OWAO 2023, Теория — Задача 8. Omega Sirius

ΩWAO 2023 ◇ Theoretical Round ◇ Problems

Условие

Estimate the distance between α\alpha and ω\omega Canis Majoris in parsecs.

α\alpha CMaω\omega CMa
Right ascension06h 45m 08.917s06^{\mathrm{h}}\,45^{\mathrm{m}}\,08.917^{\mathrm{s}}07h 14m 48.653s07^{\mathrm{h}}\,14^{\mathrm{m}}\,48.653^{\mathrm{s}}
Declination−16∘ 42′ 58.02′′-16^{\circ}\,42'\,58.02''−26∘ 46′ 21.61′′-26^{\circ}\,46'\,21.61''
Apparent magnitude in V band−1.46-1.463.82
Spectral typeA1VmB2.5Ve
Bolometric correction0−2.2-2.2
Mass, M⊙\mathfrak{M}_{\odot}2.063±0.0232.063 \pm 0.02310.1±0.710.1 \pm 0.7

Отговори

Решение

Константи
Universal
Speed of lightc=3.00⋅108 m/sc = 3.00 \cdot 10^8\ \mathrm{m/s}
Planck constanth=6.63⋅10−34 J⋅sh = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s}
Hubble constantH0=70 (km/s)/MpcH_0 = 70\ (\mathrm{km/s})/\mathrm{Mpc}
Astronomical unit1 au=149.6⋅106 km1\ \mathrm{au} = 149.6 \cdot 10^6\ \mathrm{km}
Parsec1 pc=206 265 au1\ \mathrm{pc} = 206\,265\ \mathrm{au}
Earth
RadiusR⊕=6371 kmR_{\oplus} = 6371\ \mathrm{km}
Obliquityε=23.4∘\varepsilon = 23.4^{\circ}
Surface gravityg=9.81 m/s2g = 9.81\ \mathrm{m/s^2}
Orbital periodT⊕=365.26dT_{\oplus} = 365.26^{\mathrm{d}}
Orbital eccentricitye⊕=0.0167e_{\oplus} = 0.0167
Moon
RadiusR☾=1737 kmR_{\text{☾}} = 1737\ \mathrm{km}
Orbital periodT☾=27.32dT_{\text{☾}} = 27.32^{\mathrm{d}}
Orbital inclinationi☾=5.1∘i_{\text{☾}} = 5.1^{\circ}
Sun
RadiusR⊙=6.96⋅105 kmR_{\odot} = 6.96 \cdot 10^5\ \mathrm{km}
Absolute magnitudeM⊙=4.74mM_{\odot} = 4.74^{\mathrm{m}} (bol.)
Effective temperatureT⊙=5.8⋅103 KT_{\odot} = 5.8 \cdot 10^3\ \mathrm{K}
LuminosityL⊙=3.828⋅1026 WL_{\odot} = 3.828 \cdot 10^{26}\ \mathrm{W}
Emission constants
Stefan–Boltzmannσ=5.67⋅10−8 (W/m2)/K4\sigma = 5.67 \cdot 10^{-8}\ (\mathrm{W/m^2})/\mathrm{K^4}
Wien’s displacementb=2898 μm⋅Kb = 2898\ \mu\mathrm{m} \cdot \mathrm{K}
UBV systemMean wavelengths
U bandλU=364 nm\lambda_U = 364\ \mathrm{nm}
B bandλB=442 nm\lambda_B = 442\ \mathrm{nm}
V bandλV=540 nm\lambda_V = 540\ \mathrm{nm}
Hydrogen spectrum
Lyman Lα\alphaλLα=1215.7\lambda_{L\alpha} = 1215.7 Å
Balmer Hα\alphaλHα=6562.8\lambda_{H\alpha} = 6562.8 Å

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