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%% Datos
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\title{Los puntos de Lagrange}
\institute{Centro Penitenciario de Jaén\hfill {\color{black}UNED Jaén}\\ Alcalá la Real}
\date{24, 25 de junio de 2025}
\author[Fernando Chamizo]{Fernando Chamizo {\footnotesize (UAM-ICMAT)}}
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\begin{document}


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\tableofcontents
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\section{Introducción}
\begin{frame}


\frametitle{Un poco de filosofía}



\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=I. Newton (1643--1727)\hfill{\footnotesize (artífice de la gravitación)}]
\begin{tabular}{cl}
	\begin{tabular}{c}
	\includegraphics[height=95pt]{./images/newton.jpg}
	\\[-6pt]
	{\tiny (Fuente: MacTutor)}
	\end{tabular}
&
\hspace{-21pt}
	\begin{tabular}{p{7.0cm}}
No sé cómo me verá el resto del mundo; pero yo me considero como un niño jugando en la playa que se alegra de vez en cuando al encontrar un guijarro más liso o un concha más bonita de lo normal, mientras delante de mí, sin que me diera cuenta, se extiende el vasto océano de la verdad.
	\end{tabular}
\end{tabular}
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\begin{center}
¿La ciencia es una recopilación de observaciones?
\end{center}



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\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\begin{center}
	\begin{tabular}{cc}
    \includegraphics[height=84pt]{./images/saturn.jpg}
    &
    \includegraphics[height=84pt]{./images/saturn_h.jpg}
    \\
    Saturno & Su polo norte
	\end{tabular}
\end{center}
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\begin{center}
	\begin{tabular}{ccc}
    \includegraphics[height=84pt]{./images/hyperion.jpg}
    &
    \includegraphics[height=84pt]{./images/iapetus.jpg}
    &
    \includegraphics[height=84pt]{./images/enceladus.jpg}
    \\
    Hiperión & Jápeto & Encélado
	\end{tabular}
\end{center}
\end{miblock}

\vspace{-4pt}

{\scriptsize Fuentes, APOD 20241102, 20130220, 20110227, 20230226, 20250610}

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\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\small
La ciencia, y muy especialmente la física, busca \textbf{primeros principios} que expliquen los ejemplos (experimentos) y permitan hacer predicciones.
\end{miblock}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]
\small
Uno de los grandes motores de la revolución científica fue la mecánica celeste, la explicación de los movimientos de los astros a través de la ley de gravitación universal.
La obra de Newton,  \emph{Principios matemáticos de la filosofía natural} mostraba el cosmos funcionando como un mecanismo de relojería. 
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
	\begin{tabular}{c}
	\includegraphics[height=90pt]{./images/principia1.jpg}
	\\[-6pt]
	{\tiny 1ª edición (1687)}
	\end{tabular}
	\begin{tabular}{c}
	\includegraphics[height=90pt]{./images/principia2.jpg}
	\\[-6pt]
	{\tiny Una página}
	\end{tabular}
	\begin{tabular}{c}
	\includegraphics[height=90pt]{./images/principia3.png}
	\\[-6pt]
	{\tiny Ilustración del movimiento de un cometa}
	\end{tabular}
\end{miblock}

\vspace{-10pt}

	{\tiny (Fuente: Wikipedia)}

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Las matemáticas han estado desde la antigüedad muy ligadas a la astronomía. 


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
El \emph{quadrivium} agrupaba las cuatro ``artes liberales'' relacionadas con las matemáticas:  aritmética, geometría, música y astronomía. 
\end{miblock}



\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=Galileo Galilei (1564--1642)\hfill{\footnotesize (Defensor del heliocentrismo)}]
\begin{tabular}{cl}
	\begin{tabular}{p{7.0cm}}
\lq\lq
[El universo] no puede leerse hasta que no hayamos aprendido el lenguaje y nos sean familiares los caracteres en que está escrito. Está escrito en lenguaje matemático y las letras son triángulos, círculos y otras figuras geométricas, sin las cuales es humanamente imposible entender ni una palabra''. 
	\end{tabular}
&
	\begin{tabular}{c}
	\includegraphics[height=80pt]{./images/galileo2.jpg}
	\\[-6pt]
	{\tiny (Fuente: MacTutor)}
	\end{tabular}
\end{tabular}
\end{miblock}


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Las matemáticas tienen un poder increíble para formular modelos físicos que funcionen como primeros principios y para extraer conclusiones a partir de ellos.


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=E.P. Wigner (1902--1995)\hfill{\footnotesize (Premio Nobel de Física en 1963)}]
\begin{tabular}{cl}
	\begin{tabular}{c}
	\includegraphics[height=90pt]{./images/wigner.jpg}
	\\[-6pt]
	{\tiny (Fuente: MacTutor)}
	\end{tabular}
&
\hspace{-9pt}
	\begin{tabular}{p{7.0cm}}
\lq\lq
La enorme utilidad de las matemáticas en las ciencias naturales es algo que roza el misterio y que no tiene explicación racional''.
\medskip

\lq\lq El milagro de lo apropiado que es el lenguaje de las matemáticas para 
la formulación de las leyes de la física es un regalo maravilloso que 
ni entendemos ni merecemos``.
	\end{tabular}
\end{tabular}
\end{miblock}

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\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\begin{center}
  \begin{tabular}{c}
    \includegraphics[height=150pt]{./images/rosetta2.jpg}
    \\
    {\footnotesize Trayectoria}
  \end{tabular}
  \begin{tabular}{c}
    \includegraphics[height=150pt]{./images/rosetta1.jpg}
    \\
    {\footnotesize Cometa {\scriptsize (Fuente: ESA)}}
  \end{tabular}
\end{center}
\end{miblock}


\begin{miblock}[halign=flush left,colframe=white, boxrule=1pt, coltitle=white]
\scriptsize
La sonda espacial Rosetta de la ESA aprovechó el tirón gravitacional de varios astros durante 10 años para encontrarse con el cometa 
Churyumov-Gerasimenko. 
¡Esto requiere  un montón de ecuaciones y cálculos!
\end{miblock}

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\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=R. Hamming (1915--1998)]
\begin{center}
  \begin{tabular}{c}
    \includegraphics[height=85pt]{./images/hamming.jpeg}
 \\
 {\tiny (Fuente: MacTutor)}
  \end{tabular}
  \hspace{-8pt}
  \begin{tabular}{p{7cm}}
	No es que la ciencia explique ``por qué'' las cosas son como son --la gravitación no explica por qué las cosas caen-- sino que la ciencia da tantos detalles del ``cómo'' que tenemos la sensación de que entendemos el ``porqué''.
	\\
	\ 
  \end{tabular}
\end{center}
\end{miblock}


Aspiramos a que haya pocos principios, pero por el hecho de ser principios no sabemos deducirlos, no entendemos por qué existen. 

\begin{miblock}[halign=flush left,colframe=gray, boxrule=1pt, coltitle=white]
\begin{center}
No sabemos deducir la gravitación de primeros principios
\end{center}
\end{miblock}


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\begin{miblock}[halign=flush left,colframe=green, boxrule=1pt, coltitle=white]
\begin{center}
 Las teorías científicas siempre están sujetas a revisión 
\end{center}
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\vspace{5pt}
\begin{center}
 \begin{tabular}{c|c}
  \begin{tikzpicture}[scale=0.02]
	% circles
	\draw[black, line width=1.5pt] (0,0) ellipse (100 and 98);
	\draw[red,fill=yellow] (19.89,0) circle (6.96);
	\draw[brown,fill] (-50,84.87) circle (2);
	% crosses
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};  
	% names
    \node  at (19.89,-14) {Sol};
    \node  at (-40,74.87) {Mercurio};
 \end{tikzpicture}
 &
  \begin{tikzpicture}[scale=0.02]
	% circles
	\draw[black, line width=1.5pt] (0,0) ellipse (100 and 98);
	\foreach \x in {0,20,...,100}
     {
		\draw[rotate around={\x:(19.89,0)},black, line width=1.5pt]  (0,0) ellipse (100 and 98);
     }
	\draw[red,fill=yellow] (19.89,0) circle (6.96);
	\draw[brown,fill] (-50,84.87) circle (2);
	% crosses
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};  
	% names
    \node  at (19.89,-14) {Sol};
%    \node  at (-40,74.87) {Mercurio};
 \end{tikzpicture}
 \\
 \hline
 Teoría (Newton)
 &
 Teoría (Einstein)
 \end{tabular}
\end{center}
\end{miblock}

Descontando el efecto del resto de los planetas, 
la ``circunferencia'' de la órbita de Mercurio gira alrededor del Sol como un {\sl hula hoop}.

{\footnotesize Es un efecto muy débil: un giro completo cada 3 millones de años.}


\end{frame}




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\section{Bases de mecánica}
\begin{frame}
\frametitle{Kepler nos engaña}

%\

\begin{miblock}[halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white, title=Primera ley de Kepler (1609)]
Los planetas describen órbitas elípticas {\color{gray}con el Sol en uno de los focos}. 
\end{miblock}

\begin{center}
\begin{tabular}{ccc}
% \cellcolor{white} 
    \includegraphics[height=85pt]{./images/kepler.jpeg}
  &
 \cellcolor{white} 
 \begin{tikzpicture}[scale=0.74]
    \draw[line width=2pt] (0,0) ellipse (2cm and 1cm);
    	\node at (0,-1.6) {\,};
 \end{tikzpicture}
 &
 \hspace{-10pt}   \includegraphics[height=85pt]{./images/newton.jpg}
 \\
 {\sf\small Kepler} {\tiny (Fuente: MacTutor)}
 &
 {\sf Elipse}
 &
 {\sf\small Newton} {\tiny (Fuente: MacTutor)}
\end{tabular}
\end{center}

Newton probó para las órbitas periódicas:
\vspace{-10pt}
\[
 \text{\sf\footnotesize Ley de Gravitación universal}
 \begin{array}{c}
  {\scriptstyle\Rightarrow}
  \\[-4pt]
  \Leftarrow
 \end{array}
 \text{\sf\footnotesize órbitas elípticas}
\]
\end{frame}


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\begin{frame}

\begin{miblock}[halign=flush left,colframe=red,boxrule=1pt, coltitle=white]
Esto es un poco engañoso, las órbitas de los 8 planetas son circunferencias casi perfectas.
\end{miblock}

Medida matemática de la redondez de una elipse:
\[
 \begin{array}{c}
  e=\text{excentricidad}
  \\
  e\approx 0\ \to\ \text{muy redonda},
  \qquad\ \qquad
  e\approx 1\ \to\ \text{muy oblonga}.
 \end{array}
\]


\begin{center}
\begin{tabular}{ccc}
 \cellcolor{white} 
 \begin{tikzpicture}[scale=0.75]
    \draw[line width=2pt] (0,0) ellipse (2cm and 1.9899cm);
    	\node at (0,-1.4) {\,};
 \end{tikzpicture}
  &\phantom{mmmm}&
 \cellcolor{white} 
 \begin{tikzpicture}[scale=0.75]
    \draw[line width=2pt] (0,0) ellipse (2cm and 0.62cm);
    	\node at (0,-1.8) {\,};
 \end{tikzpicture}
 \\
 $e = 0.1$
 &&
 $e = 0.95$
\end{tabular}
\end{center}


\end{frame}



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\vspace{3pt}

\begin{tcolorbox}[width=140pt,halign=flush left,colframe=azulUAM,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\noindent
{\bf\underbar{Matemáticas}}
\vspace{6pt}

Largo\,$=$\,Ancho$\sqrt{1-e^2}$
\end{tcolorbox}
\begin{tcolorbox}[width=160pt,halign=flush left,colframe=azulUAM,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\noindent
\hspace{-4pt}
{\bf\underbar{Cuenta de la vieja}} \,{\tiny ($e$ pequeño)}
\vspace{-6pt}

$50e^2\approx$\,$\%$ de reducción
\end{tcolorbox}



\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\ \ {\color{brown}\small Planetas interiores}\qquad
\begin{tabular}{l|c|c|c|c|}
&\mercury\ {\tiny Mercurio}&\venus\ {\tiny Venus}&\earth\ {\tiny Tierra}&\mars\ {\tiny Marte}
\\
\hline
e&
 0.2056
&0.0068
&0.0167
&0.0934
\\
\hline
\end{tabular}
\vspace{8pt}

\ \ {\color{blue!40!white}\small Planetas exteriores}\qquad
\begin{tabular}{l|c|c|c|c|}
&\jupiter\ {\tiny Júpiter}&\saturn\ {\tiny Saturno}&\uranus\ {\tiny Urano}&\neptune\ {\tiny Neptuno}
\\
\hline
e&
0.0484
&0.0542
&0.0472
&0.0086
\\
\hline
\end{tabular}
\end{miblock}

\vspace{7pt}


{\color{blue}Caso extremo:} Mercurio $\to$ $2\%$ 

{\color{blue}Caso medio:} Júpiter, Urano $\to$ $0.1\%$

{\color{blue}Nuestra casa:} Tierra $\to$ $0.01\%$

\


Las órbitas de los cometas sí son muy elípticas. 

{\color{blue}Cometa Halley:}  $75\%$



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\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=Órbita de Mercurio a escala\hfill{\scriptsize ¿Alguien no la ve redonda?}]
\vspace{5pt}
\begin{center}
  \begin{tikzpicture}[scale=0.03]
	% circles
	\draw[black, line width=1.5pt] (0,0) ellipse (100 and 98);
	\draw[red,fill=yellow] (19.89,0) circle (6.96);
	\draw[brown,fill] (-50,84.87) circle (2);
	% crosses
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};  
	% names
    \node  at (19.89,-14) {Sol};
    \node  at (-40,74.87) {Mercurio};
 \end{tikzpicture}
\end{center}
\end{miblock}


{\tiny Los tamaños del Sol y Mercurio no están a escala.}



\end{frame}



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\begin{frame}
\frametitle{Dos fuerzas que mueven los mundos}

\vspace{-10pt}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=Protagonistas de la mecánica]
\begin{itemize}
 \item 
 \textbf{Fuerza} $\to$ lo que causa aceleraciones
 \vspace{-6pt}
 \[
  F = m\cdot a
 \]
 \item 
 \textbf{Aceleración} $\to$ variación de la velocidad con el tiempo
 \vspace{-6pt}
 \[
  a = \dot{v}
  \qquad
  \Big(a = \frac{dv}{dt}\Big)
 \]
 \item 
 \textbf{Velocidad:} Todos sabemos qué es eso
\end{itemize}

\end{miblock}


\begin{tcolorbox}[width=145pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
{\small\begin{tabular}{l}
Son \textbf{vectoriales}:
\\
tienen una magnitud
\\
y una dirección
\end{tabular}}
\end{tcolorbox}
\quad
\begin{tcolorbox}[width=140pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\begin{tabular}{c}
  \begin{tikzpicture}[scale=0.75]
    \draw[-{Latex[length=8pt]},line width=2pt] (0,0) -- (1,1);
    \node  at (0.15,0.5) {$F$};
    \draw[-{Latex[length=8pt]},line width=2pt] (2,0) -- (2,1.41);
    \node  at (1.6,0.5) {$F$};
    \draw[-{Latex[length=8pt]},line width=2pt] (2.9,0.5) -- (2.9+1.41,0.5);
    \node  at (3.6,0.9) {$F$};
 \end{tikzpicture}
\end{tabular}
\end{tcolorbox}

\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{-10pt}

\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white, title=Ley de gravitación universal]
Masa puntual $M$ $\to$ aceleración $a$
\[
 \begin{array}{c}
   \boxed{\displaystyle a=\frac{GM}{d^2}}
   \\[20pt]
   \text{dirigida hacia la masa}
 \end{array}
 \qquad\ \qquad
 \begin{array}{c}
  \begin{tikzpicture}[scale=0.35]
	% circles
	\draw[red,fill=red] (0,0) circle (5pt);
	\draw[line width = 1pt] (4,6) node[cross=4pt,rotate=45,blue]{};  
	% arrow
%	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.15,6*0.15) -- (4*0.98,6*0.98);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4,6) -- (4*0.03,6*0.03);
	% names
	\node at (1,0) [right] {\sf\small masa $M$};
	\node at (2,3) [right] {$d$};
	\node at (4.1,5.5) [right] {\small\begin{tabular}{l}
	                                 masa de
	                                 \\
	                                 prueba
	                                \end{tabular}};
  \end{tikzpicture}
 \end{array}
\]
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
$d =$\, distancia

$G=$ constante de gravitación 
$6.67430\cdot 10^{-11}\, m^3/(kg\cdot s^2)$
\end{miblock}

Newton $\to$ la ley también es cierta cuando en vez de una masa puntual tenemos una masa esférica y consideramos la distancia al centro. 


\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{-10pt}

\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white, title=Fuerza centrífuga]
Experimentada por partículas al obligarlas a girar
\[
 \begin{array}{c}
   \boxed{\displaystyle a=\omega^2 R}
   \\[20pt]
   \text{radial hacia afuera}
 \end{array}
 \qquad\ \qquad
 \begin{array}{c}
  \begin{tikzpicture}[scale=0.35]
	% circles
	\draw[red, line width = 1pt] (0,0) circle (3);
	\draw[line width = 0.5pt] (0,0) node[cross=2pt,rotate=0,black]{};  
	\draw[line width = 0.5pt] ({3*cos(20)},{3*sin(20)}) node[cross=4pt,rotate=0,black]{};  
	\draw[line width = 0.5pt, fill] ({3*cos(20)},{3*sin(20)}) circle (4pt);  
	% arrow
	\draw[-{Latex[length=8pt]}, line width = 1pt] ({3*cos(20)},{3*sin(20)}) -- ({6.2*cos(20)},{6.2*sin(20)});
	% names
	\node at ({3*cos(20)},{3*sin(20)-0.5}) [right] {\small partícula};
	\node at ({6.2*cos(20)-2},{6.2*sin(20)}) {\small $a$};
 \end{tikzpicture}
 \end{array}
\]
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
$R =$\, radio

$\omega= 2\pi\,\text{vueltas}/\text{tiempo}$\quad (velocidad angular)
\end{miblock}


\hspace{\stretch{1}}\begin{tcolorbox}[width=280pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{3pt}
 \begin{tabular}{c}
  \begin{tikzpicture}[scale=1.2]
	% circles
	\draw[line width = 1pt, red] (-1,-1) parabola bend (0,0) (1,-1);
	\draw[line width = 1pt, dashed, blue] (0,-0.5) circle (0.5);  
	\draw[line width = 0.5pt] (0,0) node[cross=3pt,rotate=0,black]{};   
	\end{tikzpicture}
 \end{tabular}
 \begin{tabular}{p{5cm}}
  \footnotesize Para movimientos no circulares, la fuerza centrífuga se calcula aproximando por una circunferencia.
 \end{tabular}
\end{tcolorbox}\hspace{\stretch{1}}



\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{-10pt}


\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white]
¿Por qué no medimos $\omega$ en vueltas/tiempo?
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\footnotesize En matemáticas es común medir el número de vueltas multiplicadas por $2\pi$, de este modo 
\vspace{-2pt}
\[
 \text{longitud} = \text{nº vueltas}\times \text{radio}.
\]
\end{miblock}


\begin{miblock}[halign=flush left,colframe=blue, boxrule=0.5pt, coltitle=white]
\footnotesize 
  \begin{tabular}{c}
%     \includegraphics[height=85pt]{./images/pizza.jpg}
\begin{overpic}[width=85pt]{./images/pizza.jpg}
\put(82,72){\color{green}$\alpha$}
\end{overpic} 
  \end{tabular}
\begin{tabular}{c|c|c|}
 &Gente de a pie & Matemático
 \\
 \hline
 Porción $\alpha$
 &
 $1/8$ de pizza
 &
 $\pi/4$ de pizza
 \\
 \hline
 Una pizza en $1\, h$
 &
 $\omega=1/h$
 &
 $\omega=2\pi/h$
 \\
 \hline
 Longitud borde
 &
 $2\pi \alpha R$
 &
 $\alpha R$
 \\
 \hline
 Área
 &
 $\pi \alpha R^2$
 &
 $\alpha R^2/2$
 \\
 \hline
\end{tabular}

\end{miblock}



\end{frame}


%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}
\frametitle{Uno igual a dos}

\vspace{-15pt}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white, title=Problema circular de un cuerpo]
 \vspace{-10pt}
\[
 \begin{array}{c|c}
 \begin{array}{c}
  \begin{tikzpicture}[scale=0.65]
	% circles
	\draw[blue, line width = 0.5pt, dashed] (0,0) circle (3);
	\draw[red,fill=red] (0,0) circle (5pt);
	\draw[line width = 0.5pt] ({3*cos(20)},{3*sin(20)}) node[cross=4pt,rotate=0,black]{};  
	\draw[line width = 0.5pt, fill] ({3*cos(20)},{3*sin(20)}) circle (1pt);  
	% arrow
	\draw[-{Latex[length=8pt]}, line width = 1pt] ({3*cos(20)},{3*sin(20)}) -- ({5.0*cos(20)},{5.0*sin(20)});
	\draw[-{Latex[length=8pt]}, line width = 1pt] ({3*cos(20)},{3*sin(20)}) -- ({1.0*cos(20)},{1.0*sin(20)});
	% names
	\node at ({3*cos(20)},{3*sin(20)-0.5}) [right] {\small\begin{tabular}{l}
	                                 masa de
	                                 \\
	                                 prueba
	                                \end{tabular}};
	\node[rotate=20] at ({6.2*cos(20)-2},{6.2*sin(20)-0.2}) {\small centríf.};
	\node[rotate=20] at ({4.2*cos(20)-2},{4.2*sin(20)-0.2}) {\small grav.};
	\node at (0,-1) {\sf\small masa $M$};
 \end{tikzpicture}
 \end{array}
 &
 \quad
 \begin{array}{c}
 \displaystyle \frac{GM}{d^2}=\omega^2 d
  \\
  \Downarrow
  \\[5pt]
  \boxed{\displaystyle \omega^2 =\frac{GM}{d^3}}
 \end{array}
 \end{array}
\]

\vspace{-20pt}

\begin{miblock}[halign=flush left,colframe=gray, boxrule=0.5pt, coltitle=white]
\small Parte de la tercera ley de Kepler.
\end{miblock}
\end{miblock}


\vspace{-5pt}


\begin{tcolorbox}[width=303pt,halign=flush left,colframe=brown,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
{\footnotesize\begin{tabular}{l}
Estación espacial:  $d=6798\, km$ $\to$ $\omega =2\pi \cdot 15.5\, \text{vueltas}/\text{día}$.
\\
Telescopio Hubble:  $d=6918\, km$ $\to$ $\omega =2\pi \cdot 15.1\, \text{vueltas}/\text{día}$.
\\
Satélite geoestacionario:  $d=36000\, km$ $\to$ $\omega =2\pi \cdot 1\, \text{vuelta}/\text{día}$.
\end{tabular}}
\end{tcolorbox}

\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{0pt}

{\small ¿Qué ocurre al considerar dos masas, cada una ejerciendo una gravedad no despreciable?}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\begin{tabular}{c|c}
 \underbar{\textbf{Problema de un cuerpo}}
 &
 \underbar{\textbf{Problema de dos cuerpos}}
 \\[8pt]
 \begin{tikzpicture}[scale=0.5]
	% circles
	\draw[red,fill=red] (0,0) circle (30pt);
	\draw[blue,fill=blue] (4,6) circle (5pt);
	% arrow
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.15,6*0.15) -- (4*0.98,6*0.98);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4,6) -- (4*0.15,6*0.15);
	% names
	\node at (1,0) [right] {\sf\small gordo};
	\node at (4.5,6) [right] {\sf\small flaco};
	\node at (2,3) [right] {$d$};
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.5]
	% circles
	\draw[red,fill=red] (0,0) circle (20pt);
	\draw[blue,fill=blue] (4,6) circle (10pt);
%	\draw[black,fill=black] (2*0.7,3*0.7) circle (2pt);
	% arrow
	\draw[line width = 1pt, dashed] (4*0.09,6*0.09) -- (4*0.95,6*0.95);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.09+0.7,6*0.09-0.3) -- (4*0.95+0.7,6*0.95-0.3);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.95+0.7,6*0.95-0.3) -- (4*0.09+0.7,6*0.09-0.3);
	%%%%
	\draw[-{Latex[length=8pt]}, line width = 1pt] (2*0.7-1,3*0.7+0.6) -- (4*0.09-1,6*0.09+0.6);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.09-1,6*0.09+0.6) -- (2*0.7-1,3*0.7+0.6);
	%%%%
	\draw[-{Latex[length=8pt]}, line width = 1pt] (2*0.7-1,3*0.7+0.6) -- (4*0.95-1,6*0.95+0.6);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.95-1,6*0.95+0.6) -- (2*0.7-1,3*0.7+0.6);
	% names
	\node at (1,0) [right] {\sf\small medio gordo};
	\node at (1-0.5,0-0.7) [right] {\sf\small $m_1$};
	\node at (4.5,6+0.2) [right] {\sf\small medio flaco};
	\node at (4.2,6-0.5) [right] {\sf\small $m_2$};
	\node at (3,3) [right] {$d$};
%	\node at (2*0.7,3*0.7) {$\mathbf{+}$};
	\draw[line width = 1pt] (2*0.7,3*0.7) node[cross=4pt,rotate=45,black]{};  
	\node at (2*0.7,3*0.7) [left] {\small $c$};
	\node at (2*0.8-2.2,3*0.8) {$d_1$};
	\node at (2*1.7-2.2,3*1.7) {$d_2$};
 \end{tikzpicture}
\end{tabular}

\begin{miblock}[halign=flush left,colframe=blue!40!white, boxrule=1pt, coltitle=white]
\footnotesize
$c=$\, centro de masas.

$d_1,\, d_2=$\, inversamente proporcionales a $m_1,\, m_2$.
\end{miblock}
\end{miblock}

{\small Ambos problemas son ``iguales'' cambiando el centro a $c$}

\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{-5pt}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\begin{tabular}{c|c}
 \begin{tikzpicture}[scale=0.25]
	% circles
	\draw[red,fill=red] (0,0) circle (30pt);
	\draw[blue,fill=blue] (4,6) circle (5pt);
	\draw[black, dashed] (0,0) circle (7.21);
	% arrow
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.15,6*0.15) -- (4*0.98,6*0.98);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4,6) -- (4*0.15,6*0.15);
	% names
	\node at (1,0) [right] {\sf\small gordo};
	\node at (4.5,6) [right] {\sf\small flaco};
	\node at (2,3) [right] {$d$};
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.5]
	% circles
	\draw[red,fill=red] (0,0) circle (20pt);
	\draw[blue,fill=blue] (4,6) circle (10pt);
	\draw[black,dashed] (2*0.7,3*0.7) circle (4.69);
	\draw[black,dashed] (2*0.7,3*0.7) circle (2.52);
	% arrow
	\draw[line width = 1pt, dashed] (4*0.09,6*0.09) -- (4*0.95,6*0.95);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.09+0.7,6*0.09-0.3) -- (4*0.95+0.7,6*0.95-0.3);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.95+0.7,6*0.95-0.3) -- (4*0.09+0.7,6*0.09-0.3);
	%%%%
	\draw[-{Latex[length=8pt]}, line width = 1pt] (2*0.7-1,3*0.7+0.6) -- (4*0.09-1,6*0.09+0.6);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.09-1,6*0.09+0.6) -- (2*0.7-1,3*0.7+0.6);
	%%%%
	\draw[-{Latex[length=8pt]}, line width = 1pt] (2*0.7-1,3*0.7+0.6) -- (4*0.95-1,6*0.95+0.6);
	\draw[-{Latex[length=8pt]}, line width = 1pt] (4*0.95-1,6*0.95+0.6) -- (2*0.7-1,3*0.7+0.6);
	% names
	\node at (1,0) [right] {\sf\small medio gordo};
	\node at (1-0.5,0-0.7) [right] {\sf\small $m_1$};
	\node at (4.5,6+0.2) [right] {\sf\small medio flaco};
	\node at (4.2,6-0.5) [right] {\sf\small $m_2$};
	\node at (3,3) [right] {$d$};
%	\node at (2*0.7,3*0.7) {$\mathbf{+}$};
	\draw[line width = 1pt] (2*0.7,3*0.7) node[cross=4pt,rotate=45,black]{};  
	\node at (2*0.7,3*0.7) [left] {\small $c$};
	\node at (2*0.8-2.2,3*0.8) {$d_1$};
	\node at (2*1.7-2.2,3*1.7) {$d_2$};
 \end{tikzpicture}
 \\
 \hline
 &
 \\[-6pt]
 $\displaystyle\omega^2= \frac{GM}{d^3}$
 &
 $\displaystyle\omega^2= \frac{G(m_1+m_2)}{d^3}$
\end{tabular}
\end{miblock}

\begin{tcolorbox}[width=303pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
{\small\begin{tabular}{l}
Distancia mayor $\to$ frecuencia menor
\\
Masa mayor $\to$ frecuencia mayor
\end{tabular}}
{\footnotesize\begin{tabular}{l}
($d$ crece $\Rightarrow$ $\omega$ decrece)
\\
($m$ crece $\Rightarrow$ $\omega$ crece) 
\end{tabular}}
\end{tcolorbox}


\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\vspace{-5pt}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
Radio del Sol: $695\,700\, km$.

Centro de masas Sol-Júpiter: $742\,370\, km$.

Centro de masas Sol-Tierra: $449\, km$.
\end{miblock}

\begin{center}
\begin{tabular}{ccc}
  \begin{tikzpicture}[scale=0.25]
	% circles
	\draw[red,fill=yellow] (0,0) circle (6.96);
%	\draw[black,fill=black] (7.42,0) circle (10pt);
	% crosses
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};  
	\draw[line width = 1pt] (7.42,0) node[cross=4pt,rotate=0,black]{};  
 \end{tikzpicture}
 &\phantom{mmm}&
  \begin{tikzpicture}[scale=0.25]
	% circles
	\draw[red,fill=yellow] (0,0) circle (6.96);
%	\draw[black,fill=black] (7.42,0) circle (10pt);
	% crosses
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};  
	\draw[line width = 1pt] (0.00449,0) node[cross=4pt,rotate=0,black]{};  
 \end{tikzpicture}
\end{tabular}
 
\end{center}

\vspace{-5pt}

\hspace{\stretch{1}}\begin{tcolorbox}[width=250pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=blue!20!white, nobeforeafter,tcbox raise=-18pt]
\vspace{-5pt}
  \begin{tabular}{p{4.5cm}}
{\footnotesize Estrictamente, el Sol no está fijo, se mueve con respecto al centro de masas del Sistema Solar.}
  \end{tabular}
  \quad
  \begin{tabular}{c}
     \includegraphics[height=55pt]{./images/galileo.jpg}
  \end{tabular}
  \vspace{-10pt}
\end{tcolorbox}\hspace{\stretch{1}}


\end{frame}


%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\section{El problema}
\begin{frame}
\frametitle{Cuando tres no es multitud}


\vspace{-5pt}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
Estudiar tres cuerpos bajo la acción gravitatoria es complicadísimo.
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
  \begin{tabular}{c}
     \includegraphics[height=85pt]{./images/3body1_2.png}
   \\
   {\small Simulación}
   \\
   {\tiny (Fuente: Wikipedia)}
  \end{tabular}
  \hspace{-5pt}
  \begin{tabular}{c}
     \includegraphics[height=85pt]{./images/lemn.png}
   \\
   {\small Solución especial}
   \\
   {\tiny Chenciner, Montgomery \underbar{2000}}
  \end{tabular}
  \quad
  \begin{tabular}{c}
   \begin{tikzpicture}[scale=0.15]
     \draw[rotate around={53:(7.45,-13.5)},black] (7.45,-13.5) ellipse (3.6 and 1.6);
%     \draw[rotate around={33:(6.45,-14.5)},black] (6.45,-14.5) ellipse (3.04 and 1.17);
     \draw[rotate around={-12:(12.4,-13.0)},black] (12.4,-13.0) ellipse (5.26 and 1.46);
     \draw[rotate around={-89:(8.2,-8.3)},black] (8.2,-8.3) ellipse (6.2 and 2.41);
     % circles
	\draw[blue,fill=blue] (5.0,-16.3) circle (10pt);
	\draw[blue,fill=blue] (14.4,-12.0) circle (10pt);
	\draw[blue,fill=blue] (5.9,-6.1) circle (10pt);
	\draw[blue,dashed] (5.9,-6.1) -- (14.4,-12.0) -- (5.0,-16.3) -- (5.9,-6.1);
	% space
	\node at (5,-17.1) {\phantom{.}};
	\node at (5,.5) {\phantom{.}};
%	\draw[black] (5.0,-16.3) -- (5.9,-6.1);
%	\draw[rotate around={-60:(5.0,-16.3)},black] (5.0,-16.3) -- (5.9,-6.1);
%	\draw[black] (5.6,-16.3) -- (14.7,-13.0);
%	\draw[rotate around={60:(5.6,-16.3)},black] (5.6,-16.3) -- (14.7,-13.0);
   \end{tikzpicture}
   \\
   {\small Solución especial}
   \\
   {\tiny Lagrange 1772}
  \end{tabular}
\end{miblock}


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\footnotesize

Euler (1765) $\to$ soluciones masas alineadas.
\smallskip 

Lagrange (1772) $\to$ soluciones masas en triángulo equilátero.
\end{miblock}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\vspace{-5pt}

\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white, title=Problema circular restringido de tres cuerpos]
Suponemos que en el problema circular de dos cuerpos añadimos una tercera masa $m_3$ muy pequeña.
\end{miblock}


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\footnotesize

Incluso así las órbitas generales son complicadas.
\end{miblock}

\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
  \begin{tabular}{c|c}
     \includegraphics[height=63pt]{./images/ocapesc_mod.png}
     &
     \includegraphics[height=63pt]{./images/ocaps_mod.png}
   \\
   \hline
   \multicolumn{2}{c}{\color{blue!30!black}\footnotesize Algunas órbitas para un observador en la línea que une $m_1$ y $m_2$}
  \end{tabular}
\end{miblock}


{\footnotesize En general:

\begin{itemize}\setlength{\itemsep}{0pt}
 \item 
$m_1$ y $m_2$ $\to$ giro con velocidad angular $\omega$ alrededor de $c$
 \item
$m_3$ $\to$ órbitas raras si no está muy cerca de $m_1$ o $m_2$
\end{itemize}

}

\end{frame}





%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\vspace{5pt}


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
¿Es posible que $m_3$ siga una trayectoria circular con la misma $\omega$ que $m_1$ y $m_2$?
\end{miblock}


\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\textbf{Aplicación:}
Situar telescopios, observatorios, estaciones espaciales, etc. que conserven la posición relativa sin gastar combustible.
\end{miblock}

\begin{tabular}{c}
 \begin{tikzpicture}[scale=0.35]
	% circles
	\draw[yellow,fill] (0,0) circle (1.2);
	\draw[blue!70!white,fill] (10,0) circle (0.2);
	\draw[black,fill] (3,0) circle (0.11);
	\draw[-, line width = 1pt] (1.2,0) -- (9.8,0);
	% arcs
	\draw [black, line width = 1pt, dashed, domain=-10:10] plot ({10*cos(\x)}, {10*sin(\x)});
	\draw [black, line width = 1pt, domain=-25:25] plot ({3*cos(\x)}, {3*sin(\x)});
	% names
	\node at (3,-0.5) [right] {\small $m_3$};
	\node at (0,-1.8) {\small Sol $m_1$};
	\node at (10,-1.8) {\small Tierra $m_2$};
 \end{tikzpicture}
 \\
 {\footnotesize $m_3$ cerca, $\omega_3>\omega$}
\end{tabular}
\begin{tabular}{c}
 \begin{tikzpicture}[scale=0.35]
	\def\fcr{0.25} % factor
	% circles
	\draw[yellow,fill] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (10*\fcr,0) circle (0.2*\fcr);
	\draw[black,fill] (10,0) circle (0.11);
	\draw[-, line width = 1pt] (1.2*\fcr,0) -- (10,0);
	% arcs
	\draw [black, line width = 1pt, dashed, domain=-15:15] plot ({10*\fcr*cos(\x)}, {10*\fcr*sin(\x)});
	\draw [black, line width = 1pt, domain=-10:10] plot ({10*cos(\x)}, {10*sin(\x)});
	% names
	\node at (10,-0.5) [right] {\small $m_3$};
	\node at (0,-1.1) {\small Sol};
	\node at (10*\fcr,-1.1) {\small Tierra};
 \end{tikzpicture}
 \\[10pt]
 {\footnotesize $m_3$ lejos, $\omega_3<\omega$}
\end{tabular}



\begin{miblock}[halign=flush left,colframe=red!50!brown, boxrule=1pt, coltitle=white]
Desfase con la Tierra $\to$ problemas de comunicación y accesibilidad
\end{miblock}

\begin{center}
 \setlength{\tabcolsep}{15pt}
 \begin{tabular}{cccc}
 \begin{tikzpicture}[scale=0.7]
	\def\fcr{0.25} % factor
	% circles
	\draw[yellow,fill] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (4*\fcr,6*\fcr) circle (0.2*\fcr);
	\draw[black,dashed] (0,0) circle (3.6*\fcr);
	\draw[red,fill] (2*\fcr,3*\fcr) circle (0.15*\fcr);
	%%%%
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.7]
	\def\fcr{0.25} % factor
	% circles
	\draw[yellow,fill] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (4*\fcr,6*\fcr) circle (0.2*\fcr);
	\draw[black,dashed] (0,0) circle (3.6*\fcr);
	\draw[red,fill] (-1.60*\fcr,3.23*\fcr) circle (0.15*\fcr);
	%%%%
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.7]
	\def\fcr{0.25} % factor
	% circles
	\draw[yellow,fill] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (4*\fcr,6*\fcr) circle (0.2*\fcr);
	\draw[black,dashed] (0,0) circle (3.6*\fcr);
	\draw[red,fill] (1.60*\fcr,-3.23*\fcr) circle (0.15*\fcr);
	%%%%
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.7]
	\def\fcr{0.25} % factor
	% circles
	\draw[yellow,fill] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (4*\fcr,6*\fcr) circle (0.2*\fcr);
	\draw[black,dashed] (0,0) circle (3.6*\fcr);
	\draw[red,fill] (-2*\fcr,-3*\fcr) circle (0.15*\fcr);
	%%%%
 \end{tikzpicture}
 \end{tabular}

\end{center}



\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\section{La solución}
\begin{frame}

\vspace{8pt}


\begin{miblock}[halign=flush left,colframe=blue, boxrule=1pt, coltitle=white]
\footnotesize Hay cinco posiciones relativas que permiten que $m_3$ siga una órbita circular con la misma $\omega$ que $m_1$ y $m_2$. Las tres masas se moverían como unidas con barras rígidas, conservando las distancias.
\end{miblock}

\vspace{-7pt}

\hspace{\stretch{1}}\begin{tcolorbox}[width=195pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{3pt}
\smash{Estos son los \textbf{puntos de Lagrange}}
\end{tcolorbox}\hspace{\stretch{1}}

\vspace{-10pt}

{\small\color{black!90!blue}\[
 \begin{tabular}{c}
 \begin{tikzpicture}[scale=0.65]
	\def\fcr{0.25} % factor
	% line
	\draw[black,dashed, line width = 0.75pt] (-16*\fcr,0) -- (18*\fcr,0);
	\draw[black,dashed, color=green!80!black, line width = 1pt] (15*\fcr,0) -- (5.5*\fcr,16.45*\fcr) -- (-4*\fcr,0);
	\draw[black,dashed, color=green!80!black, line width = 1pt] (15*\fcr,0) -- (5.5*\fcr,-16.45*\fcr) -- (-4*\fcr,0);
	% circles
	\draw[black,dashed] (0,0) circle (15*\fcr);
	\draw[red,fill] (-4*\fcr,0) circle (1.2*\fcr);
	\draw[blue,fill] (15*\fcr,0) circle (0.5*\fcr);
	\draw[black,fill] (5.5*\fcr,16.45*\fcr) circle (0.3*\fcr); % L_4
	\draw[black,fill] (5.5*\fcr,-16.45*\fcr) circle (0.3*\fcr); % L_5
	\draw[black,fill] (12*\fcr,0) circle (0.3*\fcr); % L_1
	\draw[black,fill] (18*\fcr,0) circle (0.3*\fcr); % L_2
	\draw[black,fill] (-16*\fcr,0) circle (0.3*\fcr); % L_3
	% cross
	\draw[line width = 1pt] (0,0) node[cross=3pt,rotate=0,black]{};  
	% names
	\node at (0,-0.5) {\small $c$};
	\node at (-1,-0.75) {\small $m_1$};
	\node at (15*\fcr,-0.5) {\small $m_2$};
	\node at  (5.5*\fcr+0.5,16.45*\fcr) {\small $L_4$};
	\node at  (5.5*\fcr+0.5,-16.45*\fcr) {\small $L_5$};
	\node at  (12*\fcr,0.5) {\small $L_1$};
	\node at  (18*\fcr,0.5) {\small $L_2$};
	\node at  (-16*\fcr,0.5) {\small $L_3$};
 \end{tikzpicture}
 \end{tabular}
 \qquad 
 \setlength{\tabcolsep}{-1pt}
 \begin{array}{|c|}
 \hline
  L_1,\, L_2,\, L_3
  \\
  \Downarrow
  \\
  \begin{tabular}{c}
   \tiny ``Fáciles'' físicamente
   \\[-6pt]
   \tiny (intuitivos)
  \end{tabular}
  \\[3pt]
  \hline
  \\[-10pt]
  L_4,\, L_5
  \\
  \Downarrow
  \\
  \begin{tabular}{c}
   \tiny ``Fáciles'' matemáticamente
   \\[-6pt]
   \tiny (con ecuaciones)
  \end{tabular}
  \\
  \hline
  \multicolumn{1}{l}{\scriptstyle\triangle\, m_1\,  m_2\, L_4\rule{0pt}{18pt},\ \ \ \triangle\, m_1\,  m_2\, L_5}
  \\
  \multicolumn{1}{c}{\scriptstyle\text{triángulos equiláteros}}
 \end{array}
\]}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\frametitle{Sistema Sol-Tierra}

\vspace{-15pt}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]
La distancia de la Tierra al Sol es 150 millones de kilómetros.
\end{miblock}


\vspace{-15pt}


{\small\[
 \begin{tabular}{c}
 \begin{tikzpicture}[scale=0.72]
	\def\fcr{0.25} % factor
	% line
	\draw[black,dashed] (-16*\fcr,0) -- (18*\fcr,0);
	% circles
	\draw[black,dashed] (0,0) circle (15*\fcr);
	\draw[red,fill=yellow] (0,0) circle (1.2*\fcr);
	\draw[blue!70!white,fill] (15*\fcr,0) circle (0.5*\fcr);
	\draw[black,fill] (7.5*\fcr,13*\fcr) circle (0.3*\fcr); % L_4
	\draw[black,fill] (7.5*\fcr,-13*\fcr) circle (0.3*\fcr); % L_5
	\draw[black,fill] (12*\fcr,0) circle (0.3*\fcr); % L_1
	\draw[black,fill] (18*\fcr,0) circle (0.3*\fcr); % L_2
	\draw[black,fill] (-16*\fcr,0) circle (0.3*\fcr); % L_3
	% names
	\node at (0,-0.75) {\small Sol};
	\node at (15*\fcr,-0.5) {\small Tierra};
	\node at  (7.5*\fcr,13*\fcr-0.5) {\small $L_4$};
	\node at  (7.5*\fcr,-13*\fcr-0.5) {\small $L_5$};
	\node at  (12*\fcr,0.5) {\small $L_1$};
	\node at  (18*\fcr,0.5) {\small $L_2$};
	\node at  (-16*\fcr,0.5) {\small $L_3$};
 \end{tikzpicture}
 \end{tabular}
 \quad
 \begin{tabular}{|l|}
  \hline
  Respecto de la 
  \\
  órbita terrestre
  \\
  {\tiny\bf\color{blue}$Mkm=$ mill. de km}
  \\[4.82pt]
  {\scriptsize$\bullet$}\ 
  $L_1$: $1.5\, Mkm$
  \\
  por dentro
  \\[4.82pt]
  {\scriptsize$\bullet$}\ 
  $L_2$: $1.5\, Mkm$
  \\
  por fuera
  \\[4.82pt]
  {\scriptsize$\bullet$}\ 
  $L_3$: $0.026\, Mkm$
  \\
  por fuera
  \\[4.82pt]
  {\scriptsize$\bullet$}\ 
  $L_4,\, L_5$ en la
  \\
  órbita {\small ($150\,Mkm$)}
  \\
  \hline
 \end{tabular}
\]}

\vspace{-14pt}

{\scriptsize \underbar{Anécdota}: Un objeto en $L_3$ estaría siempre tapado por el Sol. En ciencia ficción es una excusa para una antiTierra.

}

\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}
%\vspace{-15pt}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Objetos actualmente en puntos de Lagrange Sol-Tierra]
\begin{itemize}
 \item 
 (L1) 
 \textbf{DSCOVR} (antes llamado \textbf{Triana}). Satélite climático.
 \item 
 (L1) 
 \textbf{Aditya-L1}. Observación solar.
 
 \item 
 (L2) 
 \textbf{JWST}. Telescopio espacial.
 \item 
 (L2) 
 \textbf{Euclid}. Observación de galaxias.
 
 \item 
 (L4) (alrededores) 
 \textbf{2010 TK$_7$}. Asteroide ($\approx 0.3\,km$).
 \item 
 (L4) (alrededores) 
 \textbf{2020 XL$_5$}. Asteroide ($\approx 1.2\,km$).
 
 
\end{itemize}
\end{miblock}

\


\begin{miblock}[halign=flush left,colframe=blue!70!white, boxrule=1pt, coltitle=white, title=Objetos anteriores importantes en $L_2$]
\begin{itemize}
 \item 
 \textbf{WMAP}. Medición de la radiación de fondo.
 \item 
 \textbf{Gaia}. Estudio de las estrellas.
\end{itemize}
\end{miblock}



\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


%\vspace{-15pt}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt,colback=aster, coltitle=white, title=Asteroides en el Sistema Solar interior (Wikipedia)]
\begin{center}
  \begin{tabular}{c}
     \includegraphics[height=195pt]{./images/asteroids.png}
  \end{tabular}
\end{center}
\end{miblock}



\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Explicación de $L_1$\hfill{\scriptsize(balance de aceleraciones sobre $m_3$)}]
\begin{center}
  \begin{tikzpicture}[scale=0.4]
	\draw[black,fill] (0,0) circle (0.4);
	\draw[black,fill] (10,0) circle (0.2);
	% lnes
	\draw[line width = 1pt] (7,0) node[cross=4pt,rotate=0,black]{};  
	\draw[line width = 1pt] (0,0) -- (10,0);  
	% arrows
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (0.5,0.85) -- (10.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (9.5,0.85) -- (0,0.85);
	%
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (7.5,-1.85) -- (10.0,-1.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (9.5,-1.85) -- (7.0,-1.85);
	% names
	\node at (0,-0.85) {\small $m_1$};
	\node at (7,-0.85) {\small $m_3$};
	\node at (10,-0.85) {\small $m_2$};
	\node[color=blue] at (5,1.3) {\small $D$};
	\node[color=blue] at (8.5,-1.2) {\small $d$};
 \end{tikzpicture}
\end{center}
\vspace{-15pt}

\begin{center}
 Gravedad de $m_1$ = Gravedad de $m_2$ + centrífuga
\end{center}
\[
 \frac{G m_1}{(D-d)^2}
 =
 \frac{G m_2}{d^2}
 +
 \omega^2d
\]
\end{miblock}


\hspace{\stretch{1}}\begin{tcolorbox}[width=195pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{-10pt}
\[
 \begin{array}{lcl}
  d\approx 0
  &
  \to
  &
  \text{\footnotesize gana el lado derecho}
  \\
  d\approx D
  &
  \to
  &
  \text{\footnotesize gana el lado izquierdo}
 \end{array}
\]
\end{tcolorbox}\hspace{\stretch{1}}

\vspace{-1pt}

\[
 \Downarrow
\]

\vspace{-3pt}

\hspace{\stretch{1}}\begin{tcolorbox}[width=245pt,halign=center,colframe=blue,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{2pt}
A cierta distancia ($L_1$), se compensan
\end{tcolorbox}\hspace{\stretch{1}}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Explicación de $L_2$\hfill{\scriptsize(balance de aceleraciones sobre $m_3$)}]
\begin{center}
  \begin{tikzpicture}[scale=0.4]
	\draw[black,fill] (0,0) circle (0.4);
	\draw[black,fill] (7,0) circle (0.2);
	% lnes
	\draw[line width = 1pt] (10,0) node[cross=4pt,rotate=0,black]{};  
	\draw[line width = 1pt] (0,0) -- (10,0);  
	% arrows
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (0.5,0.85) -- (7.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (6.5,0.85) -- (0,0.85);
	%
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (7.5,0.85) -- (10.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (9.5,0.85) -- (7.0,0.85);
	% names
	\node at (0,-0.85) {\small $m_1$};
	\node at (7,-0.85) {\small $m_2$};
	\node at (10,-0.85) {\small $m_3$};
	\node[color=blue] at (3.5,1.3) {\small $D$};
	\node[color=blue] at (8.5,1.3) {\small $d$};
 \end{tikzpicture}
\end{center}
\vspace{-15pt}

\begin{center}
 Gravedad de $m_1$ $+$ Gravedad de $m_2$ $=$ centrífuga
\end{center}
\[
 \frac{G m_1}{(D+d)^2}
 +
 \frac{G m_2}{d^2}
 =
 \omega^2d
\]
\end{miblock}


\hspace{\stretch{1}}\begin{tcolorbox}[width=195pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{-10pt}
\[
 \begin{array}{lcl}
  d\approx 0
  &
  \to
  &
  \text{\footnotesize gana el lado izquierdo}
  \\
  d\ \text{\footnotesize grande}
  &
  \to
  &
  \text{\footnotesize gana el lado derecho}
 \end{array}
\]
\end{tcolorbox}\hspace{\stretch{1}}

\vspace{-1pt}

\[
 \Downarrow
\]

\vspace{-3pt}

\hspace{\stretch{1}}\begin{tcolorbox}[width=245pt,halign=center,colframe=blue,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{2pt}
A cierta distancia ($L_2$), se compensan
\end{tcolorbox}\hspace{\stretch{1}}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Explicación de $L_3$\hfill{\scriptsize(balance de aceleraciones sobre $m_3$)}]
\begin{center}
  \begin{tikzpicture}[scale=0.4]
	\draw[black,fill] (0,0) circle (0.4);
	\draw[black,fill] (7,0) circle (0.2);
	\draw[black] (-7,0) circle (0.2);
	% lnes
	\draw[line width = 1pt] (-10,0) node[cross=4pt,rotate=0,black]{};  
	\draw[line width = 1pt] (-10,0) -- (7,0);  
	% arrows
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (0.5,0.85) -- (7.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (6.5,0.85) -- (0,0.85);
	%
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (-0.5,0.85) -- (-7.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (-6.5,0.85) -- (0,0.85);
	%
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (-7.5,0.85) -- (-10.0,0.85);
	\draw[-{Latex[length=8pt]}, line width = 0.75pt, color=blue] (-9.5,0.85) -- (-7.0,0.85);
	% names
	\node at (0,-0.85) {\small $m_1$};
	\node at (7,-0.85) {\small $m_2$};
	\node at (-10,-0.85) {\small $m_3$};
	\node[color=blue] at (3.5,1.3) {\small $D$};
	\node[color=blue] at (-8.5,1.3) {\small $d$};
 \end{tikzpicture}
\end{center}
\vspace{-15pt}

\begin{center}
 Gravedad de $m_1$ $+$ Gravedad de $m_2$ $=$ centrífuga
\end{center}
\[
 \frac{G m_1}{(D+d)^2}
 +
 \frac{G m_2}{(2D+d)^2}
 =
 \omega^2d
\]
\end{miblock}


\hspace{\stretch{1}}\begin{tcolorbox}[width=195pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{-10pt}
\[
 \begin{array}{lcl}
  d\approx 0
  &
  \to
  &
  \text{\footnotesize gana el lado izquierdo}
  \\
  d\ \text{\footnotesize grande}
  &
  \to
  &
  \text{\footnotesize gana el lado derecho}
 \end{array}
\]
\end{tcolorbox}\hspace{\stretch{1}}

\vspace{-1pt}

\[
 \Downarrow
\]

\vspace{-3pt}

\hspace{\stretch{1}}\begin{tcolorbox}[width=245pt,halign=center,colframe=blue,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\vspace{2pt}
A cierta distancia ($L_3$), se compensan
\end{tcolorbox}\hspace{\stretch{1}}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\frametitle{Para decir toda la verdad\dots}

\vspace{-18pt}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]
\underbar{Matemáticamente}, $L_1$, $L_2$ y $L_3$ son puntos \textbf{inestables}:
Mínimas perturbaciones de un objeto allí, acabarán alejándolo.

\medskip

En la \underbar{práctica}, son poco inestables, hay órbitas (de Lissajous, de halo) que permiten estar cerca de ellos con poco gasto de combustible.
\end{miblock}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt,colback=black, coltitle=white]
\begin{center}
  \begin{tabular}{c}
     \includegraphics[height=55pt]{./images/webb1.jpg}
     \includegraphics[height=55pt]{./images/webb2.jpg}
     \\
     {\color{white}\footnotesize  Telescopio James Webb}
  \end{tabular}
  \begin{tabular}{c}
     \includegraphics[height=55pt]{./images/webb_orbit.png}
     \\
     {\color{white}\footnotesize Órbita (Wikipedia)}
  \end{tabular}
\end{center}
\end{miblock}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]

$L_4$ y $L_5$ son puntos \textbf{estables}:
Pequeñas perturbaciones de un objeto allí, no lo alejan significativamente.
\end{miblock}


\vspace{-15pt}


\end{frame}



%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\section{Ecuaciones}
\begin{frame}


\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\begin{center}
¿Cómo sabemos que existen $L_4$ y $L_5$?
\end{center}
\end{miblock}

\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\begin{center}
¿Cómo sabemos que no existen otros puntos de Lagrange?
\end{center}
\end{miblock}


\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\begin{center}
¿Cómo simular las trayectorias (con ordenador)?
\end{center}
\end{miblock}

\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\begin{center}
¿Cómo deducir sus propiedades?
\end{center}
\end{miblock}

%\vspace{10pt}


\hspace{\stretch{1}}\begin{tcolorbox}[width=200pt,halign=flush left,colframe=ct1,boxrule=1pt, coltitle=white,enhanced,drop fuzzy shadow,colback=white, nobeforeafter,tcbox raise=-18pt]
\begin{center}
 \LARGE\bf 
 Ecuaciones
 \\
 Matemáticas
\end{center}
\end{tcolorbox}\hspace{\stretch{1}}


\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

{\footnotesize La clave para obtener ecuaciones razonables es ponerse en el lugar de un observador que está en $c$ y gira con $m_1$ y $m_2$.}

\begin{miblock}[halign=flush left,colframe=teal, boxrule=1pt, coltitle=white]
\begin{center}
\[
 \begin{array}{rl}
  a_{\text{grav}}& =\text{\footnotesize aceleración gravitatoria}
  \\[-3pt]
  & \phantom{=\ }\text{\footnotesize debida a $m_1$ y $m_2$}
  \\
  a_{\text{3}}&= \text{\footnotesize aceleración de $m_3$}
  \\
  a_{\text{3(obs)}}&= \text{\footnotesize para el observador}
  \\[8pt]
  \multicolumn{2}{l}{\text{\footnotesize Ecuación fundamental:\ }\boxed{a_3=a_{\text{grav}}}}
 \end{array}
 \quad
 \begin{tabular}{c}
  \begin{tikzpicture}[scale = 0.3]
	% lines
	\draw[line width = 0.5pt] ({-2*cos(20)},{-2*sin(20)}) --  ({8*cos(20)},{8*sin(20)});
	\draw[line width = 1pt, dashed] (0,0) -- ({4*cos(20)},{4*sin(20)}) -- ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)});
	% circles
	\draw[blue,fill] ({8*cos(20)},{8*sin(20)}) circle (0.2);
	\draw[blue,fill] ({-2*cos(20)},{-2*sin(20)}) circle (0.2);
	\draw[blue,fill] ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)}) circle (0.1);
	% cross
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};
	%
	% arc
	\draw[-latex, line width = 0.5pt] ({8*cos(20)},{8*sin(20)}) arc[radius=8, start angle=20, end angle=40];
	% names
    \node[below = 5pt]  at (0,0) {$c$};
    \node[above = 0pt,rotate=20]  at (0,0) {\includegraphics[height=25pt]{./images/observer.png}};
    \node[below = 5pt]  at ({-2*cos(20)},{-2*sin(20)}) {$m_1$};
    \node[below = 5pt]  at ({8*cos(20)},{8*sin(20)}) {$m_2$};
    \node[above = 3pt]  at ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)})     {$m_3$}; 
    \node[right = 0pt,rotate=20]  at ({4*cos(20)-3.1/2*sin(20)},{4*sin(20)+3.1/2*cos(20)}) {$y$};
    \node[above = 0pt,rotate=20]  at ({4/2*cos(20)},{4/2*sin(20)}) {$x$};
    \node[right = 2pt,rotate=20]  at ({8*cos(30)},{8*sin(30)}) {$\omega$};
 \end{tikzpicture}
 \end{tabular}
\]
\end{center}
\end{miblock}


{\footnotesize La posición de $m_3$ para el observador es $\mathbf{r}=(x,y)$. Indicando con un punto la tasa de variación con respecto del tiempo:
\[
 a_{\text{3(obs)}}
 =\dot{v}
 =(\dot{x},\dot{y})\dot{\ }
 =(\ddot{x},\ddot{y})
 =\ddot{\mathbf{r}}
\]
\textbf{Problema:} $a_3\ne  a_{\text{3(obs)}}$ porque el observador gira y también sufre aceleraciones.

}


\end{frame}


%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\begin{miblock}[halign=flush left,colframe=teal, boxrule=1pt, coltitle=white]
\begin{center}
¿Cómo afectan los giros a las aceleraciones?
\end{center}
\end{miblock}

\begin{miblock}[halign=flush left,colframe=teal, boxrule=1pt, coltitle=white]
\begin{center}
¿Qué relación hay entre $a_3$ y $a_{\text{3(obs)}}$?
\end{center}
\end{miblock}

Para estudiar este problema, definimos
\[
G_\alpha = \, \text{Giro con velocidad angular $\omega$ partiendo del ángulo $\alpha$} 
\]
y consideramos tres hechos geométricos:


\begin{miblock}[halign=flush left,colframe=white, boxrule=1pt, coltitle=white]
\[
 \begin{array}{|c|c|c|}
 \begin{tikzpicture}[scale=0.1]
	% circle
	\draw[black, line width = 1pt] (0,0) circle (10);
 	
	\foreach \x in {0,20,...,60}
     {
		\draw[-latex, line width = 1pt] (0,0) -- ({10*cos(\x)},{10*sin(\x)}) ;
		\draw[-latex, line width = 1pt, red] (0,0) -- ({-10*cos(\x)},{-10*sin(\x)}) ;
     }
	% arcs
	\draw[-latex, line width = 0.5pt] (12,0) arc[radius=12, start angle=0, end angle=30];
	\draw[-latex, line width = 0.5pt] (4,0) arc[radius=4, start angle=0, end angle=180];
	% names
    \node[right = 2pt,rotate=15]  at ({12*cos(15)},{12*sin(15)}) {$\omega$};
 	\node at (0,4) [left = 1pt] {$\scriptstyle{180^\circ}$};
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.3]
	\def\h{5} % horizontal
	\def\v{3} % vertical
	\def\al{atan(\v/\h)} % angle
	\def\r{sqrt(\v*\v+\h*\h)} % radius
	% axes
	\draw[black, line width = 0.5pt] (-1.2*\v,0) -- (1.2*\h,0);
	\draw[black, line width = 0.5pt] (0,-0.5) -- (0, 1.1*\h);
	% lines
	\draw[black, line width = 1pt, dashed] (0,0) -- (\h,0) -- (\h,\v);
	\draw[black, line width = 1pt, dashed] (0,0) -- (-\v,0) -- (-\v,\h);
	% arrows 
	\draw[-latex, line width = 1pt, black] (0,0) -- (\h,\v);
	\draw[-latex, line width = 1pt, black] (0,0) -- (-\v,\h);
	% arcs
	\draw[-latex, line width = 0.5pt, black]  (0.5*\h,0.5*\v)  arc[radius={0.5*\r}, start angle=\al, end angle={90+\al}];
	% names
    \node[above = 2pt]  at ({0.5*\r*cos(20+\al)},{0.5*\r*sin(20+\al)}) {$\scriptstyle{90^\circ}$};
    \node[below = 2pt]  at (\h/2,0) {$\scriptstyle{h}$};
    \node[right = 2pt]  at (\h,\v/2) {$\scriptstyle{v}$};
    \node[below = 2pt]  at (-\v/2,0) {$\scriptstyle{-v}$};
    \node[left = 2pt]  at (-\v,\h/2) {$\scriptstyle{h}$};
 \end{tikzpicture}
 &
 \begin{tikzpicture}[scale=0.1]
	\def\az{-30} % starting angle
	\def\v{10} % velocity
	% circle
	\draw[black, line width = 1pt] (0,0) circle (10);
 	
	\foreach \x in {\az,-15,...,60}
     {
		\draw[-latex, line width = 1pt] (0,0) -- ({10*cos(\x)},{10*sin(\x)}) ;
		\draw[-latex, line width = 1pt, blue] ({10*cos(\x)},{10*sin(\x)}) -- ({10*cos(\x)-\v*sin(\x)},{10*sin(\x)+\v*cos(\x)}) ;
     }
	% arcs
	\draw[-latex, line width = 0.5pt] (-12,0) arc[radius=12, start angle=180, end angle=210];
%	\draw[-latex, line width = 0.5pt, black]  ({10*cos(60)-\v/2*sin(60)},{10*sin(60)+\v/2*cos(60)}) -- (0,0); 
	\draw[-latex, line width = 0.5pt, black]  ({10*cos(60)-\v/2*sin(60)},{10*sin(60)+\v/2*cos(60)}) arc[radius=\v/2, start angle=150, end angle=240]; 
	% names
    \node[left = 2pt,rotate=15]  at ({12*cos(195)},{12*sin(195)}) {$\omega$};
 	\node at ({10*cos(\az)},{10*sin(\az)}) [right = 1pt] {$\scriptstyle{\alpha}$};
 \end{tikzpicture}
 \\
 \hline
 \scriptstyle{{G}_{180^\circ}=- G_{0}}
 &
 \scriptstyle{G_{90^\circ}(h,v)=(-v,h)}
 &
 \scriptstyle{\dot{G}_\alpha=\omega G_{\alpha +90^\circ}}
 \end{array}
\]
\end{miblock}

\end{frame}





%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

$G_\alpha = $\, Giro con velocidad angular $\omega$ partiendo del ángulo $\alpha$ 


\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\[
 \begin{aligned}
 (G_0\mathbf{r})\dot{\ }
 &=
 \dot{G}_{0}\mathbf{r}+ {G}_{0}\dot{\mathbf{r}}
 +
 G_0\dot{\mathbf{r}}
 \\
 &=
 \omega G_{90^\circ}\mathbf{r}
 +
 G_0\dot{\mathbf{r}}
 \end{aligned}
 \ \ 
 {\setlength{\arraycolsep}{0pt}
 \begin{array}{|c|c|}
 \hline
 \begin{tikzpicture}[scale=0.09]
	\def\az{-30} % starting angle
	\def\v{10} % velocity
	% circle
	\draw[black, line width = 1pt] (0,0) circle (10);
 	
	\foreach \x in {\az,-15,...,60}
     {
		\draw[-latex, line width = 1pt] (0,0) -- ({10*cos(\x)},{10*sin(\x)}) ;
		\draw[-latex, line width = 1pt, blue] ({10*cos(\x)},{10*sin(\x)}) -- ({10*cos(\x)-\v*sin(\x)},{10*sin(\x)+\v*cos(\x)}) ;
     }
	% arcs
	\draw[-latex, line width = 0.5pt] (-12,0) arc[radius=12, start angle=180, end angle=210];
	\draw[-latex, line width = 0.5pt, black]  ({10*cos(60)-\v/2*sin(60)},{10*sin(60)+\v/2*cos(60)}) arc[radius=\v/2, start angle=150, end angle=240]; 
	% names
    \node[left = 2pt,rotate=15]  at ({12*cos(195)},{12*sin(195)}) {$\omega$};
 	\node at ({10*cos(\az)},{10*sin(\az)}) [right = 1pt] {$\scriptstyle{\alpha}$};
 \end{tikzpicture}
 &
 \ \ 
 \begin{tikzpicture}[scale=0.09]
	% circle
	\draw[black, line width = 1pt] (0,0) circle (10);
 	
	\foreach \x in {0,20,...,60}
     {
		\draw[-latex, line width = 1pt] (0,0) -- ({10*cos(\x)},{10*sin(\x)}) ;
		\draw[-latex, line width = 1pt, red] (0,0) -- ({-10*cos(\x)},{-10*sin(\x)}) ;
     }
	% arcs
	\draw[-latex, line width = 0.5pt] (12,0) arc[radius=12, start angle=0, end angle=30];
	\draw[-latex, line width = 0.5pt] (4,0) arc[radius=4, start angle=0, end angle=180];
	% names
    \node[right = 2pt,rotate=15]  at ({12*cos(15)},{12*sin(15)}) {$\omega$};
 	\node at (0,4) [left = 1pt] {$\scriptstyle{180^\circ}$};
 \end{tikzpicture}
 \\[-3pt]
 \hline
 \scriptstyle{\dot{G}_\alpha=\omega G_{\alpha +90^\circ}}
 &
 \scriptstyle{{G}_{180^\circ}=- G_{0}}
 \\
 \hline
 \end{array}}
\]
\vspace{-10pt}
\[
 \begin{aligned}
 (G_0\mathbf{r})\ddot{\ }
 &=
 \omega (G_{90^\circ}\mathbf{r})\dot{\ }
 +
 (G_0\dot{\mathbf{r}})\dot{\ }
 \\
 &=
 \omega^2 G_{180^\circ}\mathbf{r}
 +
 \omega G_{90^\circ}\dot{\mathbf{r}}
 +
 \omega G_{90^\circ}\dot{\mathbf{r}}
 +
 G_0\ddot{\mathbf{r}}
 \\
 &=
 -\underbrace{\omega^2 G_{0}\mathbf{r}}_{\text{a. centrífuga}}
 +
 \underbrace{2\omega G_{90^\circ}\dot{\mathbf{r}}}_{\text{a. de Coriolis}}
 +
 \underbrace{G_0\ddot{\mathbf{r}}}_{a_{3\text{(obs)}}\text{ girada}}
 \end{aligned}
 \ 
 {\setlength{\arraycolsep}{0pt}
 \begin{array}{|c|}
 \hline
 \begin{tikzpicture}[scale=0.2]
	\def\h{5} % horizontal
	\def\v{3} % vertical
	\def\al{atan(\v/\h)} % angle
	\def\r{sqrt(\v*\v+\h*\h)} % radius
	% axes
	\draw[black, line width = 0.5pt] (-1.2*\v,0) -- (1.2*\h,0);
	\draw[black, line width = 0.5pt] (0,-0.5) -- (0, 1.1*\h);
	% lines
	\draw[black, line width = 1pt, dashed] (0,0) -- (\h,0) -- (\h,\v);
	\draw[black, line width = 1pt, dashed] (0,0) -- (-\v,0) -- (-\v,\h);
	% arrows 
	\draw[-latex, line width = 1pt, black] (0,0) -- (\h,\v);
	\draw[-latex, line width = 1pt, black] (0,0) -- (-\v,\h);
	% arcs
	\draw[-latex, line width = 0.5pt, black]  (0.5*\h,0.5*\v)  arc[radius={0.5*\r}, start angle=\al, end angle={90+\al}];
	% names
    \node[above = 2pt]  at ({0.5*\r*cos(20+\al)},{0.5*\r*sin(20+\al)}) {$\scriptstyle{90^\circ}$};
    \node[below = 2pt]  at (\h/2,0) {$\scriptstyle{h}$};
    \node[right = 2pt]  at (\h,\v/2) {$\scriptstyle{v}$};
    \node[below = 2pt]  at (-\v/2,0) {$\scriptstyle{-v}$};
    \node[left = 2pt]  at (-\v,\h/2) {$\scriptstyle{h}$};
 \end{tikzpicture}
 \\[-3pt]
 \hline
 \scriptstyle{G_{90^\circ}(h,v)=(-v,h)}
 \\
 \hline
 \end{array}}
\]


\end{miblock}


\begin{miblock}[halign=flush left,colframe=brown, boxrule=1pt, coltitle=white]
\[
 a_3=(G_0\mathbf{r})\ddot{\ }
 =
 -\omega^2(x,y)
 +
 2\omega(-\dot{y},\dot{x})
 +
 (\ddot{x},\ddot{y})
\]
\end{miblock}

\end{frame}






%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title= Ecuación fundamental]
\[
 {-\omega^2(x,y)
 +
 2\omega (-\dot{y},\dot{x})
 +
 (\ddot{x}, \ddot{y})
 =
 a_{\text{grav}}}
\]
\end{miblock}





\begin{miblock}[halign=flush left,colframe=white, boxrule=1pt, coltitle=white]
\begin{tabular}{cl}
 \begin{tabular}{c}
  \begin{tikzpicture}[scale = 0.35]
	% lines
	\draw[line width = 1pt] (0,0) -- ({-2*cos(20)},{-2*sin(20)}) -- ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)}) -- ({8*cos(20)},{8*sin(20)}) -- (0,0);
	\draw[line width = 2pt, dashed] (0,0) -- ({4*cos(20)},{4*sin(20)}) -- ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)});
	% circles
	\draw[blue,fill] ({8*cos(20)},{8*sin(20)}) circle (0.2);
	\draw[blue,fill] ({-2*cos(20)},{-2*sin(20)}) circle (0.2);
	\draw[blue,fill] ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)}) circle (0.1);
	% cross
	\draw[line width = 1pt] (0,0) node[cross=4pt,rotate=45,black]{};
	% names
    \node[below = 5pt]  at (0,0) {$c$};
    \node[below = 5pt]  at ({-2*cos(20)},{-2*sin(20)}) {$m_1$};
    \node[below = 5pt]  at ({8*cos(20)},{8*sin(20)}) {$m_2$};
    \node[above = 3pt]  at ({4*cos(20)-3.1*sin(20)},{4*sin(20)+3.1*cos(20)})     {$m_3$}; 
    \node[right = 0pt,rotate=20]  at ({4*cos(20)-3.1/2*sin(20)},{4*sin(20)+3.1/2*cos(20)}) {$y$};
    \node[above = 0pt,rotate=20]  at ({4/2*cos(20)},{4/2*sin(20)}) {$x$};
    \node[above = 3pt]  at ({1*cos(20)-3.1/2*sin(20)},{1*sin(20)+3.1/2*cos(20)})     {$r_1$}; 
    \node[above = 0pt]  at ({6*cos(20)-3.1/2*sin(20)},{6*sin(20)+3.1/2*cos(20)})     {$r_2$}; 
 \end{tikzpicture}
 \end{tabular}
&
	\begin{tabular}{p{4.5cm}}
	La fórmula para $a_{\text{grav}}$ es un poco complicada porque hay que sumar la debida a $m_1$ y a $m_2$.
	\end{tabular}
\end{tabular}

{\small Cambiando las unidades para que $\omega =1=\text{distancia de $m_1$ a $m_2$}$, se puede probar}
\[
 a_{\text{grav}}
 =
 \Big(
 \frac{(\mu-1)(x+\mu)}{r_1^3}
 -
 \frac{\mu(x+\mu-1)}{r_2^3}
 ,\, 
 \frac{(\mu-1)y}{r_1^3}
 -
 \frac{\mu y}{r_2^3}
 \Big) 
\]
con $\mu = (1+m_2/m_1)^{-1}$. 


\end{miblock}



\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Ecuaciones desarrolladas tras el cambio de unidades]
\[
% \boxed{
 \begin{cases}
 -x
 +
 -2\dot{y}
 +
 \ddot{x}
 &=
 \dfrac{(\mu-1)(x+\mu)}{r_1^3}
 -
 \dfrac{\mu(x+\mu-1)}{r_2^3}
 \\[15pt]
 -y
 +
 2\dot{x}
 +
 \ddot{y}
 &=
 \dfrac{(\mu-1)y}{r_1^3}
 -
 \dfrac{\mu y}{r_2^3}
 \end{cases}
%}
\]
\end{miblock}

\begin{center}
 ¿Realmente sirven de algo ecuaciones tan complicadas?
\end{center}

\begin{center}
 \bf\LARGE Sí
\end{center}

\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white]
{\small Desde el punto de vista práctico, los ordenadores pueden resolver con mucha precisión ecuaciones (diferenciales) que contienen $\dot{x}$, $\ddot{x}$, $\dot{y}$, $\ddot{y}$ esencialmente aproximando estas cantidades por incrementos.} 
\end{miblock}


\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]
\[
 \begin{cases}
 -x
 +
 -2\dot{y}
 +
 \ddot{x}
 &=
 \dfrac{(\mu-1)(x+\mu)}{r_1^3}
 -
 \dfrac{\mu(x+\mu-1)}{r_2^3}
 \\[15pt]
 -y
 +
 2\dot{x}
 +
 \ddot{y}
 &=
 \dfrac{(\mu-1)y}{r_1^3}
 -
 \dfrac{\mu y}{r_2^3}
 \end{cases}
\]
\end{miblock}


\begin{miblock}[halign=flush left,colframe=ct1, boxrule=1pt, coltitle=white]
{\small Además, sabiendo matemáticas, permiten un tratamiento teórico del problema. Por ejemplo, prueban la existencia de $L_4$ y $L_5$.} 
\end{miblock}


{\small Puntos de Lagrange $\to$ $x$, $y$ constantes $\to$  
$\dot{x}=\ddot{x}=\dot{y}=\ddot{y}=0$}


\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Ecuaciones de los puntos de Lagrange]
\[
 \dfrac{(\mu-1)(x+\mu)}{r_1^3}
 -
 \dfrac{\mu(x+\mu-1)}{r_2^3}
 +x =
 \dfrac{(\mu-1)y}{r_1^3}
 -
 \dfrac{\mu y}{r_2^3}
 +y=0
\]
\end{miblock}



\end{frame}




%%%%%%%%%%%%
% SLIDE    %
%%%%%%%%%%%%
\begin{frame}

\vspace{8pt}

{\small Si $y\ne 0$, no estamos en la situación de $L_1$, $L_2$ y $L_3$.}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white, title=Ecuaciones de los puntos de Lagrange con $y\ne 0$]
\[
 \dfrac{(\mu-1)(x+\mu)}{r_1^3}
 -
 \dfrac{\mu(x+\mu-1)}{r_2^3}
 +x =
 \dfrac{\mu-1}{r_1^3}
 -
 \dfrac{\mu}{r_2^3}
 +1=0
\]
\end{miblock}

\vspace{10pt}

\begin{miblock}[halign=flush left,colframe=ct0, boxrule=1pt, coltitle=white]
 \begin{tabular}{c}
  \begin{tikzpicture}[scale = 0.2]
	% lines
	\draw[line width = 1pt] (0,0) -- ({10*cos(20)},{10*sin(20)}) -- ({10*cos(80)},{10*sin(80)}) -- (0,0);
	\draw[line width = 1pt, dashed] (0,0) -- ({10*cos(-40)},{10*sin(-40)}) -- ({10*cos(20)},{10*sin(20)});
	% circles
	\draw[blue,fill] (0,0) circle (0.2);
	\draw[blue,fill] ({10*cos(20)},{10*sin(20)}) circle (0.2);
	\draw[blue,fill] ({10*cos(80)},{10*sin(80)}) circle (0.1);
	% names
    \node[left = 2pt]  at (0,0) {$m_1$};
    \node[right = 2pt]  at ({10*cos(20)},{10*sin(20)}) {$m_2$};
    \node[above = 1pt]  at ({5*cos(20)},{5*sin(20)}) {\small $1$};
    \node[above = 2pt]  at ({10*cos(80)},{10*sin(80)}) {$m_3$}; 
	\node[right = 3pt]  at ({8.6*cos(50)},{8.6*sin(50)})     {\small $r_2=1$}; 
    \node[left = 0pt]  at ({5*cos(80)},{5*sin(80)})     {\small $r_1=1$}; 
 \end{tikzpicture}
 \end{tabular}
 \begin{tabular}{p{6cm}}
  \small 
  Es fácil ver que con $r_1=r_2=1$ se cumplen estas ecuaciones 
  \\
  \multicolumn{1}{c}{$\Downarrow$}
  \\
  \multicolumn{1}{c}{\small
  $m_1$, $m_2$ y $m_3$ a la misma distancia}
  \\
  \multicolumn{1}{c}{$\Downarrow$}
  \\
  \multicolumn{1}{c}{%\small
  Se obtienen $L_4$ y $L_5$}
 \end{tabular}
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El estudio matemático de las ecuaciones permite extraer otras conclusiones, por ejemplo, deducir la estabilidad de $L_4$ y $L_5$ y la inestabilidad de $L_1$, $L_2$ y $L_3$.
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También permite abordar el problema de cuándo $m_3$ es ``capturada'' por alguna de las masas.

\begin{center}
\begin{tabular}{c}
	\includegraphics[height=65pt]{./images/o1confcx.png}
	\qquad
	\includegraphics[height=65pt]{./images/o1confr.png}
\\
\hline
	\includegraphics[height=65pt]{./images/ocapas.png}
\end{tabular}
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