Expert: Paul Klarreich - 1/18/2011

Question
in the question you answered here http://en.allexperts.com/q/Advanced-Math-1363/Radius-circumference.htm in 2007, you showed that the answer would be the same regardless of the original radius, because you never used an actual value. in other words, it seems as if you showed mathematically that given any sphere, one could add a meter to its circumference, invariably making its radius increase by about 6 inches. i have a hard time believing that this would hold true for a plum tomato or the universe itself, for instance. can you help me with my confusion? i have gone over the math and i just can't seem to find the place where i am misinterpreting your answer.

Answer
Questioner: Andrew
Country: United States
Category: Advanced Math
Private: No
Subject: circumference relative to radius
Question: in the question you answered here http://en.allexperts.com/q/Advanced-Math-1363/Radius-circumference.htm in 2007, you showed that the answer would be the same regardless of the original radius, because you never used an actual value. in other words, it seems as if you showed mathematically that given any sphere, one could add a meter to its circumference, invariably making its radius increase by about 6 inches. i have a hard time believing that this would hold true for a plum tomato or the universe itself, for instance. can you help me with my confusion? i have gone over the math and i just can't seem to find the place where i am misinterpreting your answer.
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It is remarkable that you happened to dig up an answer from Oct 19, 2007.  Indeed, that probably was the last one for some time, because of some history.

Anyway, although the answer might be counterintuitive, it is correct,  but not for plum tomatoes.  Those are not spherical, being fatter at one end.  Cherry tomatoes, OK.

All you need do is substitute numbers for the radius.  (You are actually dealing with a circle -- a great circle for each sphere -- not a sphere.)  

You will note that C is a linear function of r.  So any increase in r produces the same (times 2pi) increase in C.

Not so for area, of course, but that is another story.

**** answer repeated below ****

You will need the formula:  C = 2pi r

Your first band has a length equal to  2pi ER, where ER is the Earth Radius.

You are adding 1 meter to that length, so the new circumference is 2pi ER + 1.  What is the New Radius, which I will call NR?

2pi NR = 2pi ER + 1   

NR = ER + 1/(2pi)   << divided by 2pi

Now 1/(2pi) is the increase in radius, and the height of the (floating?) band of metal.  1/2pi is about 1/6.28, or about  39.37/6.28, or about 6 inches.  

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