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USAAAO 2019, II кръг — Задача 11

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Всички материали от USAAAO 2019

2019 National Astronomy Olympiad (NAO) · 25 т.

Instructions (Please Read Carefully)

(Start the exam after reading the instructions carefully and affirming that the contestant understood the instructions given.)

The top 5 eligible scorers on the NAO will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions (7 points each), Medium Questions (15 points each) and Long Questions (25 points each). The maximum score is 153 points.

The test must be completed within 2.5 hours (150 minutes). The proctor should mark the start and end time of the exam on the front page.

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as ”scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution. This exam document, solution, and all the scratch paper used should be turned in to the proctor at the end of the exam.

This is a written exam. Contestants can only use a scientific calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on March 8th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a scientific calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books, notes, laptops, mobile phones, or any other devices are allowed while taking the exam.

After reading the instructions, please make sure to sign at the bottom of this page, affirming that:

All work on this exam is mine.
I took this exam under a proctor’s supervision.
I did not receive any external aids besides the materials provided.
I am not allowed to discuss the test with others throughout the period of this examination.
Failure to follow these rules will lead to disqualification from the exam.

To be completed by the proctor:

Full name (First, Last): ______

Position (e.g. Physics Teacher at High School x): ______

Email address: ______

Student began exam at (hh:mm): ______

Student submitted the exam at (hh:mm): ______

Signature: ______ Date (mm/dd/yy): ______

To be completed by the student:

Last name: ______

First name: ______ Middle name: ______

Date of birth (mm/dd/yy): ______ Email address: ______

Signature: ______ Date (mm/dd/yy): ______

Table of Physical Constants

Constant	Symbol	Value
Electron rest mass	$m_e$	$9.109\times10^{-31}$ kg
Proton rest mass	$M_p$	$1.6726\times10^{-27}$ kg
Electronic charge	$e$	$1.6022\times10^{-19}$ C
Speed of light in free space	$c$	$2.9979\times10^8$ m s $^{-1}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}$ H m $^{-1}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}$ F m $^{-1}$
Planck’s constant	$h$	$6.626\times10^{-34}$ J s
Reduced Planck’s constant	$\hbar=h/2\pi$	$1.0546\times10^{-34}$ J s
	$\hbar c$	197.33 MeV fm
Boltzmann’s constant	$k_B$	$1.3807\times10^{-23}$ J K $^{-1}$
Gas constant	$\mathcal R=k_B/m_H$	$8.250\times10^3$ J kg $^{-1}$ K $^{-1}$
Molar gas constant	$R$	8.315 J mol $^{-1}$ K $^{-1}$
Avogadro’s number	$N_A$	$6.022\times10^{23}$ mol $^{-1}$
Standard molar volume		$22.414\times10^{-3}$ m $^3$ mol $^{-1}$
Unified atomic mass unit ( $^{12}$ C scale)	$u$	$931.5$ MeV $/c^2=1.660538\times10^{-27}$ kg
Mass of hydrogen atom	$m_H$	$1.0078u=1.6735\times10^{-27}$ kg
Bohr magneton	$\mu_B$	$9.274\times10^{-24}$ A m $^2$ or J T $^{-1}$
Nuclear magneton	$\mu_N$	$5.051\times10^{-27}$ A m $^2$ or J T $^{-1}$
Proton magnetic moment	$\mu_p$	$2.7928\mu_N$
Neutron magnetic moment	$\mu_n$	$-1.9130\mu_N$
Bohr radius	$a_0$	$5.292\times10^{-11}$ m
Fine structure constant	$\alpha=e^2/(4\pi\epsilon_0\hbar c)$	$(137.04)^{-1}$
Compton wavelength of electron	$\lambda_C=h/(m_ec)$	$2.4263\times10^{-12}$ m
Rydberg’s constant	$R_\infty$	$1.0974\times10^7$ m $^{-1}$
	$R_\infty hc$	13.606 eV
Stefan-Boltzmann constant	$\sigma$	$5.671\times10^{-8}$ W m $^{-2}$ K $^{-4}$
Radiation density constant	$a=4\sigma/c$	$7.561\times10^{-16}$ J m $^{-3}$ K $^{-4}$
Gravitational constant	$G$	$6.673\times10^{-11}$ N m $^2$ kg $^{-2}$

Rest masses of some leptons and hadrons in MeV/ $c^2$ :

$e^{\pm}$ 0.5110, $\mu^{\pm}$ 105.66, $\tau^{\pm}$ 1777, $\pi^0$ 134.98, $\pi^{\pm}$ 139.57, $K^{\pm}$ 493.7, $K^0$ 497.7, $\eta$ 547, $D^0$ 1865, $D^{\pm}$ 1869, p 938.3, n 939.6, $\Lambda^0$ 1115.7, $\Sigma^+$ 1189, $\Sigma^0$ 1193, $\Sigma^-$ 1197, $\Xi^0$ 1315, $\Xi^-$ 1321, $\Omega^-$ 1672, $Z^0$ $91.187\times10^3$ , $W^{\pm}$ $80.41\times10^3$ .

Quark	Charge	$I_3$	$S$	$C$	$B$	$T$	Mass (GeV/ $c^2$ )
u	$+2/3$	$+1/2$	0	0	0	0	$\sim0.003$
d	$-1/3$	$-1/2$	0	0	0	0	$\sim0.006$
c	$+2/3$	0	0	+1	0	0	$\sim1.25$
s	$-1/3$	0	−1	0	0	0	$\sim0.11$
t	$+2/3$	0	0	0	0	+1	174.3
b	$-1/3$	0	0	0	−1	0	4.2

Astrophysical Data — Other data and conversion factors — Trigonometrical identities — Prefixes

Astrophysical Data

Quantity	Symbol	Value
Astronomical unit	AU	$1.496\times10^{11}$ m
Parsec	pc	$3.086\times10^{16}$ m
Luminosity of Sun	$L_\odot$	$3.85\times10^{26}$ W
Mass of Sun	$M_\odot$	$1.989\times10^{30}$ kg
Radius of Sun	$R_\odot$	$6.96\times10^8$ m
Mass of Earth	$M_E$	$5.9742\times10^{24}$ kg
Radius of Earth	$R_E$	$6.3781\times10^6$ m

Other data and conversion factors

Quantity	Symbol	Value
1 ångstrom	Å	$10^{-10}$ m
1 fermi	fm	$10^{-15}$ m
1 barn	b	$10^{-28}$ m $^2$
1 pascal	Pa	1 N m $^{-2}$
1 standard atmosphere		$1.0132\times10^5$ Pa
Standard acceleration due to gravity	$g$	9.807 m s $^{-2}$
1 electron volt	eV	$1.6022\times10^{-19}$ J
	eV/ $hc$	$8.065\times10^5$ m $^{-1}$
	eV/ $k_B$	$1.1604\times10^4$ K
Wavelength of 1 eV photon		$1.2399\times10^{-6}$ m

Trigonometrical identities

\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi

\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi

\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}

\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}

\cos\alpha-\cos\beta=2\sin\frac{\alpha+\beta}{2}\sin\frac{\beta-\alpha}{2}

In a triangle ABC, $a/\sin A=b/\sin B=c/\sin C$

and $a^2=b^2+c^2-2bc\cos A$ .

Prefixes

Symbol	Prefix	Factor
T	tera	$10^{12}$
G	giga	$10^9$
M	mega	$10^6$
k	kilo	$10^3$
c	centi	$10^{-2}$
m	milli	$10^{-3}$
$\mu$	micro	$10^{-6}$
n	nano	$10^{-9}$
p	pico	$10^{-12}$
f	femto	$10^{-15}$

Archive constants sheet

The archive retains this general constants sheet alongside the USAAAO collection. The visible sheet does not identify a year or an issuing body. Its values differ from some constants explicitly used in the 2019 solutions; the archive does not establish that this was the exact sheet given to the 2019 contestants.

Условие

Cygnus X-1/HDE 226868 is a binary system consisting of a black hole Cygnus X-1 and blue supergiant HDE 226868. The mass of HDE 226868 is $30M_{\odot}$ and the period of the binary system is 5.6 days. Radial velocity data reveals that the orbital velocity of HDE 226868 is 116.68 km/s at apoapse and 123.03 km/s at periapse.

(a) Determine the eccentricity of the orbit of HDE 226868. [5 т.]

(b) Determine the length of the semimajor axis of the orbit of HDE 226868. [5 т.]

(c) Determine the mass of Cygnus X-1, to at least 3 significant figures. [5 т.]

The peak blackbody temperature of an accretion disk occurs at a distance of $r_{peak}$ and a temperature of $T_{peak}$ . One can determine the peak blackbody temperature by assuming that it corresponds to the peak in the x-ray spectrum. Due to relativistic effects, the actual peak blackbody temperature $T_{peak}$ is related to the peak color temperature $T_{color}$ derived from observed spectral data by $T_{color} = f_{GR}f_{col}T_{peak}$ , where $f_{GR} \approx 0.510$ and $f_{col} \approx 1.7$ . Three x-ray spectra of Cygnus X-1 are shown in Figure 4.

Three side-by-side panels SP1, SP2 and SP3 of x-ray spectra versus E (keV) on logarithmic axes (about 0.5 to 50 keV). Upper panels: Photons cm⁻² s⁻¹ from 0.1 to about 20, with data points and model components Total (magenta dash-dot), Thermal (red solid), Power-Law (blue dashed), Reflected (green dashed) and Line (green solid). Lower panels: Data/Model Ratio from 0.7 to 1.3 with a horizontal line at 1.0. — Figure 4: Three x-ray spectra from Cygnus X-1. From Gou et al. (2011).

(d) Using spectrum SP2, determine the peak blackbody temperature $T_{peak}$ of the accretion disk around Cygnus X-1. [4 т.]

The total luminosity of the blackbody component of the accretion disk can be estimated by $L_{disk} \approx 4\pi\sigma r_{peak}^2T_{peak}^4$ (Makishima et al. 1986). The radius $r_{last}$ of the innermost edge of the accretion disk is related to the radius $r_{peak}$ of the peak blackbody temperature by $r_{peak} = \eta r_{last}$ , where $\eta \approx 0.63$ . In 1996, the blackbody luminosity of the accretion disk around Cygnus X-1 was estimated to be $2.2\times10^{37}$ ergs/s.

(e) Determine the radius $r_{last}$ of the innermost edge of the accretion disk around Cygnus X-1. [4 т.]

Assume that the innermost edge of the accretion disk is located at the innermost stable circular orbit (ISCO), whose radius $r_{isco}$ is a function of the spin of the black hole. The relationship between $r_{isco}$ and $a_*$ , the spin parameter of the black hole, can be estimated by:

r_{isco} = \frac{GM}{c^2}\left(\sqrt{8.354\cdot[(2-a_*)^2-1]}+1\right)

(f) Determine the spin parameter $a_*$ of Cygnus X-1. [2 т.]

Отговори

Решение

Внимание: Непълно решение

The official solutions document gives no solution for part (f).

Solution:

(a) In the absence of external forces, the total angular momentum of the binary system is conserved. The angular momentum $l = mr^2\omega = mrv$ of each component must also be conserved. Therefore, $m_1r_{1,a}v_{1,a} = m_1r_{1,p}v_{1,p} \implies \frac{v_{1,p}}{v_{1,a}} = \frac{r_{1,a}}{r_{1,p}} = \frac{a(1+\epsilon)}{a(1-\epsilon)} \implies \epsilon = \frac{v_{1,p}-v_{1,a}}{v_{1,p}+v_{1,a}} = \frac{123.03-116.68}{123.03+116.68} = 0.0265$ .

(b) The area swept by the vector $\mathbf{r_1}$ (from the center of mass to the primary) over time interval $\Delta t$ is approximately a right triangle of base $r_1$ and height $v_1\Delta t$ , and therefore area $\frac{1}{2}r_1v_1\Delta t$ . Over a single orbit, the area swept is $A = \frac{1}{2}r_1v_1T = \pi a_1^2\sqrt{1-\epsilon^2}$ , for $(r_1, v_1)$ at any point in the orbit (by Kepler’s second law, the rate of area swept is constant). So, $\pi a_1^2\sqrt{1-\epsilon^2} = \frac{1}{2}a_1(1+\epsilon)v_{1,a}T \implies a_1 = \sqrt{\frac{1+\epsilon}{1-\epsilon}}\cdot\frac{v_{1,a}T}{2\pi} = \sqrt{\frac{1+0.0265}{1-0.0265}}\cdot\frac{116680\cdot5.6\cdot24\cdot3600}{2\pi} = 9.23\times10^9$ m.

(c) Kepler’s third law states that $(a_1 + a_2)^3 = \frac{G(m_1+m_2)}{4\pi^2}T^2$ , so $a_2 = \left[\frac{G(m_1+m_2)}{4\pi^2}T^2\right]^{1/3} - a_1$ . Additionally, the definition of center of mass is that $a_1m_1 = a_2m_2$ , so $a_2 = \frac{a_1m_1}{m_2}$ . Putting these two equations together yields $\left[\frac{G(m_1+m_2)}{4\pi^2}T^2\right]^{1/3} - a_1 = \frac{a_1m_1}{m_2}$ , or $\left[\frac{G(m_1+m_2)}{4\pi^2}T^2\right]^{1/3} - a_1 - \frac{a_1m_1}{m_2} = 0$ . Solving by iteration gives $m_2 = 2.40\times10^{31}$ kg, or $m_2 = 12.1M_{\odot}$ .

(d) The spectrum peaks at about 1.5 keV, or $2.4\times10^{-16}$ J. The energy of a photon is related to its wavelength by $E = \frac{hc}{\lambda}$ , which is related to the peak blackbody temperature by $T_{color} = \frac{b}{\lambda}$ , where $b$ is Wein’s constant. Therefore, $E = \frac{hcT_{color}}{b}$ , or $T_{color} = \frac{Eb}{hc} = \frac{2.4\times10^{-16}\cdot2.898\times10^{-3}}{6.626\times10^{-34}\cdot3.00\times10^8} = 3.50\times10^6$ K. Finally, $T_{peak} = \frac{T_{color}}{f_{GR}f_{col}} = 4.04\times10^6$ K.

(e) $L_{disk} = 2.2\times10^{37}$ ergs/s, or $2.2\times10^{30}$ W. $L_{disk} = 4\pi\sigma r_{peak}^2T_{peak}^4$ , so $r_{peak} = \frac{1}{T_{peak}^2}\sqrt{\frac{L_{disk}}{4\pi\sigma}} = \frac{1}{(4.04\times10^6)^2}\sqrt{\frac{2.2\times10^{30}}{4\pi\cdot5.67\times10^{-8}}} = 107.66$ km. $r_{last} = \eta r_{peak} = 67.83$ km.

Оригинал в Архива: nao_2019.pdf · официални решения: nao_2019_solution.pdf · допълнителни данни: physical-constants.pdf