Due to the Earth's rotation, what is the magnitude of the centrifugal force acting on a unit mass at the equator? What would this force be at the same latitude on an $8000 \mathrm{~m}$ high mountain? By how much would the Earth's gravitational force be reduced at the same mountain? What would be the weight of a $75 \mathrm{~kg}$ person on the same mountain?
1. $C_{f}=\frac{4 \pi^{2} r}{T^{2}}=\frac{4 \pi^{2} \cdot 6370 \cdot 10^{5}}{86164^{2}}=3,387 \mathrm{dyn}$.
2. $C_{f}^{\prime}=3,392$ dyn.
3. If the gravitational force per unit mass at the Earth's surface is $g_{0}$ and on the mountain is $g_{1}$, then

$$
g_{0}: g_{1}=(r+h)^{2}: r^{2},
$$

so

$$
g_{1}=g_{0} \frac{r^{2}}{r^{2}+2 h r+h^{2}}
$$

but since $h^{2}$ is negligible compared to the other quantities, it can be ignored; dividing by $r^{2}$ and expanding according to the binomial theorem:

$$
g_{1}=g_{0}\left(1-\frac{2 h}{r}\right)=978,90 \mathrm{dyn}
$$

and the difference in gravitational forces $=2,55$ dyn.

4. If we want to calculate the weight of the body, we need to subtract the centrifugal force from the gravitational force

$$
P=75,000 \times 975,51 \mathrm{dyn}=73163000 \mathrm{dyn} .
$$

Since on average 981000 dyn $=1 \mathrm{~kg}$ weight, therefore:

$$
P=74,54 \mathrm{~kg} \text { weight. }
$$

(László Bánó, Budapest.)

The problem was also solved by: H. Fodor, R. Földes, F. Kräuter, Gy. Jánosy, L. Pető, K. Sztrokay, O. Szilas.