/Astro540-HW

Astro540 Homework

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Astro540 Homework 1

Problem 1

  • total proper motion $\mu$

$$ \mu = \sqrt{\mu_\delta^2 + \cos^2\delta\mu_\alpha^2} $$

  • proper motion velocity component

$$ v_\tau = \mu d = \frac{1}{0.376} \mathrm{pc} \cdot 1.21'' \mathrm{a^{-1}} = 15.25 \mathrm{km\cdot s^{-1}} $$

  • total velocity

$$ v = \sqrt{v_tau^2 + v_r ^2} = \sqrt{7.6^2 + 15.25^2}\mathrm{km\cdot s^{-1}}=17.04 \mathrm{km\cdot s^{-1}} $$

  • when teh object is closest to sun, the direction of its motion is perpendicular to the line connected it and the sun. Thus the distance is caculated as

$$ d_0 = d \cdot \sin\theta = \frac{1}{0.376} \mathrm{pc} \cdot \frac{v_\tau^2}{v} = 2.38 \mathrm{pc} $$

  • the proper motion

$$ \mu = \frac{v}{d_0}1.51''\mathrm{a^{-1}} $$

Problem 2

thanks to the astropy coordinate module help to me do the coordinate conversion.

  • velocity for solar motion relative to CMB:

$$ v_{\odot-\mathrm{CMB}} = c \cdot \frac{\Delta T}{T} = 369.2 \mathrm{km\cdot s^{-1}} $$

to the direction

\begin{align*} &\alpha = 167.25^{\circ}\ &\delta = -7^{\circ} \end{align*}

  • Since

$$ \vec{v}{\odot - \mathrm{CMB}} = \vec{v}{\odot - \mathrm{MW}} + \vec{v}_{\mathrm{MW - CMB}} $$

we got

$$ \vec{v}{\mathrm{MW - CMB}} = \vec{v}{\odot - \mathrm{CMB}} - \vec{v}_{\odot - \mathrm{MW}} $$

  • in Celestial coordinate convert two motion in Cartesian representation

$$ \vec{v}{\odot - \mathrm{CMB}} = -357.44 \vec{i} + 80.88 \vec{j} + (-45.00) \vec{k} $$ $$ \vec{v}{\odot - \mathrm{MW}} = 108.70 \vec{i} + (-97.862) \vec{j} + 119.34 \vec{k} $$

  • Thus in Celestial coordinate,

$$ \vec{v}_{\mathrm{MW - CMB}} = 159.02 \vec{i} + (-22.74) \vec{j} + 541.35{k} $$

  • in spherical representation, the direction

\begin{align*} & ra = 159.02 ^{\circ}\ & dec = -22.74^{\circ} \end{align*}

with a velocity 541.3 $\mathrm{km \cdot s^{-1}}$

  • in galactic coordinate

\begin{align*} &w=-29.26\ &u=-466.20\ &v=273.60 \end{align*}

Problem 3

$$ M_{B, disk} = M_\odot - 2.5\log(1.2\times 10^{10}) =-19.72 $$ $$ M_{B, bulge} = M_\odot - 2.5\log(1.9\times 10^{9})=-17.71\\ $$

  • Luminosity density:

$$ l = l_0 \cdot \exp(-R_0/R_d) = 1280 \cdot L_\odot \mathrm{pc^{-2}}\exp(-8 / 2.7) = 62.4 L_\odot \mathrm{pc}^{-2} $$

  • Since

$$ 3.8\mathrm{Mag arcsec^{-2}} \sim 1L_\odot /(10 AU)^{-2} = 4.25\times 10 ^{8} L_\odot \mathrm{pc^{-2}} $$

$$ \mu_k - 3.8 = -2.5\log(62.4/(4.25\times 10^8)) $$ $$ \mu_k = 20.9 \mathrm{mag\cdot arcsec^{-2}} $$

Problem 4

a)

$$ \frac{f_0}{f} = 10^{-(M_V -m_v)/2.5} = 3.63\times 10^7 $$

$$ d = 10\mathrm{pc} \cdot \sqrt{\frac{f_0}{f}} = 60 \mathrm{kpc} $$

b)

  • number density of disk:

$$ n_{\mathrm{disk}} = n_0 \exp(-R/R_0) $$

  • number density of halo star:

$$ n_{\mathrm{halo}} = 0.1 n_0 \exp(-R/R_0) $$

  • At far distance

$$ \frac{n_\mathrm{disk}}{n_\mathrm{disk}}\Bigg|_{R = 60\mathrm{kpc}} = 0.1\cdot \frac{(60/3)^{-3.5}}{\exp(-60/3)} = 1356 $$

$$ P_{\mathrm{halo}} = 0.999 $$

  • At the solar neighborhood

$$ \frac{n_\mathrm{disk}}{n_\mathrm{disk}}\Bigg|_{R = 8\mathrm{kpc}} = 0.1\cdot \frac{(8/3)^{-3.5}}{\exp(-8/3)} = 0.0002 $$

$$ P_{\mathrm{halo}} = 0.0002 $$

c

$$ A_v = 3.3 E(B-V) = 1.65 \frac{d}{\mathrm{kpc}} $$

For a $M_v = 5.1$ solar type star at the distance $d$ kpc, the magnitude changes to

\begin{align*} m_v &=M_v - 2.5\log\left[\frac{(0.01\mathrm{kpc})^2}{d^2}\right]\ &=5.1 - 2.5\log\frac{0.0001}{d^2} + 1.65d \end{align*}

when $m_v = 24.0$, $d = 3.68$ kpc.