We are planning the journey of the Venus-4 spacecraft. For simplification, we consider the orbits of Earth and Venus around the Sun as circles, and the orbital planes of the two planets as identical. The ratio of the orbital radii is $1: 0.72$, and Venus completes its orbit in 224.7 days. Both planets orbit the Sun in the same direction. We can also use the ecliptic longitudes of the Sun and Venus on June 18, 1967, and September 26, 1967, which are: Sun: $86^{\circ}$ and $183^{\circ}$, Venus: $132^{\circ}$ and $148^{\circ}$. (Such a rotation brings the chosen reference direction ${ }^{3}$ to point towards the celestial body in the Earth's position on that day.)

a) Provide an approximate value for the heliocentric ecliptic longitude of Venus on the mentioned days, and on the spacecraft's arrival date, October 18 (all data are considered at $0^{h}$), as well as an approximate value for the heliocentric ecliptic longitude of Earth on the spacecraft's launch date, June 12.

b) Provide an approximate value for the distance between the spacecraft's launch and arrival positions, assuming the radius of the Earth's orbit is $150 \cdot 10^{6}$ $\mathrm{km}$, 4[^0]

[^0]:    ${ }^{2}$ See: Astronomical Yearbook for 1966, Gondolat Publishing, Budapest, 1965. p. 48. (It provides coordinates every 100 days, rounded to whole degrees).

    ${ }^{3}$ The reference direction is the vernal point, which is the direction in which the Sun is seen on March 21.

    ${ }^{4}$ To avoid the search for complex computational procedures, we note that the part of the problem that seems more difficult - but is actually just unfamiliar - only requires solving right-angled triangles.
a) Let $F$ be the Earth and $V$ be the position of Venus on June 18 and September 26, denoted as $F_{1}, F_{2}$ and $V_{1}, V_{2}$, respectively. Furthermore, let $F$'s position on March 21 be $F_{0}$, the day of the spacecraft's launch be $F_{i}$, and $V$'s position on the day of the spacecraft's arrival be $V_{\dot{e}}$. The radii of the orbits of Earth and Venus are $R$ and $r$, respectively, and the projections of the common center $N$ of the circular orbits, the Sun, on the lines of sight $F_{1} V_{1}, F_{2} V_{2}$ are $N^{\prime}$ and $N^{\prime \prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2e7a235f8726859e1c95g-1.jpg?height=921&width=854&top_left_y=347&top_left_x=619)

The angle at $F_{1}$ in the right triangle $F_{1} N N^{\prime}$ is $132^{\circ}-86^{\circ}=46^{\circ}$, so $N N^{\prime}=R \sin 46^{\circ}$, and the angle at $N$ in the right triangle $V_{1} N N^{\prime}$ is

$$
\cos V_{1} N N^{\prime} \varangle=\frac{N N^{\prime}}{N V_{1}}=\frac{R \sin 46^{\circ}}{r}=\frac{0.7193}{0.72} \approx 1.00.
$$

(Considering the rounding of the data, we cannot write more than three significant figures in any case, because the error in $46^{\circ}$ can approach $1^{\circ}$, if the rounding of $132^{\circ}$ and $86^{\circ}$ is in opposite directions and close to the maximum value, $0.5^{\circ}$.) Therefore, $V_{1} N N^{\prime} \varangle \approx 0^{\circ}$, and we consider $V_{1}$ to be the same as $N^{\prime}$, so the heliocentric ecliptic longitude of $V_{1}$ is $132^{\circ}+90^{\circ}=222^{\circ}$.

Similarly, using the right triangles $F_{2} N N^{\prime \prime}$ and $V_{2} N N^{\prime \prime}$,

$$
\cos V_{2} N N^{\prime \prime} \varangle=\frac{R \sin \left(183^{\circ}-148^{\circ}\right)}{r}=\frac{0.574}{0.72} \approx 0.797
$$

$V_{2} N N^{\prime \prime} \varangle \approx 37^{\circ}$. Since the heliocentric longitude of $N^{\prime \prime}$ is $148^{\circ}-90^{\circ}=58^{\circ}$, the longitude of $V_{2}$ is either $58^{\circ}-37^{\circ}=21^{\circ}$ or $58^{\circ}+37^{\circ}=95^{\circ}$. We choose the correct value based on the fact that in 100 days, $V$ travels about $100 / 224.7 \approx 4 / 9$ of its orbit, which is approximately $360^{\circ} \cdot 4 / 9 \approx 160^{\circ}$. Therefore, the longitude of $V_{2}$ is $222^{\circ}+160^{\circ}=382^{\circ}$, or about $22^{\circ}$. Thus, the approximate value of $21^{\circ}$ is the correct one.

For the 22 days between September 26 and October 18, a similar increase of $22 \cdot 1.6^{\circ} \approx 35^{\circ}$ occurs, so the longitude of $V_{\hat{e}}$ is $21^{\circ}+35^{\circ}=56^{\circ}$.

The heliocentric longitude of $F_{1}$ is $86^{\circ}+180^{\circ}=266^{\circ}$; its change over 100 days is $183^{\circ}-86^{\circ}=97^{\circ}$, so the longitude of $F_{1}$ increased by $6^{\circ}$ between June 12 and June 18, making the longitude of $F_{i}$ $260^{\circ}$.

b) Based on all of this, the angle at $N$ in the triangle $F_{i} V_{\hat{e}} N$ is $360^{\circ}+56^{\circ}-260^{\circ}=156^{\circ}$, and thus the side $F_{i} V_{\hat{e}}$ is 252 million km by the cosine rule.

Kecskeméty Károly (Budapest, József A. Gymnasium, III. о. t.)