OWAO 2025, Практически тур — Задача 1. Chandrasekhar Says No

ΩWAO 2025 ◇ Practical Round ◇ Problems · 20 т.

Условие

The file WD.txt contains a data table for ZZ Ceti white dwarfs (Cang et al., 2025) with the following columns:

  • Teff(K) — effective temperature in kelvin;
  • logg — base-10 logarithm of the surface gravity in cm/s2^2;
  • G(mag) — apparent magnitude in the GG band;
  • Parallax(mas) — parallax in milliarcseconds.

The file BC.txt provides bolometric corrections for the GG band (Carrasco et al., 2014):

  • Teff[K] — effective temperature in kelvin;
  • logg[cm/s2] — base-10 logarithm of the surface gravity in cm/s2^2;
  • Mbol-MG — bolometric correction.

a) Plot the dependence of the white dwarf radius (in R⊙R_\odot) on its mass (in M⊙\mathfrak{M}_\odot). [16 т.]

b) How many objects outlie the main relation? For how many objects are the parameters non-realistic? Explain. [4 т.]

Отговори

Решение

Константи

Universal

  • Gravitational constant G=6.67⋅10−11 m3kg⋅s2G = 6.67 \cdot 10^{-11}\ \frac{\mathrm{m^3}}{\mathrm{kg \cdot s^2}}
  • Speed of light c=3.00⋅108 m/sc = 3.00 \cdot 10^{8}\ \mathrm{m/s}

Astronomical

  • Astronomical unit 1 au=149.6⋅106 km1\ \mathrm{au} = 149.6 \cdot 10^{6}\ \mathrm{km}
  • Parsec 1 pc=206 265 au1\ \mathrm{pc} = 206\,265\ \mathrm{au}
  • Hubble constant H0=70 (km/s)/MpcH_0 = 70\ (\mathrm{km/s})/\mathrm{Mpc}

Emission constants

  • Stefan–Boltzmann σ=5.67⋅10−8 Wm2⋅K4\sigma = 5.67 \cdot 10^{-8}\ \frac{\mathrm{W}}{\mathrm{m^2 \cdot K^4}}
  • Wien’s displacement b=2898 μm⋅Kb = 2898\ \mathrm{\mu m \cdot K}

Earth

  • Radius R⊕=6371 kmR_\oplus = 6371\ \mathrm{km}
  • Obliquity of ecliptic ε=23.4∘\varepsilon = 23.4^{\circ}
  • Orbital period T⊕=365.26 daysT_\oplus = 365.26\ \mathrm{days}

Sun

  • Radius R⊙=6.96⋅105 kmR_\odot = 6.96 \cdot 10^{5}\ \mathrm{km}
  • Mass M⊙=1.99⋅1030 kg\mathfrak{M}_\odot = 1.99 \cdot 10^{30}\ \mathrm{kg}
  • Absolute magnitude M⊙=4.74mM_\odot = 4.74^{\mathrm{m}} (bol.)
  • Effective temperature T⊙=5.8⋅103 KT_\odot = 5.8 \cdot 10^{3}\ \mathrm{K}
  • Luminosity L⊙=3.828⋅1026 WL_\odot = 3.828 \cdot 10^{26}\ \mathrm{W}

Оригинал в Архива: OWAO_2025_Practical_Problems.pdf · решение от архива: OWAO_2025_Practical_Solutions.pdf