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Условие Отговори Решение

OWAO 2024, Теория — Задача 8. Thirty Degrees of Freedom

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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

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Start each problem on a separate Answer Sheet.
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Some problems are quite difficult. Remember that partial solutions are also scored.
A list of constants is given on page 4.

Good luck!

Условие

A spherical molecular cloud has a radius of 5 pc and a mass of $2 \cdot 10^3\,\mathfrak{M}_\odot$ . The temperature of the gas is 30 K, the cloud is homogeneous.

a) Estimate the mean free path and the mean free time (between successive collisions) of the hydrogen molecules in the cloud. The cross section radius of $\mathrm{H}_2$ is 2.7 Å.

b) Let a star pass behind the cloud along its apparent diameter with a tangential velocity of 70 km/s. For how long will the absorption be greater than $0.5^{\mathrm{m}}$ ? Absorption in the V band is related to the atomic column density by formula $A_V = 5.2 \cdot 10^{-22}\ N_H$ in magnitudes, where $N_H$ is measured in $\mathrm{cm}^{-2}$ .

Отговори

Решение

a) The mean free path of the molecules is $l_{\mathrm{H}_2} = \frac{1}{\sqrt{2}\pi d^2 n}$ , where $\pi d^2$ is the collisional cross section of a molecule, $d = 2.7$ Å is the cross section radius. The $\sqrt{2}$ coefficient arises from taking into account the motion of molecules relative to each other. Note, however, that taking the latter into account does not critically affect the estimations of the values needed for the solution.

The concentration of the molecules in the cloud

n = \frac{N}{V} = \frac{M/m_{\mathrm{H}_2}}{\frac{4}{3}\pi R^3} = \frac{2 \cdot 10^3 \times 2 \cdot 10^{30}\ \mathrm{kg}/(2 \times 1.67 \cdot 10^{-27}\ \mathrm{kg})}{\frac{4}{3}\pi \times (5 \times 206\,265 \times 1.49 \cdot 10^{11}\ \mathrm{m})^3} \approx 8 \cdot 10^7\ \mathrm{m}^{-3} = 8 \cdot 10^1\ \mathrm{cm}^{-3}.

Here the mass of $\mathrm{H}_2$ is estimated as $2m_p$ .

l_{\mathrm{H}_2} = \frac{1}{\sqrt{2}\pi \times \left(2.7 \cdot 10^{-10}\ \mathrm{m}\right)^2 \times 8 \cdot 10^7\ \mathrm{m}^{-3}} \approx 4 \cdot 10^{10}\ \mathrm{m} \approx 0.3\ \mathrm{au}.

The characteristic thermal velocity of the molecules is

v_t = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3\mathfrak{R}T}{\mu}} = \sqrt{\frac{3 \times 8.31\ \mathrm{J/(mol \cdot K)} \times 30\ \mathrm{K}}{0.002\ \mathrm{kg/mol}}} \approx 6 \cdot 10^2\ \mathrm{m/s}.

We can also estimate the characteristic velocities associated with the gravitational field of the cloud. We may use the virial theorem assuming that the cloud is not collapsing nowadays ( $2\langle E_k\rangle + \langle U\rangle = 0$ ), or calculate the circular velocity at the cloud boundary

v_g \sim \sqrt{\frac{G\mathfrak{M}}{R}} = \sqrt{\frac{G\mathfrak{M}_\odot}{a_\oplus} \cdot \frac{2 \cdot 10^3}{206\,265 \times 5}} = v_\oplus \times \sqrt{\frac{2 \cdot 10^3}{206\,265 \times 5}} = 30\ \mathrm{km/s} \times 0.044 \sim 1.3\ \mathrm{km/s}.

As a result, to estimate the mean free time we may take the value of velocity of $1.0 \div 1.5$ km/s. The mean free time is

t_{\mathrm{H}_2} = \frac{l_{\mathrm{H}_2}}{v_g} \sim \frac{0.3\ \mathrm{au}}{1\ \mathrm{km/s}} \sim 5 \cdot 10^7\ \mathrm{s} \sim 1 \div 2\ \mathrm{yr}.

b) We estimate the absorption in the direction of the cloud center, that is, the maximum possible absorption in the cloud:

\begin{aligned} \max A_V &= 5.2 \cdot 10^{-22}\ \mathrm{cm}^2 \cdot N_H = 5.2 \cdot 10^{-22}\ \mathrm{cm}^2 \cdot 2n \cdot 2R =\\ &= 5.2 \cdot 10^{-22}\ \mathrm{cm}^2 \times (2 \times 8 \cdot 10^1\ \mathrm{cm}^{-2}) \cdot (2 \times 5 \times 206\,265 \times 1.49 \cdot 10^{13}\ \mathrm{cm}) \approx 2.6^{\mathrm{m}}. \end{aligned}

The minimal path of light through the cloud for the absorption to be at least $0.5^{\mathrm{m}}$ is

Dotted circle of the cloud with centre C; chord AB near the top with its midpoint H, lines from C to A, B and H; a vertical arrow at the left marking the star’s path with dotted guide lines at the chord ends, and an arrow at the right labelled 'to observer'.

AB = 2R \cdot \frac{0.5}{2.6} \approx 2\ \mathrm{pc}.

The star has to travel the distance

2CH = 2 \cdot \sqrt{5^2 - 0.95^2} \approx 10\ \mathrm{pc},

which is approximately the cloud’s diameter.

The time needed for that travel

\tau \approx \frac{2R}{v_s} = \frac{10 \times 206\,265 \times 1.49 \cdot 10^{11}\ \mathrm{km}}{70\ \mathrm{km/s}} = 4.4 \cdot 10^{12}\ \mathrm{s} \approx 1.4 \cdot 10^5\ \mathrm{yr}.

Marking Scheme:

Question (a)
1. Concentration of the molecules — 2 pt. + 1 pt. (formula + estimation)
2. Mean free path — 2 pt. + 1 pt.
3. Characteristic velocity of molecules — 2 pt. + 1 pt.
4. Two estimations of the velocity (thermal and gravitational), conclusion — 2 pt.
5. Mean free time — 1 pt.
Question (b) 6. Absorption in the direction of the cloud center — 2 pt. + 1 pt. 7. Light travel distance corresponding to an absorption of $0.5^{\mathrm{m}}$ — 2 pt. 8. Length of the star’s path — 2 pt. 9. Star’s travel time — 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