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IOAA 2024, Теория — Задача 4. T4. White Dwarf

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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

The structure of a white dwarf is sustained against gravitational collapse by the pressure of degenerate electrons, a phenomenon explained by quantum physics and related to the Pauli Exclusion Principle for electrons. The equation of state of a gas made of non-relativistic degenerate electrons is the following:

P = \left( \frac{3}{8\pi} \right)^{2/3} \frac{h^2}{5 m_e} n_e^{5/3},

where $n_e$ is the number of electrons per unit volume, which can be expressed in terms of the mass density $\rho$ using the dimensionless factor $\mu_e$ , the number of nucleons (protons and neutrons) per unit electron. Also consider that the central pressure can be described by this equation of state.

In the condition of hydrostatic equilibrium, the pressure and gravitational forces balance each other at any distance $r$ from the centre of the star. This condition can be expressed by:

\frac{\mathrm{d}P}{\mathrm{d}r} = -\frac{G M(r) \rho(r)}{r^2},

where $M(r)$ is the mass contained in the sphere of radius $r$ , and $\rho(r)$ is the mass density of the star at a radius $r$ .

Assume that $m_p = m_n$ , the density of a white dwarf is roughly uniform, and the following approximation is valid at the surface of the star:

\left. \frac{\mathrm{d}P}{\mathrm{d}r} \right|_{r=R} \approx -\frac{P_c}{R},

where $P_c$ is the pressure at the centre of the star, and $R$ the star radius.

(a) The relationship between the mass $M$ and the radius $R$ of a white dwarf can be written in the form: $R = a \cdot M^b$ . Find the exponent $b$ and determine the coefficient $a$ in terms of physical constants and $\mu_e$ . [6 т.]

(b) Using the relationship found in the previous part, estimate the radius of a white dwarf made of fully ionised carbon ( $^{12}_{6}\mathrm{C}$ ) with a mass of $M = 1.0\,M_\odot$ . [4 т.]

Решение

Solution: Simplifying the expression given in the problem statement:

\frac{\mathrm{d}P}{\mathrm{d}r} = -\frac{G M(r)\rho}{r^2}

At $r = R$ , is is possible use the approximation provided and the expression $\rho = 3M/(4\pi R^3)$ :

-\frac{P_c}{R} = -\frac{G M\{3M/(4\pi R^3)\}}{R^2}

P_c = \frac{3GM^2}{4\pi R^4}

[1.5]

Furthermore, electron density $n_e$ can be related to mass density as:

\rho = \mu_e m_p n_e

[1.5]

Using now the equation of state:

P_c = \left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2}{5m_e}\left(\frac{\rho}{\mu_e m_p}\right)^{5/3}

= \left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2}{5m_e}\left(\frac{3M}{4\pi R^3 \mu_e m_p}\right)^{5/3}

\frac{3GM^2}{4\pi R^4} = \left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2}{5m_e}\left(\frac{3}{4\pi\mu_e m_p}\right)^{5/3} \frac{M^{5/3}}{R^5}

R = \left(\frac{4\pi}{3}\right)\left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2}{5Gm_e}\left(\frac{3}{4\pi\mu_e m_p}\right)^{5/3} M^{-1/3}

Therefore:

a = \left(\frac{4\pi}{3}\right)\left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2}{5Gm_e}\left(\frac{3}{4\pi\mu_e m_p}\right)^{5/3}

[1.5]

b = -\frac{1}{3}

[1.5]

(b) There are 2 nucleons (1 proton and 1 neutron) per unit electron for carbon, so that $\mu_e=2$ . [1.0]

Using the expression for the radius found on the previous item:

R \approx \left(\frac{4\pi}{3}\right)\left(\frac{3}{8\pi}\right)^{2/3} \frac{1}{5}\left(\frac{3}{4\pi}\right)^{5/3} \frac{h^2}{Gm_e}\,\frac{1}{(\mu_e m_p)^{5/3}}\,\frac{1}{M^{1/3}}

\approx 1.866\cdot 10^{-2}\ \frac{h^2}{Gm_e}\,\frac{1}{(\mu_e m_p)^{5/3}}\,\frac{1}{M^{1/3}}

\approx \frac{1.866 \times 10^{-2} \times (6.626 \times 10^{-34})^2}{6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times (2 \times 1.67 \times 10^{-27})^{5/3} \times (1.988 \times 10^{30})^{1/3}}

R \approx 1.44 \times 10^{6}\ \mathrm{m}

[3.0]

Separate Marking Scheme — T4

(a)

1.5 points: using hydrostatic equilibrium with linear pressure gradient to find $P_c$ in terms of M , R and constants.
1.5 points: relating $\rho$ and $n_e$ through $\rho=\mu_e m_p n_e$
1.5 points: algebraically manipulating given equation of state correctly and previous results to find a in terms of constants and $\mu_e$ .
1.5 points point: algebraically manipulating given equation of state correctly and previous results to find that $b=-1/3$ .

(b)

1 point: noticing $\mu_e\approx2$ .
3 points: using found expression and obtaining correct final answer.

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