\section*{Task 1 - V11031}

The Soviet spacecraft is expected to approach within about \(100000 \mathrm{~km}\) of Venus on May 19, 1961.

a) How many times larger would Venus appear to an observer on the spacecraft compared to the full moon as seen from Earth?

Note: Approximate the apparent sizes of the full moon and Venus using their diameters.

b) What portion of the Venusian surface could he see?

Moon diameter: 3476 km, Venus diameter: 12220 km, Earth-Moon distance: 384000 km
}

a) If the celestial bodies are considered approximately disk-shaped as required, then the spacecraft is 3.84 times closer to Venus during the flyby than the Moon is to Earth. Additionally, Venus is \(\frac{12220}{3476}\) times larger than the Moon, so it would appear \(\frac{12220}{3476} \cdot 3.84 = 13.5\) times larger to an observer in the spacecraft compared to the full moon as seen from Earth.

b) One might be tempted to answer: "the side facing it." However, the actual observation angle should be used to determine the visible area of Venus as a percentage. Unfortunately, the problem also leaves us uncertain whether the distance of \(100000 \mathrm{~km}\) refers to the distance to the center of Venus or to its surface. We assume the former:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1172.jpg?height=334&width=691&top_left_y=1169&top_left_x=248)

It is \(\cos \alpha = \frac{r}{d}\). The total surface area of Venus is

\[
A_{g} = 4 \pi r^{2}
\]

The visible part is

\[
A_{s} = 2 \pi r^{2} (1 - \cos \alpha) = 2 \pi r^{2} \left(1 - \frac{r}{d}\right)
\]

Thus, the visible fraction of the surface area is

\[
\frac{A_{s}}{A_{g}} = \frac{1 - \frac{r}{d}}{2} = \frac{d - r}{2 d}
\]

In the present case, this corresponds to approximately \(46.9\%\) of the total surface area.