#1
|
|||
|
|||
Cool math problem
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. |
#2
|
|||
|
|||
Re: Cool math problem
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.
|
#3
|
|||
|
|||
An answer
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 ? |
#4
|
|||
|
|||
Re: Cool math problem
Oops, my solution is for how high it would be if it was an equal height above the earth at all points.
|
#5
|
|||
|
|||
Re: Cool math problem
My orignal response was waaaay off. I see my error. Now I get about 400 meters.
Morgan |
#6
|
|||
|
|||
You are correct!
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/...s/May1998.html
I thought this was cool because the answer is so counter intuitive. |
#7
|
|||
|
|||
Re: You are correct!
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 |
#8
|
|||
|
|||
A more precise answer
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. |
|
|