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IOAA 2013, Теория, дълги задачи — Задача 3

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$7^{th}$ International Olympiad on Astronomy & Astrophysics · 27 July – 5 August 2013, Volos Greece · Theoretical Exam - Long Questions · 27 юли 2013 г.

Условие

Question 3

(1) Using the $\textit{virial theorem}$ for an isolated, spherical system, i.e, that $-2\langle K\rangle=\langle U\rangle$ , where “K” is the average kinetic energy and “U” is the average potential energy of the system, determine an expression for the total mass of a cluster of galaxies if we know the radial velocity dispersion, $\sigma$ , of the cluster’s galaxy members and the cluster’s radius, $R$ . Assume that the cluster is isolated, spherical, has a homogeneous density and that it consists of galaxies of equal mass.

(2) Find the $\textit{virial mass}$ , i.e. the mass calculated from the $\textit{virial theorem}$ , of the Coma cluster, which lies at a distance of 90 Mpc from us, if you know that the radial velocity dispersion of its member galaxies is $\sigma_{v_r}=1000km/s$ and that its angular diameter (on the sky) is about $4^\circ$ .

(3) From observations, the total luminosity of the galaxies comprising the cluster is approximately $L=5\times10^{12}L_{\odot}$ . If the mass to luminosity ratio, $M/L$ , of the cluster is $\sim1$ (assume that all the mass of the cluster is visible mass), this should correspond to a total mass $M\sim5\times10^{12}M_{\odot}$ for the mass of the cluster. Give the ratio of the luminous mass to the total mass of the cluster you derived in question (2).

Решение

Answer:

(1) Using the virial theorem for our isolated, spherical system of N galaxies of mass $m$ , each, we get

-2\langle K\rangle=\langle U\rangle\Rightarrow -\frac{2}{N}\sum_{1}^{N}\frac{1}{2}m_i u_i^2=\langle U\rangle\Rightarrow -\frac{m}{N}\sum_{1}^{N}u_i^2=\frac{U}{N}

(3 point)

where $\frac{1}{N}\sum_{1}^{N}u_i^2\approx\langle u^2\rangle=\langle u_r^2\rangle+\langle u_\theta^2\rangle+\langle u_\phi^2\rangle$ , where $u_r$ , $u_\theta$ and $u_\phi$ are the radial velocity and the two perpendicular velocities on the plane of the sky of the members of the cluster.

(6 points)

Assuming that $\langle u_r^2\rangle\sim\langle u_\theta^2\rangle\sim\langle u_\phi^2\rangle$ , we have

-3m\langle u_r^2\rangle=U/N=\left(-\frac{3}{5}\frac{GM^2}{R}\right)/N=-\frac{3}{5}\frac{GMm}{R}\ \Longrightarrow\ M\approx\frac{5R\langle u_r^2\rangle}{G}

(where we used the gravitational potential of a spherical homogeneous mass M enclosed within radius R)

(13 points)

Alternatively the student can give a rougher order of magnitude estimate

-2\langle K\rangle=\langle U\rangle\Longrightarrow -2\cdot\frac{1}{2}\langle u_r^2\rangle\approx-\frac{GM}{R}\quad\text{etc}

If they do not use the exact formula:

(11 Points)

(2) From the result of the previous question we have $M\approx\frac{5R\langle u_r^2\rangle}{G}$ , where

\sqrt{\langle u_r^2\rangle}=\sigma=1000\ km\ s^{-1}

(8 points)

The angular diameter of the cluster is $\varphi=4^\circ$ at a distance of $d=90\ \mathrm{Mpc}$ . Therefore the diameter D of the cluster in Mpc is calculated from:

\tan\varphi=\frac{D}{d}\Longrightarrow\varphi(\mathrm{rad})\approx\frac{D}{d}\Longrightarrow D\approx4\times\frac{\pi}{180}\times90\,\mathrm{Mpc}\approx6.3\,\mathrm{Mpc}\Longrightarrow R\approx3\,\mathrm{Mpc}

(11 points)

Therefore $M\approx\frac{5R\langle u_r^2\rangle}{G}=3.6\times10^{15}\,M_\odot$

(3 point)

(3) $\displaystyle\frac{M_{virial}}{L_{galaxies}}=\frac{3.6\times10^{15}M_\odot}{5\times10^{12}L_\odot}=720\frac{M_\odot}{L_\odot}$ . This is obviously much larger from $\displaystyle\frac{M_\odot}{L_\odot}=1$ which is found from

the visible mass of the cluster. Thus $\displaystyle\frac{M}{M_{virial}}=\frac{1}{720}$ .

(6 points)

Теми и трудност

Трудност: неоценена.

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