Съдържание

IOAA 2021, Теория — Задача 7. Menkalinan (β Aurigae) (13 points).

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2021

Theory — 14th IOAA, International Olympiad on Astronomy and Astrophysics — Q7-1 / Q7-2, English (Official) · 13 т.

Standard Table of Constants — values used in this problem

Speed of light in vacuum: $c=2.998\times10^8\ \mathrm{m\,s^{-1}}$ . Universal gravitational constant: $G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ .

Astronomical Data

1 parsec: $1\ pc = 3.086\times10^{16}m = 206\,265\ au = 3.262\ ly$

Solar System

Sun: Mass (kg) $1.988\times10^{30}$

Условие

Almost half of the stars that we see are either binary or multiple star systems. A well-known example of this is Menkalinan (Beta Aurigae), which was initially thought to be a single star, but today recognised as a binary system comprising two stars that we will refer to as Menkalinan A and B. In the following figure, a spectrum of the system (obtained by the observatory of the Universidad de los Andes, in Bogotá) is shown:

Relative flux versus wavelength in angstroms, ticks 6530–6600, with absorption minima A near 6561 and B near 6565. — Spectrum of Menkalinan binary system in the region of Hα. Y-axis is for the relative flux, and X-axis measures wavelengths. Menkalinan A is marked as A in the graph, and Menkalinan B is marked as B.

Spectrum of Menkalinan binary system in the region of $H\alpha$ . Y-axis is for the relative flux, and X-axis measures wavelengths. Menkalinan A is marked as A in the graph, and Menkalinan B is marked as B.

Answer the following questions using the plot and noting that the wavelength of $H\alpha$ line in the laboratory frame is $656.28\ \mathrm{nm}$ . Assume circular orbits, and assume that the binary system as a whole is at rest with respect to the observer.

7.1 In the spectrum, we can see the $H\alpha$ line for each star in the system. Calculate the line-of-sight velocity of each star (km/s) and determine, at the time of this observation, which of the two stars is moving towards us. [5 т.]

7.2 The binary system is located $81.1$ light years from Earth and has an orbital period of $3.96$ days. The semi-major axis for Menkalinan B (smaller star) was measured to be $3.35\,milliarcseconds$ . If the mass ratio of the two components is $1.026$ , find the total mass of the system (in solar masses). [4 т.]

7.3 Calculate the individual masses of Menkalinan $A$ and $B$ in solar masses. [2 т.]

7.4 Since Menkalinan $A$ and $B$ are main sequence stars, use the relation:

\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5}

to estimate the luminosity of each star (in solar luminosity). [2 т.]

Отговори

Решение

7.1

Observing the plot, it is possible to estimate the wavelength of both Hα lines and later find the velocity of each star by Doppler.

Official spectrum annotated with vertical lines at the two measured H-alpha wavelengths, 6561 and 6565 angstroms.

v_A=\frac{6561\ \mathrm{\mathring A}-6562.8\ \mathrm{\mathring A}}{6562.8\ \mathrm{\mathring A}}\times(2.998\cdot10^5\ \mathrm{km\cdot s^{-1}})=-82.23\ \mathrm{km\cdot s^{-1}}

[2 point]

v_B=\frac{6565\ \mathrm{\mathring A}-6562.8\ \mathrm{\mathring A}}{6562.8\ \mathrm{\mathring A}}\times(2.998\cdot10^5\ \mathrm{km\cdot s^{-1}})=+100.50\ \mathrm{km\cdot s^{-1}}

[2 point]

So, Menkalinan A is approaching and Menkalinan B is moving away from us. [1 point]

7.2

m_A>m_B\Rightarrow\alpha_B>\alpha_A

From the definition of CM:

\frac{\alpha_B}{\alpha_A}=1.026

[0.5 points]

\alpha=(\alpha_A+\alpha_B)=\alpha_B\left(1+\frac{\alpha_A}{\alpha_B}\right)=1.975\alpha_B

[0.5 points]

Applying the 3rd Kepler’s law, the total mass of the system is:

M_{\mathrm{tot}}=\frac{4\pi^2}{G}\times\frac{\alpha^3\cdot d^3}{T^2}

[1 point]

M_{\mathrm{tot}}=\frac{4\pi^2}{G}\times\frac{(0.00662''\times\frac{1\ \mathrm{rad}}{206265''})^3\cdot(81.1\ \mathrm{ly}\cdot\frac{9.46\cdot10^{15}\ \mathrm{m}}{1\ \mathrm{ly}})^3}{(3.96\ \mathrm{d}\cdot\frac{86400\ \mathrm{s}}{1\ \mathrm{d}})^2}

M_{\mathrm{tot}}=7.528\cdot10^{31}\ \mathrm{kg}=37.868M_\odot

[2 point]

7.3

\frac{m_A}{m_B}=1.026

M_{\mathrm{tot}}=m_A+m_B=1.026m_B+m_B=2.026m_B=37.868M_\odot

m_B=18.691M_\odot

[1 point]

m_A=1.026m_B=19.177M_\odot

[1 point]

7.4

Using the mass/luminosity relation given:

\frac{L_A}{L_\odot}=\left(\frac{m_A}{M_\odot}\right)^{3.5}=\left(\frac{19.177M_\odot}{M_\odot}\right)^{3.5}\quad3.088\times10^4

L_A=3.088\times10^4L_\odot

[1 point]

\frac{L_B}{L_\odot}=\left(\frac{m_B}{M_\odot}\right)^{3.5}=\left(\frac{18.691M_\odot}{M_\odot}\right)^{3.5}=2.823\times10^4

L_B=2.823\times10^4L_\odot

[1 point]

Official solution source discrepancy

The first luminosity calculation in official solution 7.4 omits the equality sign before 3.088×10⁴. Its spacing and the following explicit result are retained.

Оригинал в Архива: TQ-7-Q.pdf · официални решения: TQ7 (1).pdf · допълнителни данни: TQ-7-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf