IAO 2011, Теория, Група α — Задача 4. Photo of Jupiter

Photo geometry and identification

Source gap: the opening of this solution is missing at the top edge of scanned page 4. The visible text starts mid-sentence:

… moon (to the left) and to the shadow of the moon (to the right) the distance $L$ from the center of the moon to the center of Jupiter can be calculated using formulas of rectangular triangle:

Where $\alpha$ is the angle between the direction to the Sun (The Sun beams come a little from the left) and the direction to the Earth (from where the picture is made). We can estimate this angle in two ways.

Or we remember that the Capricorn is a zodiac constellation, and the center of this constellation (not zodiac sign, but zodiac constellation!) is passed by the Sun approximately in the end of January — the beginning of February. The angle between the Sun positions on October, 19th and, assume, on February, 3rd, is equal 3.6 months or $108^\circ$.

Or to start with that fact that the picture is made on October, 19th (a direct ascension of the Sun 26 days after an equinox we will estimate as $12^{\mathrm h}+26\cdot4^{\mathrm m}=13^{\mathrm h}44^{\mathrm m}$), Jupiter was in the middle of Capricorn constellation (not zodiac sign, but constellation!) (right ascension of this point we will estimate to $21^{\mathrm h}$), and both the Sun and Jupiter were close to the ecliptic, declination of both points nearby $-10^\circ$–$-15^\circ$. This data allow us to estimate the angle between Sun and Jupiter at the celestial sphere as about the difference in declinations, that is, $7^{\mathrm h}16^{\mathrm m}$ or $109^\circ$.

From the second drawing, knowing this angle $\varphi$, and distances to the Sun from Earth $L_E$ and from Jupiter $L_J$, it is possible to calculate the angle corner $\alpha$ necessary to us:

By measuring the values R, a and b (accordingly 48.5 mm, 59 mm and 18 mm on the author printed picture, on other scaled printed pictures other values may be obtained, but their absolute values are not important, important is only ratios that do not depend on scale) in the photo, and using the formula for a rectangular triangle, we can receive a value of $L/R$ equal to 9.64.

From the given tables it is possible to find that for Io this ratio equals to 5.90, for Europa — 9.39, for Ganimede — 14.97, for Callisto — 26.33. It is evident, that our calculations are made with some accuracy, and the proximity of values 9.64 and 9.39 (1.5% of accuracy) says that we see the moon Europa in the picture.

Transcription note: “difference in declinations” and “1.5%” follow the supplied sketch. The sketch reports distance from Jupiter's centre, while the question asks distance from its surface; a final subtraction of the radius is not explicitly supplied. The ray-diagram label at the bottom of scanned page 3 is partly cut off.