Problem: I have a rope, it fits around the equator exactly once. I add 10 cm to the rope, attach the ends, and pull up. How high off the ground can I pull the rope?

Solution: The length \(L\) of the rope before adding the 10 cm is \(L=2 \pi R\), where \(R\) is the radius of the earth and is about \(6.4 \times 10^6\) meters.

After adding the 10 cm, you give the rope a stretch in the center. When the rope is stretched to full tautness, the result will be most of the rope hugged against the equator, with 2 straight pieces, each tangent to the equator. Thus, we have the right triangle as pictured above. We are interested in solving for the height \(h\).

We proceed by considering the rope after the addition of the 10 cm which will now be denoted as \(\Delta L\). When pulled up, the rope hugs the earth outside the points of tangency. Denote the points of tangency on the earth as corresponding to an angle \(\theta\) from the vertical, and the length of one of the two sections of rope not attached to the earth as \(y\). The following relation holds:

\(L+\Delta L=2 (\pi – \theta) R+ 2 y\),

where

\(y^2=(R+h)^2-R^2=2 R h+h^2\).

Use the fact that \(L=2 \pi R\) and \(\theta=\tan^{-1} \frac{y}{R}\), and define \(w=\frac{\Delta L}{2 R}\) and \(z=\frac{y}{R}\). The above equation is then rewritten as

\(w=z-\tan^{-1}z\),

where we must solve for \(z\). Of course, this is a transcendental equation that cannot be solved exactly, but it is clear that, since \(w\) is very small, then \(z\) must also be small, and we can get places by expanding the transcendental function in a series:

\(z-\tan^{-1}z=\frac{1}{3} z^3-\frac{1}{5} z^5+\frac{1}{7} z^7-\ldots\)

The best way to use this series is to consider the first term, with all higher-order terms being some error:

\(z-\tan^{-1}z=\frac{1}{3} z^3+O(z^5)=\frac{1}{3} z^3 \left [ 1+O(z^2) \right ]\).

Then, to lowest order, we obtain the following:

\(z=\left ( 3 w \right )^{\frac{1}{3}} \left [ 1+O(w^{\frac{2}{3}}) \right ]\).

Now, to lowest order,

\(z=\sqrt{2 \frac{h}{R}} \left [ 1+O \left ( \frac{h}{R} \right ) \right ]\),

so that we now have an approximate solution and an error estimate:

\(h=\left ( \frac{9}{32} R \Delta L^2 \right )^{\frac{1}{3}}+O\left [ \left ( \frac{\Delta L^4}{R} \right )^{\frac{1}{3}} \right ]\).

Note that the solution involves the radius of the earth, which is a very large number compared with the rope extension of 10 cm. The result will then be surprisingly large; the first-order term is, using the values given above, about 26.2 meters. The error term, on the other hand, is on the order of 0.001 meters, or 0.1 cm, and can safely be called negligible. However, for larger extensions, we can simply expand the series solution further until the error estimate is within acceptable bounds.