Quote:DocYour 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.

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.Quote:ItsCalledSoccerHmmm ... 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.

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 :)

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

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:WizardNo 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:mkl654321And 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.

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.

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.

Shall I hold back in future?