Съдържание

IOAA 2023, Отборен тур — Задача 4. Southern Pole Star

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Всички материали от IOAA 2023

Group Competition

Указания

For this round, the participants will be grouped by random selection into 5-person teams, each representing a named asteroid. Each participant in the group will be from a different country. The selection will take place near the beginning of the IOAA, so that the team members can get to know each other.
Please remember the name and number of your asteroid, as this will also be used to identify your place during the Planetarium and Observation rounds.
The group competition consists of several tasks, which you will receive in a sealed envelope. Each team works together at one table under the supervision of the guides to solve the tasks. You are not allowed to communicate with participants from other teams during the round.
Dedicated answer sheets are provided for writing your answers. Enter the final answers into the appropriate boxes in the answer sheet (marked A).
Open the envelopes on the START signal given by the judges. Time is measured from this moment; the winning team will be the one which finishes in the shortest total time, after any time penalties (for example for incorrect or missing answers) are applied. The time penalties are explained in each task.
When you have solved all the problems, hand your answer sheets to the guide, who will note the total time.
The maximum time available for the round is 90 minutes. After this time any remaining answer sheets should be handed in.
The completed answer sheets will then be marked by the jury, who will apply time penalties as appropriate. The winning team will be announced at the closing ceremony.
Everything you need will be provided on the table (calculator, office supplies, geometrical instruments, paper, table of constants).
For one of the tasks, all the screens in the room will simultaneously display a video at a specified time (the video will be repeated several times).

Table of Constants

Fundamental constants
Speed of light in vacuum	$c$	$=$	$2.998 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Planck constant	$h$	$=$	$6.626 \times 10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B$	$=$	$1.381 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$=$	$5.670 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$=$	$1.602 \times 10^{-19}\ \mathrm{C}$
Universal gravitational constant	$G$	$=$	$6.674 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$
Universal electric constant	$\epsilon_0$	$=$	$8.854 \times 10^{-12}\ \mathrm{m^{-3}kg^{-1}s^{4}A^{2}}$
Universal gas constant	$R$	$=$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_\mathrm{A}$	$=$	$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$
Wien’s displacement constant	$b = \lambda_m T$	$=$	$2.898 \times 10^{-3}\ \mathrm{m\,K}$
Mass of electron	$m_\mathrm{e}$	$=$	$9.109 \times 10^{-31}\ \mathrm{kg}$
Mass of proton	$m_\mathrm{p}$	$=$	$1.673 \times 10^{-27}\ \mathrm{kg}$
Mass of neutron	$m_\mathrm{n}$	$=$	$1.675 \times 10^{-27}\ \mathrm{kg}$
Mass of Helium nucleus	$m_\mathrm{He}$	$=$	$6.645 \times 10^{-27}\ \mathrm{kg}$
Atomic mass unit (a.m.u., Dalton)		$=$	$1.661 \times 10^{-27}\ \mathrm{kg}$

Astronomical data
Hubble constant	$H_0$	$=$	$70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
North Ecliptic Pole ( $J2000.0$ )	$(\alpha_\mathrm{E},\ \delta_\mathrm{E})$		$(18^\mathrm{h}00^\mathrm{m}00^\mathrm{s}, +66^{\circ}33'39'')$
North Galactic Pole ( $J2000.0$ )	$(\alpha_\mathrm{G},\ \delta_\mathrm{G})$		$(12^\mathrm{h}51^\mathrm{m}26^\mathrm{s}, +27^{\circ}07'42'')$
1 jansky	1 Jy	$=$	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 parsec	1 pc	$=$	$3.086 \times 10^{16}\ \mathrm{m}$
			206 265 au
			3.262 ly
1 astronomical unit (au)	1 au	$=$	$1.496 \times 10^{11}\ \mathrm{m}$
1 sidereal day	$T_\mathrm{SD}$	$=$	$23.93444\ \mathrm{h}$
			$23^\mathrm{h}56^\mathrm{m}04^\mathrm{s}$
1 tropical year		$=$	365.2422 solar days
1 sidereal year		$=$	365.2564 solar days

Approximations

$(1+x)^n \approx 1 + nx$

$(1+x)(1+y) \approx 1 + x + y$ if $x \ll 1$ and $y \ll 1$

Gauss’s formulae

Spherical law of cosines: $\cos a = \cos b \cos c + \sin b \sin c \cos A$

Spherical law of sines: $\dfrac{\sin A}{\sin a} = \dfrac{\sin B}{\sin b} = \dfrac{\sin C}{\sin c}$

next page $\Longrightarrow$

The Sun
Solar luminosity	$L_\odot$	$=$	$3.826 \times 10^{26}\ \mathrm{W}$
Apparent angular diameter of Sun	$\theta_\odot$	$=$	32’
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$=$	$5778\ \mathrm{K}$
Apparent visual magnitude		$=$	$-26.75$
Absolute visual magnitude		$=$	$+4.82$
Apparent bolometric magnitude		$=$	$-26.83$
Absolute bolometric magnitude		$=$	$+4.74$
Distance of the Sun from the Galactic centre		$\approx$	$8\ kpc$

The Earth and Moon
Obliquity of the ecliptic (Earth)	$\epsilon$	$=$	$23.5'$
Platonic year (period of precession of Earth’s axis)		$=$	25 765 years
Apparent visual magnitude of full Moon		$=$	$-12.74$
Apparent angular diameter of Moon	$\theta_\mathrm{L}$	$=$	$31'$
Inclination of the lunar orbit to the ecliptic		$=$	$05^{\circ}08'43''$
Inclination of the lunar equator to its orbital plane		$=$	$6.687^{\circ}$
Lunar sidereal month	$T_\mathrm{SL}$	$=$	$27.321661\ \mathrm{d}$
			$655.71986\ \mathrm{h}$
Synodic month		$=$	$29.530589d$
Tropical month		$=$	$27.321582d$
Anomalistic month		$=$	$27.554550d$
Draconic month		$=$	$27.212221d$

The Solar System

Object	Mean radius [km]	Mass [kg]	Semimajor axis [au]	Eccentricity
Sun	695 700	$1.988 \times 10^{30}$	—	—
Mercury	2 440	$3.301 \times 10^{23}$	0.387	0.206
Venus	6 052	$4.867 \times 10^{24}$	0.723	0.007
Earth	6 378	$5.972 \times 10^{24}$	1.000	0.016 710
Moon	1 737	$7.346 \times 10^{22}$	$3.844 \times 10^{5}$ km	0.054 900 (range $0.026 - 0.077$ )
Mars	3 390	$6.417 \times 10^{23}$	1.524	0.093
Jupiter	69 911	$1.898 \times 10^{27}$	5.203	0.048
Saturn	58 232	$5.683 \times 10^{26}$	9.537	0.054
Uranus	25 362	$8.681 \times 10^{25}$	19.189	0.047
Neptune	24 622	$1.024 \times 10^{26}$	30.070	0.009

Условие

On the map of the southern sky in the answer sheet, draw the southern precession circle and determine the year, nearest to the present date, in which the star $\delta$ Velorum will become the southern pole star as a result of precession. The map is presented in the equidistant projection.

Penalties: missing or incorrect answer +10 minutes

Решение

Внимание: Непълно решение

В архива няма официално решение на тази задача.

Оригинал в Архива: team.pdf