I read on the Internet it's true, but I suspect it:

Image describing the puzzle

Take a ribbon tightly wound around the equator of the earth. Add 1 meter to that ribbon by cutting it at any point. Uniformly lift above the earth, removing the slack caused by the addition in the length of the ribbon.

The distance that the ribbon is lifted above the surface of the earth is ~16cm.

(All this is assuming that the earth is a perfect sphere.)

The first time I saw this, I was quite astonished, wondering what difference would 1 meter make, considering that the earth is huge, but anyway lets do the math. Let the radius of the sphere be R and the new radius be R', hence

$$ 2 \pi R' = 2 \pi R + 1\,m $$


$$ 2\pi(R'-R) = 1\,m$$

or, the height $R' - R$ is $$ \frac{1}{2\pi}\,m = 15.9\,cm $$

Is it really independent from the sphere's radius?

  • 3
    $\begingroup$ Yes. The relation between perimeter and radius is linear, and that means the same difference in radius always causes the same difference in perimeter and vice versa, regardless of what values radius and perimeter have. $\endgroup$ – Daniel Fischer May 7 '14 at 10:25

Yes, apart from the fact that evenly distributing this slack might be a bit problematic in practice, the theory works just the way you computed it.


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