Съдържание

IOAA 2009, Теория — Задача 17. Apparent number density of stars in the Galaxy

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2009

3rd International Olympiad on Astronomy and Astrophysics — Theoretical Competition · 45 т.

Внимание: Бележка към темата

The problem PDF's constants table has missing equation glyphs; the intact matching table and logarithm formula were recovered from original proceedings PDF 41, explicitly attributed. The full 147-page proceedings file and hash are retained, with a declared reading scope of 41 and47–67. Printed solution inconsistencies are retained and annotated: Q10 intermediate omits field factor/unit; Q12 negative sign and 3.21/3.19 mismatch; Q15 signed −1.5 and referenced but absent Figures 1/2; Q16(d) denominator minus sign; Q17 final m−5.2 versus preceding m+5.2. No missing figure or correction was invented.

Please read these instructions carefully

Each student will receive problem sheets in English and/or in his/her native language.
The available time for answering theoretical problems is 5 hours. You will have 15 short problems (Theoretical Part 1, Problem 1 to 15), and 2 long problems (Theoretical Part 2, Problem 16 and 17).
Use only the pen that has been provided on your desk.
Do Not use the back side of your writing sheets. Write only inside the boxed area.
Yellow scratch papers are not considered in marking.
Begin answering each problem in separate sheet.
Fill in the boxes at the top of each sheet of your paper with your "country name", your "student code", "problem number", and total number of pages which is used to answer to that problem.
Write the final answer for each problem in the box, labeled "Answer Sheet".
Starting and the end of the exam will be announced by ringing a bell.
The final answer in each question part must be accompanied by units, which should be in SI or appropriate units. 20% of the marks available for that part will be deducted for a correct answer without units.
The required numerical accuracy for the final answer depends on the number of significant figures given in the data values in the problem. 20% of the marks available for the final answer in each question part will be deducted for answers without required accuracy as given in the problem. Use the constant values exactly as given in the table of constants.
At the end of the exam put all papers, including scratch papers, inside the envelope and leave everything on your desk.

Table of Constants

The problem PDF's constants table has missing embedded equation glyphs. This table and useful formula are transcribed from the matching intact table on original PDF page 41 of IOAA2009_final.pdf (printed page 36), retaining the full proceedings source and its original page numbering.

(All constants are in SI)

Parameter	Symbol	Value
Gravitational constant	$G$	$6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Plank constant	$h$	$6.63\times10^{-34}\ \mathrm{J\,s}$
	$\hbar$	$1.05\times10^{-34}\ \mathrm{J\,s}$
Speed of light	$c$	$3.00\times10^8\ \mathrm{m\,s^{-1}}$
Solar Mass	$M_\odot$	$1.99\times10^{30}\ \mathrm{kg}$
Solar radius	$R_\odot$	$6.96\times10^8\ \mathrm m$
Solar luminosity	$L_\odot$	$3.83\times10^{26}\ \mathrm W$
Apparent solar magnitude (V)	$m_\odot$	$-26.8$
Solar constant	$b_\odot$	$1.37\times10^3\ \mathrm{W\,m^{-2}}$
Mass of the Earth	$M_\oplus$	$5.98\times10^{24}\ \mathrm{kg}$
Radius of the Earth	$R_\oplus$	$6.38\times10^6\ \mathrm m$
Mean density of the Earth	$\rho_\oplus$	$5\times10^3\ \mathrm{kg\,m^{-3}}$
Gravitational acceleration at sea level	$g$	$9.81\ \mathrm{m\,s^{-2}}$
Tropical year		365.24 days
Sidereal year		365.26 days
Sidereal day		86164 s
Inclination of the equator with respect to the ecliptic	$\epsilon$	$23.5^\circ$
Parsec	pc	$3.09\times10^{16}\ \mathrm m$
Light year	ly	$9.46\times10^{15}\ \mathrm m$
Astronomical Unit	AU	$1.50\times10^{11}\ \mathrm m$
Solar distance from the center of the Galaxy	$R$	$8\times10^3\ \mathrm{pc}$
Hubble constant	$H$	$75\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Mass of proton	$m_p$	$1.67\times10^{-27}\ \mathrm{kg}$
Central wavelength of V-band	$\lambda$	$550\ \mathrm{nm}$
Refraction of star light at horizon		$34'$
	$\pi$	3.1416

Useful mathematical formula:

\ln(1+x)\simeq x\quad\text{for }x\to0.

Условие

Let us model the number density of stars in the disk of Milky Way Galaxy with a simple exponential function of $n=n_0\exp(-(r-R_0)/R_d)$ , where $r$ represents the distance from the center of the Galaxy, $R_0$ is the distance of the Sun from the center of the Galaxy, $R_d$ is the typical size of disk and $n_0$ is the stellar density of disk at the position of the Sun. An astronomer observes the center of the Galaxy within a small field of view. We take a particular type of Red giant stars (red clump) as the standard candles for the observation with approximately constant absolute magnitude of $M=-0.2$ ,

a Considering a limiting magnitude of $m=18$ for a telescope, calculate the maximum distance that telescope can detect the red clump stars. For simplicity we ignore the presence of interstellar medium so there is no extinction.

b Assume an extinction of $0.7\ \mathrm{mag/kpc}$ for the interstellar medium. Repeat the calculation as done in the part (a) and obtain a rough number for the maximum distance these red giant stars can be observed.

c Give an expression for the number of these red giant stars per magnitude within a solid angle of $\Omega$ that we can observe with apparent magnitude in the range of $m$ and $m+\Delta m$ , (i.e. $\Delta N/\Delta m$ ). Red giant stars contribute fraction $f$ of overall stars. In this part assume no extinction in the interstellar medium as part (a). Assume the size of the disk is larger than maximum observable distance.

Отговори

Решение

a)

Relation between the apparent and absolute magnitude is given by

m=M+5\log\left(\frac d{10}\right).

(3 points)

where $d$ is in terms of parsec. Substituting $m=18$ and $M=-0.2$ , results in

d=4.37\times10^4\ \mathrm{pc}.

(5 points)

b)

Adding the term for the extinction, changes the magnitude distance relation as follows

m=M+0.7x+5\log(100x)

where $x$ is given in terms of kilo parsec. To have a rough value for $x$ , after substituting $m$ and $M$ , this equation reduces to

8.2=0.7x+5\log x.

(6 points)

To solve this equation we examine

x=5,5.5,6,6.5

where the best value is obtained roughly $x\simeq6.1\ \mathrm{kpc}$ .

(8 points)

c)

For a solid angle $\Omega$ , the number of observed red clump stars at the distance in the range of $x$ and $x+\Delta x$ is given by

\Delta N=\Omega x^2n(x)f\Delta x.

So the number of stars observed in $\Delta x$ is given by

\frac{\Delta N}{\Delta x}=\Omega x^2n(x)f.

(6 points)

From the relation between the distance and apparent magnitude we have

m_1=M+5\log\left(\frac x{10}\right),\qquad m_2=M+5\log\left(\frac{x+\Delta x}{10}\right).

\Delta m=5\log\left(\frac{x+\Delta x}{x}\right)=5\log\left(1+\frac{\Delta x}{x}\right).

\Delta m=\frac5{\ln10}\ln\left(1+\frac{\Delta x}{x}\right)=\frac5{\ln10}\left(\frac{\Delta x}{x}\right).

Replacing $\Delta x$ with $\Delta m$ , results in

\frac{\Delta N}{\Delta m}=\frac{\Delta N}{\Delta x}\times\frac{\Delta x}{\Delta m}.

So the number of stars for a given magnitude is obtained by

\frac{\Delta N}{\Delta m}=\frac{\Omega\ln10}{5}n(x)x^3f.

(5 points)

Finally we substituting $x$ in terms of apparent magnitude using $x=10^{(m+5.2)/5}$ . In the case of no extinction, we are able to observe the Galaxy beyond the center. So $dN/dm$ has two terms in $x<R_0$ and $x>R_0$ . The relation between $x$ and $r$ for these two cases are

x=R_0-r\quad x<R_0

and

x=R_0+r\quad x>R_0.

(6 points)

So in general we can write $\frac{\Delta N}{\Delta m}$ as

\frac{\Delta N}{\Delta m}=\frac{\Omega\ln10}{5}n_0\exp\left(\frac{10^{(m-5.2)/5}}{R_d}\right)10^{3(m-5.2)/5}f,\qquad x<R_0.

\frac{\Delta N}{\Delta m}=\frac{\Omega\ln10}{5}n_0\exp\left(\frac{2R_0}{R_d}\right)\exp\left(-\frac{10^{(m-5.2)/5}}{R_d}\right)10^{3(m-5.2)/5}f\Theta(x_0-x),\qquad x>R_0.

(6 points)

where $\Theta(x)$ is the step function and $x_0$ is the maximum observable distance.

Source note: the final formulas print $m-5.2$ , whereas the preceding substitution prints $m+5.2$ . Both are retained as printed.

Теми и трудност

Трудност: неоценена.

Галактики и космология · Светлина и астрономически измервания · Млечен път · Звездни величини · Междузвездно поглъщане

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: 2009_Theory.pdf · официални решения: IOAA2009_final.pdf