Robert Eisele
Engineer, Systems Architect and DBA

Misc: The rope around the equator

The earth has a radius of 6371km on the equator and thus a circumference of 40030km. Let us imagine that a rope is stretched along the equator. Now we extend the rope by one meter and lay it around the earth so that it has the same distance from its surface everywhere. Is it possible to push a piece of paper about one tenth of a millimeter thick between the rope and the ground?


The circumference \(U\) of a circle is \(2\pi\) times of its radius:

\[U = 2\pi r\]

If we increase the circumference by \(\Delta U\), the radius increases by \(\Delta r\) and as such:

\[ \begin{array}{rl} U + \Delta U =& 2\pi(r+\Delta r)\\ =& 2\pi r + 2\pi\Delta r \end{array} \]

We know that \(U = 2\pi r\) and can substitute it, which allows us to remove the term from both sides:

\[ \begin{array}{rrl} & \Delta U = & 2\pi\Delta r\\ \Leftrightarrow &\frac{\Delta U}{2\pi} = &\Delta r \end{array} \]

When we plug in numbers, we get \(\Delta r = \frac{1m}{2\pi}\approx 15.9cm\). That means that there is enough space for the sheet of paper, even books fit under the rope. The more interesting conclusion is that no matter what the original radius is, \(\Delta r\) always depends on \(\Delta U\) and not the original radius nor circumference. That means if you have a tennis ball, put a rope around it and increase the circumference by one meter, the added radius is also about 15.9cm.

« Back to problem overview