Circumference problem


By Rethman Chuck (M762) on Wednesday, October 11, 2000 - 02:41 am :

Okay, I heard this, and it doesn't make sense to me. Take the earth, wrap a string around it, then add a foot to the length of the string. Pull the string evenly, How far off the earth is the string? Theoretically and numerically, you get the same answer. Then, do the same to a basketball, and you get the SAME answer. This does not seem possible. Is there a radius growth factor that should be taken into account? Stumped!


By Kerwin Hui (P1312) on Wednesday, October 11, 2000 - 08:23 am :
Because circumference C is 2π×radius, so the change in circumference is 2π×[changeinradius].
By Rethman Chuck (M762) on Wednesday, October 11, 2000 - 02:26 pm :

That gives the change being r =1/2pi ., or 0.1592357 feet. Theoretically, this makes sense; intuitively it does not. What troubles me is you get the same answer using real numbers REGARDLESS of the radius--this is the answer if the radius is the earth or a basketball!!!
By the way, this was a Suisse Bank interview question.


By Michael Doré (P904) on Wednesday, October 11, 2000 - 07:38 pm :

It is very counter-intuitive but totally correct. Let's take it the other way around - if the string is one foot above the ground all the way around, then how much longer will the rope have to be than the circumference? Answer: the same for the Earth and a basketball.

One possible way of reconciling this with intuition is to note that the curvature of the Earth is very much less than that of a basketball. So the string around the Earth will have almost exactly the same curvature as the Earth (after all what difference is a mere extra foot going to make). Therefore in each metre of the Earth's circumference there will only be a very small discrepancy between the length of the rope above it and the 1 metre of ground. But then of course there are many MANY metres round the circumference so the discrepancy will build up, and turn out to be the same as the basketball in total. For a smaller sphere the curvature would be larger, so there would be more discrepancy in length above each metre of the cirumference (as the extra foot would be more important in terms of curvature), but then there would be fewer metres in total round the circumference, and the two fators cancel leaving the discrepancy in length between rope and circumference the same in each case.

I know that probably makes no rigorous sense whatsoever, but I'm trying to give an intuitive rather than mathematical feel for the question. (Kerwin has already done the mathematical part).