OWAO 2023, Практически тур — Задача 1. Kholshevnikov Metric

ΩWAO 2023 ◇ Practical Round ◇ Problems

Условие

Consider two points in 3-dimensional space, defined by their radius vectors r⃗1\vec{r}_1 and r⃗2\vec{r}_2. According to the Pythagorean theorem, the distance between these points d12=(r⃗1−r⃗2)2d_{12} = \sqrt{(\vec{r}_1 - \vec{r}_2)^2}. The concept of “distance” can be generalized to arbitrary spaces.

In order to determine which meteor shower a meteoroid belongs to, it is required to determine the “distance” between the orbits — the so-called metric in the space of orbital parameters. One example of metric is Kholshevnikov metric, described in detail below.

Let aa, ee, ii, ω\omega, Ω\Omega be the semi-major axis, eccentricity, inclination, argument of the pericentre and the longitude of the ascending node of the meteoroid’s orbit, respectively (Keplerian elements).

We introduce two vectors u⃗\vec{u} and v⃗\vec{v}:

u⃗:=(sin⁡i a(1−e2) sin⁡Ω−sin⁡i a(1−e2) cos⁡Ωcos⁡i a(1−e2)),v⃗:=(e a(1−e2)(cos⁡ωcos⁡Ω−cos⁡isin⁡ωsin⁡Ω)e a(1−e2)(cos⁡ωsin⁡Ω+cos⁡isin⁡ωcos⁡Ω)esin⁡i a(1−e2) sin⁡ω),\vec{u} := \begin{pmatrix} \sin i\,\sqrt{a(1-e^2)}\,\sin\Omega \\ -\sin i\,\sqrt{a(1-e^2)}\,\cos\Omega \\ \cos i\,\sqrt{a(1-e^2)} \end{pmatrix}, \quad \vec{v} := \begin{pmatrix} e\,\sqrt{a(1-e^2)}(\cos\omega\cos\Omega - \cos i\sin\omega\sin\Omega) \\ e\,\sqrt{a(1-e^2)}(\cos\omega\sin\Omega + \cos i\sin\omega\cos\Omega) \\ e\sin i\,\sqrt{a(1-e^2)}\,\sin\omega \end{pmatrix},

Let us define a distance ρ\rho — Kholshevnikov metric — between two orbits in the space of Keplerian elements by the formula

ρ=(u⃗1−u⃗2)2+(v⃗1−v⃗2)2.\rho = \sqrt{(\vec{u}_1 - \vec{u}_2)^2 + (\vec{v}_1 - \vec{v}_2)^2}.

Let a0a_0, e0e_0, i0i_0, ω0\omega_0, Ω0\Omega_0 be the Keplerian elements of the mean orbit of the meteor shower. If the distance ρ\rho between this orbit and the orbit of a given meteoroid is smaller than a limiting value, we may suppose the meteoroid to belong to this meteor shower.

The file catalog2018.csv contains the observation times and equatorial coordinates of meteoroids, as well as the Keplerian elements of the orbit of each meteoroid. The data was taken from the archives of the project SonotaCo Network.

YearMonthDayα\alpha, ∘^{\circ}δ\delta, ∘^{\circ}aa, aueeii, ∘^{\circ}ω\omega, ∘^{\circ}Ω\Omega, ∘^{\circ}
201811.4241883.569001.114001.9670.579508.8111055.40330100.69660
201811.48810107.8190024.129001.9730.752901.44970280.02610280.76860
…………………………

We are looking for meteoroids related to the Geminids meteor shower. The mean orbit of the shower corresponds to a0=1.31a_0 = 1.31 au, e0=0.889e_0 = 0.889, i0=22.9∘i_0 = 22.9^{\circ}, ω0=324.3∘\omega_0 = 324.3^{\circ}, Ω0=261.7∘\Omega_0 = 261.7^{\circ}. For this problem, let us set the limiting value log⁡10ρ=−1.0\log_{10}\rho = -1.0.

a) Plot all meteoroids on the graph in the coordinates (λ;β)(\lambda; \beta) where λ\lambda and β\beta are the ecliptic longitude and latitude of the meteoroid, respectively.

b) How many meteoroids belong to the Geminids meteor shower? Use the limiting value of the log⁡10ρ\log_{10}\rho specified above and consider only meteoroids with elliptic orbits. Mark these meteoroids on the graph in the coordinates (λ;β)(\lambda; \beta).

c) Plot histograms for semi-major axes and log⁡10ρ\log_{10}\rho values for meteoroids belonging to the Geminids meteor shower and the histogram for days of observing these meteoroids. On what date was the largest number of meteoroids from the Geminids meteor shower observed?

d) Determine the median ecliptic and equatorial coordinates of the meteoroids belonging to the Geminids meteor shower.

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_Practical_Round_Problems.pdf · официални решения: OWAO_2023_Practical_Round_Solutions.pdf