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.

Quote:WizardYes, 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:WizardYes, 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?

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Quote:weaselmanBy 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?

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? :)

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.Quote:rdw4potusThis 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.

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:weaselman4,200,000 (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:WizardI 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:weaselmanWhat 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.