OWAO 2025, Практически тур — Задача 2. At the World’s Edge

ΩWAO 2025 ◇ Practical Round ◇ Problems · 20 т.

Условие

Cape Fligely (81∘ 51′ N;  59∘ 14′ E81^{\circ}\,51'\,\mathrm{N};\ \ 59^{\circ}\,14'\,\mathrm{E}) is located on the northern coast of Rudolf Island in Franz Josef Land, Russian Federation. It is the northernmost point of Russia, Europe, and the Eurasian continent as a whole.

a) Find the distance from the cape to the North Pole of the Earth. [2 т.]

b) At what solar declination does the polar day begin at the cape? Assume that atmospheric refraction at the horizon is 35′35'. [2 т.]

You are provided with a table eclipses.xlsx of lunar eclipses of the 21st century. The table lists the eclipse date, Saros number, gamma1^1, the moments of the Moon’s disk contacts with the Earth’s shadow (see Fig. 1), and the time of greatest eclipse.

c) How many total lunar eclipses will occur in the 21st century? partial? penumbral? [4 т.]

Point Ξ (0; 59∘ 14′ E)\Xi\,(0;\ 59^{\circ}\,14'\,\mathrm{E}) lies on the equator at the same longitude, somewhere in the Indian Ocean.

d) Determine how many total lunar eclipses will be visible from both Cape Fligely and point Ξ\Xi. List the corresponding event numbers.

Count an eclipse as visible if any part of totality occurs above the horizon. Ignore events visible only in partial phases. [12 т.]

1^1Gamma is the distance of the center of the Moon’s disk from the center of the Earth’s umbral shadow, expressed in units of Earth’s equatorial radius. It is defined at the moment of greatest eclipse, when its absolute value reaches a minimum. For a lunar eclipse, it shows whether the Moon passes north or south of the center of the Earth’s shadow; a positive value means the Moon passes to the north of the center.

Diagram of a central total lunar eclipse: a large light-grey disk labelled Penumbra with a darker concentric disk labelled Umbra; the Moon's path is a straight arrow from upper right to lower left, with Moon disks marking the contacts P1, U1, U2 (entering) and U3, U4, P4 (leaving); U2 and U3 are drawn red inside the umbra.
Figure 1: Designation of contacts illustrated with the example of a central total lunar eclipse

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