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Условие Отговори Решение

OWAO 2024, Теория — Задача 5. Almost Circumpolar

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Всички материали от OWAO 2024

ΩWAO 2024 ◇ Theoretical Round ◇ Problems · 20 т.

Указания

The theoretical round is 5 hours long and accounts for 40 % of the total grade.
The maximum score for each problem is 20 points. The round total is 180 points.
There are Answer Sheets to write down detailed solutions.

On each Answer Sheet, please fill in

Student code
Problem number
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Start each problem on a separate Answer Sheet.
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Draft papers will not be graded.
Some problems are quite difficult. Remember that partial solutions are also scored.
A list of constants is given on page 4.

Good luck!

Условие

Let’s consider stars for which it is true at the same time: (i) in their diurnal motion they cross the horizon, and (ii) their upper culmination occurs to the north of the zenith.

a) Express the fraction of such stars as a function of latitude $\varphi$ . Assume that stars are distributed uniformly over the celestial sphere.

b) Determine what is the maximum possible fraction of such stars and at what latitude this maximum is achieved.

Отговори

Решение

a) Let us determine the location of suitable stars on the diagram $(\varphi;\ \delta)$ . The upper culmination of the star occurs to the north of the zenith if $\delta > \varphi$ .

A star does not cross the horizon in two cases: either it is circumpolar or it is non-rising. A star is circumpolar if its lower culmination is above the horizon: $h_{\min} = |\varphi + \delta| - 90^{\circ} \geqslant 0^{\circ} \Longrightarrow |\varphi + \delta| \geqslant 90^{\circ}$ . A star is non-rising if its upper culmination is below the horizon: $h_{\max} = 90^{\circ} - |\varphi - \delta| \leqslant 0^{\circ} \Longrightarrow |\varphi - \delta| \geqslant 90^{\circ}$ . Considering $\delta > \varphi$ , we find 4 different cases:

\begin{array}{llcr} \text{Case 1:} & \varphi \leqslant -45^{\circ} & \Longrightarrow & -(90^{\circ} + \varphi) \leqslant \delta \leqslant 90^{\circ} + \varphi;\\ \text{Case 2:} & -45^{\circ} \leqslant \varphi \leqslant 0^{\circ} & \Longrightarrow & \varphi \leqslant \delta \leqslant 90^{\circ} + \varphi;\\ \text{Case 3:} & 0^{\circ} \leqslant \varphi \leqslant 45^{\circ} & \Longrightarrow & \varphi \leqslant \delta \leqslant 90^{\circ} - \varphi;\\ \text{Case 4:} & \varphi \geqslant 45^{\circ} & \Longrightarrow & \delta \in \varnothing. \end{array}

Diagram in the (φ; δ) plane from −90° to +90° on both axes: the line φ = δ, dashed diagonals, shaded triangles labelled 'non-rising stars' and 'circumpolar stars' (two each), a hatched region below the line φ = δ, and the admissible white regions numbered 1, 2 and 3. — Figure 4: Selecting areas that meet the conditions of the problem

The fraction of the area of a sphere enclosed between small circles with declinations from $\delta_{\min}$ to $\delta_{\max}$ can be calculated using the spherical segment area formula $S_{\delta \geqslant \delta_0} = 2\pi R^2(1 - \sin\delta_0)$ :

\eta = \frac{2\pi R^2(1 - \sin\delta_{\min}) - 2\pi R^2(1 - \sin\delta_{\max})}{4\pi R^2} = \frac{\sin\delta_{\max} - \sin\delta_{\min}}{2}\,.

Next, we obtain the fractions for the each case stated above:

\begin{array}{ll} \text{Case 1:} & \eta_1 = \dfrac{\sin(90^{\circ} + \varphi) - \sin(-90^{\circ} - \varphi)}{2} = \cos\varphi;\\[2ex] \text{Case 2:} & \eta_2 = \dfrac{\sin(90^{\circ} + \varphi) - \sin\varphi}{2} = \dfrac{\cos\varphi - \sin\varphi}{2} = \dfrac{\sin(45^{\circ} - \varphi)}{\sqrt{2}}\,;\\[2ex] \text{Case 3:} & \eta_3 = \dfrac{\sin(90^{\circ} - \varphi) - \sin\varphi}{2} = \dfrac{\cos\varphi - \sin\varphi}{2} = \dfrac{\sin(45^{\circ} - \varphi)}{\sqrt{2}}\,. \end{array}

Finally,

\eta(\varphi) = \begin{cases} \cos\varphi, & \varphi \leqslant -45^{\circ};\\ \frac{1}{\sqrt{2}}\sin(45^{\circ} - \varphi), & -45^{\circ} \leqslant \varphi \leqslant 45^{\circ};\\ 0, & \varphi \geqslant 45^{\circ}. \end{cases}

Graph of the fraction of stars η (vertical axis, 0 to 0.8) against latitude φ in degrees (horizontal axis, −90 to 90): the curve rises linearly-looking from 0 at −90° to about 0.71 at −45° (marked by a vertical dashed line), then falls to 0 at 45° and stays 0 up to 90°.

b) Obviously, the maximum of $\eta$ is achieved at $\varphi = -45^{\circ}$ and equals

\cos(-45^{\circ}) = \frac{\sin(45^{\circ} - (-45^{\circ}))}{\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.71.

Marking Scheme:

Question (a)
1. The presence of a graph or formulas for culminations — 3 pt.
2. Indication of 4 possible cases — 4 × 1 pt. = 4 pt.
3. Relationship of latitude and declination for each case — 4 × 1 pt. = 4 pt.
4. Calculation of the spherical area — 2 pt.
5. Providing answers for three possible cases — 3 × 1 pt. = 3 pt.
Question (b) 6. Discussion of the maximum value — 2 pt. ( $[-45^{\circ};\ 45^{\circ}]$ ) + 1 pt. ( $[-90^{\circ};\ -45^{\circ}]$ ) 7. Correct final answer — 1 pt.

Константи

Universal
Speed of light	$c = 3.00 \cdot 10^8\ \mathrm{m/s}$
Planck constant	$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s}$
Boltzmann constant	$k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Gas constant	$\mathfrak{R} = 8.314\ \mathrm{J/(mol \cdot K)}$
Proton mass	$m_p = 1.673 \cdot 10^{-27}\ \mathrm{kg}$
Astronomical
Astronomical unit	$1\ \mathrm{au} = 149.6 \cdot 10^6\ \mathrm{km}$
Parsec	$1\ \mathrm{pc} = 206\,265\ \mathrm{au}$
Hubble constant	$H_0 = 70\ \mathrm{(km/s)/Mpc}$
Earth
Radius	$R_\oplus = 6371\ \mathrm{km}$
Mass	$\mathfrak{M}_\oplus = 5.97 \cdot 10^{24}\ \mathrm{kg}$
Obliquity of ecliptic	$\varepsilon = 23.4^{\circ}$
Surface gravity	$g = 9.81\ \mathrm{m/s^2}$
Orbital period	$T_\oplus = 365.26^{\mathrm{d}}$
Orbital eccentricity	$e_\oplus = 0.0167$
Hydrogen spectrum
Lyman L $\alpha$	$\lambda_{\mathrm{L}\alpha} = 1215.7$ Å
Balmer H $\alpha$	$\lambda_{\mathrm{H}\alpha} = 6562.8$ Å
Jupiter
Radius	$R_J = 6.99 \cdot 10^4\ \mathrm{km}$
Mass	$\mathfrak{M}_J = 1.90 \cdot 10^{27}\ \mathrm{kg}$
Orbital radius	$a_J = 5.20\ \mathrm{au}$
Orbital period	$T_J = 11.86\ \mathrm{yr}$
Sun
Radius	$R_\odot = 6.96 \cdot 10^5\ \mathrm{km}$
Mass	$\mathfrak{M}_\odot = 1.99 \cdot 10^{30}\ \mathrm{kg}$
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	$\sigma = 5.67 \cdot 10^{-8}\ \mathrm{(W/m^2)/K^4}$
Wien’s displacement	$b = 2898\ \mathrm{\mu m \cdot K}$
UBV system	Mean wavelengths
U band	$\lambda_U = 364\ \mathrm{nm}$
B band	$\lambda_B = 442\ \mathrm{nm}$
V band	$\lambda_V = 540\ \mathrm{nm}$

Number of Siriuses in the Solar system $\geqslant 1$

Оригинал в Архива: OWAO-2024-prob-T.pdf · официални решения: OWAO-2024-sol-T.pdf