A string that is 25,000 miles long...?

Question

A man buys a string that is 25,000 miles long and he sets out to stretch is around the circumference of the Earth. When he reaches his starting point, he finds that the string is 25,000 miles and one yard long. Rather than cut the string he decides to tie the two ends of the string together and distribute the extra 36 inches evenly around the globe. Disregarding the amount of string used to tie the knot, how far does the string stand out from the Earth because of the extra length?

According to my calculations:

C[of Earth]=25,000 miles

C[of String]=25,000 miles and 1 yard

R[of Earth]= (C[of Earth])/(2π)

R[of String]= (C[of String])/(2π)

25,000 Miles=1,584,000,000 inches

R[of Earth]= (1,584,000,000")/(2π)=252,101,429.858"

R[of String]= (1,584,000,036")/(2π)=252,101,435.587"

252,101,435.587"

- 252,101,429.858"

_______________

5.729"

Is this correct?

Answer 1

I believe you've done it right....though it is only theoretical and not really possible to do this without spending billions.

Answer 2

there is a much simpler way to think of this...

the circumference of a circle is 2 pi R

if the string is greater than the circumference of the Earth by an amount 36 inches, then we know that the circle created by the string will have a circumference 36 inches greater than the circumfernece of the earth

so, if we think of the string's circumference as C+C', where C is the circ of the earth and C' is the additional 36 inches, we have:

C+c'=2pi R +2 pi R'

or C'=2 pi R'

or the additional radius R'=36/6.28=5.73 inches

you are correct...but thought you might be interested to see a simpler way to do this

Answer 3

....Or an even simpler way using only (a little) calculus:

C = 2πR

dC = 2πdR

dR = dC/2π

Since dC = 1yd, it is a sufficiently small fraction of C to qualify as a differential unit.........