TITLE: A String Tied Around The Earth
QUESTION [6 upvotes]: Say you're standing on the equator and you have a string below you tied around the equator (40,075 km) that is the length of the equator + 1 meter (40,075.001 km). What is the maximum height you can you lift the string off the ground? Can you create a function of both circumference of the circle (earth) and string to output the distance between the two if pulled tight?
Assumptions:

For illustration, the result would be pulled from a single point, making a triangle until it met with the earth, in which it would follow the curvature of the earth. Similar to a snow-cone or O>
The string does not stretch
The earth can be assumed to be a perfect sphere

REPLY [2 votes]: Draw a picture. Let $P$ be the peak of the stretched string, and let $C$ be the centre of the Earth. Let $G$ be the point on the Earth's surface which is on the line $PC$, and let $T$ and $T'$ be the points of tangency of the string with the Equator after the string has been pushed upwards to full height $PG$.
Let $r$ be the radius of the Earth, and let $\epsilon r$ be the excess amount of string in addition to $2\pi r$. Note that $\epsilon$, in your example, is very small.
Let $\theta=\angle TCG$ Note that $\theta$ is small. The key equation is
$$\epsilon  r=2r\tan\theta -2r\theta.$$
This holds because $2r\tan\theta$ is the sum of the lengths of the two tangent segments $PT$ and $PT'$, while $2 r\theta$ is the amount of string saved because it no longer covers the minor arc $TGT'$. The difference is equal to the amount $\epsilon r$ of extra string we have available.
Using the fact that $\tan\theta\approx \theta +\frac{\theta^3}{3}$, we get
$$\theta\approx \left(\frac{3\epsilon}{2}\right)^{1/3}.$$ 
Now that we know $\theta$, we can find the height $PG$ of the peak of the string. From the diagram, we can see that $PG=r(\sec\theta-1)$. Since $\theta$ is small, this is well approximated by $\frac{\theta^2}{2}$, and we end up with
$$PG\approx \frac{r}{2}\left(\frac{3\epsilon}{2}\right)^{2/3}.$$