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IOAA 2024, Теория — Задача 9. T9. Physics of Accretion

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Всички материали от IOAA 2024

IOAA 2024 — Theory Examination, English (Official) · 35 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

The accretion of matter onto compact objects, such as neutron stars and black holes, is one of the most efficient ways to produce radiant energy in astrophysical systems. Consider an element of gas of mass $\Delta m$ in a stationary and geometrically thin disc of matter with a maximum radius of $R_{max}$ and minimum stable orbital radius of $R_{min}$ (with $R_{min}/R_{max} \ll 1$ ), in rotation around a compact object of mass $M$ and radius $R$ .

(a) Assuming that an element of gas in the disc follows an approximately Keplerian circular orbit, find an expression for the total mechanical energy per unit mass $\frac{\Delta E}{\Delta m}$ released by this gas from when it starts orbiting at a radius $R_{max}$ until it reaches an orbital radius of $r \ll R_{max}$ . This process occurs very slowly, transforming kinetic energy into internal energy of the gas disc through viscous dissipation. Note: Ignore the gravitational interaction between particles within the accretion disc and give your final answer in terms of $G$ , $M$ , and $r$ . [6 т.]

(b) Considering that the disc receives mass at an average rate of $\dot{M}$ , and assuming that all the mechanical energy lost is ultimately converted into radiation, find an expression for the total luminosity of the disc. [5 т.]

(c) Consider now the ring composed of all mass elements with radii between $r$ and $r + \Delta r$ . In this scenario, find an expression for the luminosity generated by the disc over its small width $\Delta r$ at this radius; that is, find the expression for $\frac{\Delta E}{\Delta t \Delta r}$ . [8 т.]

(d) Assuming that the gravitational energy released in this ring is locally emitted by the surface of the ring in the form of black-body radiation, find an expression for the surface temperature $T$ of the ring. [10 т.]

(e) Consider that the central object is a stellar black hole with a mass of $3M_{\odot}$ and a rate of accretion of $\dot{M} = 10^{-9}\ M_{\odot}/\ \mathrm{year}$ . Consider also that $R_{min} = 3R_{sch}$ , where $R_{sch}$ is the Schwarzschild radius of the black hole. Determine the luminosity of the disc and the peak wavelength of emission of its innermost part. Ignore gravitational redshift effects and assume that the emission from the innermost part of the ring dominates the total emission. [3 т.]

(f) Now, considering another accretion system with $\dot{M} = 1\ M_{\odot}/\ \mathrm{year}$ and a peak emission wavelength of $\lambda = 6 \times 10^{-8}\ \mathrm{m}$ , estimate the mass of this black hole. [3 т.]

Решение

(T9) Physics of Accretion (35 points)
The accretion of matter onto compact objects, such as neutron stars and black holes, is one of the most efficient ways to produce radiant energy in astrophysical systems. Consider an element of gas of mass $\Delta m$ in a stationary and geometrically thin disk of matter with a maximum radius of $R_{max}$ and minimum stable orbital radius of $R_{min}$ (with $R_{min}/R_{max} \ll 1$ ), in rotation around a compact object of mass $M$ and radius $R$ .

(a) (6 points) Assuming that an element of gas in the disk follows an approximately Keplerian circular orbit, find the expression of the total mechanical energy per unit of mass $\frac{\Delta E}{\Delta m}$ released by this gas from the moment it starts orbiting at a radius $R_{max}$ until the moment it reaches an orbit of radius $r \ll R_{max}$ . This process occurs very slowly, transforming kinetic energy into internal energy of the gas disk through viscous dissipation.
Note: Disregard the gravitational interaction between particles within the accretion disk and give your final answer in terms of $G$ , $M$ and $r$

Solution: For a Keplerian orbit: [2.0]

\Delta m\frac{v_k^2}{r} = \frac{GM\Delta m}{r^2} \Rightarrow v_k^2 = \frac{GM}{r}

Total energy: [2.0]

E = E_{kin} + E_{pot} \Rightarrow E(r) = \frac{1}{2}\Delta m v_k^2 - \frac{GM\Delta m}{r} \Rightarrow E(r) = -\frac{1}{2}\frac{GM\Delta m}{r}

E(R_{max}) - E(r) = \frac{1}{2}GM\Delta m\left(\frac{1}{r} - \frac{1}{R_{max}}\right) = \frac{GM\Delta m}{2r}\left(1 - \frac{r}{R_{max}}\right) \approx \frac{GM\Delta m}{2r}

[1.0]

Therefore:

\frac{\Delta E}{\Delta m} \approx \frac{GM}{2r}

[1.0]

(b) (5 points) Considering that the disk receives mass at an average rate of $\dot{M}$ , and assuming that all the mechanical energy lost is ultimately converted into radiation, find an expression for the total luminosity of the disk.

Solution: Variation in energy:

\Delta E_{Tot} = E(R_{max}) - E(R_{min})

Luminosity: [3.0]

L_{Tot} = \frac{\Delta E_{Tot}}{\Delta t} = \frac{\Delta E_{Tot}}{\Delta m} \times \dot{M}

L_{Tot} = \frac{GM\dot{M}}{2R_{min}}

[2.0]

Where the result from (a) was used, allied with $R_{min} \ll R_{max}$ .

(c) (8 points) Consider now the ring composed of all mass elements from radius between $r + \Delta r$ to $r$ . In this scenario, find an expression of the luminosity generated by the disc over its small length $\Delta r$ at this radius, that is, find the expression for $\frac{\Delta E}{\Delta t\Delta r}$ .

Solution: Variation in the energy of the tiny mass: [5.0]

\Delta E = E(r + \Delta r) - E(r) = \frac{1}{2}GM\Delta m\left(\frac{1}{r} - \frac{1}{r + \Delta r}\right) \approx \frac{1}{2}GM\Delta m\left(\frac{\Delta r}{r^2(1 + \Delta r/r)}\right)

\Rightarrow \frac{\Delta E}{\Delta m} \approx \frac{1}{2}GM\Delta r\,\frac{1}{r^2}

Multiplying this expression by the rate of mass variation per time, it is possible to obtain an expression for $\frac{\Delta E}{\Delta t\Delta r}$ :

\frac{\Delta E}{\Delta t} = \frac{\Delta E}{\Delta m} \times \dot{M} = \frac{GM\dot{M}\Delta r}{2r^2}

\Rightarrow \frac{\Delta E}{\Delta t\Delta r} = \frac{GM\dot{M}}{2r^2}

[3.0]

(d) (10 points) Assuming that the gravitational energy released in this ring is locally emitted by the surface of the ring in the form of blackbody radiation, find an expression for the surface temperature $T$ of the ring.

Solution: Surface area of the ring: [4.0]

A = 2\left[\pi(r + \Delta r)^2 - \pi r^2\right] = 2(\pi r^2 + 2\pi r\Delta r + \pi \Delta r^2 - \pi r^2)

\Rightarrow A \approx 4\pi r\Delta r

Using Stefan-Boltzmann's law: [2.0]

A\sigma T^4 = L

[2.0]

4\pi r\Delta r\sigma T^4 = \frac{\Delta E}{\Delta t} \Rightarrow 4\pi r\sigma T^4 = \frac{\Delta E}{\Delta r\Delta t}

4\pi r\sigma T^4 = \frac{GM\dot{M}}{2r^2}

\Rightarrow T = \left(\frac{GM\dot{M}}{8\pi\sigma r^3}\right)^{1/4}

[2.0]

(e) (3 points) Consider that the central object is a stellar black hole with a mass of $3M_\odot$ and a rate of accretion of $\dot{M} = 10^{-9}\ M_\odot/\mathrm{year}$ . Consider also that $R_{min} = 3R_{sch}$ , where $R_{sch}$ is the Schwarzschild radius of the black hole. Determine the luminosity of the disk and the peak wavelength of emission of its innermost part. Disregard gravitational redshift effects and assume that the emission from the innermost part of the ring dominates the total emission.

Solution: Internal radius of the ring: [1.5]

R_{sch} = \frac{2GM}{c^2} \Rightarrow R_{min} = 3R_{sch} = \frac{6GM}{c^2}

R_{min} = \frac{6GM}{c^2} = \frac{6 \cdot 6{,}674 \cdot 10^{-11} \cdot 3 \cdot 1{,}988 \cdot 10^{30}}{(2{,}998 \cdot 10^8)^2} = 2{,}7 \times 10^4\,\mathrm{m}

Total luminosity:

L_{Tot} = \frac{GM\dot{M}}{2R_{min}} = \frac{\dot{M}c^2}{12}

\Rightarrow L_{Tot} = \frac{10^{-9} \cdot 1{,}988 \cdot 10^{30} \cdot (2{,}998 \cdot 10^8)^2}{12 \cdot 365{,}2422 \cdot 24 \cdot 60 \cdot 60}

\Rightarrow L_{Tot} = 5 \times 10^{29}\ W

Surface temperature:

T=\left(\frac{GM\dot M}{8\pi\sigma R_{min}^3}\right)^{1/4}=\left(\frac{6.674\cdot10^{-11}\cdot3\cdot10^{-9}\cdot(1.988\cdot10^{30})^2}{8\pi\cdot5.670\times10^{-8}\cdot(2.7\cdot10^4)^3\cdot365.2422\cdot24\cdot60\cdot60}\right)^{1/4}

T=5.5\times10^6\ K

Using Wien's law:

\lambda=\frac{b}{T}=\frac{2.898\cdot10^{-3}}{5.5\cdot10^6}

\Rightarrow\ \boxed{\lambda=5\times10^{-10}\ m}

[1.5]

(f) (3 points) Now, considering another accretion system with $\dot M=1\ M_\odot/\mathrm{year}$ and peak wavelength of emission is $\lambda=6\times10^{-8}\,\mathrm{m}$ , estimate the mass of this black hole.

Solution: Using Wien's law:

\lambda=\frac{b}{T}\Rightarrow T=\frac{b}{\lambda}=\frac{2.898\cdot10^{-3}}{6\times10^{-8}}

\Rightarrow T=4.8\cdot10^4\ K

[1.0]

Considering this temperature to be that of the innermost of the accretion disk, and using $R_{\min}=3R_{\mathrm{sch}}$ :

T=\left(\frac{GM\dot M}{8\pi\sigma R_{min}^3}\right)^{1/4}\Rightarrow T^4=\left(\frac{GM\dot M}{8\pi\sigma(3R_{\mathrm{sch}})^3}\right)=\left(\frac{GM\dot M}{8\pi\sigma(6GM/c^2)^3}\right)

\Rightarrow M=\left(\frac{\dot M}{8\pi\sigma6^3}\right)^{1/2}\ \frac{c^3}{GT^2}

\Rightarrow M=\frac{1}{(8\pi)^{1/2}}\ \frac{1}{6^{3/2}}\ \left\{\frac{1.988\cdot10^{30}/(3.15\cdot10^7)}{5.670\cdot10^{-8}}\right\}^{1/2}\ \frac{(2.998\cdot10^8)^3}{6.674\cdot10^{-11}\cdot(4.8\cdot10^4)^2}

\boxed{M\approx2.5\times10^{39}\ kg\approx1.3\times10^{9}\ M_\odot}

[2.0]

Separate Marking Scheme — T9

(a)

2 points: evaluating the centripetal force expression to calculate the velocity $v_k$ .
2 points: obtaining the correct expression for the total energy.
1 point: using the fact that $r\ll R_{\max}$ to simplify the expression for $E(R_{\max})-E(r)$ .
1 point: correct final expression.

(b)

3 points: obtaining an expression for $L_{\mathrm{tot}}$ .
2 points: correct final value.

(c)

5 points: evaluating the expression for energy using reasonable approximations, such as the binomial approximation.
3 points: correct final expression.

(d)

4 points: obtaining the expression for the area of the ring. 2 points should be deducted if the student misses the factor 2.
2 points: using Stefan-Boltzmann’s Law to relate temperature and area.
2 points: writing the thermodynamic equilibrium expression.
2 points: correct final expression.

(e)

1.5 points: correctly calculating the luminosity.
1.5 points: correctly calculating the wavelength.

(f)

1 point: correctly calculating the temperature.
2 points: correctly calculating the mass.

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