In the logbook of a ship's captain, we found the formula $d=p \sqrt{h}$ for determining the distance of the visible horizon, but the number in place of $p$ is not clearly readable. In the formula, $d$ represents the distance of the visible horizon in kilometers, and $h$ represents the height of the observer's eye above sea level in meters. Determine the value of $p$ so that we get a usable formula. (Take the radius of the Earth as 6370 km.)
Solution. Based on the problem statement, let's create a diagram of the Earth and the interpretation of the horizon ($r=$ $6370 \mathrm{~km}, d=p \sqrt{h}$). The horizon can be interpreted as the tangent line to the Earth at a point $\left(E_{1}, E_{2}\right)$, where the emission point is the observer point $M$. Since the line from the center of the circle $\left(O E_{1}, O E_{2}\right)$ to the tangent point is perpendicular to the tangent at that point, we get right triangles: $O M E_{1}, O M E_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7c172c0794aab8f9d3e6g-1.jpg?height=513&width=501&top_left_y=356&top_left_x=801)

Using the Pythagorean theorem, we can write: $r^{2}+d^{2}=(r+h)^{2}$, where $h$ is given in kilometers. The expression $d=p \sqrt{h}$ can be substituted into the Pythagorean theorem, but we need to be careful that in this expression, $h$ is given in meters:

$$
r^{2}+p^{2} h=r^{2}+h^{2}+2 r h, \quad \text { i.e., } \quad p^{2} h=h^{2}+2 r h
$$

To avoid discrepancies due to unit conversions, the left side of the equation must be multiplied by $1000$ ($1 \mathrm{~km}=1000 \mathrm{~m}$): $1000 p^{2} h=h^{2}+2 r h, 1000 p^{2}=h+2 r$. From this,

$$
p=\sqrt{\frac{h+2 r}{1000}}
$$

Since $\frac{h}{1000}$ is negligible compared to $r$, we have

$$
p \approx \sqrt{\frac{2 \cdot 6370}{1000}} \approx 3.57
$$

The sought value: 3.57.