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IOAA 2024, Теория — Задача 2. T2. Galaxy Cluster

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Всички материали от IOAA 2024

IOAA 2024 — Theory Examination, English (Official) · 10 т.

Внимание: Бележка към темата

The printed instructions refer to blank Detailed Worksheets and a summary Answer Sheet. Those forms are absent from the available 2024 archive packet. The complete question booklet, solution booklet and separate marking scheme are included. Printed source inconsistencies are identified in the relevant tasks and retained without silently changing the source.

T2: The solution first gives $R=2.3143\times10^{23}\ \mathrm m$ and later substitutes $3.857\times10^{23}\ \mathrm m$ , while retaining its boxed $3.9\times10^{44}\ \mathrm{kg}\simeq2\times10^{14}M_\odot$ . The main double-sum potential expression also omits the factor $1/2$ used for unique pairs in the appendix. These printed expressions are retained; the marking scheme allows valid alternative estimates.

T6: The question names Palomar 4; the solution's repeated question names Palomar 6. The solution gives $V_{sky}=12.9$ mag but substitutes $12.8$ in the following surface-brightness line. The native question and printed calculation are retained.

T7: The printed expansion on solution page 13 has $-\sin(a)\cos(\phi)$ where the preceding cosine-law equation would give $-\sin(a)\sin(\phi)$ . It is retained as printed; the subsequent sunrise calculation sets $a=0$ .

T10: Solution page 25 prints the unit of $|\vec M|^2-R_\oplus^2=1.288\times10^{17}$ as m rather than $\mathrm m^2$ . In (j), the final vector prints 77.59 m/s after previously using 77.76 m/s; the magnitude uses 77.76 m/s. These printed values are retained.

T11: The question gives Maceió's longitude as $35.7^\circ W$ , while solution (e) uses $36.4^\circ W$ . The solution's printed margin marks total 11 points in (g), versus 12 points in the question and separate marking scheme; its (j) heading and marks total 7 points, versus 6 in the question and separate marking scheme. All three sources are retained with their original point values.

Common solar-system table: The native solar-system table labels the semi-major-axis column AU, while explicitly giving the Moon's entry in km; its printed unit is retained.

Указания

The theoretical competition will be 5 hours in duration and is marked out of a total of 300 points.
There are Detailed Worksheets for carrying out detailed work / rough work. On each of the Detailed Worksheets, please fill in Student Code, Question No., Page no. and total number of pages.
Start each problem on a new page of the Detailed Worksheets. 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 marked, cross it out.
There is a summary Answer Sheet with your student ID code for your final answers.
Please remember that the graders may not understand your language. As far as possible, write your solutions only using mathematical expressions and numbers. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your exam desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Detailed Worksheets, etc.), please put up your hand to signal the invigilator.
The beginning and end of the competition will be indicated by a long sound signal. Additionally, there will be a short sound signal fifteen minutes before the end of the competition (before the final long sound signal).
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the competition room.
A list of constants for this competition is given on the next page.

Fundamental Constants

Constant	Value
Speed of light in a vacuum	$c=2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h=6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e=1.602\times10^{-19}\ \mathrm C$
Universal Gravitational constant	$G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement law	$\lambda_mT=2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of the electron	$m_e=9.11\times10^{-31}\ \mathrm{kg}$
Mass of the proton	$m_p=1.67\times10^{-27}\ \mathrm{kg}$
Mass of the neutron	$m_n=1.67\times10^{-27}\ \mathrm{kg}$

Astronomical Data and Solar Magnitudes

Astronomical Data	Value
1 parsec	$1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm m=206\,265\ \mathrm{AU}=3.262\ \mathrm{ly}$
1 Astronomical Unit (AU)	$1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm m$
1 Jansky	$1\ \mathrm{Jy}=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
Hubble constant	$H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Solar luminosity	$L_\odot=3.826\times10^{26}\ \mathrm W$
Angular diameter of the Sun (from Earth)	$32'$
Effective temperature of the Sun	$T_{\mathrm{eff}}=5778\ \mathrm K$
Obliquity of the ecliptic	$23.5^\circ$
Inclination of the Moon's orbit (to ecliptic)	$5^\circ08'43''$
Magnitude of full Moon	$-12.74$
Coordinates of North Ecliptic Pole	$(18^{\mathrm h}00^{\mathrm m}00^{\mathrm s},+66^\circ33'39'')$
Coordinates of North Galactic Pole	$(12^{\mathrm h}51^{\mathrm m}26^{\mathrm s},+27^\circ07'42'')$
Sidereal day	$23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$
Tropical year	$365.2422$ solar days
Sidereal year	$365.2564$ solar days

Solar Magnitudes

Apparent Visual	Absolute Visual	Apparent Bolometric	Absolute Bolometric
$-26.75$	$+4.82$	$-26.83$	$+4.74$

Solar System, Spherical Trigonometry and Calculus

Solar System Data

Object	Radius (km)	Mass (kg)	Semi-major axis (AU)	Orbital eccentricity	Albedo
Sun	695500	$1.988\times10^{30}$	n/a	n/a	n/a
Mercury	2440	$3.301\times10^{23}$	0.387	0.206	0.088
Venus	6052	$4.867\times10^{24}$	0.723	0.007	0.76
Earth	6378	$5.972\times10^{24}$	1.000	0.016710	0.31
Moon	1737	$7.346\times10^{22}$	$3.844\times10^5\ \mathrm{km}$	0.054900	0.11
Mars	3390	$6.417\times10^{23}$	1.524	0.093	0.25
Jupiter	69911	$1.898\times10^{27}$	5.203	0.048	0.51
Saturn	58232	$5.683\times10^{26}$	9.537	0.054	0.34
Uranus	25362	$8.681\times10^{25}$	19.189	0.047	0.30
Neptune	24622	$1.024\times10^{26}$	30.070	0.009	0.29

Spherical Trigonometry

Cosine formula:

\cos a=\cos b\cos c+\sin b\sin c\cos A

Sine formula:

\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}

Four parts formula:

\cot b\sin a=\cos a\cos C+\sin C\cot B

Native common spherical-trigonometry reference sphere and triangle A, B, C with sides a, b, c.

Calculus

Chain Rule:

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

Product Rule:

\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

\int x^r\,dx=\frac{x^{r+1}}{r+1}+C\qquad(r\ne-1)

\frac{d}{dx}\ln(x)=\frac1x,\qquad\frac{d}{dx}\tan(x)=\sec^2(x)

Условие

An astrophysical survey mapped all the galaxies in a small region of the sky, of angular diameter $\Delta\theta = 0.01\ \mathrm{rad}$ , where many galaxies seemed to be concentrated around the central area of the image. When the positions and redshifts of all the galaxies in this cluster were measured, an interesting distribution emerged, which is shown in the plot below.

Galaxy redshift distribution and cluster image, with redshift 0.7, dispersion 0.0005 and angular diameter 0.01 rad.

Using these observations, estimate the total mass of the galaxy cluster and express your answer in solar masses. Assume that this galaxy cluster is in dynamical equilibrium, with a root-mean-square redshift dispersion $\sigma_z = \sqrt{\langle (z - 0.7)^2 \rangle} = 0.0005$ . Feel free to make reasonable approximations when considering the average velocities, masses, and spatial distribution of the galaxies.

Consider that the distance to $\bar{z} = 0.7$ in the standard cosmological model is $D_A = 1500\ \mathrm{Mpc}$ . Ignore cosmological effects on the distance.

Решение

If $U$ and $K$ are the total gravitational potential energy and the total kinetic energy of the cluster, respectively, the virial theorem says that:

U + 2K = 0

[2.0]

Let's say that there are $N$ galaxies of masses $m_i$ , where $i = 1, ..., N$ , moving with velocities $\vec{u}_i = \vec{v}_i - \vec{v}_0$ with respect to the cluster, where $\vec{v}_0$ is the velocity of the cluster as a whole. Therefore, in the cluster frame, $K$ is given by

K = \sum_{i=1}^{N} \frac{1}{2} m_i u_i^2

As an estimation, consider that all galaxies have mass equal to the average mass $\langle m \rangle = M_T/N$ :

K = \frac{1}{2} \frac{M_T}{N} \sum_{i=1}^{N} u_i^2 = \frac{1}{2} M_T \frac{\sum u_i^2}{N} = \frac{1}{2} M_T \sigma_v^2

[2.0]

Where $\sigma_v = \sqrt{\langle u_i^2 \rangle}$ is the root-mean-square galaxy speed, also known as velocity dispersion. Notice, however, that only the dispersion in redshift is given, which translates into a radial velocity dispersion:

\sigma_{v_r} = c \cdot \sigma_z = 2.998 \times 10^8\ \mathrm{m/s} \times 0.0005 = 1.499 \times 10^5\ \mathrm{m/s}

[1.0]

There is no penalization for students who use that $\sigma_v = c\sigma_z$ . For completeness of the solution, however, we relate $\sigma_v$ and $\sigma_{v_r}$ by assuming three-dimensional isotropy for velocities, which yields $\sigma_v^2 = 3\sigma_{v_r}^2$ . $K$ is then given by:

K = \frac{3}{2} M_T \sigma_{v_r}^2

The gravitational potential energy, on the other hand, is given by:

U = -\sum_{i \neq j} \frac{G m_i m_j}{r_{ij}}

This can be estimated in different ways. Usually, the spatial distribution of the large number of galaxies may be approximated as uniform, so that one estimates $U$ as homogeneous spherical mass distribution, for which $U$ is given by:

U = -\frac{3}{5} \frac{G M_T^2}{R}

[2.0]

And, assuming the cluster is spherical, its radius is:

R = \frac{D_A \cdot \Delta\theta}{2} = \frac{1500 \times 10^6 \times 206265 \times 1.496 \times 10^{11} \times 10^{-2}}{2} = 2.3143 \times 10^{23}\ \mathrm{m}

[1.0]

Now, using the virial theorem, it is possible to calculate the total mass:

\therefore -\frac{3}{5} G M_T^2 \frac{1}{R} + 2 \cdot \frac{3}{2} M_T \sigma_{v_r}^2 = 0

\Rightarrow M_T = \frac{5 R \sigma_{v_r}^2}{G}

M_T = \frac{5 \times 3.857 \times 10^{23} \times (1.499 \times 10^5)^2}{6.67 \times 10^{-11}}\ \mathrm{kg}

\boxed{M_T \approx 3.9 \times 10^{44}\ \mathrm{kg} = 2.0 \times 10^{14}\ \mathrm{M}_\odot}

[2.0]

Appendix on calculation of $U$ : Other valid methods for estimating $U$ include explicitly writing the sum as:

U = -\frac{1}{2} G \left( \sum_i m_i \right)^2 \left\langle \frac{1}{r_{ij}} \right\rangle

U = -\frac{1}{2} \frac{G M_T^2}{R}

where the factor of $1/2$ comes from counting all the unique pairs, and $\langle r_{ij}^{-1} \rangle$ was taken (as an approximation) as $R^{-1}$ .

Or, alternatively, using dimensional analysis to argue that:

U = -\frac{G M_T^2}{R}

Separate Marking Scheme — T2

2 points: correctly making use of the Virial Theorem to relate U and K. Deduction of -0.5 points if the student uses that $\lvert U\rvert=K$ .
2 points: obtain a correct expression for K in terms of $M_T$ and $\sigma_v$ . Students who only write the definition of kinetic energy as a sum for all galaxies without relating it to the velocity dispersion will only receive 0.5 points.
2 points: obtain a correct expression for U in terms of $M_T$ , R and G. Full marks are awarded as long as the student uses an acceptable method of estimation. For U in the form $-\alpha GM_T^2/R$ , we suggest answers with factors $\alpha=[1/2,3/5,1]$ are considered correct, provided they use a valid method. Valid methods include, but are not necessarily limited to: dimensional analysis ( $\alpha=1$ ); homogeneous sphere ( $\alpha=3/5$ ); $\langle r_{ij}^{-1}\rangle=R^{-1}$ approach ( $\alpha=1/2$ ).
If the student writes right away that $M_T\sim R\sigma_v^2/G$ (For instance, using dimensional analysis or using circular orbits), award 2.5 points out of the 6 points totaled by the first three criteria.
1 point: writing $\sigma_{v_r}=c\sigma_z$ . It is fine if the student just assumes that $\sigma_{v_r}=\sigma_v$ , or does not realize that $\sigma_z$ is directly related to $\sigma_{v_r}$ , not $\sigma_v$ .
1 point: correctly calculating the physical size (radius or diameter) of the cluster using $D_A$ and $\Delta\theta$ . Numerical value not needed.
2 points: correct estimate for the mass. Regardless of the method, answers inside the range $3\times10^{13}M_\odot\le M_T\le3\times10^{14}M_\odot$ obtain full marks. No points given for this criterion if outside the range.

General Rules for Marking

These are general guidelines for the grading of all problems. All of the rules listed below refer to deductions on specific items in the marking scheme.

Case	Marking
Using absurd physical or mathematical concepts (despite correct answers)	No points given
Giving correct answer but the method is unclear	Deduct 50% of the marks for final answer
Minor mistakes in the calculations, e.g., wrong signs, numerical constants	Deduct 20% of the marks for that part, unless there is a physical incoherence
Too many or too few significant figures ( $n\pm1$ where $n$ is the expected number of significant figures)	Deduct 50% of the marks for final answer
Correct answer but missing or wrong units	No points given for final answer
For numerical values, the answer is not within a 5% precision range	No points given for final answer, unless differently stated in the problem or marking scheme
Error resulting from an earlier part error, for which the student already lost marks, if the answer is physically reasonable	No double penalization, full points given
Error resulting from an earlier part error, if the answer is physically unreasonable	No points given

First example: If due to an error in an earlier item, the student calculates the speed of a satellite as $4.0\times10^4\ \mathrm{m/s}$ instead of $4.5\times10^4\ \mathrm{m/s}$ , they will only lose marks for the earlier item. However, if, for the same reason, a student calculates the velocity of the satellite as $4.5\times10^{12}\ \mathrm{m/s}$ , they should realize this is wrong (higher than the speed of light) and thus should lose some marks for this item as well.

Second example: Consider a question that requires a student to determine the temperature of a planet that orbits a star with luminosity $L$ at a distance $r$ . To answer correctly, the student must calculate the star's radiant flux, the power incident on the planet, the power radiated by the planet, and then, using the condition of thermal equilibrium, the planet's temperature. If the student only calculates the radiant flux and arrives at the correct final answer without clearly explaining the other steps, they will receive marks only for the flux calculation, but not for the other steps, neither for the final answer.

Оригинал в Архива: theory_examination_304b6eb66a8940d79fccc404cf1ba9ab_preview.pdf · официални решения: theory solutions.pdf · допълнителни данни: theory marking scheme.pdf