# The earth and the pea0817

$$r_1 = 6370km$$ at the earth (as a sphere with $$r_1 = 6370km$$ ) and a pea (as a sphere with $$r_2 = 2mm$$ ) and $$r_2 = 2mm$$ a rope over the equator so that it lies tightly on the surface. Now you lengthen both ropes by one meter each. Both ropes should now lie fully extended over the equator again - they no longer lie completely on the surface, but hover over the equator. How high above the surface does the rope float above the earth, how high above the pea?

The two ropes have the initial length:

$$l_1 = 2\cdot 6370 km \cdot \pi \Leftrightarrow r_1 = 6370 km = \frac{l_1}{2 \cdot \pi}$$

and

$$l_2 = 2 \cdot 2mm \cdot \pi \Leftrightarrow r_2 = 2mm = \frac{l_2}{2 \cdot \pi}.$$

But now, after the extra time.

$$r_{1 NEU} = \frac{l_1 + 1m}{2\cdot \pi}$$

and

$$r_{2 NEU} = \frac{l_2 + 1m}{2\cdot \pi}.$$

Now, amazingly enough.

$$r_{1 NEU} - r_1 = \frac{l_1 + 1m}{2\cdot \pi} - \frac{l_1}{2\cdot \pi} = \frac{l_1 + 1m - l_1}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m$$

and

$$r_{2 NEU} - r_2 = \frac{l_2 + 1m}{2\cdot \pi} - \frac{l_2}{2\cdot \pi} = \frac{l_2 + 1m - l_2}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m.$$

Thus the distance from the surface is independent of $$l_1$$ or $$l_2$$ , i.e. independent of the radii $$r_1$$ or $$r_2$$ the spheres. The astonishing answer is thus: Both ropes float at the same height $$0.159m$$ ) above the surface.

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