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
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:
$$ d\approx \sqrt[3]{\frac{9R}{32}}\approx \color{red}{121.5\,m}.$$
Half a meter.  You take the slack extra meter and lift it half a meter high.