Съдържание

IOAA 2008, Анализ на данни — Задача 2. Determination of stellar masses in a visual binary system

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Всички материали от IOAA 2008

The 2nd International Olympiad on Astronomy and Astrophysics — Practical Competition: Data Analysis · 21 август 2008 г. · 100 т.

Please read this carefully:

Every student receives problem sheets in English and/or in his/her native language, answer sheets, millimeter block papers, and scratch sheets.
The time available is five hours for the data analysis and observation competitions. There are three data analysis problems, and two observation problems.
Use only the materials provided.
Fill in the boxes at the top of each sheet of paper with your country code and your student code.
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 answer sheets and the millimeter block papers 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.
At the end of the exam place the answer sheets and the millimeter block papers inside the envelope and leave everything on your desk.

Astronomical and Physical Constants

Quantity	Value
Astronomical unit (AU)	149,597,870.691 km
Light year (ly)	$9.4605 \times 10^{17}$ cm = 63,240 AU
Parsec (pc)	$3.0860 \times 10^{18}$ cm = 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 km
Earth mass ( $\mathfrak{M}_{\oplus}$ )	$5.9736 \times 10^{27}$ g
Earth's mean radius	6,371.0 km
Earth's mean velocity in orbit	29.783 km/s
Moon’s mass (𝔐☽)	$7.3490 \times 10^{25}$ g
Moon’s mean radius	1,738 km
Sun mass ( $\mathfrak{M}_{\odot}$ )	$1.9891 \times 10^{33}$ g
Mean Earth radius	$6.3710 \times 10^{6}$ cm
Sun radius ( $R_{\odot}$ )	$6.96 \times 10^{10}$ cm
Sun luminosity ( $L_{\odot}$ )	$3.96 \times 10^{33}$ erg s $^{-1}$
Sun effective temperature ( $T_{eff\odot}$ )	5 800 °K
Sun apparent magnitude ( $m_{\odot}$ )	-26.8
Sun bolometric magnitude ( $m_{bol\odot}$ )	-26.79
Sun absolute magnitude ( $M_{\odot}$ )	4.82
Sun absolute bolometric magnitude ( $M_{bol\odot}$ )	4.72
Speed of light ( $c$ )	$2.9979 \times 10^{10}$ cm/s
Gravitational constant ( $G$ )	$6.6726 \times 10^{-8}$ dyne cm $^{2}$ g $^{-2}$
Boltzmann constant ( $k$ )	$1.3807 \times 10^{-16}$ erg. K $^{-1}$
Stefan-Boltzmann constant ( $\sigma$ )	$5.6705 \times 10^{-5}$ erg cm $^{-2}$ K $^{-4}$ s $^{-1}$
Planck constant ( $h$ )	$6.6261 \times 10^{-27}$ erg 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
Proton mass ( $m_p$ )	1.007276470 amu
Neutron mass ( $m_n$ )	1.008664904 amu
Deuterium nucleus mass ( $m_d$ )	2.013553214 amu
Hydrogen mass	1.00794 amu
Helium mass	4.002603 amu
Carbon mass	12.01070 amu

Conversion table

1 Å	0.1 nm
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)	$1.6606 \times 10^{-24}$ g
1 atm (atmosphere)	101,325 Pa = 1.01325 bar
1 dyne	$10^{-5}$ N

Points

300 points for 3 problems, 100 points for each problem

Условие

The star α-Centauri (Rigel Kentaurus) is a triple star which consists of two main-sequence stars α-Centauri A and α-Centauri B representing visual binary system, and the third star, called Proxima Centauri, which is smaller and fainter than the other two stars. The angular distance between α-Centauri A and α-Centauri B is 17.59”. The binary system has an orbital period of 79.24 years. The visual magnitudes of α-Centauri A and α-Centauri B are -0.01 and 1.34 respectively. Their color indices are 0.65 and 0.85 respectively. Use the data below to answer the following questions.

Data for main-sequence stars

$(B-V)_0$	$T_{\mathrm{eff}}$	$BC$
-0.25	24500	2.30
-0.23	21000	2.15
-0.20	17700	1.80
-0.15	14000	1.20
-0.10	11800	0.61
-0.05	10500	0.33
0.00	9480	0.15
0.10	8530	0.04
0.20	7910	0
0.30	7450	0
0.40	6800	0
0.50	6310	0.03
0.60	5910	0.07
0.70	5540	0.12
0.80	5330	0.19
0.90	5090	0.28
1.00	4840	0.40
1.20	4350	0.75

$BC$ =Bolometric Correction, $(B-V)_0$ =Intrinsic Color

Questions:

1. Plot the curve $BC$ versus $(B-V)_0$ .

2. Determine the apparent bolometric magnitudes of α-Centauri A and α-Centauri B using the corresponding curve.

3. Calculate the mass of each star.

Notes:

Bolometric correction (BC) is a correction that must be made to the apparent magnitude of an object in order to convert an object's visible magnitude to its bolometric magnitude:

BC = m_v - m_{bol} \quad \text{or} \quad BC = M_v - M_{bol}

Luminosity mass relation : $M_{bol} = -10.2\log\left(\frac{\mathrm{M}}{\mathrm{M}_{\odot}}\right) + 4.9$

Отговори

Решение

Scatter plot of BC (vertical axis, 0 to 2.5) against B - V (horizontal axis, -0.40 to 1.40) on a fine grid, with diamond markers for the tabulated main-sequence data points.

Smooth curve of BC against B-V through the tabulated points (crosses), axes from -0.40 to 1.40 and -0.5 to 2.5; dashed horizontal lines labelled α-Cent A and α-Cent B meet the curve above arrows at B-V = 0.65 and 0.85. — Main Sequence Stars

(20%)

⊙ For α Cen. A. , its colour index is given by $B - V = 0.65$ . Then, by taking the line that intersecting the above curve we find $BC = 0.10$ . (10%)

⊙ For α Cen. B. , its colour index is given by $B - V = 0.85$ . Then, by taking the line that intersecting the above curve we find $BC = 0.25$ . (10%)

From the relation $m_{\mathrm{v}} - m_{\mathrm{bol}} = BC$ it is then obtained,

⊙ For α Cen. A, $m_{\mathrm{v}} = -0.01$ , we get $m_{\mathrm{bol}} = -0.01 - 0.10 = -0.11$ (10%)

⊙ For α Cen. B, $m_{\mathrm{v}} = 1.34$ , we get $m_{\mathrm{bol}} = 1.34 - 0.25 = 1.09$ (10%)

Periodic orbit of α Centauri star is $P = 79.24$ years and its angle of semi major axis is $\alpha = 17.59''$ . From the result 2a we find that the apparent bolometric magnitude of α Centauri A is $m_{\mathrm{bolA}} = -0.11$ , while the apparent bolometric magnitude of α Centauri B is $m_{\mathrm{bolB}} = 1.09$ . (two decimal places)

1st iteration :

Step 1 : Let $\mathcal{M}_{\mathrm{A}}$ and $\mathcal{M}_{\mathrm{B}}$ be the mass of Centauri A and Centauri B stars, respectively. Then, we firstly take $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}} = 2$

Step 2 : Inserting the values of $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}}$ , $P$ , and $\alpha$ to the third Kepler equation namely,

\left(\frac{\alpha}{p}\right)^3 = \left(\mathcal{M}_A + \mathcal{M}_B\right)P^2

One gets $p$ , that is

p = \left[\frac{\alpha^3}{\left(\mathcal{M}_A + \mathcal{M}_B\right)P^2}\right]^{1/3} = \left[\frac{(17.59)^3}{2(79.24)^2}\right]^{1/3} = \left[0.4334\right]^{1/3} = 0.7568''

Step 3 : We determine the magnitude of absolute bolometric using Pogson equation,

m_{bol} - M_{bol} = -5 - 5\log p \Longrightarrow M_{bol} = m_{bol} + 5 + 5\log p

For α Centauri A:

M_{bolA} = m_{bolA} + 5 + 5\log p = -0.11 + 5 + 5\log(0.7568) = -0.11 + 5 - 0.6051 = 4.2849

For α Centauri B :

M_{bolB} = m_{bolB} + 5 + 5\log p = 1.09 + 5 + 5\log(0.7568) = 1.09 + 5 - 0.6051 = 5.4849

Step 4 : We determine the mass of both stars using luminosity mass relation,

M_{bol} = -10.2\log\left(\frac{\mathcal{M}}{\mathcal{M}_{\odot}}\right) + 4.9 \Longrightarrow \mathcal{M} = 10^{\left(\frac{4.9 - M_{bol}}{10.2}\right)}\mathcal{M}_{\odot}

For α Centauri A:

\mathcal{M}_{\mathrm{A}} = 10^{\left(\frac{4.9 - M_{bolA}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 4.2849}{10.2}\right)}\mathcal{M}_{\odot} = 1.1490\mathcal{M}_{\odot}

For α Centauri B:

\mathcal{M}_{\mathrm{B}} = 10^{\left(\frac{4.9 - M_{bolB}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 5.4949}{10.2}\right)}\mathcal{M}_{\odot} = 0.8763\mathcal{M}_{\odot}

(30%)

2nd iteration Repeating the step 1 until 4 and using the previous resulting mass in the first iteration:

Step 1 : $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}} = 1.1490 + 0.8763 = 2.0253$

Step 2 :

p = \left[\frac{\alpha^3}{\left(\mathcal{M}_A + \mathcal{M}_B\right)P^2}\right]^{1/3} = \left[\frac{(17.59)^3}{(2.0253)(79.24)^2}\right]^{1/3} = 0.7536

Step 3 : For α Centauri A:

M_{bolA} = m_{bolA} + 5 + 5\log p = -0.11 + 5 + 5\log(0.7536) = 4.2757

For α Centauri B :

M_{bolB} = m_{bolB} + 5 + 5\log p = 1.09 + 5 + 5\log(0.7536) = 5.4757

Step 4 : For α Centauri A:

\mathcal{M}_{\mathrm{A}} = 10^{\left(\frac{4.9 - M_{bolA}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 4.2757}{10.2}\right)}\mathcal{M}_{\odot} = 1.1513\mathcal{M}_{\odot}

For α Centauri B:

\mathcal{M}_{\mathrm{B}} = 10^{\left(\frac{4.9 - M_{bolB}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 5.4757}{10.2}\right)}\mathcal{M}_{\odot} = 0.8781\mathcal{M}_{\odot}

(10%)

3 rd Iteration Repeating the step 1 until 4 and using the previous resulting mass in the second iteration:

Step 1 : $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}} = 1.1513 + 0.8781 = 2.0294$

Step 2 :

p = \left[\frac{\alpha^3}{\left(\mathcal{M}_A + \mathcal{M}_B\right)P^2}\right]^{1/3} = \left[\frac{(17.59)^3}{(2.0294)(79.24)^2}\right]^{1/3} = 0.75314

Step 3 : For α Centauri A:

M_{bolA} = m_{bolA} + 5 + 5\log p = -0.11 + 5 + 5\log(0.7531) = 4.2751

For α Centauri B :

M_{bolB} = m_{bolB} + 5 + 5\log p = 1.09 + 5 + 5\log(0.7531) = 5.4751

Step 4 : For α Centauri A:

\mathcal{M}_{\mathrm{A}} = 10^{\left(\frac{4.9 - M_{bolA}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 4.2751}{10.2}\right)}\mathcal{M}_{\odot} = 1.1515\mathcal{M}_{\odot}

For α Centauri B:

\mathcal{M}_{\mathrm{B}} = 10^{\left(\frac{4.9 - M_{bolB}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 5.4751}{10.2}\right)}\mathcal{M}_{\odot} = 0.8782\mathcal{M}_{\odot}

(0%)

4 th Iteration Repeating the step 1 until 4 and using the previous resulting mass in the third iteration:

Step 1 : $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}} = 1.1515 + 0.8782 = 2.0297$

Step 2 :

p = \left[\frac{\alpha^3}{\left(\mathcal{M}_A + \mathcal{M}_B\right)P^2}\right]^{1/3} = \left[\frac{(17.59)^3}{(2.0297)(79.24)^2}\right]^{1/3} = 0.7531

Step 3 : For α Centauri A:

M_{bolA} = m_{bolA} + 5 + 5\log p = -0.11 + 5 + 5\log(0.7531) = 4.2753

For α Centauri B :

M_{bolB} = m_{bolB} + 5 + 5\log p = 1.09 + 5 + 5\log(0.7531) = 5.4743

Step 4 : For α Centauri A:

\mathcal{M}_{\mathrm{A}} = 10^{\left(\frac{4.9 - M_{bolA}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 4.2753}{10.2}\right)}\mathcal{M}_{\odot} = 1.1517\mathcal{M}_{\odot}

For α Centauri B:

\mathcal{M}_{\mathrm{B}} = 10^{\left(\frac{4.9 - M_{bolB}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 5.4743}{10.2}\right)}\mathcal{M}_{\odot} = 0.8784\mathcal{M}_{\odot}

(0%)

5 th Iteration Repeating the step 1 until 4 and using the previous resulting mass in the fourth iteration:

Step 1 : $\mathcal{M}_{\mathrm{A}} + \mathcal{M}_{\mathrm{B}} = 1.1517 + 0.8784 = 2.0301$

Step 2 :

p = \left[\frac{\alpha^3}{\left(\mathcal{M}_A + \mathcal{M}_B\right)P^2}\right]^{1/3} = \left[\frac{(17.59)^3}{(2.0301)(79.24)^2}\right]^{1/3} = 0.7530''

Step 3 : For α Centauri A:

M_{bolA} = m_{bolA} + 5 + 5\log p = -0.11 + 5 + 5\log(0.7530) = 4.2740

For α Centauri B :

M_{bolB} = m_{bolB} + 5 + 5\log p = 1.10 + 5 + 5\log(0.7530) = 5.4740

Step 4 : For α Centauri A:

\mathcal{M}_{\mathrm{A}} = 10^{\left(\frac{4.9 - M_{bolA}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 4.2740}{10.2}\right)}\mathcal{M}_{\odot} = 1.1518\mathcal{M}_{\odot}

For α Centauri B:

\mathcal{M}_{\mathrm{B}} = 10^{\left(\frac{4.9 - M_{bolB}}{10.2}\right)}\mathcal{M}_{\odot} = 10^{\left(\frac{4.9 - 5.4740}{10.2}\right)}\mathcal{M}_{\odot} = 0.8785\mathcal{M}_{\odot}

(0%)

The results of the fifth iteration equal the results of the fourth iteration, so

The mass of α Centauri A : $\mathcal{M}_{\mathrm{A}} = 1.1518\ \mathcal{M}_{\odot}$

The mass of α Centauri B : $\mathcal{M}_{\mathrm{B}} = 0.8785\ \mathcal{M}_{\odot}$

Оригинал в Архива: 2008_Data_Analysis.pdf · официални решения: dataan_sol.doc