Съдържание

Earth Moon Sun Jupiter Saturn Physical constants Условие Решение

IOAA 2017, Теория — Задача 9. Galactic Outflow

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Всички материали от IOAA 2017

IOAA 2017 — Theoretical Examination · 20 т.

Original Examination Instructions

The theoretical examination lasts for 5 hours and is worth a total of 300 marks.
Dedicated IOAA Summary Answer Sheets are provided for writing your answers. Enter the final answers into the appropriate boxes in the corresponding Summary Answer Sheet. On each Summary Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
There are Answer Sheets for carrying out detailed work/rough work. On each Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
- Question no.
- Page no. and total number of pages.
Start each problem on a separate Answer Sheet. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Answer Sheets, etc.), please draw the attention of the proctor using the Help card.
The beginning and end of the examination will be indicated by a long sound signal. Additionally, there will be a buzzer sound, fifteen minutes before the end of the examination (before the final sound signal).
At the end of the examination you must stop writing immediately. Sort and put your sheets in separate stacks, a) Stack 1: Summary Answer Sheets, Answer Sheets of part 1
b) Stack 2: Summary Answer Sheets, Answer Sheets of part 2
c) Stack 3: Summary Answer Sheets, Answer Sheets of part 3
d) Stack 4: question papers and paper sheets you do not want to be graded.
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and a table of the mark distribution for this exam are given on the next two pages.

Original List of Constants

Earth

Quantity	Value
Mass ( $M_\oplus$ )	$5.98\times10^{24}$ kg
Radius ( $R_\oplus$ )	$6.38\times10^6$ m
Acceleration of gravity ( $g$ )	$9.8 \mathrm{m\cdot s^{-1}}$
Obliquity of Ecliptic	$23^\circ27'$
Length of Tropical Year	365.2422 mean solar days
Length of Sidereal Year	365.2564 mean solar days
Albedo	0.39

Moon

Quantity	Value
Mass (M with subscript ☾)	$7.35\times10^{22}$ kg
Radius (R with subscript ☾)	$1.74\times10^6$ m
Mean distance from Earth	$3.84\times10^8$ m
Orbital inclination with the Ecliptic	$5.14^\circ$
Albedo	0.14
Apparent magnitude (mean full moon)	−12.74

Sun

Quantity	Value
Mass ( $M_\odot$ )	$1.99\times10^{30}$ kg
Radius ( $R_\odot$ )	$6.96\times10^8$ m
Luminosity ( $L_\odot$ )	$3.83\times10^{26}$ W
Absolute Magnitude ( $\mathcal M_\odot$ )	4.80 mag
Angular diameter	0.5 degrees
Rotational velocity in the Galaxy	$220 \mathrm{km\,s^{-1}}$
Distance from Galactic centre	8.5 kpc

Jupiter

Quantity	Value
Mass	$1.89\times10^{27}$ kg
Orbital semi-major axis	5.20 au
Orbital period	11.86 year

Saturn

Quantity	Value
Mass	$5.68\times10^{26}$ kg
Orbital semi-major axis	9.58 au
Orbital period	29.45 year

Physical constants

Quantity	Value
1 au	$1.50\times10^{11}$ m
1 pc	206,265 au
Gravitational constant ( $G$ )	$6.67\times10^{-11} \mathrm{N\cdot m^2\cdot kg^{-2}}$
Planck constant ( $h$ )	$6.62\times10^{-34} \mathrm{J\cdot s}$
Boltzmann constant ( $k_B$ )	$1.38\times10^{-23} \mathrm{J\cdot K^{-1}}$
Stefan-Boltzmann constant ( $\sigma$ )	$5.67\times10^{-8} \mathrm{W\cdot m^{-2}\cdot K^{-4}}$
Hubble constant ( $H_0$ )	$67.8 \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Speed of light in vacuum ( $c$ )	$299,792,458 \mathrm{m\cdot s^{-1}}$
Proton mass	$938.27 \mathrm{MeV\cdot c^{-2}}$
Deuterium mass	$1875.60 \mathrm{MeV\cdot c^{-2}}$
Neutron mass	$939.56 \mathrm{MeV\cdot c^{-2}}$
Helium-3 mass	$2808.30 \mathrm{MeV\cdot c^{-2}}$
Helium-4 mass	$3727.40 \mathrm{MeV\cdot c^{-2}}$

Original Mark Distribution

Problem number	Marks
T1	10
T2	10
T3	10
T4	10
T5	10
T6	15
T7	20
T8	20
T9	20
T10	25
T11	50
T12	40
T13	60
Total	300

Условие

Cannon et al. (2004) conducted an HI observation of a disk starburst galaxy, IRAS 0833+6517, with the Very Large Array (VLA). The galaxy is located at a distance of 80.2 Mpc with an approximate inclination angle of 23 degrees. According to the HI velocity map, IRAS 0833+6517 appears to be undergoing regular rotation with an observed radial velocity of the HI gas of roughly $5850 \mathrm{km\,s^{-1}}$ at a distance of 7.8 kpc from the centre (the left panel of the figure below).

Gas outflow from IRAS 0833+6517 is traced by using the blueshifted interstellar absorption lines observed against the backlight of the stellar continuum (the right panel of the figure). Assuming that this galaxy is gravitationally stable and all the stars are moving in circular orbits,

a) Determine the rotational velocity ( $v_{\mathrm{rot}}$ ) of IRAS 0833+6517 at the observed radius of HI gas. [5 т.]

b) Calculate the escape velocity for a test particle in the gas outflow at the radius of 7.8 kpc. [9 т.]

c) Determine if the outflowing gas can escape from the galaxy at this radius by considering the velocity offset of the C II $\lambda1335$ absorption line, which is already corrected for the cosmological recessional velocity. (The central rest-frame wavelength of the CII absorption line is 1335 Å.) (YES / NO) [6 т.]

Both complete original galaxy/observer geometry and CII absorption spectrum, all axes, ticks and full native curve.

Решение

The original key heading prints “(T2) Galactic Outflow”; its complete prompt, diagram and three requests identify this as question T9.

a) Solution

The systemic or recessional velocity of IRAS 0833+6517 can be determined by the Hubble’s law

v_{\mathrm{sys}}=H_0d=67.8\ \mathrm{km\,s^{-1}\,Mpc^{-1}}\times80.2\ \mathrm{Mpc}=5437.6\ \mathrm{km\,s^{-1}}.

[2]

The observed radial velocity can be written in terms of systemic velocity and rotational velocity of a galaxy as

v_{\mathrm{rad}}=v_{\mathrm{sys}}+v_{\mathrm{rot}}\sin i,

[1]

where $v_{\mathrm{rad}}$ , $v_{\mathrm{sys}}$ , and $v_{\mathrm{rot}}$ are the radial, systemic, and rotational velocity of the galaxy and $i$ is an inclination angle.

The rotational velocity of the galaxy can thus be obtained by

v_{\mathrm{rot}}=\frac{v_{\mathrm{rad}}-v_{\mathrm{sys}}}{\sin i}=\frac{5850-5437.6\ \mathrm{km\,s^{-1}}}{0.39}=1055\ \mathrm{km\,s^{-1}}.

[2]

Answers between 1050 to 1060 are acceptable for full marks.

b) Solution — Method 1

In order to determine the escape velocity, we need to know the dynamical mass of the system. In case of the rotation dominated galaxy, a dynamical mass can be estimated as

M_{\mathrm{dyn}}(<R)=v_{\mathrm{rot}}^2R/G,

[2]

where $M_{\mathrm{dyn}}(<R)$ is the dynamical mass with in the radius $R$ .

So the dynamical mass of IRAS 0833+6517 within the radius of 7.8 kpc is

M_{\mathrm{dyn}}(<R)=\frac{(1055\times10^3\ \mathrm{m\,s^{-1}})^2\times7.8\times10^3\ \mathrm{pc}\times206265\ \mathrm{au/pc}\times1.50\times10^{11}\ \mathrm{m/au}}{6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}}=4.0\times10^{42}\ \mathrm{kg},

[2]

or $M_{\mathrm{dyn}}(<R)=2.0\times10^{12}M_{\square}$ .

Assuming the virial theorem, the escape velocity can be written as

v_{\mathrm{esc}}=\sqrt{\frac{2GM}R}.

[2]

Therefore, the escape velocity is

v_{\mathrm{esc}}=\sqrt{\frac{2\times6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\times4.0\times10^{42}\ \mathrm{kg}}{7.8\times10^3\ \mathrm{pc}\times206265\ \mathrm{au/pc}\times1.50\times10^{11}\ \mathrm{m/au}}}=1492\ \mathrm{km\,s^{-1}}.

[3]

b) Solution — Method 2

In order to determine the escape velocity, we need to know the dynamical mass of the system. In case of the rotation dominated galaxy, a dynamical mass can be estimated as

M_{\mathrm{dyn}}(<R)=v_{\mathrm{rot}}^2R/G,

[2]

where $M_{\mathrm{dyn}}(<R)$ is the dynamical mass with in the radius $R$ . According to the virial theorem, the escape velocity is

v_{\mathrm{esc}}=\sqrt{\frac{2GM}R}.

[2]

Substituting the dynamical mass into the above equation, we get

v_{\mathrm{esc}}=\sqrt{\frac{2GM}R}=\sqrt{2v_{\mathrm{rot}}^2}

[3]

=\sqrt{2\times(1055\ \mathrm{km\,s^{-1}})^2}=1492\ \mathrm{km\,s^{-1}}.

[2]

Answers between 1487 to 1497 are acceptable for full marks.

c) Solution

To examine whether or not outflowing gas can escape from the galaxy, one needs to derive the outflow velocity of the gas from the blueshifted absorption line.

The outflow velocity or the velocity offset of any emission or absorption lines can be calculated by the following equation:

\frac{\Delta v}c=\frac{\Delta\lambda}{\lambda}.

[2]

Full native CII key graph with red acceptable-wavelength limits and blue arrow, all axes/ticks and absorption curve.

Figure 1 The CII absorption line with the acceptable range of the blueshifted wavelengths.

For the C II $\lambda1335$ absorption line (Figure 1), we obtain

\Delta v_{\mathrm{obs}}(\mathrm{C\ II}\lambda1335)=299792.458\ \mathrm{km\,s^{-1}}\left(\frac{1332-1335}{1335}\right)=-673.7\ \mathrm{km\,s^{-1}}.

[2]

Answers between $-300\ \mathrm{km\,s^{-1}}$ to $-900\ \mathrm{km\,s^{-1}}$ are acceptable with full marks.

The velocity offsets of the C II $\lambda1335$ absorption line is $-673.7\ \mathrm{km\,s^{-1}}$ . It means that the outflowing gas is going out of the galaxy at this speed, which is smaller than the escape velocity obtained in (b).

We conclude that the outflowing gas CANNOT escape from IRAS 0833+6517. [2]

The printed solar-mass subscript in the method-1 mass result is a missing-glyph box, retained as $M_{\square}$ .

Бележки за изданието

The original question and key contain several differing expressions. The Galactic-arm question uses a flat curve beyond 5 kpc and its worked key uses 2 kpc. The HI survey key prints differing flux-threshold exponents and an unusual Hubble-constant unit. The Bethlehem key has differing multiplication and remainder values. The Galactic Outflow key heading says T2 while its full prompt matches T9. GOTO assigns a different pair of marks to parts a) and d) in the question and key. The Shipwreck key prints differing sidereal-time values in its main and alternative methods. Visible missing-font glyph squares and the Exomoon key’s grading dependency are preserved. The original constants table prints the acceleration unit as m·s^-1. These printed forms remain literal.

Оригинал в Архива: Theory.pdf · официални решения: Theory_Solution.pdf