Съдържание

IAO 2013, Теория — Задача 9. αβ-6. Остатък от взрив на свръхнова

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2013

XVIII Международная астрономическая олимпиада. XVIII International Astronomy Olympiad. Литва, Вильнюс, 6 – 14. IX. 2013, Vilnius, Lithuania. Round Theo · 6 септември 2013 г. · 8 т.

Внимание: Бележка към темата

Ръкописни български условия за групи α, β и общи αβ; решенията и критериите са на английски. В αβ-6 ръкописът дава 3400 km, а английският оригинал и ключът — 3400 pc. Източниковите разминавания са запазени и отбелязани.

Общо указание

Общо указание. Възможно е не във всички задачи въпросите да са поставени коректно. Някои въпроси (възможно е това да е основният въпрос в задачата, а може и да е някой подвъпрос) може да нямат смисъл. В такива случаи в отговора трябва да се напише (на руски или на английски): „ситуация невозможна - impossible situation“. Разбира се, отговорът трябва да е подкрепен с пресмятания или логически разсъждения.

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Общо указание

Данните от таблиците (Слънчева система, звезди, константи) могат да се използват във всички задачи.

Отговорите „Да - Yes“ или „Нет - No“ трябва да бъдат написани на руски или на английски.

Parameters of orbits and physical characteristics

The bilingual source sheet gives the following values. Blank cells reproduce blank source cells.

Body / planet	Average distance (AU)	Average distance ( $10^6$ km)	Sidereal period (tropical years)	Sidereal period (days)	Orbital inclination $i$	Eccentricity $e$	Equatorial diameter (km)	Mass ( $10^{24}$ kg)	Average density (g/cm³)	Gravity at surface (m/s²)	Axial tilt	Maximum magnitude from Earth	Albedo
Sun	$1.6\cdot10^9$	$2.5\cdot10^{11}$	$2.2\cdot10^8$	$8\cdot10^{10}$			1392000	1989000	1.409			−26.74
Mercury	0.387	57.9	0.241	87.969	7.00°	0.206	4879	0.3302	5.43	3.70	0.01°		0.06
Venus	0.723	108.2	0.615	224.7007	3.40	0.007	12104	4.8690	5.24	8.87	177.36		0.78
Earth	1.000	149.6	1.000	365.2564	0.00	0.017	12756	5.9742	5.515	9.81	23.44		0.36
Moon	0.00257	0.38440	0.0748	27.3217	5.15	0.055	3475	0.0735	3.34	1.62	6.7	−12.7	0.07
Mars	1.524	227.9	1.880	686.98	1.85	0.093	6794	0.6419	3.94	3.71	25.19	−2.0	0.15
Jupiter	5.204	778.6	11.862	4332.59	1.30	0.048	142984	1899.8	1.33	24.86	3.13	−2.7	0.66
Saturn	9.584	1433.7	29.458	10759.20	2.48	0.054	120536	568.50	0.70	10.41	26.73	0.7	0.68

For outer planets and Moon — in mean opposition.

Some constants and formulae

Constant or formula	Value
Speed of light in vacuum, c (m/s)	299792458
Constant of gravitation, G (N·m²/kg²)	$6.674\cdot10^{-11}$
Solar constant, A (W/m²)	1367
Hubble parameter, $H_0$ (km/s/Mpc), mean value	71
Hubble parameter, diapason of values	50–100
Planck constant, h (J·s)	$6.626\cdot10^{-34}$
Charge of electron, e (C)	$1.602\cdot10^{-19}$
Mass of electron, $m_e$ (kg)	$9.109\cdot10^{-31}$
Proton-to-electron ratio	1836.15
Faraday constant, F (C/mol)	96485
Magnetic constant, $\mu_0$ (H/m)	$1.257\cdot10^{-6}$
Universal gas constant, R (J/mol/K)	8.314
Boltzmann constant, k (J/K)	$1.381\cdot10^{-23}$
Stefan–Boltzmann constant, $\sigma$ (W/m²/K⁴)	$5.670\cdot10^{-8}$
Wien’s displacement constant, b (m·K)	0.002897
Laboratory wavelength of Hα (Å)	6562.81
Tropical year length, T (days)	365.242199
Standard atmosphere (Pa)	101325
Visible light extinction by the terrestrial atmosphere in zenith (minimum)	19%, 0.23 mag
Refractive index of water for 20°C, n	1.334
Moment of inertia of a solid ball	$I=\frac25MR^2$
Volume of ball	$V=\frac43\pi R^3$
Area of sphere	$S=4\pi R^2$
$\pi$	3.14159265
e	2.71828183
Golden ratio, $\varphi$	1.61803399

Data of some stars

Star	Designation	RA	DEC	Parallax $p$	Apparent magnitude $m$	Spectral class	Mass ( $\mathcal M_{\odot}$ )
Sun	⊙	0h–24h	−23°26′ to +23°26′	8.794″	vis −26.74; bol −26.82	G2	1
Proxima Centauri	V645 Cen, α Cen C	14h 29m 43s	−62°40′46″	0.769″	11.05	M5.5	0.123
Alpha Centauri A	α Cen A	14h 39m 37s	−60°50′02″	0.747″	−0.01	G2	1.1
Alpha Centauri B	α Cen B	14h 39m 35s	−60°50′14″		1.34	K1	0.9
Beta Centauri	β Cen	14h 03m 49s	−60°22′23″	0.009″	0.61	B1	21
Epsilon Eridani	ε Eri	03h 32m 56s	−09°27′30″	0.311″	3.74	K2	0.82
Gliese 581	HO Lib	15h 19m 27s	−07°43′20″	0.16″	vis 10.57; bol 8.0	M3V	0.31

The Alpha Centauri parallax is printed once beside the pair; the second cell is left blank here as in the source.

Historical units

Units of power

1 horse-power (hp) = 735.49875 W.

Local units of length in the end of XVIII century

1 arshin (arsh) = 0.711187 m
1 span (sp) = 1/4 arshin
1 vershok (vrsh) = 1/4 span
1 sajene (sj) = 3 arshin
1 verst (vrst) = 500 sajene

Условие

С помощта на космическия телескоп „Чандра“ е получено рентгеново изображение на остатъка от взрива на свръхновата (SNR) в Cas A, намиращ се от нас на разстояние $d = 3400\ \mathrm{km}$ . Негативът на тази снимка е даден на фиг. SNR. С кръг са отбелязани границите на областта на SNR. В горния ляв ъгъл на фигурата е даден мащаб. Точката, разположена близо до центъра на кръга, е неутронната звезда (ядрото на избухналата звезда). Правоъгълните знаци извън кръга са реперните точки, послужили за построяване на центъра на кръга. Да предположим, че количеството енергия, освободена при взрива на свръхновата, е било от порядъка на $E_{SN} \approx 10^{46}\ \mathrm{J}$ . 1% от тази енергия движи разширяващата се материя. Средната плътност на веществото в SNR е от порядъка на $\rho \approx 10^{-21}\ \mathrm{kg/m^3}$ .

SNR Cas A: 100 arcsecond scale, boundary circle and external reference marks

6.1. Оценете възрастта на SNR Cas A. [4 т.]

6.2. Определете средната скорост на движение на неутронната звезда от центъра на SNR. [2 т.]

Отговори

Решение

Официалното решение и критериите по-долу са на английски, както в източника.

Official English solution sketch. The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Разминаване между източниците: българският ръкопис отпечатва $d=3400\ \mathrm{km}$ ; оригиналните английски условия и официалното решение използват $d=3400\ \mathrm{pc}$ . Българската единица в условието е запазена.

6.1. Age of the supernova remnant

Assume that the SNR is a sphere with the linear radius R. Using a ruler we accurately measure the diameter of the circle and the scale length (e.g. in mm) in Fig. SNR (see the figure below, the initial figure was almost 2 times larger). From these measurements we get the radius of the circle $\rho=48.5\ \mathrm{mm}$ and 100″ scale length $l=28.5\ \mathrm{mm}$ . With these data we can calculate the angular radius of the SNR

\theta=100^{\prime\prime}\rho/l,

\theta=100^{\prime\prime}\times48.5/28.5\approx170^{\prime\prime}.

If the radius R of the SNR is measured in au, the distance d in pc, and its angular radius $\theta$ in arcsec, then

R=d\times\theta,

R=170^{\prime\prime}\times3400=578000\ \mathrm{AU}\approx8.6\cdot10^{16}\ \mathrm m.

The mass of the SNR is confined within the sphere of the radius R.

M=\rho\cdot\frac43\pi R^3,

M=10^{-21}\cdot4/3\cdot3.14\cdot(8.5\cdot10^{16})^3\approx2.7\cdot10^{30}\ \mathrm{kg}.

The expansion speed of the SNR is calculated using the expression of the kinetic energy of the SNR

E_{\mathrm{kin}}=\tfrac12Mv^2,

v=(2E_{\mathrm{kin}}/M)^{1/2}=(2\cdot0.01E_{\mathrm{SN}}/M)^{1/2},

v=(2\cdot0.01\cdot10^{46}\ \mathrm J/(2.7\cdot10^{30}\ \mathrm{kg}))^{1/2}\approx8.6\cdot10^6\ \mathrm{m/s}\approx10^7\ \mathrm{m/s}.

As we see from the image, the darkest parts of it are distributed at distances about 60% of its radius. So we may propose that most mass of the SNR is distributed at these distances. Thus, by the above formulae we have found the average speed of matter at 0.6 R.

Assuming that the expansion velocity was constant we get the age of the SNR:

t=0.6R/v,

t=0.6\times8.6\cdot10^{16}\ \mathrm m/(8.6\cdot10^6\ \mathrm{m/s})\approx6\cdot10^9\ \mathrm s\approx190\ \mathrm{yr}\approx200\ \mathrm{yr}.

Taking into account that most data in the problem conditions was done very roughly, with maximum 1 significant digit (and even with zero significant digits as only order of magnitude is done: the energy was about $10^{46}$ J, 1% of which drives the expansion; the average density is $10^{-21}$ kg/m³), the answer should be expressed also with only one significant digit.

So the answer is: ~200 years.

Source note: the key uses $\rho$ both for the measured image radius and for the matter density, and uses $8.5\cdot10^{16}$ m in the mass calculation after $8.6\cdot10^{16}$ m above. These are retained.

Cas A image with reference circle and measurement lines in the official solution

6.2. Neutron-star velocity

Using a ruler we measure the distance of the neutron star from the center of circle in Fig. SNR. Say that we get $\sigma=3\ \mathrm{mm}$ . Then the angular distance of the neutron star from the center of the SNR is

\delta=100^{\prime\prime}\sigma/l,

\delta=100^{\prime\prime}\times3/28.5\approx10.5^{\prime\prime}.

The linear distance of the neutron star from the center of the SNR is

S_N=\sigma\times d,

S_N=10.5\times3400\approx35700\ \mathrm{AU}\approx5.3\cdot10^{12}\ \mathrm{km}.

The velocity of the motion of the neutron star is

u=S_N/t,

u=5.3\cdot10^{12}\ \mathrm{km}/(6\cdot10^9\ \mathrm s)\approx900\ \mathrm{km/s}\approx10^3\ \mathrm{km/s}.

Also taking into account that most data was done very roughly, the answer should be expressed also as an order of magnitude only or with only one significant digit.

The answer is: ~ $10^3$ km/s.

Source note: the key prints $S_N=\sigma d$ although the substituted 10.5 is the angular distance $\delta$ ; its notation is preserved.

Official basic criteria

6.1. Totally 4 pt, including:

Correct measurements of distances — 0.5 pt.
Finding necessary distances and mass — 1.5 pt.
Energy, finding typical speeds — 1 pt.
Final the time, correct calculations — 1 pt.

6.2. Totally 2 pt, including:

Correct measurements of distance from the center — 0.5 pt.
Calculations and correct answer — 1.5 pt.

6.common. Totally 2 pt, including:

The most data in the problem conditions was done very roughly, with maximum 1 significant digit, so the answer should be expressed also with only one significant digit, follow this principle — 2 pt.

Official solution sketches: note for jury and team leaders

The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Basic criteria: note for jury

Jury members should evaluate the students’ solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto using these positions.

Оригинал в Архива: bgtheo.pdf · официални решения: theosol.pdf · допълнителни данни: theodata.pdf · допълнителни данни: theo_cr_1.pdf · допълнителни данни: theo_cr_2.pdf · допълнителни данни: theo_cr_3.pdf · допълнителни данни: constants.pdf · допълнителни данни: hrd.pdf · допълнителни данни: alphatheo.pdf · допълнителни данни: betatheo.pdf · допълнителни данни: theo_cr_1.png · допълнителни данни: theo_cr_2.png · допълнителни данни: theo_cr_3.png