Consider the earth (as a sphere with \(r_1 = 6370km\) ) and a pea (as a sphere with \(r_2 = 2mm\) ) and stretch a rope across the equator so that it lies tightly on the surface. Now extend both ropes by one meter each. Both cables should now lie completely extended above the equator - but they are no longer completely on the surface, but float above the equator. How high above the surface does the rope float above the earth, how high above the pea?
The two ropes initially have the 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.