#LyX 2.0.0beta3 created this file. For more info see http://www.lyx.org/
\lyxformat 410
\begin_document
\begin_header
\textclass article
\begin_preamble
\newcommand{\bb}{\\}
\end_preamble
\use_default_options true
\maintain_unincluded_children false
\language catalan
\inputencoding auto
\fontencoding global
\font_roman default
\font_sans default
\font_typewriter default
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100
\font_tt_scale 100

\graphics default
\default_output_format default
\output_sync 0
\bibtex_command default
\index_command default
\paperfontsize default
\spacing single
\use_hyperref true
\pdf_bookmarks true
\pdf_bookmarksnumbered false
\pdf_bookmarksopen false
\pdf_bookmarksopenlevel 1
\pdf_breaklinks false
\pdf_pdfborder false
\pdf_colorlinks false
\pdf_backref false
\pdf_pdfusetitle true
\papersize default
\use_geometry true
\use_amsmath 1
\use_esint 1
\use_mhchem 1
\use_mathdots 1
\cite_engine basic
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\use_refstyle 0
\index Índex
\shortcut idx
\color #008000
\end_index
\leftmargin 2cm
\topmargin 2cm
\rightmargin 2cm
\bottommargin 3cm
\headheight 2cm
\headsep 2cm
\footskip 2cm
\secnumdepth 3
\tocdepth 3
\paragraph_separation skip
\defskip smallskip
\quotes_language english
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
\html_math_output 0
\html_be_strict false
\end_header

\begin_body

\begin_layout Title
Solucions als problemes d'Astronomia
\end_layout

\begin_layout Author
Sergi Blanco Cuaresma
\end_layout

\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents

\end_inset


\end_layout

\begin_layout Section
Astronomia de posició
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
\begin_inset CommandInset label
LatexCommand label
name "Recordatori:1.1"

\end_inset

Recordatori
\end_layout

\begin_layout Plain Layout
Transformació coordenades Equatorials - Horitzontals
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\cos\left(\delta\right)\sin\left(H\right)=\cos\left(h\right)\sin\left(\alpha\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\cos\left(h\right)\cos\left(\alpha\right)=\sin\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)-\sin\left(\delta\right)\cos\left(\phi\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Angle horari / Temps sideri:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\theta=\theta_{0}+1.00273791\cdot t+\lambda$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\theta=H+\alpha$
\end_inset


\end_layout

\begin_layout Plain Layout
Transformació coordenades Horitzontals - Equatorials
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\sin\left(\delta\right)=\sin\left(\phi\right)\sin\left(h\right)-\cos\left(\phi\right)\cos\left(h\right)\cos\left(a\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\cos\left(\delta\right)\sin\left(H\right)=\cos\left(h\right)\sin\left(\alpha\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\cos\left(\delta\right)\cos\left(H\right)=\sin\left(\phi\right)\cos\left(h\right)\cos\left(a\right)+\sin\left(h\right)\cos\left(\phi\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
A on:
\end_layout

\begin_layout Description
\begin_inset Formula $t$
\end_inset

 Temps universal d'observació
\end_layout

\begin_layout Description
\begin_inset Formula $\theta_{0}$
\end_inset

 Temps sideri del punt àries (
\begin_inset Formula $\Upsilon$
\end_inset

) a Greenwich a les 0:00 de temps universal
\end_layout

\begin_layout Description
\begin_inset Formula $\theta$
\end_inset

 Temps sideri local del punt àries (
\begin_inset Formula $\Upsilon$
\end_inset

)
\end_layout

\begin_layout Description
\begin_inset Formula $h,a$
\end_inset

 Coordenades horitzontals
\end_layout

\begin_layout Description
\begin_inset Formula $\delta,H$
\end_inset

 Coordenades horàries
\end_layout

\begin_layout Description
\begin_inset Formula $\delta,\alpha$
\end_inset

 Coordenades equatorials
\end_layout

\begin_layout Description
\begin_inset Formula $\phi,\lambda$
\end_inset

 Coordenades de l'observador
\end_layout

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="2">
<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $h$
\end_inset

 Alçada [
\begin_inset Formula $-90^{o}$
\end_inset

,
\begin_inset Formula $+90^{o}$
\end_inset

]
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $a$
\end_inset

 Azimut [
\begin_inset Formula $0^{o}$
\end_inset

,
\begin_inset Formula $360^{o}$
\end_inset

]
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\delta$
\end_inset

 Declinació [
\begin_inset Formula $-90^{o}$
\end_inset

,
\begin_inset Formula $+90^{o}$
\end_inset

]
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H$
\end_inset

 Angle horari [
\begin_inset Formula $0^{H}$
\end_inset

,
\begin_inset Formula $24^{H}$
\end_inset

]
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\alpha$
\end_inset

 Ascensió recta [
\begin_inset Formula $0^{H}$
\end_inset

,
\begin_inset Formula $24^{H}$
\end_inset

]
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\phi$
\end_inset

 Latitud [
\begin_inset Formula $-90^{o}$
\end_inset

,
\begin_inset Formula $+90^{o}$
\end_inset

]
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda$
\end_inset

 Longitud [
\begin_inset Formula $0^{H}$
\end_inset

,
\begin_inset Formula $24^{H}$
\end_inset

]
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Graphics
	filename Crepuscle-Twilight.gif
	scale 50

\end_inset


\end_layout

\begin_layout Standard
El radi angular del sol des de la Terra és de 16' (
\begin_inset Formula $R_{\odot}=16^{m}$
\end_inset

).
 Per tant, la definició més minimalista de crepuscle podria ser el temps
 que triga el sol en ocultarse (
\begin_inset Formula $dh_{\odot}=32^{m}$
\end_inset

).
 Com estem tractant amb alçades molt petites 
\begin_inset Formula $h\sim0$
\end_inset

, es podran fer les següents aproximacions: 
\begin_inset Formula $\sin\left(h\right)\sim0$
\end_inset

 i 
\begin_inset Formula $\cos\left(h\right)\sim1$
\end_inset

.
\end_layout

\begin_layout Standard
La clau d'aquest problema es entendre que la diferència en angle horari
 
\begin_inset Formula $\Delta H$
\end_inset

 correspon exactament al temps que triga el moviment.
\end_layout

\begin_layout Standard
A partir de l'equació (
\begin_inset Formula $\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)$
\end_inset

) aïllem i aproximem:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(H\right)=\frac{\sin\left(h\right)-\sin\left(\phi\right)\sin\left(\delta\right)}{\cos\left(\phi\right)\cos\left(\delta\right)}\simeq\frac{\sin\left(\phi\right)\sin\left(\delta\right)}{\cos\left(\phi\right)\cos\left(\delta\right)}=-\tan\left(\phi\right)\tan\left(\delta\right)\label{eq:cos_H_derivada}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
L'aproximació només es vàlida si 
\begin_inset Formula $h\sim0$
\end_inset

 (en aquest cas el centre del sol estarà a 16') i per tant podem dir que
 
\begin_inset Formula $\sin\left(h\right)\sim0$
\end_inset

.
 A partir d'aquesta equació es pot derivar (elevem tot al quadrat, i utilitzem
 
\begin_inset Formula $\cos^{2}\left(H\right)=1-\sin^{2}\left(H\right)$
\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\sin\left(H\right)=\pm\sqrt{1-\tan^{2}\left(\phi\right)\tan^{2}\left(\delta\right)}\label{eq:sin_H_derivada}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El signe positiu correspon a la posta de sol, mentre que el negatiu a la
 sortida.
 
\end_layout

\begin_layout Standard
Per altra banda, a partir de l'equació (
\begin_inset Formula $\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)$
\end_inset

) podem derivar:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{d}{dh}\sin\left(h\right)=\cos\left(h\right)\Rightarrow\sin\left(h\right)=\cos\left(h\right)dh
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{d}{dH}\left(\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)\right)=-\cos\left(\phi\right)\cos\left(\delta\right)\sin\left(H\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(h\right)dh=-\cos\left(\phi\right)\cos\left(\delta\right)\sin\left(H\right)dH\label{eq:cos_hdh}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
A on 
\begin_inset Formula $dh$
\end_inset

 és la diferencia en alçada del sol i 
\begin_inset Formula $dH$
\end_inset

 és la diferencia en angle horari, que equival al temps que trigarà en recorre
 
\begin_inset Formula $dh$
\end_inset

.
 A continuació, substituïm l'equació (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:sin_H_derivada"

\end_inset

) a (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:cos_hdh"

\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(h\right)dh=\pm\cos\left(\phi\right)\cos\left(\delta\right)\sqrt{1-\tan^{2}\left(\phi\right)\tan^{2}\left(\delta\right)}dH
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Aïllem el temps que trigarà:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
dH=\pm\frac{\cos\left(h\right)dh}{\cos\left(\phi\right)\cos\left(\delta\right)\sqrt{1-\tan^{2}\left(\phi\right)\tan^{2}\left(\delta\right)}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Simplifiquem:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\left|dH\right|=\frac{\left|dh\right|}{\sqrt{\cos^{2}\left(\phi\right)\cos^{2}\left(\delta\right)-\sin^{2}\left(\phi\right)\sin^{2}\left(\delta\right)}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Si avaluem l'equació a l'equador 
\begin_inset Formula $\phi=0$
\end_inset

, llavors 
\begin_inset Formula $\cos\left(\phi\right)=1$
\end_inset

, 
\begin_inset Formula $\sin\left(\phi\right)=0$
\end_inset

 i per tant la durada només dependrà de la declinació del sol (
\begin_inset Formula $\delta$
\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\left|dH\right|=\frac{\left|dh\right|}{\cos\left(\delta\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Quan la declinació és zero (
\begin_inset Formula $\delta=0\Rightarrow\cos\left(\delta\right)=1$
\end_inset

) correspon a l'equinox i la posta de sol serà la més curta.
 En canvi a la declinació màxima/mínima estem al solstici d'estiu/tardor
 (
\begin_inset Formula $\delta=\pm23.45\Rightarrow\left|\cos\left(\delta\right)\right|<1$
\end_inset

) i el temps de la posta de sol serà més llarga.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
\begin_inset CommandInset label
LatexCommand label
name "Recordatori:1.2"

\end_inset

Recordatori
\end_layout

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Graphics
	filename Esfera celest.png
	lyxscale 75
	scale 25

\end_inset


\end_layout

\begin_layout Description
Zona
\begin_inset space ~
\end_inset

tropical: lloc del planeta a on el sol passa pel zenit com a mínim un cop
 a l'any.
\end_layout

\begin_layout Description
Zona
\begin_inset space ~
\end_inset

polar: lloc del planeta a on al menys un cop l'any el sol té com alçada
 màxima l'horitzó
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Per a que el sol pugui ser-hi al zenit al menys un cop a l'any, la banda
 del sol ha de coincidir amb el punt marcat com a zenit al dibuix anterior.
 Això implica que l'equador celest no pot estar més lluny de la obliqüitat
 eclíptica 
\begin_inset Formula $\varepsilon$
\end_inset

 del zenit del lloc, es a dir (per equivalència d'angles), la latitud ha
 de ser igual o inferior a 
\begin_inset Formula $\varepsilon$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\left|\phi\right|\leq\varepsilon
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per altra banda, per a que el sol estigui sobre l'horitzó al menys un cop
 l'any, la obliqüitat eclíptica ha de ser com a mínim igual a 90 graus menys
 la latitud de l'observador tal i com es pot visualitzar al dibuix anterior:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
90-\phi\leq\varepsilon\Rightarrow\phi\geq90-\varepsilon
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Si combinem els dos límits:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
90-\varepsilon\leq\left|\phi\right|\leq\varepsilon
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, com a mínim és necessari que l'obliqüitat de l'eclíptica sigui:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
90-\varepsilon\leq\varepsilon\Rightarrow\varepsilon=45
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Cal destacar que si fos a l'hemisferi sud del planeta, el limit variaria:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\phi\leq-\left(90-\varepsilon\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Veure recordatori del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.2"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
A partir del dibuix del 1.2 i sabent que el pol nord galàctic (PNG) es troba
 a:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left(\alpha_{G},\delta_{G}\right)=\left(12^{H}51^{m}26^{s},27^{o}.4\right)
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Horitzó
\end_layout

\begin_layout Standard
Aliniem l'equador galàctic amb l'horitzó,es a dir, fem coincidir el pol
 galàctic amb el zenit del lloc i obtenim que el pla equatorial està a 27º.4
 del zenit i per tant l'eix de rotació respecte a l'horitzó (que equival
 a la declinació del lloc) està a:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\phi=90-(90-27.4)=90-62.6=27.4
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Circumpolars
\end_layout

\begin_layout Standard
Les estrelles circumpolars són aquelles que es troben sempre sobre l'horitzó.
 Sabem que des de l'eix de rotació fins al pol nord galàctic hi ha:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
90-\left(distancia\, PNG\, al\, pla\, equador\, celest\right)=90-\delta_{G}=90-27^{o}.4=62^{o}.6
\]

\end_inset


\end_layout

\begin_layout Standard
Per tant, s'ha de trobar un lloc a on siguin visibles els objectes que estiguin
 a 
\begin_inset Formula $\pm62^{o}.4$
\end_inset

 de l'eix de rotació (que equival a la declinació del lloc), això implica
 que:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\phi\geq62^{0}.4
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
1er i 2on grups de Bessel
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\begin{array}{ccc}
\sin\left(a\right)\sin\left(B\right)=\sin\left(b\right)\sin\left(A\right) & | & \cos\left(a\right)=\cos\left(b\right)\cos\left(c\right)+\sin\left(b\right)\sin\left(c\right)\cos\left(A\right)\\
\sin\left(a\right)\sin\left(C\right)=\sin\left(c\right)\sin\left(A\right) & | & \cos\left(b\right)=\cos\left(a\right)\cos\left(c\right)+\sin\left(a\right)\sin\left(c\right)\cos\left(B\right)\\
\sin\left(b\right)\sin\left(C\right)=\sin\left(c\right)\sin\left(B\right) & | & \cos\left(c\right)=\cos\left(a\right)\cos\left(b\right)+\sin\left(a\right)\sin\left(b\right)\cos\left(C\right)
\end{array}$
\end_inset


\end_layout

\begin_layout Plain Layout
3er grup de Bessel
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\begin{array}{c}
\sin\left(a\right)\cos\left(B\right)=\cos\left(b\right)\sin\left(c\right)-\sin\left(b\right)\cos\left(c\right)\cos\left(A\right)\\
\sin\left(a\right)\cos\left(C\right)=\cos\left(c\right)\sin\left(b\right)-\sin\left(c\right)\cos\left(b\right)\cos\left(A\right)\\
\sin\left(b\right)\cos\left(A\right)=\cos\left(a\right)\sin\left(c\right)-\sin\left(a\right)\cos\left(c\right)\cos\left(B\right)\\
\sin\left(b\right)\cos\left(C\right)=\cos\left(c\right)\sin\left(a\right)-\sin\left(c\right)\cos\left(a\right)\cos\left(B\right)\\
\sin\left(c\right)\cos\left(A\right)=\cos\left(a\right)\sin\left(b\right)-\sin\left(a\right)\cos\left(b\right)\cos\left(C\right)\\
\sin\left(c\right)\cos\left(B\right)=\cos\left(b\right)\sin\left(a\right)-\sin\left(b\right)\cos\left(a\right)\cos\left(C\right)
\end{array}$
\end_inset


\end_layout

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Graphics
	filename Spherical_triangle_3d-Bessel.png
	lyxscale 50
	scale 30

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Els angles coneguts:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
a=47^{g}12^{\text{m}}39,^{s}9=47^{g}+12^{m}\frac{1^{g}}{60^{m}}+39,^{s}9\frac{1^{m}}{60^{s}}\frac{1^{g}}{60^{m}}=47,^{g}211\frac{\pi rad}{180^{g}}=0.82399rad
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
b=61^{g}46^{m}9,^{s}1=61^{g}+46^{m}\frac{1^{g}}{60^{m}}+9,^{s}1\frac{1^{m}}{60^{s}}\frac{1^{g}}{60^{m}}=61,^{g}079\frac{\pi rad}{180^{g}}=1.066rad
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
c=43^{g}59^{m}17,^{s}6=43^{g}+59^{m}\frac{1^{g}}{60^{m}}+17,^{s}69\frac{1^{m}}{60^{s}}\frac{1^{g}}{60^{m}}=43,^{g}005\frac{\pi rad}{180^{g}}=0.75058rad
\]

\end_inset


\end_layout

\begin_layout Standard
Per tant, aplicant les fórmules de Bessel:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
A=\arccos\left(\frac{\cos\left(a\right)-\cos\left(b\right)\cos\left(c\right)}{\sin\left(b\right)\sin\left(c\right)}\right)=0,99385rad\frac{180^{g}}{\pi rad}=56,^{g}943
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
B=\arccos\left(\frac{\cos\left(b\right)-\cos\left(a\right)\cos\left(c\right)}{\sin\left(a\right)\sin\left(c\right)}\right)=1.5971rad\frac{180^{g}}{\pi rad}=91,^{g}507
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
C=\arccos\left(\frac{\cos\left(c\right)-\cos\left(a\right)\cos\left(b\right)}{\sin\left(a\right)\sin\left(b\right)}\right)=0.89303rad\frac{180^{g}}{\pi rad}=51,^{g}167
\]

\end_inset


\end_layout

\begin_layout Standard
Per a la resta de triangles es segueix una estratègia similar per obtenir
 les incògnites utilitzant les equacions de Bessel.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
A la posta de sol, el punt més baix del sol es troba a una alçada aparent
 de zero graus (
\begin_inset Formula $h_{aparent}=0$
\end_inset

) i per tant a una distància zenital de 90 (
\begin_inset Formula $z_{mesurat}=90^{g}$
\end_inset

).
 Segons la taula, per aquesta distància s'ha d'aplicar un factor correctiu
 
\begin_inset Formula $R=2123^{s}=35^{m}23^{s}$
\end_inset

.
 Per tant, l'alçada real:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
z_{real}=z_{mesurat}+R=90^{g}35^{m}23^{s}\Rightarrow h_{real}=90^{g}-z_{real}=-35^{m}23^{s}
\]

\end_inset


\end_layout

\begin_layout Standard
Com que el diàmetre del sol es de 32 minuts, es troba totalment sota l'horitzó.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Veure recordatori del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.1"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Per tal de considerar la sortida o posta del sol volem que el sol es trobi
 a alçada zero (
\begin_inset Formula $h=0$
\end_inset

), per tant 
\begin_inset Formula $\cos\left(h\right)=1$
\end_inset

 i 
\begin_inset Formula $\sin\left(h\right)=0$
\end_inset

 i en conseqüència podem trobar equacions per l'azimut i l'angle horari
 que depenen de la declinació i latitud del lloc:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(\delta\right)=\sin\left(\phi\right)\sin\left(h\right)-\cos\left(\phi\right)\cos\left(h\right)\cos\left(a\right)\Rightarrow\sin\left(\delta\right)=-\cos\left(\phi\right)\cos\left(a\right)\Rightarrow\cos\left(a\right)=-\frac{\sin\left(\delta\right)}{\cos\left(\phi\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)\Rightarrow\cos\left(H\right)=\frac{-\sin\left(\phi\right)\sin\left(\delta\right)}{\cos\left(\phi\right)\cos\left(\delta\right)}\Rightarrow\cos\left(H\right)=-\tan\left(\phi\right)\tan\left(\delta\right)
\]

\end_inset


\end_layout

\begin_layout Standard
La declinació del sol al solsticis és 
\begin_inset Formula $\delta=0$
\end_inset

 i als equinox 
\begin_inset Formula $\delta=\varepsilon=\pm23,^{g}44$
\end_inset

 i la latitud del lloc s'indica al problema 
\begin_inset Formula $\phi=41,^{g}4$
\end_inset

.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Veure recordatori del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.1"

\end_inset

 i del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.2"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Si volem considerar la diferencia en angle horari utilitzant la següent
 expressió:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
H+\alpha=\theta=\theta_{0}+1.00273791\cdot t+\lambda
\]

\end_inset


\end_layout

\begin_layout Standard
Podem observar que durant 24 hores tots els termes seran constants excepte
 el temps, el qual volem que sigui de mínim 18 hores per sobre de l'horitzó:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\Delta H\geq1.00273791\cdot\Delta t=1.00273791\cdot18^{h}=18,^{h}049
\]

\end_inset


\end_layout

\begin_layout Standard
Per altra banda, l'angle horari té com a punt de referència (
\begin_inset Formula $H=0^{h}$
\end_inset

) l'equador celest a on talla amb el meridià del lloc (cercle que va de
 nord a sud i passa pel zenit).
 El temps que l'estrella estarà des de la sortida fins arribar a 
\begin_inset Formula $H=0^{h}$
\end_inset

 és igual al temps des de 
\begin_inset Formula $H=0^{h}$
\end_inset

 fins la posta del sol, per tant podem dir que l'alçada de l'estrella a
 la posta de sol es just la meitat del temps que la volem per sobre de l'horitzó
:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
H_{posta}\geq\frac{\Delta H}{2}=9,^{h}0246=9^{h}01^{m}29^{s}\frac{360^{g}}{24^{h}}=135^{g}22^{m}10^{s}\frac{\pi rad}{180^{g}}=2,3626rad
\]

\end_inset


\end_layout

\begin_layout Standard
A partir de les fórmules de transformació:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)\Rightarrow\cos\left(H\right)=\frac{-\sin\left(\phi\right)\sin\left(\delta\right)}{\cos\left(\phi\right)\cos\left(\delta\right)}\Rightarrow\cos\left(H\right)=-\tan\left(\phi\right)\tan\left(\delta\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Utilitzant l'hora de la posta i la declinació 
\begin_inset Formula $\delta=-30^{g}\frac{\pi rad}{180}=-0,52360rad$
\end_inset

, podem trobar la latitud del lloc:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\phi\leq\arctan\left(\frac{-\cos\left(H_{posta}\right)}{\tan\left(\delta\right)}\right)=0,889919rad=-50,9747^{g}
\]

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Veure recordatori del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.1"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Observador:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\left(\phi,\lambda\right)=\left(40^{o},3^{o}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Objecte:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\left(\delta,\alpha\right)=(-10^{o},5^{o})
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Moment: 25 de febrer de 1994 a les 22:00 (temps universal) amb temps sideri
 local (
\begin_inset Formula $\theta_{0}=10^{H}18^{m}31^{s}$
\end_inset

)
\end_layout

\begin_layout Standard
Transformen les coordenades equatorials en horitzontals:
\end_layout

\begin_layout Itemize
Angle horari
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
H+\alpha=\theta_{0}+1.0029377t+\lambda
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\alpha=5^{o}\frac{24^{H}}{360^{o}}=0^{H}.33
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\theta_{0}=10^{H}+18^{m}\frac{1^{H}}{60^{m}}+31\frac{1^{m}}{60^{s}}\frac{1^{H}}{60^{m}}=10^{H}.3086
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\lambda=3^{o}\frac{24^{H}}{360^{o}}=0^{H}.2
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
H=10^{H}.3086+1.0029377\left(22\right)+0^{H}.2-0^{H}.33=32^{H}.24
\end{equation}

\end_inset


\end_layout

\begin_layout Itemize
Alçada
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\phi=40^{o}\frac{\pi}{180^{0}}=0.70rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\delta=-10^{o}\frac{\pi}{180^{0}}=-0.17rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
H=32^{o}.24\frac{\pi}{180^{0}}=0.56rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
h=\sin^{-1}\left(\sin\left(0.70\right)\sin\left(-0.17\right)+\cos\left(0.70\right)\cos\left(-0.17\right)\cos\left(0.56\right)\right)=0,56rad\frac{180^{0}}{\pi}=31^{o}.94
\end{equation}

\end_inset


\end_layout

\begin_layout Itemize
Azimut
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\sin\left(a\right)=\frac{\cos\left(\delta\right)\sin\left(H\right)}{\cos\left(h\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
a=\sin^{-1}\left(\frac{\cos\left(-0.17\right)\sin\left(0.56\right)}{\cos\left(0.56\right)}\right)=\sin^{-1}\left(0.62\right)=0.66rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(a\right)=\frac{\sin\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)-\sin\left(\delta\right)\cos\left(\phi\right)}{\cos\left(h\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El sinus es positiu (0.62) per tant, l'angle pot ser-hi als quadrants superiors.
 Calculem l'alçada també amb l'altra fórmula del cosinus:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
a=\cos^{-1}\left(\frac{\sin\left(0.70\right)\cos\left(-0.17\right)\cos\left(0.56\right)-\sin\left(-0.17\right)\cos\left(0.70\right)}{\cos\left(0.56\right)}\right)=\cos^{-1}\left(0.79\right)=0.66rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El cosinus també és positiu (0.79) per tant l'angle es troba al primer quadrant
 i no hem de fer cap ajustament:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
a=0.66rad\frac{180^{0}}{\pi}=38^{o}.11
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
A continuació, passem a avaluar si l'objecte serà visible.
 L'obstacle es troba a 3 metres i té 3 metres d'alçada, si miréssim en línia
 perpendicular a la paret el punt més alt es troba a un angle:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\tan\left(\beta\right)=\frac{alçada}{distancia}=\frac{3}{3}\label{eq:tan_angle}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\beta=\tan^{-1}\left(1\right)=0.785\, rad=44.94^{o}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Però com l'estrella no es troba en línia recta cap el sud (
\begin_inset Formula $a=38^{o}.11$
\end_inset

), llavors l'obstacle s'interposa a una distància diferent.
 Com la paret esta orientada al sud, l'angle en el que mirem correspondrà
 a l'azimut de l'estrella (
\begin_inset Formula $a$
\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(a\right)=\frac{distancia\, perpendicular}{distancia\, real}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
distancia\, real=\frac{dist\, perpendicular}{\cos\left(a\right)}=\frac{3}{\cos\left(0.66\right)}=3.80m
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, el punt més alt de la paret en aquesta direcció es troba en un
 angle de:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\beta=\tan^{-1}\left(3.80\right)=1.31\, rad=75^{o}.15
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per a que l'objecte sigui visible, la seva alçada ha de ser superior a l'angle
 que representa la paret 
\begin_inset Formula $a>\beta$
\end_inset

 i en aquest cas no es compleix.
 No és possible veure l'estrella.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Veure recordatori del problema 
\begin_inset CommandInset ref
LatexCommand vref
reference "Recordatori:1.1"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Les ascensions rectes i declinacions de Sírius i Arcturus:
\end_layout

\begin_layout Standard
\begin_inset Formula $\alpha_{sirius}=6^{h}34^{m}9^{s}=6,^{h}5692\frac{360^{g}}{24^{h}}\frac{\pi rad}{180^{g}}=1,7198rad$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta_{sirius}=-16^{g}42^{m}58^{s}=-16,^{g}716\frac{\pi rad}{180^{g}}=-0,29175rad$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\alpha_{arcturus}=14^{h}15^{m}40^{s}=14,^{h}261\frac{360^{g}}{24^{h}}\frac{\pi rad}{180^{g}}=3,7335rad$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta_{arcturus}=19^{g}10^{m}57^{s}=19,^{g}183\frac{\pi rad}{180^{g}}=0,33481rad$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\phi_{bcn}=41,^{g}4\frac{\pi rad}{180^{g}}=0,72257rad$
\end_inset


\end_layout

\begin_layout Standard
Quan Sírius es posa, la seva alçada és zero (
\begin_inset Formula $h=0$
\end_inset

) i per tant 
\begin_inset Formula $\cos\left(h\right)=1$
\end_inset

 i 
\begin_inset Formula $\sin\left(h\right)=0$
\end_inset

.
 En conseqüència:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(h\right)=\sin\left(\phi\right)\sin\left(\delta\right)+\cos\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)\Rightarrow\cos\left(H\right)=\frac{-\sin\left(\phi\right)\sin\left(\delta\right)}{\cos\left(\phi\right)\cos\left(\delta\right)}\Rightarrow\cos\left(H\right)=-\tan\left(\phi\right)\tan\left(\delta\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Per tant, podem obtenir l'angle horari:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
H_{sirius}=\arccos\left(-\tan\left(\phi_{bcn}\right)\tan\left(\delta_{sirius}\right)\right)=1,3028rad
\]

\end_inset


\end_layout

\begin_layout Standard
I el temps sideri local:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\theta=H_{sirius}+\alpha_{sirius}=1,3028+1,7198=3,0226
\]

\end_inset


\end_layout

\begin_layout Standard
Amb aquest, podem realitzar la conversió de coordenades equatorials a horitzonta
ls de Arcturus:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
H_{arcturus}=\theta-\alpha_{arcturus}=3,0226-3,7335=-0,71087rad
\]

\end_inset


\end_layout

\begin_layout Standard
Calculem l'alçada:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(h_{arcturus}\right)=\sin\left(\phi_{bcn}\right)\sin\left(\delta_{arcturus}\right)+\cos\left(\phi_{bcn}\right)\cos\left(\delta_{arcturus}\right)\cos\left(H_{arcturus}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
h_{arcturus}=0,85438rad\frac{180^{0}}{\pi}=48{}^{o}.953
\]

\end_inset


\end_layout

\begin_layout Standard
Calculem l'azimut:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(a_{arcturus}\right)=\frac{\cos\left(\delta_{arcturus}\right)\sin\left(H_{arcturus}\right)}{\cos\left(h_{arcturus}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
a_{arcturus}=\arcsin\left(-0,93844\right)=-1,2181rad
\]

\end_inset


\end_layout

\begin_layout Standard
El sinus es negatiu (-0,938) per tant, l'angle pot ser-hi als quadrants
 inferiors.
 Calculem l'alçada també amb l'altra fórmula del cosinus:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\cos\left(a\right)=\frac{\sin\left(\phi\right)\cos\left(\delta\right)\cos\left(H\right)-\sin\left(\delta\right)\cos\left(\phi\right)}{\cos\left(h\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
a=\cos^{-1}\left(0,34542\right)=\cos^{-1}\left(0.79\right)=1,2181rad
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El cosinus és positiu (0.79) per tant l'angle es troba al quadrant de la
 dreta i no hem de fer cap ajustament:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
a_{arcturus}=-1,2181rad\frac{180^{o}}{\pi rad}=-69,^{g}792
\]

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Transformació de coordenades horitzontals a horàries:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula \[
\overrightarrow{r_{horàries}}=R_{1}\left(\varepsilon\right)\cdot\overrightarrow{r_{horitzontals}}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula \[
\left(\begin{array}{c}
\cos\left(\delta\right)\cos\left(H\right)\\
\cos\left(\delta\right)\sin\left(H\right)\\
\sin\left(\delta\right)
\end{array}\right)=\left(\begin{array}{ccc}
\sin\left(\phi\right) & 0 & \cos\left(\phi\right)\\
0 & 1 & 0\\
-\cos\left(\phi\right) & 0 & \sin\left(\phi\right)
\end{array}\right)\left(\begin{array}{c}
\cos\left(h\right)\cos\left(a\right)\\
\cos\left(h\right)\sin\left(a\right)\\
\sin\left(h\right)
\end{array}\right)
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Per convertir coordenades equatorials a eclíptiques:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\overrightarrow{r_{ecliptiques}}=R_{1}\left(\varepsilon\right)\cdot\overrightarrow{r_{equatorials}}
\]

\end_inset


\end_layout

\begin_layout Standard
Considerem 
\begin_inset Formula $\lambda$
\end_inset

 com la distància des del punt àries i 
\begin_inset Formula $\beta$
\end_inset

 com l'alçada des de la eclíptica:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left(\begin{array}{c}
\cos\left(\beta\right)\cos\left(\lambda\right)\\
\cos\left(\beta\right)\sin\left(\lambda\right)\\
\sin\left(\beta\right)
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & \cos\left(\varepsilon\right) & \sin\left(\varepsilon\right)\\
0 & -\sin\left(\varepsilon\right) & \cos\left(\varepsilon\right)
\end{array}\right)\left(\begin{array}{c}
\cos\left(\delta\right)\cos\left(\alpha\right)\\
\cos\left(\delta\right)\sin\left(\alpha\right)\\
\sin\left(\delta\right)
\end{array}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Considerem que la declinació del zenit és la latitud del lloc (
\begin_inset Formula $\delta_{z}=\phi$
\end_inset

 ) i l'ascensió recta del zenit és l'hora sideri local (
\begin_inset Formula $\alpha_{z}=\theta$
\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left(\begin{array}{c}
\cos\left(\beta_{z}\right)\cos\left(\lambda_{z}\right)\\
\cos\left(\beta_{z}\right)\sin\left(\lambda_{z}\right)\\
\sin\left(\beta_{z}\right)
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & \cos\left(\varepsilon\right) & \sin\left(\varepsilon\right)\\
0 & -\sin\left(\varepsilon\right) & \cos\left(\varepsilon\right)
\end{array}\right)\left(\begin{array}{c}
\cos\left(\phi\right)\cos\left(\theta\right)\\
\cos\left(\phi\right)\sin\left(\theta\right)\\
\sin\left(\phi\right)
\end{array}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
A partir de la última línia obtenim la relació entre 
\begin_inset Formula $\beta_{z}$
\end_inset

 i la latitud - temps sideri local - obliqüitat de l'eclíptica:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sin\left(\beta_{z}\right)=0\cdot\left(\cos\left(\phi\right)\cos\left(\theta\right)\right)+\left(-\sin\left(\varepsilon\right)\right)\cdot\left(\cos\left(\phi\right)\sin\left(\theta\right)\right)+\cos\left(\varepsilon\right)\cdot\sin\left(\phi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Per altra banda, si desenvolupem la primera línia:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\cos\left(\beta_{z}\right)\cos\left(\lambda_{z}\right)=\cos\left(\phi\right)\cos\left(\theta\right)\Rightarrow\cos\left(\beta_{z}\right)=\frac{\cos\left(\phi\right)\cos\left(\theta\right)}{\cos\left(\lambda_{z}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
Per combinar-la amb la segona:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\cos\left(\beta_{z}\right)\sin\left(\lambda_{z}\right)=0\cdot\left(\cos\left(\phi\right)\cos\left(\theta\right)\right)+\cos\left(\varepsilon\right)\cdot\left(\cos\left(\phi\right)\sin\left(\theta\right)\right)+\sin\left(\varepsilon\right)\cdot\sin\left(\phi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
\frac{\cos\left(\phi\right)\cos\left(\theta\right)}{\cos\left(\lambda_{z}\right)}\sin\left(\lambda_{z}\right)=\cos\left(\varepsilon\right)\cdot\cos\left(\phi\right)\sin\left(\theta\right)+\sin\left(\varepsilon\right)\cdot\sin\left(\phi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Obtenim la relació entre 
\begin_inset Formula $\lambda_{z}$
\end_inset

 i la latitud - temps sideri local - obliqüitat de l'eclíptica:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\tan\left(\lambda_{z}\right)=\frac{\cos\left(\varepsilon\right)\cdot\cos\left(\phi\right)\sin\left(\theta\right)+\sin\left(\varepsilon\right)\cdot\sin\left(\phi\right)}{\cos\left(\phi\right)\cos\left(\theta\right)}=\cos\left(\varepsilon\right)\tan\left(\theta\right)+\frac{\sin\left(\varepsilon\right)\tan\left(\phi\right)}{\cos\left(\theta\right)}
\]

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Definició de parsec: 
\begin_inset Formula $1pc=\frac{1UA}{1\, segon\, d'arc}=\frac{1UA}{1^{s}\frac{1^{o}}{3600^{s}}\frac{2\pi rad}{360^{o}}}=206.265UA$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Tenim una estrella de massa 
\begin_inset Formula $M_{\star}=2\cdot10^{33}g=M_{\odot}$
\end_inset

, amb un planeta de massa 
\begin_inset Formula $M_{\diamond}=1.9\cdot10^{30}g=M_{J}$
\end_inset

 girant en orbita circular a 5.2 AU.
 Per tant, prenent l'estrella com a punt 0 AU i el planeta es troba a un
 radi 5.2 AU, el centre de masses (CDM) es troba a:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
x_{CDM}=\frac{0\cdot M_{\star}+r\cdot M_{\diamond}}{M_{\star}+M_{\diamond}}=\frac{5.2\cdot1.9\cdot10^{30}}{2\cdot10^{33}+1.9\cdot10^{30}}=0.0049AU
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
L'estrella es mourà al voltant del centre de masses i recorrerà una distancia
 de 
\begin_inset Formula $2\cdot x_{CDM}=0.0099AU$
\end_inset

, que traduït a distància angular des de la terra i utilitzant l'aproximació
 de la tangent (donat que 
\begin_inset Formula $\theta_{max}\sim0$
\end_inset

) dependrà de la distància:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\tan\left(\theta_{max}\right)=\frac{2\cdot x_{CDM}}{distancia_{max}}\simeq\theta_{max}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Com que només podem distingir angles de com a mínim 0
\begin_inset Quotes erd
\end_inset

.01 (resolució límit), la distància angular pot ser com a molt tan petita
 com aquesta resolució límit.
 Això implica que la distancia màxima a la que pot ser-hi el sistema planetari
 és:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\theta_{max}=\frac{2\cdot x_{CDM}}{distancia_{max}}\geq\theta_{resol}\Rightarrow distancia_{max}\leq\frac{2\cdot x_{CDM}}{\theta_{resol}}=\frac{0.0099UA}{0".01}=0.99pc
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Estels
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Lluminositat solar: 
\begin_inset Formula $L_{\odot}=3,86\cdot10^{26}W=3,86\cdot10^{26}\frac{J}{s}$
\end_inset


\end_layout

\begin_layout Plain Layout
Unitat d'energia: 
\begin_inset Formula $1erg=10^{-7}J$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Tenim una estrella amb lluminositat 
\begin_inset Formula $L_{\star}=10^{3}L_{\odot}$
\end_inset

 i distància 
\begin_inset Formula $r=1kpc=1000\cdot3,0857\cdot10^{16}m$
\end_inset

.
 Per tant l'energia que ens arriba a nosaltres segueix la llei de la inversa
 del quadrat.
 El nostre telescopi te un diàmetre 
\begin_inset Formula $\phi=2m$
\end_inset

 que limita la quantitat d'energia recollida, l'utilitzem durant 
\begin_inset Formula $\Delta t=10^{H}=10\cdot60\cdot60=36.000$
\end_inset

.
\end_layout

\begin_layout Standard
Per tant, l'energia total recollida correspon a l'energia de l'estrella
 (
\begin_inset Formula $L_{\star}$
\end_inset

) per l'esfera (
\begin_inset Formula $\frac{1}{4\pi r^{2}}$
\end_inset

) per l'àrea col·lectora (
\begin_inset Formula $A=r^{2}\pi=\left(\frac{\phi}{2}\right)^{2}\pi$
\end_inset

) pel temps total (
\begin_inset Formula $\Delta t$
\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
E_{Total}=L_{\star}\frac{1}{4\pi r^{2}}\left(\frac{\phi}{2}\right)^{2}\pi\Delta t
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
E_{Total}=10^{3}\cdot3,86\cdot10^{26}\frac{J}{s}\frac{1}{4\pi\left(3.085,7\cdot10^{16}m\right)^{2}}\left(\frac{2m}{2}\right)^{2}\pi36.000s=36.49\cdot10^{-7}J
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
L'energia cinètica d'un floquet de neu de massa 
\begin_inset Formula $m=0.1g=0.0001kg$
\end_inset

 que cau a velocitat 
\begin_inset Formula $v=10\frac{cm}{s}=0.1\frac{m}{s}$
\end_inset

 és de:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
E_{c}=\frac{1}{2}mv^{2}=\frac{1}{2}0.0001kg\cdot0.1^{2}\frac{m^{2}}{s^{2}}=5\cdot10^{-7}J
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Subsubsection
Solució 1
\end_layout

\begin_layout Standard
Considerem que la diferencia fa referència a magnituds absolutes.
 A partir d'aquesta, podem obtenir un coeficient de lluminositat:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M_{1}=-2,5\log\left(\frac{L_{1}}{L_{2}}\right)+M_{2}\Rightarrow\frac{L_{1}}{L_{2}}=10^{\frac{M_{2}-M_{1}}{2,5}}=10^{\frac{0,5}{2,5}}=1.5849
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Donat que la lluminositat també és igual a 
\begin_inset Formula $L=4\pi R^{2}F$
\end_inset

 i el flux a la superfície no variarà:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{L_{1}}{L_{2}}=\frac{4\pi R_{1}^{2}F}{4\pi R_{2}^{2}F}=\left(\frac{R_{1}}{R_{2}}\right)^{2}=1.2589
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant la diferencia de radi ve determinada per un factor:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{R_{1}}{R_{2}}=\sqrt{\frac{L_{1}}{L_{2}}}=\sqrt{1.5849}=1.2589
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Solució 2
\end_layout

\begin_layout Standard
Considerem que la diferencia fa referència a magnituds aparents.
 Desenvolupem la relació entre magnituds aparents i flux:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m_{1}-m_{2}=2,5\log_{10}\frac{f_{1}}{f_{2}}=2,5\log_{10}\frac{F\cdot\left(\frac{r_{1}}{d}\right)^{2}}{F\cdot\left(\frac{r_{2}}{d}\right)^{2}}=2,5\log_{10}\frac{\sigma\text{·}T^{4}\cdot\left(\frac{r_{1}}{d}\right)^{2}}{\sigma\text{·}T^{4}\cdot\left(\frac{r_{2}}{d}\right)^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
I considerarem que la temperatura no varia encara que canvii la magnitud
 (equivalent a dir que no varia el flux a la superfície), llavors la relació
 entre radis és:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m_{1}-m_{2}=2,5\log_{10}\frac{\left(r_{1}\right)^{2}}{\left(r_{2}\right)^{2}}=0,5\Rightarrow\frac{r_{1}}{r_{2}}=\sqrt{10^{\frac{0,5}{2,5}}}=1,2589
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Magnitud aparent màxima que pot registra l'ull humà: 
\begin_inset Formula $m_{v}=6$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Existeixen dos tipus d'estels de la seqüència principal tenen magnituds
 
\begin_inset Formula $M_{V,A0V}=0.6$
\end_inset

 i 
\begin_inset Formula $M_{V,G2V}=4.7$
\end_inset

, així mateix hi ha 1 A0V per cada 10 G2V (p.ex.
 el sol) per unitat de volum en la nostra proximitat (
\begin_inset Formula $\frac{\rho_{A0V}}{\rho_{G2V}}=\frac{1}{10}$
\end_inset

).
 Si l'ull humà només pot veure estrelles de magnitud aparent 6, calculem
 la distancia màxima a la que podrem veure una estrella tipus A0V utilitzant
 la relació 
\begin_inset Formula $M=m-5\log_{10}\left(d_{L}\right)+5$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
d_{Lmax,A0V}=10^{\frac{m-M+5}{5}}=10^{\frac{6-0.6+5}{5}}=120.23pc
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
I la distància d'una estrella tipus G2V:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
d_{Lmax,G2V}=10^{\frac{m-M+5}{5}}=10^{\frac{6-4.7+5}{5}}=18.197pc
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per saber qui predominarà, si el nombre d'estrelles A0V o G2V fem el seu
 rati i així podem substituir el rati de densitats conegut:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{N_{A0V}}{N_{G2V}}=\frac{\frac{4}{3}\pi\left(d_{Lmax,A0V}\right)^{3}}{\frac{4}{3}\pi\left(d_{Lmax,G2V}\right)^{3}}\frac{\rho_{A0V}}{\rho_{G2V}}=\frac{\frac{4}{3}\pi\left(120.23\right)^{3}}{\frac{4}{3}\pi\left(18.197\right)^{3}}\frac{1}{10}=28.843
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Es poden observar 28 A0V per cada G2V.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
Hipòtesis: Suposem que el flux emès per Júpiter és nul.
\end_layout

\begin_layout Standard
Les lluminositats del Sol quan no es troba Júpiter al davant:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{\odot}=4\pi R^{2}F_{\odot}=4\pi d_{L}^{2}f_{\odot}\Rightarrow f_{\odot}=\frac{4\pi R^{2}F_{\odot}}{4\pi d_{L}^{2}}=\frac{R^{2}F_{\odot}}{d_{L}^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
I quan Júpiter està al davant (hem de restar el flux del sol que tapa l'àrea
 del planeta):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{\odot,J}=4\pi R^{2}F_{\odot}-4\pi R_{J}^{2}F_{\odot}=4\pi d_{L}^{2}f_{\odot,J}\Rightarrow f_{\odot,J}=\frac{4\pi R^{2}F_{\odot}-4\pi R_{J}^{2}F_{\odot}}{4\pi d_{L}^{2}}=\frac{R^{2}F_{\odot}-R_{J}^{2}F_{\odot}}{d_{L}^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Si ara utilitzem la relació flux - magnitud aparent per trobar la variació
 en magnitud aparent:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m_{\odot}-m_{\odot,J}=2,5\log_{10}\frac{\frac{R^{2}F_{\odot}}{d_{L}^{2}}}{\frac{R^{2}F_{\odot}-R_{J}^{2}F_{\odot}}{d_{L}^{2}}}=2,5\log_{10}\frac{R^{2}F_{\odot}}{R^{2}F_{\odot}-R_{J}^{2}F_{\odot}}=2,5\log_{10}\frac{R^{2}}{R^{2}-R_{J}^{2}}=2,5\log_{10}\frac{1}{1-\left(\frac{R_{J}}{R}\right)^{2}}\label{eq:variacio_mag}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El problema ens indica que les densitats són semblants:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\rho_{\odot}=\frac{M_{\odot}}{\frac{4}{3}\pi R_{\odot}^{3}}\simeq\rho_{J}=\frac{M_{J}}{\frac{4}{3}\pi R_{J}^{3}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
I donat que la massa de Júpiter 
\begin_inset Formula $1.000\cdot M_{J}=M_{\odot}$
\end_inset

, llavors:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{1.000\cdot M_{J}}{\frac{4}{3}\pi R_{\odot}^{3}}=\frac{M_{J}}{\frac{4}{3}\pi R_{J}^{3}}\Rightarrow10^{3}=\frac{R_{\odot}^{3}}{R_{J}^{3}}\Rightarrow10=\frac{R_{\odot}}{R_{J}}\Rightarrow\frac{R_{J}}{R_{\odot}}=\frac{1}{10}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, podem substituir a la formula (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:variacio_mag"

\end_inset

):
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m_{\odot}-m_{\odot,J}=2,5\log_{10}\frac{1}{1-\left(\frac{R_{J}}{R}\right)^{2}}=2,5\log_{10}\frac{1}{1-\left(\frac{1}{10}\right)^{2}}=0.011
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
Per trobar la magnitud integrada no es poden sumar directament les magnituds
 individuals, sinó que necessitem obtenir el flux de cada una de les 7 estrelles.
 A partir de la definició 
\begin_inset Formula $m_{i}=-2,5\log f_{i}+ct$
\end_inset

 obtenim:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
f_{i}=10^{-\frac{m_{i-ct}}{2,5}}=10^{-0.4m_{i}+0.4ct}=10^{-0.4m_{i}}10^{0.4ct}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, el flux total serà:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
f_{total}=\sum f_{i}=10^{0.4ct}\sum_{i=1}^{7}10^{-0.4m_{i}}=10^{0.4ct}10^{-0,6627}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Finalment, la magnitud integrada:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m_{integrada}=-2,5\log f_{total}+ct=-2,5\log\left(10^{-0,6627}\right)=1,657
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
La constant del flux total s'anul·la amb la constant d'aquesta última equació.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
La constant solar 
\begin_inset Formula $f=1,36\cdot10^{6}erg\, s^{-1}cm^{-2}$
\end_inset

 que correspon a la potència rebuda per unitat de superfície a la posició
 de la Terra no és més que el flux.
\end_layout

\begin_layout Standard
Sabent el diàmetre angular del sol 
\begin_inset Formula $\theta_{d}=32^{m}$
\end_inset

 des de la terra (radi angular 
\begin_inset Formula $\theta_{r}=16^{m}\frac{1^{g}}{60^{m}}\frac{2\pi rad}{360^{g}}=0,0046542rad$
\end_inset

) i la distància 
\begin_inset Formula $d=1UA\simeq1,5\cdot10^{13}cm$
\end_inset

, podem obtenir el radi 
\begin_inset Formula $\tan\left(0,0046542\right)\simeq0,0046542=\frac{R}{d}\Rightarrow R=6.,9813\cdot10^{10}cm$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Lluminositat
\end_layout

\begin_layout Standard
\begin_inset Formula $L=4\Pi d_{L}^{2}f=4\pi\cdot\left(1,5\cdot10^{13}\right)^{2}\cdot1,36\cdot10^{6}=3.845\cdot10^{33}$
\end_inset


\end_layout

\begin_layout Subsubsection
Temperatura efectiva
\end_layout

\begin_layout Standard
A partir del Flux a la superfície:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L=4\Pi R^{2}F\Rightarrow F=\frac{L}{4\Pi R^{2}}=\frac{3.845\cdot10^{33}}{4\pi\left(6,9813\cdot10^{10}\right)^{2}}=6,2779\cdot10^{10}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Podem aplicar la relació amb la temperatura:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
F=\sigma T_{eff}^{4}\Rightarrow T_{eff}=\sqrt[4]{\frac{6,2779\cdot10^{10}}{5,67\text{·}10^{-5}}}=5.768,4K
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Longitud d'ona màxima de l'espectre solar (cos negre)
\end_layout

\begin_layout Standard
\begin_inset Formula $\lambda_{max}=\frac{b}{T}=\frac{2,8977685\cdot10^{-3}mK}{5.768,4K}=5,0235\cdot10^{-7}$
\end_inset

 
\end_layout

\begin_layout Subsubsection
Fracció de la lluminositat interceptada per la terra
\end_layout

\begin_layout Standard
La lluminositat interceptada per la terra correspon al flux que arriba per
 l'àrea de la terra, sabent que el seu radi és 
\begin_inset Formula $R_{terra}=6.400km$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula $L_{interceptada}=f\cdot\pi R_{terra}^{2}=\left(1,36\cdot10^{6}erg\, s^{-1}cm^{-2}\right)\pi\left(6.400\cdot10^{5}cm\right)^{2}=1,75\cdot10^{24}erg\, s^{-1}$
\end_inset


\end_layout

\begin_layout Standard
Per tant, la fracció o percentatge de lluminositat interceptada:
\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{L_{interceptada}}{L_{\odot}}=\frac{1,75\cdot10^{24}}{3.845\cdot10^{33}}=4.5515\cdot10^{-10}$
\end_inset


\end_layout

\begin_layout Subsubsection
Magnitud absoluta del sol
\end_layout

\begin_layout Standard
Si la magnitud aparent és 
\begin_inset Formula $m=-26.8$
\end_inset

 i la distància en parsec 
\begin_inset Formula $d=1UA\frac{1pc}{206.265UA}=4,848\cdot10^{-6}pc$
\end_inset

, l'absoluta és:
\end_layout

\begin_layout Standard
\begin_inset Formula $M=m-5\log_{10}\left(d_{L}\right)-5=-26.8-5\log_{10}\left(4,848\cdot10^{-6}pc\right)-5=4,7722$
\end_inset


\end_layout

\begin_layout Standard
Cal destacar que si la magnitud aparent és al visual, l'absoluta també ho
 serà.
 Per altra banda, si la magnitud aparent és bolomètrica, l'absoluta també.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Lluminositat del sol: 
\begin_inset Formula $L_{\odot}=3,84\cdot10^{33}erg\, s^{-1}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="3">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Estrella
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Magnitud
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Distància
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Sírius
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m=-1,46$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d=2,7pc$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Arcturus
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m=-0,04$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d=11pc$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Algol
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $M=-0,20$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d=29pc\frac{3,0857\cdot10^{18}cm}{1pc}=8,9485\cdot10^{19}cm$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Standard
Calculem la magnitud absoluta per Sírius i Arcturus:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M_{Sirius}=m-5\log_{10}\left(d_{L}\right)+5=-1,46-5\log_{10}\left(2.7\right)+5=1,3832
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M_{Arcturus}=m-5\log_{10}\left(d_{L}\right)+5=-0,04-5\log_{10}\left(11\right)+5=-0.24696
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Addicionalment, sabem que el flux rebut a la terra d'Algol és 
\begin_inset Formula $f_{Algol}=3,769\cdot10^{-6}erg\, s^{-1}cm^{-2}$
\end_inset

.
 Per tant, la seva lluminositat:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Algol}=4\pi d_{L}^{2}f=4\pi\cdot\left(8,9485\cdot10^{19}cm\right)^{2}\cdot3,769\cdot10^{-6}erg\, s^{-1}cm^{-2}=3,793\cdot10^{35}erg\, s^{-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Algol}=3,793\cdot10^{35}erg\, s^{-1}\frac{1L_{\odot}}{3,84\cdot10^{33}erg\, s^{-1}}=98,776L_{\odot}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, podem comparar la lluminositat d'Algol amb la de Sírius o Arcturus
 per obtenir-la amb la relació 
\begin_inset Formula $M_{1}=-2,5\log\left(\frac{L_{1}}{L_{2}}\right)+M_{2}\Longrightarrow\frac{L_{1}}{L_{2}}=10^{\frac{M_{2}-M_{1}}{2,5}}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Sirius}=L_{Algol}\cdot10^{\frac{M_{Algol}-M_{Sirius}}{2,5}}=3,793\cdot10^{35}\cdot10^{\frac{-0,2-1,3832}{2,5}}=8,8248\cdot10^{34}erg\, s^{-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Arcturus}=L_{Algol}\cdot10^{\frac{M_{Algol}-M_{Arcturus}}{2,5}}=3,793\cdot10^{35}\cdot10^{\frac{-0,2-3,793}{2,5}}=3.96\cdot10^{35}erg\, s^{-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Sirius}=8,8248\cdot10^{34}erg\, s^{-1}\frac{1L_{\odot}}{3,84\cdot10^{33}erg\, s^{-1}}=22,98L_{\odot}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{Arcturus}=9,5892\cdot10^{33}erg\, s^{-1}\frac{1L_{\odot}}{3,84\cdot10^{33}erg\, s^{-1}}=103,14L_{\odot}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
Radi del sol: 
\begin_inset Formula $R_{\odot}=700.000km$
\end_inset


\end_layout

\begin_layout Plain Layout
Lluminositat del sol: 
\begin_inset Formula $L_{\odot}=3,86\cdot10^{26}W$
\end_inset


\end_layout

\begin_layout Plain Layout
Magnitud bolomètrica del sol: 
\begin_inset Formula $M_{Bolometrica,\odot}=4,75mag$
\end_inset


\end_layout

\begin_layout Plain Layout
Constant solar a la terra (flux): 
\begin_inset Formula $f=1,36\cdot10^{6}erg\, s^{-1}cm^{-2}=1,36\cdot10^{3}W\, m^{-2}$
\end_inset

 
\end_layout

\begin_layout Plain Layout
Massa del sol: 
\begin_inset Formula $M_{\odot}=2\cdot10^{30}kg$
\end_inset


\end_layout

\begin_layout Plain Layout
Constant de Stefan-Boltzmann: 
\begin_inset Formula $\sigma=\frac{2\text{·}K_{B}^{4}\text{·}\Pi^{5}}{c^{2}\text{·}L^{3}\text{·}15}=5,67\text{·}10^{-8}Wm^{-2}K^{-4}=5,67\text{·}10^{-5}erg\, cm^{-2}K^{-4}$
\end_inset


\end_layout

\begin_layout Plain Layout
Constant universal de gravitació: 
\begin_inset Formula $G=6.6726\cdot10^{-11}N\, m^{2}/kg^{2}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Tenim una estrella supergegant amb lluminositat 
\begin_inset Formula $L_{B0}=2,5\cdot10^{5}L_{\odot}=9,65\cdot10^{31}W$
\end_inset

 i radi 
\begin_inset Formula $R_{B0}=20R_{\odot}=1,4\cdot10^{10}m$
\end_inset


\end_layout

\begin_layout Subsubsection
Magnitud absoluta i temperatura efectiva
\end_layout

\begin_layout Standard
A partir de la relació 
\begin_inset Formula $L=4\pi R^{2}F$
\end_inset

 tenim que el Flux a la superfície és:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
F=\frac{L}{4\pi R^{2}}=\frac{9,65\cdot10^{31}}{4\pi\left(1,4\cdot10^{10}\right)^{2}}=3,918\cdot10^{10}Wm^{-2}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, la seva temperatura efectiva:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
F=\sigma T_{eff}^{4}\Rightarrow T_{eff}=\sqrt[4]{\frac{3,918\cdot10^{10}}{5,67\text{·}10^{-8}}}=28.832K
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
I per comparació amb el sol podem extreure la magnitud absoluta:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M=-2,5\log\left(\frac{L}{L_{\odot}}\right)+M_{Bolometrica,\odot}=-2,5\log\left(\frac{9,65\cdot10^{31}W}{3,86\cdot10^{26}W}\right)+4,75=-8,7449
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Distancia planeta per a tenir la mateixa constant solar que la terra
\end_layout

\begin_layout Standard
Sabem que la constant solar (flux) a la terra és 
\begin_inset Formula $f=1,36\cdot10^{3}W\, m^{-2}$
\end_inset

, per tant per calcular la distància utilitzarem:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L=4\pi d_{L}^{2}f\Rightarrow d_{L}=\sqrt{\frac{L}{4\pi f}}=\sqrt{\frac{9,65\cdot10^{31}W}{4\pi\left(1,36\cdot10^{3}W\, m^{-2}\right)}}=7,3926\cdot10^{13}m\frac{1UA}{150\cdot10^{9}m}=492.84
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Període orbital del planeta
\end_layout

\begin_layout Standard
Per obtenir la massa de l'estrella utilitzem la relació 
\begin_inset Formula $\log\left(\frac{L}{L_{\odot}}\right)\simeq3,45\log\left(\frac{M}{M_{\odot}}\right)$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M=M_{\odot}10^{\frac{1}{3,45}\log\left(\frac{L}{L_{\odot}}\right)}=2\cdot10^{30}\cdot10^{\frac{1}{3,45}\log\left(\frac{9,65\cdot10^{31}}{3,86\cdot10^{26}}\right)}=7,3392\cdot10^{31}kg=36,696M_{\odot}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, el període orbital considerant la distància calculada a l'apartat
 anterior:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
T=\sqrt{\frac{4\pi^{2}r^{3}}{G\left(M_{1}\right)}}=\sqrt{\frac{4\pi^{2}\left(7,3926\cdot10^{13}m\right)^{3}}{6.6726\cdot10^{-11}kg\, m^{3}s^{-2}kg^{-2}\cdot7,3392\cdot10^{31}kg}}=5,7069\cdot10^{10}segons=1809,6anys
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Magnitud aparent de l'estrella des del planeta
\end_layout

\begin_layout Standard
\begin_inset Formula $m=M+5\log_{10}\left(d_{L}\right)-5=-8,7449+5\log_{10}\left(7,3926\cdot10^{13}m\frac{1pc}{3,0857\cdot10^{16}m}\right)-5=-26,803$
\end_inset

 
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Subsubsection
Pressió al centre
\end_layout

\begin_layout Standard
Com prenem la hipòtesi de que la densitat és uniforme, no dependrà del radi:
 
\begin_inset Formula $\rho\left(r\right)=constant=\rho_{\star}$
\end_inset


\end_layout

\begin_layout Standard
Mitjançant la conservació de la massa 
\begin_inset Formula $dm=4\pi r^{2}\rho\left(r\right)dr=4\pi r^{2}\rho_{\star}dr$
\end_inset

 deduïm:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\int_{m\left(r\right)}^{m\left(R\right)=0}dm=m\left(r\right)=\int_{0}^{r}4\pi r^{2}\rho_{\star}dr=\frac{4\pi r^{3}}{3}\rho_{\star}-0
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, la gravetat en funció del radi 
\begin_inset Formula $g\left(r\right)=\frac{Gm\left(r\right)}{r^{2}}=G\cdot\frac{4\pi r}{3}\rho_{\star}$
\end_inset

 i l'equació de l'equilibri hidrostàtic 
\begin_inset Formula $\frac{dP}{dr}=-g\left(r\right)\rho\left(r\right)=-g\left(r\right)\rho_{\star}=-G\cdot\frac{4\pi r}{3}\rho_{\star}^{2}$
\end_inset


\end_layout

\begin_layout Standard
Com a condició de contorn sabem que la pressió a la superfície és cero 
\begin_inset Formula $P\left(R\right)=0$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\intop_{P\left(r\right)}^{P\left(R\right)=0}dP=P\left(r\right)=\int_{r}^{R}-G\cdot\frac{4\pi r}{3}\rho_{\star}^{2}dr=-G\cdot\frac{4\pi R^{2}}{2\cdot3}\rho_{\star}^{2}--G\cdot\frac{4\pi r^{2}}{2\cdot3}\rho_{\star}^{2}=-G\cdot\frac{2\pi}{3}\rho_{\star}^{2}\cdot\left(R^{2}-r^{2}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Al centre 
\begin_inset Formula $r=0$
\end_inset

, per tant la pressió al centre és:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
P_{c}=-G\cdot\frac{2\pi}{3}\rho_{\star}^{2}\cdot R^{2}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Temperatura central
\end_layout

\begin_layout Standard
La llei del gas ideal és 
\begin_inset Formula $P=\frac{n}{V}kT$
\end_inset

 a on la constant 
\begin_inset Formula $k=8,314J\text{·}K^{-1}mol^{-1}$
\end_inset

 i 
\begin_inset Formula $n=\frac{m}{\mu\cdot m_{H}}$
\end_inset

 amb 
\begin_inset Formula $m$
\end_inset

 com a massa total, 
\begin_inset Formula $\mu=1\cdot10^{-3}kg/mol=1g/mol$
\end_inset

 i 
\begin_inset Formula $m_{H}$
\end_inset

 la massa atòmica de l'hidrogen.
 Sabem que la densitat és constant i que la massa mitja per partícula en
 unitats de massa d'hidrogen és
\begin_inset Formula $\mu\cdot m_{H}=0.6m_{H}$
\end_inset

, per tant:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{n}{V}=\frac{m}{V}\cdot\frac{1}{\mu\cdot m_{H}}=\rho\left(r\right)\cdot\frac{1}{\mu\cdot m_{H}}=\frac{\rho_{\star}}{0.6m_{H}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
En conseqüència:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
P_{c}=\frac{\rho_{\star}}{0.6m_{H}}\cdot k\cdot T_{c}=-G\cdot\frac{2\pi}{3}\rho_{\star}^{2}\cdot R^{2}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
T_{c}=-\frac{G\cdot\frac{2\pi}{3}\rho_{\star}^{2}\cdot R^{2}}{\frac{\rho_{\star}}{0.6m_{H}}\cdot k}=-\frac{0.6m_{H}\cdot G\cdot2\pi\rho_{\star}\cdot R^{2}}{3\cdot k}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Aplicar al sol
\end_layout

\begin_layout Standard
A les fórmules anteriors es considerarà el radi del sol 
\begin_inset Formula $R=R_{\odot}=7\cdot10^{8}m$
\end_inset

, per tant el volum 
\begin_inset Formula $V_{\odot}=\frac{4}{3}\pi R_{\odot}^{3}$
\end_inset

 i la densitat 
\begin_inset Formula $\rho_{\star}=\frac{M_{\odot}}{V_{\odot}}=\frac{2\cdot10^{30}kg}{\frac{4}{3}\pi\left(7\cdot10^{8}\right)^{3}}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
P_{c}=-G\cdot\frac{2\pi}{3}\rho_{\star}^{2}\cdot R^{2}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
T_{c}=-\frac{0.6m_{H}\cdot G\cdot2\pi\rho_{\star}\cdot R^{2}}{3\cdot k}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Recordatori
\end_layout

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="1">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Tabular
<lyxtabular version="3" rows="12" columns="2">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Filter letter
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Description
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
U
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Ultraviolet
\end_layout

\end_inset
</cell>
</row>
<row>
<cell multicolumn="1" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Visible
\end_layout

\end_inset
</cell>
<cell multicolumn="2" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
B
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Blue
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
V
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Visual
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
G
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Green
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
R
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Red
\end_layout

\end_inset
</cell>
</row>
<row>
<cell multicolumn="1" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Near-Infrared
\end_layout

\end_inset
</cell>
<cell multicolumn="2" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
I
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Infrared
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Z, Y, J, H, K, L
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Others
\end_layout

\end_inset
</cell>
</row>
<row>
<cell multicolumn="1" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Mid-Infrared
\end_layout

\end_inset
</cell>
<cell multicolumn="2" alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
M, N, Q
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Others
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Diagrama HR:
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Graphics
	filename DiagramaHR-Hertzsprung-Russel.jpg
	scale 50

\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Les estrelles en la seva fase de gegant vermella emeten energia principalment
 en la banda vermella (com el seu nom indica).
 Per altra banda, les nanes blanques poden emetre en un rang ampli en funció
 de les seves característiques (p.ex.
 temperatura efectiva), no obstant aquest rang es troba per sobre de la
 banda vermella (més aviat blau-blanc).
\end_layout

\begin_layout Standard
En conseqüència, un eclipse de la gegant vermella per la nana blanca tindrà
 els següents efectes:
\end_layout

\begin_layout Itemize
Observació amb filtre U: Quan la nana passi pel davant de la gegant s'incrementa
rà l'energia rebuda a l'ultraviolat, donat que la gegant vermella gairebé
 no emet en aquesta banda i la nana si.
\end_layout

\begin_layout Itemize
Observació amb filtre R: Decaurà l'energia rebuda al vermell quan la nana
 passi pel davant de la gegant degut a que la nana absorbirà/taparà i la
 seva lluminositat intrínseca no contribueix en aquesta banda.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Subsubsection
Longitud d'ona i radi
\end_layout

\begin_layout Standard
Donades les fórmules 
\begin_inset Formula $\mbox{\ensuremath{\lambda=\frac{1}{1,09737312\cdot10^{5}cm^{-1}\text{·}\left(\frac{1}{m_{in}^{2}}-\frac{1}{n_{out}^{2}}\right)}}}$
\end_inset

 i 
\begin_inset Formula $r_{n}=0.5292\cdot10^{-10}n^{2}m$
\end_inset

 obtenim:
\end_layout

\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="7">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Destí (
\begin_inset Formula $m_{in}$
\end_inset

)
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Serie
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Banda
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=1$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=2$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=3$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=4$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Lyman
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
UV
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=1,215\cdot10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=1,0252\cdot10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=9,72\cdot10^{-6}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
2
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Balmer
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
V
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=6,56\cdot10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=4,86\cdot10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
3
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Paschen
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
IR
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\lambda=1,87\cdot10^{-4}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
4
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Brackett
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
-
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{1}=5,29\cdot10^{-11}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{2}=2,117\cdot10^{-10}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{3}=4,764\cdot10^{-10}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{4}=8,47\cdot10^{-10}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Subsubsection
Nombre d'àtoms
\end_layout

\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="4">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Serie Balmer amb 
\begin_inset Formula $n_{max}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Radi màxim
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Distància interatòmica
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Estrelles B
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=19$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{19}=1,91\cdot10^{-8}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d_{19}=1,91\cdot10^{-7}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Estrelles Be
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=42$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{42}=9,335\cdot10^{-8}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d_{42}=9,335\cdot10^{-7}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Nana blanca
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $n_{out}=6$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $r_{6}=1,9\cdot10^{-9}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $d_{6}=1,9\cdot10^{-8}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Standard
Per tant, considerant que la densitat 
\begin_inset Formula $\rho=\frac{M}{\frac{4}{3}\pi R^{3}}$
\end_inset

, llavors:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\rho_{B}=\frac{1atom}{\frac{4}{3}\pi\left(1,91\cdot10^{-7}\cdot\frac{100cm}{1m}\right)^{3}}=\frac{3,4262\cdot10^{13}atom}{1cm^{3}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\rho_{Be}=\frac{1atom}{\frac{4}{3}\pi\left(9,335\cdot10^{-7}\cdot\frac{100cm}{1m}\right)^{3}}=\frac{2,9347\cdot10^{11}atom}{1cm^{3}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\rho_{Nanes}=\frac{1atom}{\frac{4}{3}\pi\left(1,9\cdot10^{-8}\cdot\frac{100cm}{1m}\right)^{3}}=\frac{3,480\cdot10^{16}atom}{1cm^{3}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Les nanes blanques tenen una densitat molt superior a una estrella B o Be.
 Així com les B tenen una densitat superior a les Be.
\end_layout

\begin_layout Subsubsection
Transicions
\end_layout

\begin_layout Standard
Transformem les freqüències 
\begin_inset Formula $\nu_{1}=4.874,157Mhz$
\end_inset

 i 
\begin_inset Formula $\nu_{2}=4.897,339Mhz$
\end_inset

 a longitud d'ones:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\lambda_{1}=\frac{299.792.458m/s}{4.874,157\cdot10^{6}Hz}=0,0615m\frac{100cm}{1m}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\lambda_{2}=\frac{299.792.458m/s}{4.897,339\cdot10^{6}Hz}=0,061215m\frac{100cm}{1m}
\end{equation}

\end_inset


\end_layout

\begin_layout Paragraph
Lambda 1
\end_layout

\begin_layout Standard
Pel primer cas, podem utilitzar 
\begin_inset Formula $\mbox{\ensuremath{\frac{1}{\lambda}=R\text{·}\left(\frac{1}{m_{in}^{2}}-\frac{1}{n_{out}^{2}}\right)}}$
\end_inset

 i el fet de que 
\begin_inset Formula $n_{out}=m_{in}+1$
\end_inset

 per determinar la transició:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\mbox{\ensuremath{\frac{1}{\lambda}=R\text{·}\left(\frac{1}{m_{in}^{2}}-\frac{1}{\left(m_{in}+1\right)_{out}^{2}}\right)=R\text{·}\left(\frac{\left(m_{in}+1\right)_{out}^{2}-m_{in}^{2}}{\left(m_{in}+1\right)_{out}^{2}\cdot m_{in}^{2}}\right)=R\text{·}\left(\frac{m_{in}^{2}+2m_{in}+1-m_{in}^{2}}{\left(m_{in}+1\right)^{2}\cdot m_{in}^{2}}\right)=R\text{·}\left(\frac{2m_{in}+1}{\left(m_{in}+1\right)^{2}\cdot m_{in}^{2}}\right)}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\mbox{\ensuremath{\frac{\left(m_{in}+1\right)^{2}\cdot m_{in}^{2}}{2m_{in}+1}=2\cdot R\cdot\lambda}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Com 
\begin_inset Formula $r_{n}=0.5292\cdot10^{-10}n^{2}m=0.5292\cdot10^{-10}\left(m_{in}+1\right)^{2}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\mbox{\ensuremath{\frac{\left(m_{in}+1\right)^{2}\cdot m_{in}^{2}}{2m_{in}+1}=2\cdot\lambda\cdot0.5292\cdot10^{-10}\left(m_{in}+1\right)^{2}}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\mbox{\ensuremath{\frac{m_{in}^{2}}{2m_{in}+1}=2\cdot\lambda\cdot0.5292\cdot10^{-10}}}
\end{equation}

\end_inset


\end_layout

\begin_layout Paragraph
Lambda 2
\end_layout

\begin_layout Standard
Pel segon cas, podem utilitzar 
\begin_inset Formula $\mbox{\ensuremath{\frac{1}{\lambda}=R\text{·}\left(\frac{1}{m_{in}^{2}}-\frac{1}{n_{out}^{2}}\right)}}$
\end_inset

 i el fet de que 
\begin_inset Formula $n_{out}=m_{in}+2$
\end_inset

 per determinar la transició
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\mbox{\ensuremath{\frac{1}{\lambda}=R\text{·}\left(\frac{1}{m_{in}^{2}}-\frac{1}{\left(m_{in}+2\right)_{out}^{2}}\right)=R\text{·}\left(\frac{\left(m_{in}+2\right)_{out}^{2}-m_{in}^{2}}{\left(m_{in}+2\right)_{out}^{2}\cdot m_{in}^{2}}\right)=R\text{·}\left(\frac{m_{in}^{2}+2m_{in}+4-m_{in}^{2}}{\left(m_{in}+2\right)^{2}\cdot m_{in}^{2}}\right)=R\text{·}\left(\frac{2m_{in}+4}{\left(m_{in}+2\right)^{2}\cdot m_{in}^{2}}\right)}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{\left(m_{in}+2\right)^{2}\cdot m_{in}^{2}}{2m_{in}+4}=2\cdot R\cdot\lambda
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Com 
\begin_inset Formula $r_{n}=0.5292\cdot10^{-10}n^{2}m=0.5292\cdot10^{-10}\left(m_{in}+2\right)^{2}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{\left(m_{in}+2\right)^{2}\cdot m_{in}^{2}}{2m_{in}+4}=2\cdot\lambda\cdot0.5292\cdot10^{-10}\left(m_{in}+2\right)^{2}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{m_{in}^{2}}{2m_{in}+4}=2\cdot\lambda\cdot0.5292\cdot10^{-10}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Formació i evolució estel·lar
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
Per demostrar que el temps de caiguda lliure d'un núvol de gas esfèric de
 radi R i massa M és 
\begin_inset Formula $t_{cl}=\frac{\pi}{2\sqrt{2}}\left(\frac{R^{3}}{GM}\right)^{\frac{1}{2}}$
\end_inset

:
\end_layout

\begin_layout Enumerate
Suposem simetria axial al núvol de gas esfèric
\end_layout

\begin_layout Enumerate
Considerem una partícula de la superfície (
\begin_inset Formula $r=R$
\end_inset

) inicialment en repòs (
\begin_inset Formula $v\left(r\right)=0$
\end_inset

).
\end_layout

\begin_layout Enumerate
No existeix equilibri hidrostàtic, només actual la força de la gravetat
 
\begin_inset Formula $F_{g}=-\frac{G\cdot M\cdot m}{r^{2}}$
\end_inset

 on M és la massa total i 
\begin_inset Formula $m$
\end_inset

 la massa d'una partícula a la superfície.
\end_layout

\begin_layout Standard
Aplicant la segona llei de Newton 
\begin_inset Formula $F=m\cdot a=m\cdot\frac{dv}{dt}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m\cdot\frac{dv}{dt}=-\frac{G\cdot M\cdot m}{r^{2}}\Rightarrow\frac{dv}{dt}=-\frac{G\cdot M}{r^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Calculem la velocitat en funció del radi:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{dv}{dt}=\frac{dv}{dr}\cdot\frac{dr}{dt}=\frac{dv}{dr}\cdot v
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Combinem les dos fórmules anteriors e integrem:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
v\cdot\frac{dv}{dr}=-\frac{G\cdot M}{r^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\int_{v(r)}^{v(R)=0}v\cdot dv=\frac{v\left(r\right)^{2}}{2}=\int_{r}^{R}-\frac{G\cdot M}{r^{2}}dr\Rightarrow\frac{v\left(r\right)^{2}}{2}=\frac{G\cdot M}{R}-\frac{G\cdot M}{r}=G\cdot M\left(\frac{1}{R}-\frac{1}{r}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
v\left(r\right)=-\left(2\cdot\frac{G\cdot M}{R}\left(\frac{R}{r}-1\right)\right)^{\frac{1}{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Però volem conèixer el temps que triga en col·lapsar 
\begin_inset Formula $v\left(r\right)=\frac{dr}{dt}\Rightarrow dt=\frac{dr}{v\left(r\right)}$
\end_inset

, per tant:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\int_{t=0}^{t=t_{cl}}dt=\int_{r=R}^{r=0}\frac{1}{v\left(r\right)}dr
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
t_{cl}=\int_{r=R}^{r=0}\frac{1}{-\left(2\cdot\frac{G\cdot M}{R}\left(\frac{R}{r}-1\right)\right)^{\frac{1}{2}}}dr=-\left(\frac{R}{2\cdot G\cdot M}\right)^{\frac{1}{2}}\int_{r=R}^{r=0}\frac{1}{\left(\frac{R}{r}-1\right)^{\frac{1}{2}}}dr
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Invertim el periode d'integració per poder canviar el punt de referència
 i el signe:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
t_{cl}=\left(\frac{R}{2\cdot G\cdot M}\right)^{\frac{1}{2}}\int_{r=0}^{r=R}\frac{1}{\left(\frac{R}{r}-1\right)^{\frac{1}{2}}}dr
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per integrar, fem canvi de variable 
\begin_inset Formula $q=\frac{R}{r}$
\end_inset

, per tant 
\begin_inset Formula $r=\frac{R}{q}$
\end_inset

 i 
\begin_inset Formula $dr=-\frac{R}{q^{2}}dq$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\int_{r=0}^{r=R}\frac{1}{\left(\frac{R}{r}-1\right)^{\frac{1}{2}}}dr=\int_{q=\infty}^{q=1}\frac{1}{\left(q-1\right)^{\frac{1}{2}}}\cdot\left(-\frac{R}{q^{2}}dq\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
A partir del llistat d'integrals del llibre 
\emph on

\begin_inset Quotes eld
\end_inset

Schaum's Mathematical handbook of Formulas and Tables
\begin_inset Quotes erd
\end_inset


\emph default
: 
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
t_{cl}=\sqrt{\frac{R^{3}}{2GM}}\left[\frac{\sqrt{q-1}}{q}+\arctan\sqrt{q-1}\right]_{1}^{\infty}=\frac{\pi}{2\sqrt{2}}\left(\frac{R^{3}}{GM}\right)^{\frac{1}{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Cas del sol
\end_layout

\begin_layout Standard
\begin_inset Formula $R=6.,96\cdot10^{10}cm$
\end_inset

, 
\begin_inset Formula $M=1,98\cdot10^{33}g$
\end_inset

 i 
\begin_inset Formula $G=6,6732\cdot10^{-8}$
\end_inset

 per tant 
\begin_inset Formula $t_{d}=1.775s=29,6minuts$
\end_inset


\end_layout

\begin_layout Subsubsection
Cas d'un núvol molecular
\end_layout

\begin_layout Standard
\begin_inset Formula $R=10pc=3,086\cdot10^{19}cm$
\end_inset

, 
\begin_inset Formula $M=10^{3}M_{\odot}=1,98\cdot10^{36}g$
\end_inset

 i 
\begin_inset Formula $G=6,6732\cdot10^{-8}$
\end_inset

 per tant 
\begin_inset Formula $t_{d}=5,24\cdot10^{14}s=16,6\cdot10^{6}anys$
\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
La llei de conservació de la massa tenint en compte que la densitat ha de
 ser del tipus 
\begin_inset Formula $\rho=ar^{b}$
\end_inset

 és 
\begin_inset Formula $\frac{dm}{dr}=4\pi r^{2}\rho\left(r\right)=4\pi ar^{b+2}$
\end_inset

.
 Per tant, la massa en funció del radi:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m\left(r\right)=\intop_{0}^{r}dm=\int_{0}^{r}4\pi ar^{b+2}dr=\frac{4\pi ar^{b+3}}{b+3}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per altra banda, l'equilibri hidrostàtic es compleix si 
\begin_inset Formula $\frac{dP}{dr}=-g\left(r\right)\rho\left(r\right)=-\frac{Gm\left(r\right)}{r^{2}}\cdot a\cdot r^{b}=-\frac{G\cdot4\pi ar^{b+3}}{r^{2}\cdot\left(b+3\right)}\cdot a\cdot r^{b}=-\frac{G\cdot4\pi a}{\left(b+3\right)}\cdot\frac{r^{b+3}\cdot r^{b}}{r^{2}}$
\end_inset


\end_layout

\begin_layout Standard
Com estem considerant que és un gas ideal (
\begin_inset Formula $P\cdot V=nkT\Rightarrow P=\frac{m}{V\cdot\mu\cdot m_{H}}kT=\rho\left(r\right)\frac{k}{\mu\cdot m_{H}}T=ar^{b}\frac{k}{\mu\cdot m_{H}}T$
\end_inset

) i la temperatura es constant 
\begin_inset Formula $T\left(r\right)=ct$
\end_inset

, llavors la variació de la pressió respecte al radi serà:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{dP}{dr}=\frac{d}{dr}\left(a\cdot r^{b}\frac{k}{\mu\cdot m_{H}}T\right)=a\cdot b\cdot r^{b-1}\cdot\frac{k}{\mu\cdot m_{H}}T
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Combinant les dos anteriors equacions:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
a\cdot b\cdot r^{b-1}\cdot\frac{k}{\mu\cdot m_{H}}T=\frac{G\cdot4\pi a}{\left(b+3\right)}\cdot\frac{r^{b+3}\cdot r^{b}}{r^{2}}\Rightarrow b\cdot\frac{k}{\mu\cdot m_{H}}T\cdot r^{b-1}=-\frac{G\cdot4\pi a}{\left(b+3\right)}\cdot r^{2b+1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, igualant els exponents i els coeficients:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
2b+1=b-1\Rightarrow b=-2
\end{equation}

\end_inset

 
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
-2\cdot\frac{k}{M}T=-\frac{G\cdot4\pi a}{\left(-2+3\right)}\Rightarrow a=\frac{kT}{2\pi\mu\cdot m_{H}G}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $a$
\end_inset

 i 
\begin_inset Formula $b$
\end_inset

 són valors reals, la funció de densitat proposada pot complir l'equació
 d'equilibri hidrostàtic.
\end_layout

\begin_layout Standard
La massa respecte al radi creix de forma proporcional:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
m\left(r\right)=\frac{4\pi ar^{b+3}}{b+3}=\frac{2\cdot k\cdot T}{M\cdot G}\cdot r
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
La cadena protó-protó al sol allibera 
\begin_inset Formula $27MeV\frac{1,6\cdot10^{-6}erg}{1MeV}=4,32\cdot10^{-5}erg$
\end_inset

, donat que la lluminositat és 
\begin_inset Formula $L_{\odot}=3,9\cdot10^{33}erg/s$
\end_inset

, es produeixen 
\begin_inset Formula $n_{p-p}=\frac{3,9\cdot10^{33}erg/s}{4,32\cdot10^{-5}erg}=9,027\cdot10^{37}$
\end_inset

 reaccions p-p per segon.
\end_layout

\begin_layout Standard

\emph on
Hipòtesis
\emph default
: considerem que el sol esta compost únicament per H i no es produeixen
 reaccions nuclears més enllà del Heli.
\end_layout

\begin_layout Standard
En una reacció p-p desapareixen 4 àtoms d'H, cadascun amb una massa 
\begin_inset Formula $m_{H}\simeq m_{protó}=1,673\cdot10^{-24}g$
\end_inset

.
 Si el sol té una massa 
\begin_inset Formula $M_{\odot}=2\cdot10^{33}g$
\end_inset

, llavors existeixen 
\begin_inset Formula $n_{H}=\frac{2\cdot10^{33}g}{1,673\cdot10^{-24}g}=1,1955\cdot10^{57}$
\end_inset

 àtoms d'hidrogen que es consumeixen a un ritme de 
\begin_inset Formula $4\cdot9,027\cdot10^{37}$
\end_inset

 per segon, per tant la durada del sol:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
t=\frac{1,1955\cdot10^{57}atoms}{4\cdot9,027\cdot10^{37}atoms/s}=3,3108\cdot10^{18}s\frac{1h}{3600s}\frac{1d}{24h}\frac{1any}{365.25d}=1\cdot10^{11}anys
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
La lluminositat del sol és 
\begin_inset Formula $L_{\odot}=3,9\cdot10^{33}erg/s\cdot\frac{1g\text{·}cm^{2}/s^{2}}{1erg}$
\end_inset

 i la relació entre energia i massa segons Einstein 
\begin_inset Formula $E=m\cdot c^{2}$
\end_inset

, llavors:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{dE}{dt}=c^{2}\cdot\frac{dm}{dt}\Rightarrow\frac{dm}{dt}=\frac{\frac{dE}{dt}}{c^{2}}=\frac{L_{\odot}}{c^{2}}=\frac{3,9\cdot10^{33}g\text{·}cm^{2}/s}{\left(299.792.458\cdot10^{2}cm/s\right)^{2}}=4,3393\cdot10^{12}g/s
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Si la durada del sol dintre de la seqüència principal és 
\begin_inset Formula $\Delta t=10^{10}anys\frac{365.25d}{1any}\frac{24h}{1d}\frac{3600s}{1h}=3,1558\cdot10^{17}$
\end_inset

 i la variació de la massa depèn de 
\begin_inset Formula $\Delta m=\frac{dm}{dt}\cdot\Delta t$
\end_inset

, llavors el percentatge de massa perduda:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{\Delta m}{M_{\odot}}=\frac{4,3393\cdot10^{12}g/s\cdot3,1558\cdot10^{17}s}{2\cdot10^{33}g}=0,0007=0,07\%
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
La funció inicial de massa (IMF) estableix el nombre d'estel per intervals
 de massa: 
\begin_inset Formula $n\left(M\right)dM\propto M^{-2,3}dM$
\end_inset

.
\end_layout

\begin_layout Standard
Sigui 
\begin_inset Formula $M_{1}$
\end_inset

la massa d'una estrella poc massiva i 
\begin_inset Formula $M_{2}$
\end_inset

 la massa d'una estrella molt massiva a on 
\begin_inset Formula $M_{1}\ll M_{2}$
\end_inset

, podem calcular el nombre d'estrelles en l'interval:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
N_{total}=\int_{M_{1}}^{M_{2}}n\left(M\right)dM\propto\int_{M_{1}}^{M_{2}}M^{-2,3}dM=\frac{-0.769}{M_{2}^{1,3}}-\frac{-0.769}{M_{1}^{1,3}}\approx\frac{0.769}{M_{1}^{1,3}}
\end{equation}

\end_inset

 donat que 
\begin_inset Formula $\frac{-0.769}{M_{2}^{1,3}}\sim0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
M_{total}=\int_{M_{1}}^{M_{2}}M\cdot n\left(M\right)dM\propto\int_{M_{1}}^{M_{2}}M\cdot M^{-2,3}dM=\int_{M_{1}}^{M_{2}}M^{-1,3}dM=\frac{-3.3}{M_{2}^{0,3}}-\frac{-3.3}{M_{1}^{0,3}}\approx\frac{3.3}{M_{1}^{0,3}}
\end{equation}

\end_inset

 donat que 
\begin_inset Formula $\frac{-3.3}{M_{2}^{0,3}}\sim0$
\end_inset


\end_layout

\begin_layout Standard
Per tant, la 
\begin_inset Formula $N_{total}\left(M\right)\cdot M_{total}$
\end_inset

 per cada franja de masses seria similar a:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Graphics
	filename Massa vs quantitat.jpg
	lyxscale 50
	scale 35

\end_inset


\end_layout

\begin_layout Standard
Clarament la major part de la massa es troba en estrelles de massa petita.
\end_layout

\begin_layout Standard
Per altra banda, respecte a la lluminositat 
\begin_inset Formula $L\propto M^{3.45}$
\end_inset

, per tant:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
L_{total}=\int dL=\int_{M_{1}}^{M_{2}}L\left(M\right)\cdot n\left(M\right)\cdot dM\propto\int_{M_{1}}^{M_{2}}M^{3.45}\cdot M^{-2,3}\cdot dM=\frac{1}{2.15}\cdot\left(M_{2}^{2.15}-M_{1}^{2.15}\right)\approx\frac{1}{2.15}\cdot\left(M_{2}^{2.15}\right)
\end{equation}

\end_inset

 donat que 
\begin_inset Formula $M_{1}^{2.15}\sim0$
\end_inset


\end_layout

\begin_layout Standard
Per tant, la 
\begin_inset Formula $N_{total}\left(M\right)\cdot L_{total}$
\end_inset

 per cada franja de masses seria similar a:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Graphics
	filename Massa vs lluminositat.jpg
	lyxscale 50
	scale 35

\end_inset


\end_layout

\begin_layout Standard
És obvi que la major part de la lluminositat prové d'estrelles amb masses
 grans.
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
Per tal que una partícula estigui gravitacionalment lligada a un sistema
 
\begin_inset Formula $E_{total}=E_{cinètica}+E_{potencial}<0$
\end_inset

.
 Per tant, el lligam es trenca si l'energia total és igual o superior a
 0.
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
E_{total}=E_{c}+E_{p}\geq0\Rightarrow E_{c}=-E_{p}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Utilitzant l'energia potencial 
\begin_inset Formula $E_{p}=-\frac{G\cdot M\cdot m}{r}$
\end_inset

 i 
\begin_inset Formula $E_{c}=\frac{1}{2}m\cdot v^{2}$
\end_inset

, llavors la velocitat d'escapament d'una partícula és:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{1}{2}m\cdot v_{esc}^{2}=\frac{G\cdot M\cdot m}{r}\Rightarrow v_{esc}=\sqrt{\frac{2\cdot G\cdot M}{r}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tal que l'àtom estigui lligat:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{3}{2}kT<\frac{1}{2}m\cdot v_{esc}^{2}=\frac{1}{2}m_{H}\cdot\frac{2\cdot G\cdot M}{r}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
El radi de l'estrella el podem deduir a partir de 
\begin_inset Formula $L=4\pi R^{2}\sigma T^{4}$
\end_inset

 i 
\begin_inset Formula $L=1,3\cdot10^{6}L_{\odot}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
R=\sqrt{\frac{L}{4\pi\sigma T^{4}}}=\sqrt{\frac{1,3\cdot10^{6}\cdot3,84\cdot10^{33}erg\, s^{-1}}{4\pi\cdot5,67\text{·}10^{-5}erg\, cm^{-2}K^{-4}\cdot\left(2500K\right)^{4}}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Per tant, la massa mínima per a que una partícula no escapi de l'atracció
 gravitatòria:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
\frac{3}{2}kT<\frac{1}{2}m_{H}\cdot\frac{2\cdot G\cdot M}{\sqrt{\frac{L}{4\pi\sigma T^{4}}}}\Rightarrow M>\frac{3}{2}\frac{k}{G\cdot m_{H}}\sqrt{\frac{L}{4\pi\sigma T^{4}}}
\end{equation}

\end_inset

a on la constant de Boltzman és 
\begin_inset Formula $k=1.3806\cdot10^{-16}1g\text{·}cm^{2}\cdot s^{-2}\cdot K^{-1}$
\end_inset

, la constant gravitacional 
\begin_inset Formula $G=6.6726\cdot10^{-11}N\, m^{2}/kg^{2}$
\end_inset

, la massa de l'hidrogen 
\begin_inset Formula $m_{H}=1,637\cdot10^{-24}g$
\end_inset

, la lluminositat 
\begin_inset Formula $L=1,3\cdot10^{6}L_{\odot}=1,3\cdot10^{6}\cdot3,84\cdot10^{33}erg\, s^{-1}$
\end_inset

 i la temperatura 
\begin_inset Formula $T=2500K$
\end_inset

.
\end_layout

\begin_layout Standard
Substituint aquestes darrers valors obtenim 
\begin_inset Formula $M>1.97\cdot10^{33}g$
\end_inset


\end_layout

\begin_layout Section
La Galàxia
\end_layout

\begin_layout Subsection
Problema
\end_layout

\begin_layout Standard
Entre els 5 kpc i els 8.5 kpc es troba un anell amb núvols moleculars, que
 suposarem que giren en órbits circulars de radi 
\begin_inset Formula $R_{0}=5kpc$
\end_inset

.
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Graphics
	filename La galaxia - above.png
	lyxscale 50
	scale 35

\end_inset


\begin_inset Graphics
	filename La galaxia - side.png
	lyxscale 50
	scale 35

\end_inset


\end_layout

\begin_layout Standard
L'acceleració centrifuga respecte al sistema de referència que gira amb
 velocitat angular 
\begin_inset Formula $w_{0}$
\end_inset

en la posició 
\begin_inset Formula $R_{0}$
\end_inset

 ve donada per l'expressió 
\begin_inset Formula $a_{c}=4A\left(A-B\right)\Delta R$
\end_inset

 a on es prendran com a constant d'Oort 
\begin_inset Formula $A=22km\cdot s^{-1}\cdot kpc^{-1}$
\end_inset

 i 
\begin_inset Formula $B=-22km\cdot s^{-1}\cdot kpc^{-1}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Massa mínima
\end_layout

\begin_layout Standard
Per a que un núvol de radi 
\begin_inset Formula $r=\frac{D}{2}$
\end_inset

 sobrevisqui als efectes de la rotació diferencial de la galàxia, la força
 de la gravetat ha de compensar l'acceleració centrífuga:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Graphics
	filename La galaxia - nuvol.png
	lyxscale 75
	scale 50

\end_inset


\end_layout

\begin_layout Standard
Per tant, la força de la gravetat pròpia del núvol 
\begin_inset Formula $F_{g}=\frac{GM}{r^{2}}=\frac{4GM}{D^{2}}$
\end_inset

 ha de ser superior o igual a l'acceleració centrífuga 
\begin_inset Formula $4A\left(A-B\right)\frac{D}{2}\leq\frac{4GM}{D^{2}}$
\end_inset

 i la massa mínima requerida per a la seva supervivència: 
\begin_inset Formula $M_{min}\left(D\right)\geq A\left(A-B\right)\frac{D^{3}}{2G}$
\end_inset


\end_layout

\begin_layout Subsubsection
Efectes de la rotació diferencial
\end_layout

\begin_layout Standard
Els diàmetres dels núvols estan entre 22 i 80pc amb una llei de distribució
 del tipus 
\begin_inset Formula $N\left(D\right)=5,3\cdot10^{5}D^{-2,3}$
\end_inset

.
 Per tant, la massa mínima del tots els núvols moleculars que sobreviuen:
\end_layout

\begin_layout Standard
\begin_inset Formula $M_{tot}=\int_{D=22}^{D=80}N\left(D\right)M_{min}\left(D\right)dD=5,2\cdot10^{7}M_{\odot}$
\end_inset


\end_layout

\begin_layout Standard
Aquesta massa mínima representa una petita fracció de la massa total real
 estimada (
\begin_inset Formula $2-4\cdot10^{9}M_{\odot}$
\end_inset

), per tant els núvols moleculars són prou estables com per superar els
 efectes de la rotació galàctica.
 El trencament d'aquests núvols ha de tenir un origen diferent, com la formació
 d'estrelles als seu nucli o les supernoves.
\end_layout

\end_body
\end_document
