-jump off of a tall building (altitude 1,000 ft) or out of an airplane (4,000 feet), this is certainly the case - the only "side-to-side" is provided by some initial (provided by velocity of the plane or outward push away from the building) or in-flight change (eg, parachute slows it) in tangential velocity.

In this case (assuming the stack is rigid), the top of the stack from which he jumps would be moving at some tangential speed, and when he's on the top, he's moving at the same tangential speed. When he jumps off, he's still moving at that same tangential speed. But as he proceeds down the stack, the bills that make it up are going at a slower and slower tangential speed (but the same angular velocity).

If I recall kinematics-in-a-vacuum, then the "slip" he experiences when he's up high - the earth spinning underneath him and not taking him along with it, before he reaches the "base-jump level" or "skydiving level" or whatever altitude it is before this "slip" becomes infinitessimal - is exactly made up by the difference in his original (and continuing) tangential velocity and the decreasing tangential velocity of the stack (reference point) as he descends.

I don't know the math for this, but I seem to recall some basic classroom experiments that confirmed this, like dropping a stuffed bear from a tall standard, at the same time shooting a marble originally aimed at the bear, and by the time the marble had traveled the horizontal distance to the bear-drop, it would ALWAYS hit the bear, no matter the original velocity of the marble (assuming it was enough to reach the bear in the first place).

Quote: ItsCalledSoccer

You don't have to be connected to the earth to turn with it.

Mostly true. However, it is my understanding that very long-range rifle shots are affected by the movement of the earth under the path of the bullet, and will result in being slightly off target if not adjusted for properly. I'm not sure why, but here is a link about it.

Quote: Wizard

...it is my understanding that very long-range rifle shots are affected by the movement of the earth under the path of the bullet, and will result in being slightly off target if not adjusted for properly. I'm not sure why, but here is a link about it.

My initial reaction is skepticism to the amount adjustment that article suggests is necessary for Coriolis effect on a rifle bullet. I would like to see some more details on those calculations. On the other hand, as I suggested above, there is indeed a significant adjustment necessary for a missile traveling a few thousand miles.

Quote: Doc

On the other hand, as I suggested above, there is indeed a significant adjustment necessary for a missile traveling a few thousand miles.

Wouldn't the same principle apply to the bullet?

Quote: Wizard

Mostly true. However, it is my understanding that very long-range rifle shots are affected by the movement of the earth under the path of the bullet, and will result in being slightly off target if not adjusted for properly. I'm not sure why, but here is a link about it.

I think that's right but in that case, there's a great deal of tangential velocity ... which, by the way, is what keeps satellites in orbit. I guess I was assuming that tangential velocity was 0, at least that component that's not attributable to the ever-present angular velocity * radial length. After all, in our daily life, we're moving at one rotation per day (not counting the speed from the orbit around the sun), and it doesn't seem like we're moving at all!

Quote: ItsCalledSoccer

I don't know the math for this, but I seem to recall some basic classroom experiments that confirmed this, like dropping a stuffed bear from a tall standard, at the same time shooting a marble originally aimed at the bear, and by the time the marble had traveled the horizontal distance to the bear-drop, it would ALWAYS hit the bear, no matter the original velocity of the marble (assuming it was enough to reach the bear in the first place).

Your stuffed bear experiment does generally work that way, but it is very different geometry for the problem with the tower. If there were no rotation at all, when Bill stepped off the tower, he will just drop straight down, landing next to the base (ignoring the burning up on re-entry issue). With the rotation, things are different.

Suppose the base of the tower was at 40 deg North latitude. That point on the surface of the earth is rotating in a circle north of the equator. The top of the tower would also be traveling in a circle, but it would not be in the same plane as the circle of the base. The peak would be traveling in a circle even farther north. In fact, my quick calculations indicate the plane of that circle would be almost 1,000 miles above the north pole. Suppose for a moment that the velocity at the peak were such that when Bill stepped off he would maintain orbit at that altitude. He would not stay next to the tower platform. Instead he would begin to orbit around the center of the earth. Since the velocity would not really be adequate to maintain orbit (I don't think), he would fall, but not next to the tower.

Quote: Wizard

Wouldn't the same principle apply to the bullet?

The same principle applies, but it's a matter of whether the effect is significant over small distances. It is very much like the comments from your father that you noted in your blog.

Quote: Doc

The same principle applies, but it's a matter of whether the effect is significant over small distances. It is very much like the comments from your father that you noted in your blog.

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.

I did ask my dad about this, but have not heard back. Coincidentally, I have bothered him twice about the coriolis effect lately. For those who care, here is the blog entry Doc referred to: My father's report from the equator.

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. ...

I agree. That's why I said I would be interested in seeing more details on the calculations.

Quote: Doc

Your stuffed bear experiment does generally work that way, but it is very different geometry for the problem with the tower. If there were no rotation at all, when Bill stepped off the tower, he will just drop straight down, landing next to the base (ignoring the burning up on re-entry issue). With the rotation, things are different.

Suppose the base of the tower was at 40 deg North latitude. That point on the surface of the earth is rotating in a circle north of the equator. The top of the tower would also be traveling in a circle, but it would not be in the same plane as the circle of the base. The peak would be traveling in a circle even farther north. In fact, my quick calculations indicate the plane of that circle would be almost 1,000 miles above the north pole. Suppose for a moment that the velocity at the peak were such that when Bill stepped off he would maintain orbit at that altitude. He would not stay next to the tower platform. Instead he would begin to orbit around the center of the earth. Since the velocity would not really be adequate to maintain orbit (I don't think), he would fall, but not next to the tower.

Hmmm ... that doesn't sound right. If the tower is straight, then the entire tower is along the same radius (that is, the imaginary line starting at the center of the earth, going through the first bill in the stack at the surface, and continuing through all the bills in the entire stack). Since gravity only acts radially, then that's the only direction that matters. Non-radial projections onto the surface have no force component. And, you don't have to be attached to the earth to rotate with it.

I think the example you give would be the same as saying, someone at 40-lat can stand straight up-and-down, but someone standing at, say, 70-lat can't because gravity doesn't act radially there.

At least that's how I'm reading it ... feel free to correct that.

Quote: ItsCalledSoccer

Hmmm ... that doesn't sound right. If the tower is straight, then the entire tower is along the same radius (that is, the imaginary line starting at the center of the earth, going through the first bill in the stack at the surface, and continuing through all the bills in the entire stack). Since gravity only acts radially, then that's the only direction that matters. Non-radial projections onto the surface have no force component. And, you don't have to be attached to the earth to rotate with it.

Once the billionaire takes the leap and until there is resistance from the atmosphere, I agree that the only force acting on him is gravity, and it is acting radially. The resulting acceleration is of course toward the center of the earth, but his motion is not.

The complicating factor is that there is an initial velocity that is not radial but tangential and at a greater radius than the base of the tower. The resulting motion is a decaying orbit. Suppose the tower were at the 40 N latitude that I suggested, or anywhere other than the equator or poles. The base would be on rigid ground circling the earth's axis. It would therefore be experiencing a centripetal acceleration toward the earth's axis, not toward the center of the earth but in the plane of the 40th parallel.

The falling Bill Gates would have no forces to accelerate him in that direction, so by the time his charred remains reached the ground, he would be nowhere close to the base of the tower. Did I make it any clearer that time? It would be much easier if I could draw some figures, but I think that would be even more difficult here than the equations whose absence I have lamented before.

Quote: ItsCalledSoccer

Hmmm ... that doesn't sound right. If the tower is straight, then the entire tower is along the same radius (that is, the imaginary line starting at the center of the earth, going through the first bill in the stack at the surface, and continuing through all the bills in the entire stack). Since gravity only acts radially, then that's the only direction that matters. Non-radial projections onto the surface have no force component. And, you don't have to be attached to the earth to rotate with it.

I think, Doc's point is that the tangential component of the Bill's velocity is not orthogonal to the gravitational force. If the tower was on the equator, your reasoning would apply, but at a latitude you'll still be rotating when you leave the tower, but at an angle to the original plane of rotation - you'll now be rotating at the plane of a great circle (the plane, containing your original velocity vector and the center of the Earth)

Edit: oops, he beat me to it, and actually said it better :)

I just want to interject that I am loving this discussion. It is getting too deep for my level of physics, but I'm honored to have such intelligent comments, about such a ridiculous question, on my forum. No wonder we don't seem to have many women here. Okay, please get back to the discussion, I don't want to hijack the thread.

Quote: Wizard

... I'm honored to have such intelligent comments, about such a ridiculous question, on my forum. ...

Please recall that in my first post, back on page #2, I said that if this was too geeky and if no one wanted to suggest more absurdities, we could go back to discussing your issue of 5%/second interest rates. Would that be a more productive line of thought? And I had tried to keep a degree of levity in the commentary.

And now that we have "gently redirected" the pile-of-money-and-interest-rates thread to discuss gravity and orbital motion, Wizard don't you dare hijack it. Thread hijacking is a violation of the new rules!

Quote: Wizard

No wonder we don't seem to have many women here..

We do, though, per another poster who shall remain nameless, have lots of "girly men".

And are you saying we don't have women here because only guys would discuss something this silly at such great length?

Quote: mkl654321

And are you saying we don't have women here because only guys would discuss something this silly at such great length?

Yes! Now, please, back to gravity and orbital motion.

There are certain facts that you have to ignore. First, the pile of bills is rigid. Bill and the pile of bills is moving at the same velocity at the start of the experiment. The earth is rotating but Bill is moving with the rotation. When Bill starts to plummet to the ground, because he is not attached to the bills, he is not a rigid object. He would continue to have the same moments of velocity relative to the ground. Remove the effect of friction or wind.

Given that the radius of the earth is 6.378 million meters, when you are standing on the earth, and the length of a day is 23 hours, 56 hours and 4.09 seconds, your velocity on the ground is 2 pi r x cosine (latitude) / (length of a day). At the equator, your velocity is 465.1 meters per second. At the latitude of Las Vegas which is 36 degrees north, your velocity on the ground is 465.09 x cos (36) =376.3 meters per second. At the north pole, your velocity on the ground is 0.

Now when you are 4,200 meters higher, your latitude remains the same (since the bills are perpendicular to the ground), but your velocity is different, because your radius is now 10,578,000 meters.

Your velocity is therefore 2 x pi x 10,578,000 / 86,164.09 meters per second = 771.4 meters / second. Therefore, as you fall, there is nothing stopping you from changing that angular velocity. The force of gravity is down and would not affect your velocity. Therefore, you would NEVER hit the floor. You would fly away from that stack of bills and land quite a distance away from it. Further, if you were north or south of the equator, you would fly towards the equator due to coriolis force.

There are two parts of the equation to solve. The rate of growth is 42,000,000,000 x .05 / (86400 x 365) = 66.59056 dollars per second. The height of the pile is increasing by 0.0659 meters per second, which is miniscule. Compounding interest has no effect.

Therefore 55 t = 4,200,000 + .0659056t.

54.99334t = 4,200,000
t = 76,372.9 seconds

In that length of time, my calculation shows that you will land 11,695km away from the dollars at the equator.

Excellent work, boymimbo! Except I didn't understand the comment, "Therefore you would NEVER hit the floor."

Now (if we are to keep this going so as to satisfy the Wizard), would you like to solve this for the latitude of Las Vegas (or the random 40 N I suggested earlier)? Or maybe it's time to start considering the non-Newtonian aspects associated with the high velocity and the variable gravitational field. Or maybe we should just recognize that we have run this one into the ground. I think I vote for that one.

"ridiculous question"?

Shall I hold back in future?

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.

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.

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?

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.

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 ...

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).

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?

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.

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?

According the long range calculators, yes, but only if you are shooting due east or due west at the equator.