How long would it take for the Earth to complete its axial rotation so that the centrifugal force at the equator would cancel out the weight of objects? The question also needs to be solved by trial and error: at the current rotational speed, at what distance from the Earth would the centrifugal force be equal to the gravitational force?
$$
\text { 1. } \frac{4 \pi^{2} r}{T^{2}}=g ; T=2 \pi \sqrt{\frac{r}{g}}=5062 \text{ sec}=1 \text { day } 24 \text{ h } 22 \text{ min }
$$

2. If the distance from the center of the Earth is $x$, and if the gravitational force acting on a unit mass at the other end of this distance is $g_{1}$, then

$$
g: g_{1}=x^{2}: r^{2}
$$

from which

$$
g_{1}=g \frac{r^{2}}{x^{2}}
$$

This must be equal to the centrifugal force, so

$$
\begin{gathered}
g \frac{r^{2}}{x^{2}}=\frac{4 \pi^{2} x}{T^{2}} \\
x^{3}=\frac{g r^{2} T^{2}}{4 \pi^{2}} \\
x=44500 \text{ km}
\end{gathered}
$$

(Alfréd Haar, Budapest.)

The problem was also solved by: L. Bánó, V. Gráf, J. Csada, H. Fodor, R. Földes, F. Kräuter, Gy. Jánosy, P. Merse, L. Pető, O. Szilas.