String around the earth

31.Dec.2012

I heard this problem on a TV show, where Girl_A asks Girl_B this question. Girl_A was threatened by Girl_B and wanted to ask her a ‘tough’ question to make her look bad (it worked, Girl_B wasn’t able to answer it) I found this odd because while watching the show, I had the answer in my head almost instantly.

A few weeks back a friend was visiting me, and on a long (boring) road trip, just for the heck of it, I decided to ask my friend the same question. Here goes:

“I take a string, and wrap it around the earth at the equator. Then, once I’ve done that, I add one meter to the string. What will be the gap between the surface of the earth and the string?” I then promoted the assumption that for simplicity, assume the earth is a sphere.

You’re right! the answer is 0.15 meters. Unfortunately, my friend did not reach the same conclusion. I then tried to explain how to get to that answer, but was unable to do so. She was just not able to comprehend that this value (0.15) is independent of the radius of the sphere and we decided to ‘agree to disagree’.

The whole incident gave me deep insight into how some other people think and function. There’s a spatial way of thinking (something as big as the earth, how will 1 meter of anything affect it) and a factual way of thinking (what’s the formula for circumference, and how will it change if ‘r’ radius changes). I would imagine most people think about the formula, but it turns out, most people think spatially.

I knew this was the case when I wasn’t asked about the formula for circumference, but rather the radius of the earth!
 Anyway, here’s how to think about the problem and get the answer in a few seconds:

Start with understanding that wrapping a string around a sphere effectively makes it a circle (you can also get to this from the assumption/hint promoted at the end of the question). Then (if you don’t know it already –you should– the formula of circumference and transform the verbal problem and the two stages of it into mathematical equations.

c = 2πr -> (1)
c+1 = 2π(r+x) -> (2)

(1) represents the act of wrapping a string along the equator

(2) represents the the act of adding 1 (m) to the string

The hard part is done. All you need to do now is solve for x!
So,
(1) ÷ (2)

c / (c+1) =  (2πr) / (2π(r+x))
c+1 / c = (r+x) / r
1 / c = x / r
c = r / x
2πr = r / x
x = 1 / 2π = 0.15

hence, x = 0.15 independent of radius.

Simple, right? You don’t need to know the radius of the earth (who remembers the exact value anyway?) because it doesn’t matter!

Cheers, and Happy New Year!

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