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Mano
03-11-2003, 02:35 AM
Not a probability question, but I thought it was cool, so I will post it here for your enjoyment:

If a belt were placed around the Earth's equator, and then had six meters of length added to it, and you grabbed it at a point and lifted it until all the slack was gone, how high above Earth's surface would you be?

Assume the radius of the Earth is 6,400 km.

BruceZ
03-11-2003, 04:54 AM
New circumference = 2pi(6,400,000) + 6 = 2pi(6,400,000 +6/2pi) = 2pi(r) so the radius is increased by 6/2pi or 0.95 meters.

irchans
03-11-2003, 08:28 AM
I got h = (9 x^2 r / 32 )^(1/3) where x is the excess amount of rope and r is the radius of the earth. ( x = 6 meters, r = 6.336*10^6 meters )

h is approximately 400 meters.

400 meters seems to be rather high. Is it right ?

BruceZ
03-11-2003, 12:01 PM
Oops, my solution is for how high it would be if it was an equal height above the earth at all points.

morgan
03-11-2003, 02:07 PM
My orignal response was waaaay off. I see my error. Now I get about 400 meters.

Morgan

Mano
03-11-2003, 03:02 PM
Looks like you nailed it. I got this one off of IBM researches website. Their answer was 402 meters. For their solution go to: http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/solutions/May1998.html

I thought this was cool because the answer is so counter intuitive.

Bozeman
03-13-2003, 12:53 AM
Using the first term of the taylor expansion for tan(x)-x=x^3/3, I got 401.7 m. Error should be considerably less than 1 m.

Craig

irchans
03-13-2003, 09:06 AM
The formula I gave for h was really an approximation derived using Taylor series expansion of trig functions. If we just use a computer to numerically solve the exact equations, we get

h = 401.66733241529...

for a radius of 6400 km.