## Task 14/61

During its flight around the Moon, the Soviet space station Luna 3 approached the Earth's satellite to within about $7000 \mathrm{~km}$. For the following calculations, assume the Moon's radius $r$ to be $r \approx 1750 \mathrm{~km}$, and the surface area $F$ of a spherical cap with the sphere's radius $r$ and cap height $h$ is $F=2 \pi r h$, where $\pi \approx \frac{22}{7}$ is used.

a) How large is the area of the Moon that could be seen from this distance?

b) What percentage of the Moon's surface does this represent?

c) Under what viewing angle $\varphi$ would the Moon be observable from this distance?

d) How wide must an object be to be seen under the same viewing angle from a distance of $100 \mathrm{~m}$?
a) The area of the moon that is to be overlooked is the area of a spherical cap (figure). The radius $r$ of the spherical cap is known; it is equal to the radius $r$ of the moon. To calculate the area $F$, only the height $h$ of the spherical cap is still needed. This can be calculated by multiple applications of the Pythagorean theorem or with the help of the altitude theorem or with the help of the theorem of Euclid.

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-016.jpg?height=465&width=780&top_left_y=1612&top_left_x=638)

1. Calculation using the Pythagorean theorem: The triangles $S T_{1} M, T_{1} Q M$ and $S Q T_{1}$ are right-angled. Therefore, $u^{2}+r^{2}=(H+h)^{2}(1), s^{2}+(r-h)^{2}=r^{2}(2)$ and $(H+h)^{2}+s^{2}=u^{2}$ (3).

From this, if we substitute (3) into (1) and solve (2) for $s$,

$$
(H+h)^{2}+s^{2}+r^{2}=(H+r)^{2} \quad \text { and } \quad s^{2}=r^{2}-(r-h)^{2}
$$

If we substitute the equations into each other, we get $(H+h)^{2}+r^{2}-(r-h)+r^{2}=(H+r)^{2}$. This equation contains the height $h$ of the spherical cap as the only unknown. We solve it for $h$:

$$
h=\frac{H r}{H+r}
$$

Thus, the area $F$ of the spherical cap is $F=2 \pi r h=\frac{2 \pi r^{2} H}{H+r}$. Since in our case $H=2 r$, the calculation becomes particularly simple numerically:

$$
F=\frac{2 \pi r^{2} H}{H+r}=\frac{4 \pi r^{2}}{3}
$$

With $r=1750 \mathrm{~km}, \pi=\frac{22}{7}$, we get $F \approx 13000000 \mathrm{~km}^{2}$.

b) The percentage $p$ of the moon's surface area $F$ that can be overlooked relative to the total moon's surface area $O$ is obtained by dividing $F$ by $O$ and multiplying by $100 \%$:

$$
p=\frac{2 \pi r^{2} H \cdot 100}{4 \pi r^{2}(r+H)} \%=\frac{100 r}{3 r} \% \approx 33.3 \%
$$

We see that the percentage $p$ is independent of the radius $r$ if the height $H$ of the observation point is expressed as a multiple of $r$: $H=k r$. Then the size $r$ cancels out in the fraction.

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-017.jpg?height=246&width=359&top_left_y=708&top_left_x=263)

c) As can be seen from the figure, $\sin \frac{\varphi}{2}=\frac{r}{r+H}$.

Since $H=2 r$, it follows that $\sin \frac{\varphi}{2}=\frac{1}{3}$, and thus for the angle $\varphi \approx 39.94^{\circ}$.

d) If $b$ is the width of the object and $a$ is the distance of the observer (figure), then the following must hold:

$$
\frac{A C}{C D}=\frac{B C}{C D}=\frac{b}{2 a}=\tan \frac{\varphi}{2} \rightarrow b=2 a \tan \frac{\varphi}{2} \approx 70.70 \mathrm{~m}
$$

## Addition by Erich Schiffner:

The question of the relative size of the area that can be overlooked from a distance $a$ from the surface is best solved generally. Let $a=k \cdot r$. Then,

$$
(r-h)(k+1) r=r^{2} \quad \text { from which it follows } \quad h=\frac{k}{k+1} r ; \quad F=2 \pi r^{2} \frac{k}{k+1}
$$

For the ratio of $F$ to $H$ ( $H=$ half-sphere surface area $)$, we have

$$
\frac{F}{H}=\frac{k}{k+1}
$$

For $k=2$ (as in the problem), $\frac{F}{H}=\frac{2}{3}$, and for $k=1$, $\frac{F}{H}=\frac{1}{2}$. If $k=\frac{1}{n}<1$, we get

$$
\frac{F}{H}=\frac{1}{n+1}
$$

For $k \approx \frac{1}{30}$, i.e., $n=30$ (Vostok I and II), we get $\frac{F}{H}=\frac{1}{31} \approx 0.032$. The two cosmonauts could therefore overlook about $3 \%$ of the Earth's hemisphere surface from each point of their orbit.