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OWAO 2023, Теория — Задача 5. Hide and Seek

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Всички материали от OWAO 2023

ΩWAO 2023 ◇ Theoretical Round ◇ Problems

Условие

At the 50 $^{\mathrm{th}}$ parallel ( $50^{\circ}$ N), some star rises just before Antares ( $\alpha_A = 16^{\mathrm{h}}\,29^{\mathrm{m}};\ \delta_A = -26^{\circ}\,26'$ ) disappears behind the horizon, and it sets just when Sirius ( $\alpha_S = 6^{\mathrm{h}}\,45^{\mathrm{m}};\ \delta_S = -16^{\circ}\,43'$ ) appears. The star is brighter than $+1.5^{\mathrm{m}}$ . Estimate the equatorial coordinates of the star. What is the name of this star? Consider the Earth to be perfectly spherical with no atmosphere.

Отговори

Решение

First we determine hour angle $t_A$ of Antares setting. Using spherical law of cosines for the spherical triangle with vertices in the north celestial pole, zenith and Antares — navigational triangle, we have [2 pt]

\cos z_A = \sin\varphi\sin\delta_A + \cos\varphi\cos\delta_A\cos t_A. \qquad (2)

We assume $z_A = 90^{\circ}$ because we neglect the refraction. Therefore, [1 pt]

\cos t_A = -\tan\varphi\tan\delta_A = 0.592 \quad\Rightarrow\quad t_A = 53.7^{\circ}.

Antares setting sidereal time is [1 pt]

s_1 = t_A + \alpha_A = \frac{53.7^{\circ}}{15^{\circ}/\mathrm{h}} + 16^{\mathrm{h}}\,29^{\mathrm{m}} = 20.06^{\mathrm{h}}.

Similarly, we estimate hour angle $t_S$ and sidereal time $s_2$ at the time of Sirius rising above the horizon [1 pt]:

\cos t_S = -\tan\varphi\tan\delta_S = 0.359 \quad\Rightarrow\quad t_S = -69.0^{\circ},

s_2 = \frac{-69^{\circ}}{15^{\circ}/\mathrm{h}} + 6^{\mathrm{h}}\,45^{\mathrm{m}} = 2.15^{\mathrm{h}}.

The sidereal time of the star’s upper culmination is equal to its right ascension [2 pt]:

\alpha_{\star} = s_c = \frac{s_1 + s_2}{2} \approx 23^{\mathrm{h}}.

The hour angle of the star setting is

t_{\star} = s_c - s_1 = 23.1^{\mathrm{h}} - 20.06^{\mathrm{h}} = 3.04^{\mathrm{h}}.

Therefore, we may estimate the declination [2 pt] by the formula (2):

\cos t_{\star} = -\tan\varphi\tan\delta_{\star} \quad\Rightarrow\quad \tan\delta_{\star} = -\frac{\cos t_{\star}}{\tan\varphi} = -0.587 \quad\Rightarrow\quad \delta_{\star} \approx -30^{\circ}.

These coordinates are quite similar to the coordinates of Fomalhaut ( $\alpha$ Piscis Austrinus; $\alpha = 22^{\mathrm{h}}\,57^{\mathrm{m}};\ \delta = -29^{\circ}\,37'$ ) [1 pt].

Константи

Universal
Speed of light	$c = 3.00 \cdot 10^8\ \mathrm{m/s}$
Planck constant	$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s}$
Hubble constant	$H_0 = 70\ (\mathrm{km/s})/\mathrm{Mpc}$
Astronomical unit	$1\ \mathrm{au} = 149.6 \cdot 10^6\ \mathrm{km}$
Parsec	$1\ \mathrm{pc} = 206\,265\ \mathrm{au}$

Earth
Radius	$R_{\oplus} = 6371\ \mathrm{km}$
Obliquity	$\varepsilon = 23.4^{\circ}$
Surface gravity	$g = 9.81\ \mathrm{m/s^2}$
Orbital period	$T_{\oplus} = 365.26^{\mathrm{d}}$
Orbital eccentricity	$e_{\oplus} = 0.0167$

Moon
Radius	$R_{\text{☾}} = 1737\ \mathrm{km}$
Orbital period	$T_{\text{☾}} = 27.32^{\mathrm{d}}$
Orbital inclination	$i_{\text{☾}} = 5.1^{\circ}$

Sun
Radius	$R_{\odot} = 6.96 \cdot 10^5\ \mathrm{km}$
Absolute magnitude	$M_{\odot} = 4.74^{\mathrm{m}}$ (bol.)
Effective temperature	$T_{\odot} = 5.8 \cdot 10^3\ \mathrm{K}$
Luminosity	$L_{\odot} = 3.828 \cdot 10^{26}\ \mathrm{W}$

Emission constants
Stefan–Boltzmann	$\sigma = 5.67 \cdot 10^{-8}\ (\mathrm{W/m^2})/\mathrm{K^4}$
Wien’s displacement	$b = 2898\ \mu\mathrm{m} \cdot \mathrm{K}$

UBV system	Mean wavelengths
U band	$\lambda_U = 364\ \mathrm{nm}$
B band	$\lambda_B = 442\ \mathrm{nm}$
V band	$\lambda_V = 540\ \mathrm{nm}$

Hydrogen spectrum
Lyman L $\alpha$	$\lambda_{L\alpha} = 1215.7$ Å
Balmer H $\alpha$	$\lambda_{H\alpha} = 6562.8$ Å

Оригинал в Архива: OWAO_2023_Theoretical_Round_Problems.pdf · официални решения: OWAO_2023_Theoretical_Round_Solutions.pdf