
\input ../povetron
\input ../kurzmac

\def\\{\hfil\break}
\let\rho=\varrho
\let\phi=\varphi
\def\beginitem{\begingroup\leftskip=12pt}
\def\enditem{\endgroup}
\mathcode`\,="613B  % anglicka desetinna carka v matematice  

\tracingall
\tracingonline=0

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\sect* 3.~kapitola o~probl�mu dvou t�les


\myitem{Newtonovy pohybov� z�kony:}

\beginitem
\item{1.} existuj� inerci�ln�%
%
\myfootnote{Tyto soustavy souvis� spolu {\em Galileiho transformac�\/} sou�adnic $\vec r' = \vec r - \vec v t$, kde $\vec v$~ozna�uje jejich vz�jemnou rychlost.}
%
\ vzta�n� soustavy, v~nich� se voln�%
%
\myfootnote{Nep�sob� na n� jin� s�ly ne� gravitace vzd�len�ch hmot.}
%
\ hmotn� body pohybuj� rovnom�rn� p��mo�a�e nebo jsou v~klidu;

\item{2.} o~��inku s�ly: $\vec F = m_{\rm setrva�n�}\,\vec a = {\d p\over\d t}$;

\item{3.} akce a~reakce: $\vec F' = -\vec F$.

\enditem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{Gravita�n� z�kon} (pro hmotn� body):

$$\vec F_{12} = -G {m_1 m_{2\,{\rm gravita�n�}}\over r^2} {\vec r_{12}\over r}\label{eq:gravitacni_zakon}$$

m��en� $G = 6{,}6726(5)\,{\rm N}\cdot{\rm kg}^{-2}\cdot{\rm m}^2$ (viz Bro� \& Roskovec 1988):

Adjustace fyzik�ln�ch konstant nen� u~$G$ pou�iteln�,
neexistuje toti� vztah $G$ a~jin�ch konstant
(pat�� tedy mezi m�lo p�esn� konstanty).%
%
\myfootnote{%
Bouguer (1738)   --- �chylka svislice u~hory Chimborasso v~Peru $7''$ (odhad $m_{\rm hory}$ z~$V$ a~$\rho$);
Carlini (1824)   --- zm�na $g$ p�i v�stupu na horu (sestupu do dolu);
Cavendish (1797) --- laboratorn� torzn� v�hy (v�chylky v~bodech obratu);
Jolly (1879)	 --- dvouramenn� v�hy (t�leso nad/pod hmotnou koul�) $\sim\!0{,}15\,{\rm mg}$;
Boys (1895)      --- torzn� kmity (nutno m��it rozm�ry a~momenty setrva�nosti t�les).
}


homogenn� pole:%
%
\myfootnote{P�d kuli�ky: $s = 1\,{\rm m}$, $t = 0{,}45\,{\rm s}$, $g = {2s\over t^2} \doteq 10\,{\rm m}\cdot{\rm s}^{-2}$
a~hmotnost Zem� $M_\zeme = {g R_\zeme^2\over G} \doteq 6\cdot10^{24}\,{\rm kg}$.}
%
\ $F = m g = G {M_\zeme m\over R_\zeme^2}$


gravita�n� pole {\em vn�\/} homogenn� sf�ry: jako u~hmotn�ho bodu!\,%
%
\myfootnote{%
Co� bychom ov��ili integrac� koule \uv{p�es prst�nky} souos� s~$\vec r$:
$\vec\Delta = \vec r-\vec r'$,
$\Delta^2 = r^2 - 2rr'\cos\phi + r'^2$ (kos�nov� v�ta).
%
Jednodu��� je pou��t skal�rn� potenci�l~$U$ m�sto vektorov� s�ly $\vec F = \nabla U$:
$$U = \int_{\rm kouli}\!\! -{G\,\d m\over\Delta}
= -G \int_V {\rho\,\d V\over\Delta}
= -G \int_{r'=\,0}^R \int_{\phi=\,0}^\pi {\rho(r')\over\Delta}\, \overbrace{2\pi r'\!\sin\phi}^{\rm obvod}\,\, r'\d\phi\,\, \d r'\,.$$
%
Trik: $\d\phi$ vyj�d��m z~$\Delta = (r^2 - 2rr'\cos\phi + r'^2)^{1\over2}$
a zam�n�m integra�n� meze $(0, \pi)$ za $(r-r', r+r')$.
%
$${\d\Delta\over\d\phi} = {1\over2} (r^2 - 2rr'\cos\phi + r'^2)^{-{1\over2}} (-2rr')(-\sin\phi)
= {rr'\sin\phi\over\Delta}\,,$$
%
odkud $\d\phi = {\Delta\over rr'\sin\phi} \d\Delta$ a
%
$$\eqalign{
U &= -G \int_0^R\!\!\! \int_{r-r'}^{r+r'}\!\! {\rho(r')\over\Delta}\, 2\pi r'\!\sin\phi\, {\Delta\over rr'\sin\phi} \d\Delta\, \d r'
= -{2\pi G\over r} \int_0^R\!\!\! \int_{r-r'}^{r+r'}\!\!\! \rho(r') r' \d\Delta\, \d r' \cr
&= -{2\pi G\over r} \int_0^R \rho(r') r' 2r' \d r'
= -{G\over r} \overbrace{\int_0^R 4\pi r'^2 \rho(r')\, \d r'}^{\hbox{celkov� hmotnost}}
= -{G M\over r}\,,
}$$
%
tj.~p�esn� jako hmotn� bod.}

OBR vektor� $\vec r$, $\vec r'$, $\vec\Delta$, koule a prst�nk�

uvnit�: $M(r) = {4\over 3} \pi r^3 \rho$, $F = G {M m\over r^2} = {4\over 3} \pi\, G \rho\,m \cdot r$
\=> line�rn� r�st

OBR sf�ry a~plochy uvnit�, $S\propto R^2$, $F\propto R^{-2}$ \=> $F_1=F_2$

OBR $|\vec F|(r)$ uvnit� a~vn�

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{Keplerovy z�kony:}

\beginitem
\item{1.} elipsy, ohnisko $\slunce$;

\item{2.} stejn� plochy opsan� pr�vodi�em za stejnou dobu;

\item{3.} $\displaystyle {a^3\over T^2} = {\rm konst.}$ (ne zcela p�esn�!).

\enditem

Jak se m�n� rychlost se vzd�lenost�?
$v = {2\pi r\over T} = 2\pi\sqrt{\rm konst.} {1\over\sqrt{r}}$

OBR srovnej s~plochou rota�n� k�ivkou galaxie,\\
pozn.~o~temn� hmot� v~galaktick�m halu

Pozorovat mohu sp��e {\em synodickou\/} dobu ob�hu, pot�ebuj� p�evod $t_{\rm syn}\-> t_{\rm siderick�}$:

$\omega_1 t_{\rm syn} = 2\pi + \phi$,
$\omega_2 t_{\rm syn} = \phi$,
$\omega_1 = {2\pi\over t_{{\rm sid}1}}$,
$\omega_2 = {2\pi\over t_{{\rm sid}2}}$ \=>

$${1\over t_{\rm sid1}} = {1\over t_{\rm syn}} + {1\over t_{\rm sid2}}\label{eq:t_sid}$$

OBR kru�nic a~�hlov�ch rychlost�

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\eject

\myitem{Probl�m dvou t�les} (Murray \& Dermott 1999):

inerci�ln� soustava sou�adnic, $\vec r = \vec r_2 - \vec r_1$, pou�it� {\em v�ech\/} t�� NZ:%
%
\myfootnote{Pozn�mka o~derivac�ch pro ty, kte�� o~nich je�t� nesly�eli:
%
sklon fce $\tg\alpha = {\Delta y\over\Delta x}$ popisuje \uv{jak moc se fce m�n�};
%
{\em derivace\/} funkce $y(x)$ v~bod� $x_0$ je definov�na jako limita:
%
$$y' \equiv {\d y\over\d x} \equiv \lim_{\Delta x\to 0} {y(x_0 + \Delta x) - y(x_0)\over \Delta x}\,.$$
%
P��kladem prvn� derivace sou�adnic podle �asu je okam�it� rychlost
($\vec v \equiv {\d\vec r\over\d t}$, ozna�uje se t� $\dot r$);
zrychlen� je prvn� derivace rychlosti
nebo t� druh� derivace sou�adnic ($\vec a \equiv {\d\vec v\over\d t} \equiv \dot\vec v
= {\d^2\vec r\over\d t^2} \equiv \ddot r$).
%
Uve�me derivace n�kter�ch element�rn�ch fc�:
%
$C' = 0$,
$x' = 1$,
$ax' = a$,
$(x^n)' = n x^{n-1}$,
$(\sin x)' = \cos x$,
$(\cos x)' = -\sin x$.
%
Derivace sou�tu $(f+g)' = f' + g'$,
sou�inu $(f\cdot g)' = f'g + fg'$,
slo�en� fce $(f(g(x)))' = f'(g(x)) g'(x)$.
%
Opa�n�m �kolem k~derivov�n� je hled�n� {\em primitivn� funkce\/} (neboli integrov�n�).}


$$\eqalignno{
\vec F_1 &= +G {m_1 m_2\over r^3} \vec r = m_1\ddot\vec r_1\,, & \label{eq:F_1} \cr
\vec F_2 &= -G {m_1 m_2\over r^3} \vec r = m_2\ddot\vec r_2\,, & \label{eq:F_2} \cr
}$$

tzn.~{\em celkem 12~nezn�m�ch skal�rn�ch fc� �asu\/}:
$\vec r_1(t)$, $\dot\vec r_1(t)$, $\vec r_2(t)$, $\dot\vec r_2(t)$.


P�i �e�en� t�chto diferenci�ln�ch rovnic pou�iji t�i triky:

\beginitem
\item{1.} budu studovat {\em relativn�\/} pohyb, nikoli p��mo $\vec r_1$, $\vec r_2$;

\item{2.} nejprve najdu {\em tvar\/} trajektorie $r(\phi)$, nikoli $\vec r(t)$;

\item{3.} pou�iji {\em substituce\/} za ${1\over r}$.

\enditem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{integr�ly pohybu (3~pro t�i�t� a~3~pro hybnosti):}

snadno najdu prvn�ch 6~integr�l� (celkem jich je 10), kter� �lohu podstatn� zjednodu��
(hypoteticky, kdybych na�el 12~{\em prvn�ch\/} integr�l�,
t�lesa by \uv{tr�ela} na m�st� nebo se pohybovala rovnom�rn� p��mo�a�e):

sou�et rovnic~\ref{eq:F_1} a~\ref{eq:F_2}: $m_1\ddot\vec r_1 + m_2\ddot\vec r_2 = \vec 0$,\\
prvn� neur�it� integr�l: $m_1\dot\vec r_1 + m_2\dot\vec r_2 = \vec a$,\\
druh�: $m_1\vec r_1 + m_2\vec r_2 = \vec a\,t + \vec b$,
kde $\vec a$, $\vec b$ jsou konstantn� vektory (ony integr�ly pohybu;
proto�e $\vec b$ je a� {\em druh�\/} integr�l, vyskytuj� se zde konstantn� rychlosti,
neboli line�rn� z�vislost sou�adnic na �ase).

\=> t�i�t� $\vec T = {m_1\vec r_1 + m_2\vec r_2\over m_1+m_2}$
je bu� v~klidu, nebo se rovnom�rn� p��mo�a�e pohybuje
(a~v�bec nez�le�� na tom, �e $F_{\rm G} \propto {1\over r^2}$;
sta��, �e s�ly jsou stejn� velik� a~opa�n�ho sm�ru).

OBR vektory $\vec r_1$, $\vec r_2$, $\vec r$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{rovnice relativn�ho pohybu (a~3~integr�ly momentu hybnosti):}

nyn� vy�et��me relativn� pohyb $m_2$ vzhledem k~$m_1$ --- ode�teme~\ref{eq:F_2} od~\ref{eq:F_1}\\
(p�ejdeme do {\em neinerci�ln�\/} sou�adnicov� soustavy):
%
$$\ddot\vec r = \ddot\vec r_2 - \ddot\vec r_1 = -\overbrace{G(m_1+m_2)}^{\textstyle\mu} {\vec r\over r^3}$$
%
$${\ddot\vec r} + \mu{\vec r\over r^3} = 0\label{eq:relativni_pohyb}$$

dal�� integr�ly pohybu najdeme vektorov�m n�soben�m rce~$\vec r$:
%
$$0 = \ddot\vec r\times\vec r + {\mu\over r^3}\overbrace{\vec r\times\vec r}^{\vec 0}
= \ddot\vec r\times\vec r\,,$$
%
proto�e vektorov� sou�in rovnob�n�ch vektor� je v�dy $\vec 0$.
Neur�it� integr�l je
%
$$\vec r\times\dot\vec r = \vec h = {\rm konst.}\label{eq:integral_h}$$
%
neb $(\vec r\times\dot\vec r - \vec h)' = \overbrace{\dot\vec r\times\dot\vec r}^{\vec 0} +\, {\vec r\times\ddot\vec r} - \vec 0 = \vec 0$;
$\vec h$~je~\uv{n�co jako} moment hybnosti%
%
\myfootnote{Rozhodn� to {\em nen�\/} celkov� moment hybnosti v~inerci�ln� soustav�
$\vec L = \vec r_1\times m_1\dot\vec r_1 + \vec r_2\times m_2\dot\vec r_2$;
v~$\vec h$ jsou p�itom vektorov� sou�iny \uv{pom�chan�}:
$\vec h = (\vec r_1-\vec r_2)\times(\dot\vec r_2 - \dot\vec r_1)
= \vec r_1\times\dot\vec r_2 - \vec r_1\times \dot\vec r_1 + \dots$}
%
\ na jednotku hmotnosti~$\mu$.

\=> pohyb prob�h� {\em v~rovin�\/} a~plo�n� rychlost se zachov�v� \=> {\bf II.~KZ}\\
(op�t nevy�aduje $F_{\rm G} \propto {1\over r^2}$,
$\vec F_{\rm G}$ pouze mus� sm��ovat pod�l spojnice t�les)

OBR $\vec r\times\vec v$ a~plo�n� rychlost

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{p�echod do pol�rn�ch sou�adnic v~rovin�:}

sou�adnou soustavu orientuji tak, �e rovina~$xy$ je kolm� na~$\vec h$ a~dosad�m%
%
\myfootnote{Ov��en� vztahu pro 2.~derivaci v~pol�rn�ch sou�adnic�ch:
%
b�zov� vektory
$\hat r = (\cos\phi, \sin\phi)$,
$\hat\phi = (-\sin\phi, \cos\phi)$;
%
$\dot{\hat r} = (-\sin\phi, \cos\phi) \dot\phi = \dot\phi\hat\phi$,
$\dot{\hat\phi} = -\hat r\dot\phi$;
%
$\vec r = r\hat r$,
%
$\dot\vec r = \dot r\hat r + r\dot{\hat r} = \dot r\hat r + r\dot\phi\hat\phi$,
%
$\ddot\vec r = \ddot r\hat r + \dot r\dot\phi\hat\phi + \dot r\dot\phi\hat\phi + r\ddot\phi\hat\phi + r\dot\phi(-\hat r\dot\phi)
= (\ddot r - r\dot\phi^2) \hat r + (2\dot r\dot\phi + r\ddot\phi) \hat\phi
= (\ddot r - r\dot\phi^2) \hat r + \left[{1\over r}{\d\over\d t} (r^2\dot\phi)\right] \hat\phi$.}%
%
\ za $\vec r = r\hat r$,
$\ddot\vec r = (\ddot r - r\dot\phi^2)\,\hat r + \left[{1\over r}{\d\over\d t} (r^2\dot\phi)\right]\hat\phi$
do vektorov� rce~\ref{eq:relativni_pohyb} \=> skal�rn� rce (p��slu�n� slo�ce~$\hat r$):
%
$$\ddot r - r\dot\phi^2 = -{\mu\over r^2}\,,\label{eq:polarni} $$
%
Rovnice ${1\over r}{\d\over\d t} (r^2\dot\phi) = 0$ pro slo�ku $\hat\phi$ 
je na�t�st� spln�na automaticky, proto�e to je vlastn� vyj�d�en� integr�lu~$\vec h$
v~pol�rn�ch sou�adnic�ch:
%
$$\vec h = \vec r\times\dot\vec r = r\hat r \times (\dot r\hat r + r\dot\phi\hat\phi) = r^2\dot\phi\hat z\label{eq:h_z}$$
%
a~velikost tohoto vektoru je evidentn�:
%
$$h = r^2\dot\phi = {\rm konst.}\label{eq:h}$$
%
Jak �e�it rovnici~\ref{eq:polarni} pro {\em dv�\/} nezn�m� fce �asu $r(t)$, $\phi(t)$?!

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{eliminace �asu pomoc� substituce $u = {1\over r}$ a~integr�lu~$h$}:

Do rovnice~\ref{eq:polarni} mus�m dosadit jednak za $r = {1\over u}$,
a~jednak za druhou derivaci podle �asu~$\ddot r$. Nejprve spo�tu prvn�:
%
$$\dot r  = -{1\over u^2} {\d u\over\d t} = -{1\over u^2} {\d u\over\d\phi}\!\overbrace{\d\phi\over\d t}^{\hbox to0pt{\hss\kern1.65cm$\scriptstyle\equiv\,\dot\phi\,=\,{h\over r^2}\,=\,h u^2$\hss}} = h\,{\d u\over\d\phi}\,;\label{eq:dot_r}$$
%
zde jsem elegantn� derivoval slo�enou fci $u(\phi(t))$
a~vyu�il integr�l~$h$~\ref{eq:h}. Druh� je
%
$$\ddot r = -h\,{\d^2u\over\d\phi^2}\,\dot\phi = -h^2 u^2{\d^2u\over\d\phi^2}\,.\label{ddot_r}$$
%
Vyjde $-h^2 u^2 {\d^2 u\over\d\phi^2} - {1\over u} h^2 u^4 = -\mu u^2\,$ a~po kr�cen�:
%
$${\d^2 u\over\d\phi^2} + u = {\mu\over h^2}\,,\label{eq:u}$$
%
tj.~oby�ejn� line�rn� diferenci�ln� rovnice (ODE) 2.~��du pro funkci $u(\phi)$.
%
Jej� obecn� �e�en� \uv{uh�dnu}:%
%
\myfootnote{%
Mohu se p�esv�d�it jeho derivov�n�m
${\d u\over\d\phi} = -{\mu\over h^2} e\sin(\phi-\varpi)$,
${\d^2 u\over\d\phi^2} = -{\mu\over h^2} e\cos(\phi-\varpi)$,
�e rovnici~\ref{eq:u} opravdu spl�uje
a~bez d�kazu \uv{v���m}, �e neexistuje n�jak� jin� �e�en�.}
%
$$u = {\mu\over h^2} [1 + e\cos(\phi-\varpi)]\,,\label{eq:u_reseni}$$
%
kde m�m dv� integra�n� konstanty: $e$, $\varpi$%
%
\myfootnote{$\varpi$~je zvl�tn� tvar �eck�ho p�smene~$\pi$, nikoli~$\omega$;
v~\TeX u se p��e \verbtt|\backsl varpi|.}
%
\ (plus onen p�edchoz� integr�l~$h$ a~hmotnost~$\mu$,
kter� ozna��m dohromady jako jeden parametr~$p$:
%
$$p = {h^2\over\mu}\,.\label{eq_p}$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{�e�en� probl�mu 2~t�les --- rovnice ku�elose�ky:}

zp�tn� substituce d�v� fci $r(\phi)$:
%
$$r = {p\over 1+e\cos(\phi-\varpi)}\,,\label{eq:r_polarni}$$
%
co� je {\em obecn� rce ku�elose�ky\/} s~ohniskem v~po��tku
(t�lese~1; 2.~ohnisko je pr�zdn�) \=> {\bf I.~KZ}

$e$~\dots excentricita,
$\varpi$~\dots d�lka pericentra,
$p$~\dots parametr.

M��eme rozli�it 4~p��pady:
%
\table{\halign{%
#\hfil &\ #\hfil &\ #\hfil\cr
kru�nice  & $e = 0$	& $p = a$ \cr
elipsa    & $0 < e < 1$ & $p = a\,(1-e^2)$ \cr
parabola  & $e = 1$     & $p = 2q$ \cr
hyperbola & $e > 1$     & $p = a\,(e^2-1)$ \cr
}}
%
kde $a$~ozna�uje {\em velkou poloosu\/}
a~$q$~vzd�lenost pericentra.

OBR ��znut�ho ku�ele a~elipsy

Pol�rn� sou�adnice $\phi$ je zv�na {\em prav� d�lka\/} (m��en� od osy~$x$),
�asto u��v�me {\em pravou anom�lii\/}~$f$:
%
$$f = \phi-\varpi\,,\label{eq:f}$$
%
m��enou od pericentra.
Potom pro elipsu je
%
$$r = {a\,(1-e^2)\over 1+e\cos f}\label{eq:elipsa}$$
%
a~kdybychom m�li osu~$x$ orientovanou p��mo do pericentra, bylo by:
%
$$x = r\cos f\,,\quad y = r\sin f\,.$$
%
U�ite�n� je znalost vztah� pro vzd�lenosti pericentra~$q$ a~apocentra~$Q$ od ohniska:
%
$$\eqalignno{%
q &= a\,(1-e)\,, & \label{eq:q}\cr
Q &= a\,(1+e)\,, & \label{eq:Q}\cr
}$$
%
a~tak� integr�lu~$h$ vyj�d�en�ho v~orbit�ln�ch elementech:
%
$$h = \sqrt{p\mu} = \sqrt{a(1-e^2)\mu}\,;\label{eq:h_elmts}$$
%
oboj� plyne z~geometrie elipsy.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{Jak� je vztah mezi periodou ob�hu a~rozm�rem dr�hy?}

plocha~$A$ elipsy opsan� za celou orbit�ln� periodu
%
$$A = \pi a b = \pi a^2 \sqrt{1-e^2} = {1\over 2} r^2\dot\phi\,T = h\,{T\over 2} = \sqrt{a(1-e^2)\mu}\,{T\over 2}\,,$$
%
odtud
%
$${a^3\over T^2} = {G(m_1+m_2)\over 4\pi^2}\label{eq:3KZ}$$
%
\=> {\bf 3. KZ}

Kepler ho neznal p�esn� (zanedbal $m_2\ll m_1$)

dal�� u�ite�n� tvary 3.~KZ:
%
$$n^2 a^3 = \mu\,,\label{eq:stredni_pohyb}$$
%
kde $n = {2\pi\over T}$ je {\em st�edn� pohyb\/}; nebo
%
$${[a]_{\rm AU}^3\over [T]_{\rm yr}^2} = [M]_{M_\slunce} = 1\,,\label{eq:3KZc}$$
%
kter�to jedni�ka plat� v~na�� slune�n� soustav�.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{m��en� hmotnosti planet(k)y se satelitem:}

${a^3\over T^2} = {G(m_\odot + m_{\rm planety})\over 4\pi^2}$ (ve slune�n� soustav�)

${a'^3\over T'^2} = {G(m_{\rm planety} + m_{\rm satelitu})\over 4\pi^2}$ (syst�m se satelitem)

$$\left({a\over a'}\right)^3 \left({T'\over T}\right)^2
= {m_\odot+m_{\rm p}\over m_{\rm p}+m_{\rm s} }
\simeq {m_\odot\over m_{\rm p}}\label{eq:satelity}$$

zanedb�n� planety v��i Slunci a~satelitu v��i planet�

pro n�kter� velk� planetky je mo�n� m��it $m$ p��mo
z~perturbac�, kter� p�sob� na planety
(Standish \& Hellings 1989)

Ida a~Dactyl z~Galilea (Belton \etal\ 1995):
$\rho = (2{,}6\pm 0{,}5)\,{\rm kg}\cdot{\rm m}^{-3}$

pozn. o~nejednozna�n� ur�en� trajektorii Dactylu\\
a~pravd�podobn� vysok�ch porozit�ch asteroid�

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\eject

\myitem{Jak� je rychlost v~dr�ze? (1~integr�l energie)}

Je�t� existuje des�t� integr�l pohybu, kter� n�m umo�n� snadno ur�ovat
velikost rychlosti v~dr�ze (zat�m zn�m jen $r(\phi)$ a~II.~KZ).
Rci relativn�ho pohybu~\ref{eq:relativni_pohyb} $\ddot\vec r + \mu {\vec r\over r^3} = \vec 0$
n�sob�m skal�rn� rychlost�~$\dot\vec r$:%
%
\myfootnote{%
Proto�e
$\vec r=r\hat r$,
$\dot\vec r = \dot r\hat r + r\dot\phi\hat\phi$
a~$\hat r\cdot\hat\phi = 0$,
je~${\vec r\over r^3}\cdot\dot\vec r = {r\hat r \cdot \dot r\hat r \over r^3} + {r\hat r \cdot r\dot\phi\hat\phi \over r^3}
= {\dot r\over r^2}$.}
%
$$\ddot\vec r\cdot\dot\vec r + \mu {\dot r\over r^2} = \vec 0\,,$$
%
prvn� neur�it� integr�l je:%
%
\myfootnote{%
Proto�e $v^2 = \dot\vec r\cdot\dot\vec r$
a~$(\dot\vec r\cdot\dot\vec r)' = 2\ddot\vec r\cdot\dot\vec r$.}
%
$${1\over2} v^2 - {\mu\over r} = C\,,\label{eq:v2}$$
%
kde $C$ je integra�n� konstanta; tj.~integr�l \uv{�iv� s�ly} (ZZE na jednotku hmotnosti).

Kolik je p�esn� $C$?
%
Do v�razu pro $v^2$ v~pol�rn�ch sou�adnic�ch
dosad�me z~pol�rn� rce elipsy
a~pou�ijeme je�t� integr�l momentu hybnosti~$h$
a~III.~KZ (vlastn� to odvod�me znovu jinak):
%
$$v^2 = \dot\vec r\cdot\dot\vec r = \dot r\cdot\dot r + r^2\dot\phi^2 = \dot r^2 + (r\dot f)^2\,,\label{eq:v2rphi}$$
%
proto�e $\phi = f+\varpi$, $\varpi = {\rm konst.}$ a~$\dot\phi = \dot f$.
Jen�e tam se krom� $r$ vyskytuje je�t� $\dot r$ a~$\dot f$.
%
Proto ze zn�m� rovnice elipsy~\ref{eq:elipsa}
%
$r = {a(1-e^2)\over 1+e\cos f(t)}$
%
vypo��t�me derivaci~$\dot r$:
%
$$\dot r = -{a(1-e^2)\over (1+e\cos f)^2} \cdot e(-\sin f)\dot f\,,$$
%
p�i�em� $\dot f$ vyj�d��me z~integr�lu $h$~\ref{eq:h_elmts}:
%
$$\dot f = {na^2\sqrt{1-e^2}\over r^2}$$
%
a~vyjde:
%
$$\dot r = {na\over\sqrt{1-e^2}}\,e\sin f\,,$$
%
$$r\dot f = {na\over\sqrt{1-e^2}}(1+e\cos f)\,.$$
%
V�e dosad�me do~\ref{eq:v2} a~uprav�me:
%
$$\eqalignno{
v^2 &= {n^2a^2\over 1-e^2}\,(e^2\sin^2 f + 1 + 2e\cos f + e^2\cos^2 f)
     = {n^2a^2\over 1-e^2}\,(e^2 + 1 + 2e\overbrace{\cos f}^{\hbox to0pt{\hss$\scriptstyle=\,\bigl({a(1-e^2)\over r} - 1\bigr){1\over e}$}}) = \cr
    &= {n^2a^2\over 1-e^2} \left({2a(1-e^2)\over r} - (1-e^2)\right)
     = n^2a^3\left({2\over r} - {1\over a}\right)
     = \mu\left({2\over r} - {1\over a}\right) & \label{eq:v2_C}\cr
}$$
%
Vid�me, �e integra�n� konstanta v~\ref{eq:v2} je $C = -{\mu\over 2a}$ a~nez�vis� na $e$;
pro parabolickou dr�hu $C = 0$, pro hyperbolickou $C = +{\mu\over 2a}$.

Maxim�ln� rychlost je v~pericentru, kde $r = q = a(1-e)$:
%
$$v_{\rm p}^2 = \mu\left({2\over a(1-e)} - {1\over a}\right) = {\mu\over a} {2 - (1-e)\over 1-e} = n^2a^2 {1+e\over 1-e}\,,$$
%
$$v_{\rm p} = na\sqrt{1+e\over 1-e}\,,\label{eq:v_p}$$
%
minim�ln� v~apocentru, p�i $r = Q = a(1+e)$:
%
$$v_{\rm a} = na\sqrt{1-e\over 1+e}\,.\label{eq:v_a}$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{Jak� je vztah mezi $f$ a~$t$?}

Doposud jsme m�li toti� pouze z�vislost $r(\phi)$, nikoli $r(t)$.
%
Zjist�me to oklikou:

\beginitem
\item{1.} najdeme vztah mezi {\em pravou\/} anom�li�~$f$ a~{\em excentrickou\/} anom�li�~$E$;

OBR opsan� kru�nice, $E$, $f$

\item{2.} mezi $E$ a~{\em st�edn�\/} anom�li�~$M = n(t-\tau)$;

$M$~{\em nem�\/} jednoduchou geometrickou reprezentaci!\\
$\tau$~ozna�uje okam�ik pr�chodu pericentrem

\enditem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Ad 1) Z~geometrie elipsy hned vid�me, �e kart�zsk� sou�adnice:
%
$$\eqalignno{
x &= a\cos E - ae        = r\cos f\,, & \label{eq:xEf} \cr
y &= a\sqrt{1-e^2}\sin E = r\sin f\,, & \label{eq:yEf} \cr
}$$
%
pak
%
$$\eqalignno{
r &= \sqrt{x^2+y^2}
   = \sqrt{a^2 \cos^2 E - 2 a^2 e \cos E + a^2 e^2 + a^2 (1-e^2) \sin^2 E} \cr
  &= a \sqrt{1 - 2e\cos E + e^2(1-\sin^2 E)}
   = a (1-e\cos E) & \label{eq:rE} \cr
}$$

proto�e $\cos f = {x\over r} = {\cos E-e\over 1-e\cos E}$
%
\=>
%
\myfootnote{%
Vyu�ijeme vztah� pro polovi�n� �hel:
$\sin^2{f\over 2} = {1-\cos f\over 2}$,
$\cos^2{f\over 2} = {1+\cos f\over 2}$
a~vypo�tu si
$1-\cos f = {1-e\cos E - \cos E+e\over 1-e\cos E} = {(1+e)(1-\cos E)\over 1-e\cos E}$,
$1+\cos f = {(1-e)(1+\cos E)\over 1-e\cos E}$.}
%
$$\tg{f\over 2} = \sqrt{1+e\over 1-e} \tg{E\over 2}\,.\label{eq:fE}$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Ad 2) Nap��eme �asov� derivace sou�adnic $(x,y)$:

$$\eqalign{
\dot x &= -a\sin E\dot E\, \cr
\dot y &= a\sqrt{1-e^2}\cos E\dot E\, \cr
}$$
%
a~vyu�ijeme \uv{asi post�} integr�lu~$h$~\ref{eq:h}:
%
$$\eqalign{
h &= na^2\sqrt{1-e^2}
   = |\vec r\times\dot\vec r|
   = |(x,y,0)\times(\dot x,\dot y,0)|
   = |(0,0,y\dot x - x\dot y)| \cr
  &= a^2\sqrt{1-e^2}\sin^2 E\dot E + a(\cos E-e) a\sqrt{1-e^2}\cos E\dot E \cr
}$$
%
odkud dostaneme diferenci�ln� rovnici pro funkci $E(t)$:
%
$$\dot E = {n\over1-e\cos E}\,,\label{eq:dE}$$
%
neboli
%
$$\d E\,(1-e\cos E) = n\,\d t\,,$$
%
jej�� neur�it� integr�l je:
%
$$E - e\sin E = n t + C\,.$$
%
Hrani�n� podm�nka: pro $E = 0$ je $t = \tau$ \=> $C = -n\tau$.
%
$$M = E - e\sin E\label{eq:keplerova}$$
%
{\em Keplerova rce\/} $\leftarrow$ transcendentn� :( nezn�me algebraick� vyj�d�en� $E(M)$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myitem{Jak �e�it Keplerovu rovnici?}

\beginitem
\item{1.} itera�n� metodou:

$$E_{i+1} = M + e\sin E_i\,,\quad E_0 = M\label{eq:iteracni}$$

$E_1 = M + e\sin M$,
$E_2 = M + e\sin E_1$,
\dots,
dokud $|E_{i+1} - E_i| > \epsilon$ mal�.

lep�� po��te�n� odhad $E_0 = M + \sgn(\sin M)\,k e$,
kde $k\in(0,1]$, nap�.~$k = 0{,}85$.

\item{2.} Newton--Raphsonovou metodou (obecn� pro hled�n� ko�en� fce $f(x) = 0$):

$$f(E) = E - e\sin E - M\,,$$
%
hled�me tedy $f(E) = 0$ postupn�mi iteracemi
%
$$E_{i+1} = E_i - {f(E_i)\over f'(E_i)}\,,\label{eq:raphson}$$
%
kde
%
$$f'(E) = 1 + e\cos E$$
%
je derivace $f(E)$ podle~$E$.

OBR Newton--Raphsonovy metody

\item{3.} analyticky, Fourierov�m rozvojem:

Vych�z� z~1.~itera�n� metody a~pou�ije sou�tov� vzorec
%
$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$
%
a~Taylorovy rozvoje goniometrick�ch funkc�
%
$\sin x = x - {1\over 6} x^3 + o(x^5)\,,$
$\cos x = 1 - {1\over 2} x^2 + o(x^4)\,.$

$$\eqalign{
E_1 &= M + e\sin M \cr
%
E_2 &= M + e\sin (M + e\sin M) \cr
    &\simeq M + e\sin M \overbrace{\cos(e\sin M)}^{\doteq\,1} + e\cos M \overbrace{\sin(e\sin M)}^{\doteq\,e\sin M} \cr
    &\simeq M + e\sin M + {1\over 2}e^2\sin 2M \cr
%
E_3 &= M + e\sin(M + e\sin M + {1\over 2}e^2\sin 2M) \cr
    &\simeq M + \left(e-{1\over 8}e^3\right) \sin M + {1\over 2} e^2 \sin 2M + {3\over 8} e^3\sin 3M \cr
}$$
%
\=> Fourierova �ada
%
$$E-M = \sum_{s=1}^\infty b_s(e)\sin s M\,,\label{eq:EM_fourier}$$
%
kde koeficienty
%
$$b_s(e) = {2\over s} J_s(se)$$
%
jsou Besselov�mi funkcemi 1.~druhu
%
$$J_s(se) = {1\over\pi} \int_0^\infty \cos(sE - se\sin E)\d E\,.$$
%
Pozor! �ada konverguje pouze do ur�it�ho maxim�ln�ho $e \doteq 0{,}66$.

\enditem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end


