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IOAA 2008, Теория — Задача 17. Theoretical Part 2 — Question 2

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Всички материали от IOAA 2008

The 2nd International Olympiad on Astronomy and Astrophysics, Bandung, Indonesia, Saturday, 23 August 2008 — Theoretical Competition · 23 август 2008 г. · 100 т.

Внимание: Бележка към темата

The official key places the final short question 10 in long_sol.doc as question 1 and the final long question 1 in short_sol.doc as question 10. Solutions are matched by their statements. For short question 5 the key uses dark-energy density 6.7 × 10⁻³⁰ g/cm³, whereas the final question specifies 7.1 × 10⁻³⁰ g/cm³; the supplied solution is retained without recalculation. Printed numerical, unit and notation discrepancies in the original keys are preserved.

Please read this carefully:

Every student receives problem sheets in English and/or in native language, an answer book and a scratch book.
The time available is 5 hours for the theoretical competition. There are fifteen short questions (Theoretical Part 1), and three long questions (Theoretical Part 2).
Use only Black or dark blue pen
Use only the front side of answer sheets. Write only inside the boxed area.
Begin answering each question on a separate sheet.
Numerical results should be written with as many digits as are appropriate.
Write on the blank answer sheets whatever you consider is required for the solution of each question. Please express your answer primarily in term of equations, numbers, figures, and plots. If necessary provide your answers with concise text. Full credit will be given to correct answer with detailed steps for each question. Underline your final result.
Fill in the boxes at the top of each sheet of paper with your country code and your student code.
At the end of the exam place the books inside the envelope and leave everything on your desk.

Astronomical and Physical Constants

Quantity	Value
Astronomical unit (AU)	149 597 870 691 m
Light year (ly)	$9.4605 \times 10^{15}$ m = 63,240 AU
Parsec (pc)	$3.0860 \times 10^{16}$ m = 206,265 AU
Sidereal year	365.2564 days
Tropical year	365.2422 days
Gregorian year	365.2425 days
Sidereal month	27.3217 days
Synodic month	29.5306 days
Mean sidereal day	$23^{\mathrm{h}}56^{\mathrm{m}}4^{\mathrm{s}}.091$ of mean solar time
Mean solar day	$24^{\mathrm{h}}3^{\mathrm{m}}56^{\mathrm{s}}.555$ of sidereal time
Mean distance, Earth to Moon	384 399 000 m
Earth mass ( $M_{\oplus}$ )	$5.9736 \times 10^{24}$ kg
Earth mean radius	6 371 000 m
Earth mean velocity in orbit	29 783 m/s
Moon mass (M☽)	$7.3490 \times 10^{22}$ kg
Moon mean radius	1 738 000 m
Sun mass ( $M_{\odot}$ )	$1.9891 \times 10^{30}$ kg
Sun radius ( $R_{\odot}$ )	$6.96 \times 10^{8}$ m
Sun luminosity ( $L_{\odot}$ )	$3.96 \times 10^{26}$ J s $^{-1}$
Sun effective temperature ( $T_{eff\odot}$ )	5 800 $^{\mathrm{o}}$ K
Sun apparent magnitude ( $m_{\odot}$ )	-26.8
Sun absolute magnitude ( $M_{\odot}$ )	4.82
Sun absolute bolometric magnitude ( $M_{bol\odot}$ )	4.72
Speed of light ( $c$ )	$2.9979 \times 10^{8}$ m/s
Gravitational constant ( $G$ )	$6.6726 \times 10^{-11}$ N m $^{2}$ kg $^{-2}$
Boltzmann constant ( $k$ )	$1.3807 \times 10^{-23}$ J K $^{-1}$
Stefan-Boltzmann constant ( $\sigma$ )	$5.6705 \times 10^{-8}$ J s $^{-1}$ m $^{-2}$ K $^{-4}$
Planck constant ( $h$ )	$6.6261 \times 10^{-34}$ J s
Electron charge ( $e$ )	$1.602 \times 10^{-19}$ C = $4.803 \times 10^{-10}$ esu
Electron mass ( $m_e$ )	$5.48579903 \times 10^{-4}$ amu = $9.11 \times 10^{-31}$ kg
Proton mass ( $m_p$ )	1.00727647 amu = $1.67268 \times 10^{-27}$ kg
Neutron mass ( $m_n$ )	1.008664904 amu = $1.67499 \times 10^{-27}$ kg
Deuterium nucleus mass ( $m_d$ )	2.013553214 amu = $3.34371 \times 10^{-27}$ kg
Hydrogen mass	1.00794 amu = $1.67379 \times 10^{-27}$ kg
Helium mass	4.002603 amu = $1.646723 \times 10^{-27}$ kg

Conversion table

1 Å	0.1 nm = $10^{-10}$ m
1 barn	$10^{-28}$ m $^{2}$
1 G	$10^{-4}$ T
1 erg	$10^{-7}$ J = 1 dyne cm
1 esu	$3.3356 \times 10^{-10}$ C
1 amu (atomic mass unit)	$1.6606 \times 10^{-27}$ kg
1 atm (atmosphere)	101,325 Pa = 1.01325 bar
1 dyne	$10^{-5}$ N

THEORETICAL PART 1

(300 points for 15 Theoretical Part-1 questions, 20 points for each question)

Show your method of solution step by step in the answer sheets completely as your final answer. The scratch sheet is to be used for your personal calculation and will not be marked. Partial credits will be given for answers without showing method of solution.

THEORETICAL PART 2

(300 points for 3 Theoretical Part-2, 100 points for each question)

Условие

A UBV photometric (UBV Johnson’s) observation of a star gives $U = 8.15$ , $B = 8.50$ , and $V = 8.14$ . Based on the spectral class, one gets the intrinsic color $(U - B)_{\mathrm{o}} = -0.45$ . If the star is known to have radius of $2.3\ R_{\odot}$ , absolute bolometric magnitude of -0.25, and bolometric correction (BC) of -0.15, determine:

a. the intrinsic magnitudes $U$ , $B$ , and $V$ of the star (take, for the typical interstellar matters, the ratio of total to selective extinction (color excess) $R_V = 3.2$ ),

b. the effective temperature of the star,

c. the distance to the star in pc.

Note: The relation between color excess of $U$ - $B$ and of $B - V$ is $E(U - B) = 0.72\ E(B - V)$ .

Let $A_v$ be the interstellar extinction and $R = 3.2$ , then $A_v = 3.2\ E(B\text{-}V)$ .

Решение

$U=8.15$ , $B=8.50$ , and $V=8.14$ .

$(U-B)_o=-0.45$ , $R=2.3R_\odot=1.60\times10^{11}$ cm,

$M_{bol}=-0.25$ , $BC=-0.15$

a)

$U-B=8.15-8.50=-0.35$

The color excess for U-B:

E(U-B)=(U-B)-(U-B)_o=-0.35-(-0.45)=0.10

The relation between color excess of U-B and of B-V:

E(U-B)=0.72E(B-V)\Longrightarrow E(B-V)=0.10/0.72=0.14\qquad(15\%)

Let $A_v$ be the interstellar extinction and $R=3.2$ , then

A_v=3.2E(B-V)=3.2(0.14)=0.45

V-V_o=A_v\Longrightarrow V_o=V-A_v=8.14-0.45=7.69

E(B-V)=(B-V)-(B-V)_o\Longrightarrow(B-V)_o=(B-V)-E(B-V)

=(8.50-8.14)-0.14=0.22\qquad(15\%)

(B-V)_o=B_o-V_o=0{,}22\Longrightarrow B_o=0.22+V_o=0.22+7.69=7.91\qquad(10\%)

(U-B)_o=U_o-B_o=-0{,}45\Longrightarrow U_o=B_o-0.45=7.91-0.45=7.46\qquad(10\%)

b)

Luminosity - absolute magnitude relation:

M_{bol}-M_{bol\odot}=-2.5\log L/L_\odot\Longrightarrow L/L_\odot=10^{-(M_{bol}-M_{bol\odot})/2.5}

=10^{-(-0.25-4.75)/2.5}=100\qquad(10\%)

L=100L_\odot=3.90\times10^{35}\text{ erg/det}

L=4\pi\sigma R^2T_{ef}^4

T_{ef}=\left(\frac L{4\pi\sigma R^2}\right)^{1/4}=\left(\frac{3.90\times10^{35}}{4\pi(5.67\times10^{-5})(1.60\times10^{11})^2}\right)^{1/4}=12092\text{ K}\qquad(15\%)

c)

M_v-M_{bol}=BC\Longrightarrow M_v=M_{bol}+BC=-0.25-0.15=-0.40\qquad(10\%)

m_v-M_v=-5+5\log d+A_v\Longrightarrow5\log d=m_v-M_v+5-A_v

=8.14+0.40+5-0.45=13.09

d=10^{13.09/5}=10^{2.62}=414.9\text{ pc}\qquad(15\%)

Оригинал в Архива: 2008_Theory.pdf · официални решения: short_sol.doc · допълнителни данни: long_sol.doc · допълнителни данни: constants.doc