Съдържание

IAO 2013, Теория — Задача 8. αβ-5. Пладне – XXI век

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2013

XVIII Международная астрономическая олимпиада. XVIII International Astronomy Olympiad. Литва, Вильнюс, 6 – 14. IX. 2013, Vilnius, Lithuania. Round Theo · 6 септември 2013 г. · 8 т.

Внимание: Бележка към темата

Ръкописни български условия за групи α, β и общи αβ; решенията и критериите са на английски. В αβ-6 ръкописът дава 3400 km, а английският оригинал и ключът — 3400 pc. Източниковите разминавания са запазени и отбелязани.

Общо указание

Общо указание. Възможно е не във всички задачи въпросите да са поставени коректно. Някои въпроси (възможно е това да е основният въпрос в задачата, а може и да е някой подвъпрос) може да нямат смисъл. В такива случаи в отговора трябва да се напише (на руски или на английски): „ситуация невозможна - impossible situation“. Разбира се, отговорът трябва да е подкрепен с пресмятания или логически разсъждения.

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Общо указание

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Parameters of orbits and physical characteristics

The bilingual source sheet gives the following values. Blank cells reproduce blank source cells.

Body / planet	Average distance (AU)	Average distance ( $10^6$ km)	Sidereal period (tropical years)	Sidereal period (days)	Orbital inclination $i$	Eccentricity $e$	Equatorial diameter (km)	Mass ( $10^{24}$ kg)	Average density (g/cm³)	Gravity at surface (m/s²)	Axial tilt	Maximum magnitude from Earth	Albedo
Sun	$1.6\cdot10^9$	$2.5\cdot10^{11}$	$2.2\cdot10^8$	$8\cdot10^{10}$			1392000	1989000	1.409			−26.74
Mercury	0.387	57.9	0.241	87.969	7.00°	0.206	4879	0.3302	5.43	3.70	0.01°		0.06
Venus	0.723	108.2	0.615	224.7007	3.40	0.007	12104	4.8690	5.24	8.87	177.36		0.78
Earth	1.000	149.6	1.000	365.2564	0.00	0.017	12756	5.9742	5.515	9.81	23.44		0.36
Moon	0.00257	0.38440	0.0748	27.3217	5.15	0.055	3475	0.0735	3.34	1.62	6.7	−12.7	0.07
Mars	1.524	227.9	1.880	686.98	1.85	0.093	6794	0.6419	3.94	3.71	25.19	−2.0	0.15
Jupiter	5.204	778.6	11.862	4332.59	1.30	0.048	142984	1899.8	1.33	24.86	3.13	−2.7	0.66
Saturn	9.584	1433.7	29.458	10759.20	2.48	0.054	120536	568.50	0.70	10.41	26.73	0.7	0.68

For outer planets and Moon — in mean opposition.

Some constants and formulae

Constant or formula	Value
Speed of light in vacuum, c (m/s)	299792458
Constant of gravitation, G (N·m²/kg²)	$6.674\cdot10^{-11}$
Solar constant, A (W/m²)	1367
Hubble parameter, $H_0$ (km/s/Mpc), mean value	71
Hubble parameter, diapason of values	50–100
Planck constant, h (J·s)	$6.626\cdot10^{-34}$
Charge of electron, e (C)	$1.602\cdot10^{-19}$
Mass of electron, $m_e$ (kg)	$9.109\cdot10^{-31}$
Proton-to-electron ratio	1836.15
Faraday constant, F (C/mol)	96485
Magnetic constant, $\mu_0$ (H/m)	$1.257\cdot10^{-6}$
Universal gas constant, R (J/mol/K)	8.314
Boltzmann constant, k (J/K)	$1.381\cdot10^{-23}$
Stefan–Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, b (m·K)	0.002897
Laboratory wavelength of Hα (Å)	6562.81
Tropical year length, T (days)	365.242199
Standard atmosphere (Pa)	101325
Visible light extinction by the terrestrial atmosphere in zenith (minimum)	19%, 0.23 mag
Refractive index of water for 20°C, n	1.334
Moment of inertia of a solid ball	$I=\frac25MR^2$
Volume of ball	$V=\frac43\pi R^3$
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265
e	2.71828183
Golden ratio, $\varphi$	1.61803399

Data of some stars

Star	Designation	RA	DEC	Parallax $p$	Apparent magnitude $m$	Spectral class	Mass ( $\mathcal M_{\odot}$ )
Sun	⊙	0h–24h	−23°26′ to +23°26′	8.794″	vis −26.74; bol −26.82	G2	1
Proxima Centauri	V645 Cen, α Cen C	14h 29m 43s	−62°40′46″	0.769″	11.05	M5.5	0.123
Alpha Centauri A	α Cen A	14h 39m 37s	−60°50′02″	0.747″	−0.01	G2	1.1
Alpha Centauri B	α Cen B	14h 39m 35s	−60°50′14″		1.34	K1	0.9
Beta Centauri	β Cen	14h 03m 49s	−60°22′23″	0.009″	0.61	B1	21
Epsilon Eridani	ε Eri	03h 32m 56s	−09°27′30″	0.311″	3.74	K2	0.82
Gliese 581	HO Lib	15h 19m 27s	−07°43′20″	0.16″	vis 10.57; bol 8.0	M3V	0.31

The Alpha Centauri parallax is printed once beside the pair; the second cell is left blank here as in the source.

Historical units

Units of power

1 horse-power (hp) = 735.49875 W.

Local units of length in the end of XVIII century

1 arshin (arsh) = 0.711187 m
1 span (sp) = 1/4 arshin
1 vershok (vrsh) = 1/4 span
1 sajene (sj) = 3 arshin
1 verst (vrst) = 500 sajene

Условие

Както е известно, в Литовската република (вижте картата) се използва време $\mathrm{UT} + 02^{\mathrm{h}}$ през зимата и $\mathrm{UT} + 03^{\mathrm{h}}$ през лятото. Пресметнете и направете извод:

Map of Lithuania with geographic coordinates and towns

Equation of time through the year, using the sign convention supplied in the source

5.1. Има ли в Литва такива места, където днес (8 септември 2013 г.) Слънцето ще се намира точно на юг в момента, когато часовниците на жителите на тези места ще показват 12:00? („да-yes“ или „нет-no“). [3 т.]

5.2. А въобще в други дни от годината, могат ли да се намерят такива места? („да-yes“ или „нет-no“). Ако да, пресметнете на кои дати, ако не, обосновете това с изчисления. [5 т.]

Отговори

Решение

Официалното решение и критериите по-долу са на английски, както в източника.

Official English solution sketch. The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

5.1. September 8

Today (on September 8) the summer time is in operation, so the watches of citizens of Lithuania ahead the universal time by 3 hours. Local mean solar time coincide with this time on the meridian 45° of East longitude, located too far to the east (at the latitude of Lithuania it is in the Nizhny Novgorod region of Russia).

The adding of the equation of time (see figure) is in the range of −16 – +14 minutes, and this effect may replace the meridian not more than 4° in longitude. But even with this addition such areas are far beyond the borders of Lithuania.

Answer: «нет-no».

5.2. Other days of the year

To answer the second question of the problem you need to consider two periods: summer and winter time. For any of the day during summer time operation the situation is similar to that we discussed in sub-point 1.

The territory of Lithuania is approximately (we can measure it with an accuracy of a quarter degree) between 21°00′ and 26°45′ of East longitude. At the period when the winter time is in operation, the situation is queerer. Local mean solar time correspond to the meridian 30° East. Thus, in the winter time period mean local solar noon also does not occur before 12:00 of Lithuanian time anywhere in Lithuania.

However, there is a question about the upper culmination of the sun in the problem condition, as a rule it does not coincide with the mean solar culmination at 12:00 of mean solar time. The difference between the apparent (or true) solar time ( $T_A$ ) and mean solar time ( $T_M$ ) is defined by the equation of time,

T_M=T_A+\eta,

where $\eta$ is parameter of the equation of time, which is hereinafter referred to as simply “equation of time”, as it is accepted by astronomers.

The meridian 30° East passes very close to the eastern border of Lithuania (about this meridian St. Petersburg, Nevel, Vitebsk, Orsha, Mogilev and Kiev are located), so one should pay attention to the equation of time. Winter time is in operation from the end of October to the end of March, and within this range the values of the equation of time may be both positive and negative.

Positive values of the equation of time move the apparent solar noon to later time (by the watches of inhabitants), and the negative to earlier time. Using the graph of the equation of time one can determine that the adding is minimal on November 3 and its value in a minimum is −16.4 minutes. The corresponding meridian (for the apparent culmination of the Sun at 12:00 Lithuanian time) is shifted by 4°06′ West and by almost a degree enters the territory of Lithuania, till the longitude $\lambda=25^\circ54^{\prime}$ .

Thus, the apparent noon will occur at 12:00 of Lithuanian time somewhere at the territory of the country, if the values of the equation of time were below −3°45′ in degrees, and so the values of $\eta$ were below −15 minutes. And $\eta$ is below −15 minutes from October 18 to November 17 (approximately, as we can find from the graph).

Answer: «да-yes», October 18 – November 17 (part of this period when winter time is in operation).

Note. There is no universally accepted definition of the sign of the equation of time. This solution is written in accordance with the historically European definition of the sign of the equation of time (positive values of $\eta$ in January–March, as it is displayed in the supplement page). However, in the British and American (and also translated from English) literature, the definition is inverse (negative values of $\eta$ in January–March). The members of Jury must consider that solutions using both alternatives of definition are correct (of course, the answer does not depend on the choice of the alternative, turning “+” and “−” at the graph leads to turn the sign of $\eta$ in the equation (1)).

Source note: the stated eastern border 26°45′ would differ from 30° by 3°15′, whereas the key subsequently uses 3°45′ and 15 minutes. The printed values and resulting date range are preserved.

Official basic criteria

5.1. Totally 3 pt, including:

Summer time, meridian 45°, far to the east, so not possible — 2 pt.
Mention that adding of the equation of time — correction not more than 4° in longitude — 1 pt.

5.2. Totally 5 pt, including:

For any day during summer time the situation is similar #5.1. — 1 pt.
Finding longitudes of Lithuania, meridian 30° is not too far — 1 pt.
Correct equation of time using — 1 pt.
Values of $\eta$ were below −15 minutes — 1 pt.
Finding the period from October 18 to November 17 (approximately) — 1 pt.

Official solution sketches: note for jury and team leaders

The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Basic criteria: note for jury

Jury members should evaluate the students’ solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto using these positions.

Оригинал в Архива: bgtheo.pdf · официални решения: theosol.pdf · допълнителни данни: theodata.pdf · допълнителни данни: theo_cr_1.pdf · допълнителни данни: theo_cr_2.pdf · допълнителни данни: theo_cr_3.pdf · допълнителни данни: constants.pdf · допълнителни данни: hrd.pdf · допълнителни данни: alphatheo.pdf · допълнителни данни: betatheo.pdf · допълнителни данни: theo_cr_1.png · допълнителни данни: theo_cr_2.png · допълнителни данни: theo_cr_3.png