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Math fun

Illustration 1 Ok, we have the earth. The earth as a circumference of 40.000 km. Now we put a rope around the aequator. The rope lies directly on the surface of our earth (and our earth has the shape of a perfect sphere...)

Illustration 2

Now we have a guy, who cuts this rope and adds 1 m to it. What is the distance between the earth and the rope, when we assume that the distance is equal around the whole aequator?

Illustration 3 Ok, remember your time in school... This should be quite easy. The distance is the difference between the radius from the earth and the rope.

\begin{eqnarray*}
40000\, km & = & 2\cdot \pi \cdot r_{1}\\
r_{1} & = & \frac{40000\, km}{2\cdot \pi }\\
r_{1} & = & 6366,19772\, km
\end{eqnarray*}

\begin{eqnarray*}
40000,001\, km & = & 2\cdot \pi \cdot r_{2}\\
r_{2} & = & \frac{40000,001\, km}{2\cdot \pi }\\
r_{2} & = & 6366,19788\, km
\end{eqnarray*}

\begin{eqnarray*}
distance & = & r_{2}-r_{1}\\
& = & 1,5915\cdot 10^{-4}\, km\\
& = & 15,915\, cm
\end{eqnarray*}

Ok, this looks quite reasonable. But what do you think will happen, if we do the same with a tennis ball? What is the distance between the tennis ball and the rope when we lengthen it by 1 m?

Believe me, it's also 15,915 cm. You don't belive it? Ok, let's calculate it. Let assume our ball has a circumference of 20 cm.

\begin{eqnarray*}
20\, cm & = & 2\cdot \pi \cdot r_{1}\\
r_{1} & = & \frac{20\, cm}{2\cdot \pi }\\
r_{1} & = & 3,1830\, cm
\end{eqnarray*}

\begin{eqnarray*}
120\, cm & = & 2\cdot \pi \cdot r_{2}\\
r_{2} & = & \frac{120\, cm}{2\cdot \pi }\\
r_{2} & = & 19,0985\, cm
\end{eqnarray*}

\begin{eqnarray*}
distance & = & r_{2}-r_{1}\\
& = & 15,915\, cm
\end{eqnarray*}

This distance remains the same, whatever object you coose. You may lengthen the circumference of a atom or of the galaxy by 1 m, the distance will always be 15,915 cm.

The explanation is quite simple - it's just math! Let's assume our initial circumference is C.

\begin{eqnarray*}
C & = & 2\cdot \pi \cdot r_{1}\\
r_{1} & = & \frac{C}{2\cdot \pi }
\end{eqnarray*}

\begin{eqnarray*}
C+1m & = & 2\cdot \pi \cdot r_{2}\\
r_{2} & = & \frac{C+1\, m}{2\cdot \pi }
\end{eqnarray*}

\begin{eqnarray*}
distance & = & r_{2}-r_{1}\\
& = & \frac{C-1\, m}{2\cdot \pi...
...pi }\\
& = & \frac{1\, m}{2\pi }\\
distance & = & 15,915\, cm
\end{eqnarray*}

No matter what C you choose, your distance will always be \( \frac{1\, m}{2\pi } \) , and thus be 15,915 cm.

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