A String Tied Around The Earth 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

 A: We have to solve (with respect to $d$) the trascendental equation:
$$ 2\sqrt{(R+d)^2-R^2}+2R\arcsin\frac{R}{R+d}=\pi R+1 $$
or solve (with respect to $x=\frac{d}{R}$) the trascendental equation:
$$ 2\sqrt{x^2+2x}+2\arcsin\frac{1}{1+x}=\pi+\frac{1}{R} $$
that is equivalent to:
$$ \sqrt{x^2+2x}-\arctan\sqrt{x^2+2x}=\frac{1}{2R}.$$
In order to solve $z-\arctan z=\frac{1}{2R}$ we may apply Newton's method, then $x=-1+\sqrt{z^2+1}$.
Since $z$ is close to zero, $z-\arctan z\approx\frac{z^3}{3}$, hence:
$$ d\approx \sqrt[3]{\frac{9R}{32}}\approx \color{red}{121.5\,m}.$$
A: 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
$$PG\approx \frac{r}{2}\left(\frac{3\epsilon}{2}\right)^{2/3}.$$
A: EDIT 1:
Let $\Delta C$ be the extra length of loose string.
$$ r = \frac{C}{2 \pi} ;\  \Delta r = \frac{\Delta C}{2 \pi} = h= \frac{1000\, mm}{2 \pi}  = 159. 155   \; {mm \,!} $$  
The relation/ratio is the same on Mars or Saturn or any earth based large round cylinder. 
In this problem you want $ h $ for given input $ (R, \Delta C) $
We calculate tangent length $T$ shallow horizon range using right triangle property :
$$ T^2 =  h ( 2 R -h ) \tag{1} $$
$$ T = R \tan \theta \tag{2}$$
And eliminate $T$ to solve for $ \tan \theta $
$$ \tan \theta= \sqrt{ (h/a) (2-h/a) } \tag{3}$$
From the diagram the slack above involute is indicated by red line,
$$ = IE =\tan \theta - \theta = \Delta C/ (2 h) $$
or,
$$ \sqrt { (h/a) (2-h/a) } -\tan^{-1}{\sqrt{ (h/a) (2-h/a) } } = \Delta C/(2 R) \tag {4} $$
Plugging in numerical values to solve for $h$
$$ R = 40075 * 10^3/( 2 \pi) \, meters ; \, \Delta C =1 m   \rightarrow h = 121.508 m \tag{5}$$
$\because h <<  R$ an approximation using Madhava-Gregory-Leibnitz series 
$$ \tan u - \tan ^{-1}u \approx  u^3/3 $$
gives us $$ h^3 =\frac {9}{32}\cdot  {\Delta C}^2 \cdot R \tag {6}$$
giving only a bit different value 
$ h= 121.505 m $.

A: Half a meter.  You take the slack extra meter and lift it half a meter high.
