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Na pervuyu stranicu << 2.3. Sfericheskaya sistema koordinat | Oglavlenie | 3. Astronomicheskie sistemy koordinat >>

2.4. Osnovnye formuly sfericheskoi geometrii

Rassmotrim sfericheskii treugol'nik $ ABC$ na nebesnoi sfere, prichem tochka $ A$ yavlyaetsya polyusom, a tochka $ B$ lezhit v ploskosti $ Oxz$ (ris. 2.6).

Ris. 2.6. K vyvodu formul sinusov i kosinusov

Dekartovy koordinaty edinichnyh vektorov $ {\bf r}_A,{\bf r}_B, {\bf r}_C$ opredelyayutsya soglasno (2.20) pri $ r=1$:
$\displaystyle {\bf r}_A$ $\displaystyle =$ $\displaystyle (0,0,1),$  
$\displaystyle {\bf r}_B$ $\displaystyle =$ $\displaystyle (\sin c, 0, \cos c),$  
$\displaystyle {\bf r}_C$ $\displaystyle =$ $\displaystyle (\sin b\cos A, \sin b\sin A, \cos b).$  

Po opredeleniyu skalyarnogo proizvedeniya imeem $ {\bf r}_B\cdot {\bf r}_C=\cos a$. Eto zhe proizvedenie v dekartovyh koordinatah imeet vid:

$\displaystyle {\bf r}_B\cdot {\bf r}_C=\sin c\sin b \cos A+ \cos c\cos b. $

Perepishem eti formuly v sleduyushem vide:

$\displaystyle \cos a=\cos b\cos c + \sin b\sin c \cos A.$ (2.35)

My dokazali teoremu:"Kosinus storony sfericheskogo treugol'nika raven proizvedeniyu kosinusov dvuh drugih ego storon plyus proizvedenie sinusov etih storon na kosinus ugla mezhdu nimi". Obychno sootnoshenie (2.35) nazyvayut formuloi kosinusov. S pomosh'yu ciklicheskoi perestanovki mozhno napisat' formuly kosinusov dlya dvuh drugih storon:

$\displaystyle \cos b$ $\displaystyle =\cos a\cos c + \sin a\sin c \cos B,$    
$\displaystyle \cos c$ $\displaystyle =\cos a\cos b + \sin a\sin b \cos C.$    

Teper' vychislim vektornoe proizvedenie $ {\bf r}_C \times {\bf r}_B$. Soglasno (2.16) poluchim:

$\displaystyle \vert{\bf r}_C \times {\bf r}_B\vert= \sin a. $

Dopustim, chto vektor $ {\bf r}_C \times {\bf r}_B$ napravlen v tochku $ D$ (ris. 2.6), to est'

$\displaystyle {\bf r}_C \times {\bf r}_B={\bf r}_D \sin a,$ (2.36)

$ {\mathbf{r}}_D$ -- edinichnyi vektor. Ispol'zuya (2.17), zapishem levuyu chast' (2.36) v vide:

\begin{displaymath}\begin{split}&{\bf r}_C \times {\bf r}_B = \begin{pmatrix}{\bf i}& {\bf j}& {\bf k}\\ \sin b \cos A& \sin b\sin A & \cos b \\ \sin c & 0 & \cos c \end{pmatrix} =\\ &=(\sin b \sin A\cos c)\ {\bf i}+ (\cos b\sin c-\sin b \cos A\cos c)\ {\bf j} + (-\sin b \sin c\sin A)\ {\bf k}. \end{split}\end{displaymath} (2.37)

Pravuyu chast' (2.36) soglasno (2.5) mozhno zapisat' v vide:

$\displaystyle {\bf r}_D \sin a=\sin a\left[{\bf i}(\sin AD\cos BAD) +{\bf j}(\sin AD \sin BAD)+ {\bf k}\cos AD \right].$ (2.38)

V treugol'nike $ BAD$ storona $ \widehat{BD}=90^\circ$, i ploskost' $ OBD$ perpendikulyarna ploskosti $ OBC$ ( $ \angle DBC=90^\circ$). Poetomu $ \angle ABD=90^\circ + B$. Po formule kosinusov imeem

$\displaystyle \cos \widehat{AD}=\cos 90^\circ \cos c + \sin 90^\circ \sin c \cos (90^\circ+B) $

ili

$\displaystyle \cos \widehat{AD}=- \sin c \sin B. $

Priravnivaya $ z$-komponenty v formulah (2.37) i (2.38), poluchim:

$\displaystyle -\sin b \sin c \sin A=-\sin a \sin c \sin B $

ili

$\displaystyle \frac{\sin b}{\sin B} =\frac{\sin a}{\sin A}. $

Po analogii iz treugol'nika $ CAD$ poluchim:

$\displaystyle \cos \widehat{AD}=- \sin b \sin C, $

chto privodit k vyrazheniyu:

$\displaystyle -\sin b \sin c \sin A=-\sin a \sin b \sin C. $

V rezul'tate my mozhem zapisat', chto

$\displaystyle \frac{\sin a}{\sin A}=\frac{\sin b}{\sin B} =\frac{\sin c}{\sin C}.$ (2.39)

Eti sootnosheniya izvestny kak formuly sinusov. Sformuliruem poluchennyi rezul'tat kak teoremu: "V sfericheskom treugol'nike otnoshenie sinusa storony k sinusu protivolezhashego ugla est' velichina postoyannaya".

Dlya vyvoda sleduyushei gruppy sootnoshenii mezhdu storonami i uglami sfericheskogo treugol'nika zapishem formulu sinusov dlya treugol'nika $ ABD$:

$\displaystyle \frac{\sin AD}{\sin (90^\circ +B)} =\frac{\sin 90^\circ}{\sin BAD}, $

ili: $ \cos B=\sin AD \sin BAD$. Sravnivaya $ y$-komponenty v uravneniyah (2.37) i (2.38), poluchim:

$\displaystyle \sin a\cos B=\cos b\sin c - \sin b\cos c \cos A.$ (2.40)

Sformuliruem sleduyushuyu teoremu: "Proizvedenie sinusa storony sfericheskogo treugol'nika na kosinus prilezhashego ugla ravno proizvedeniyu kosinusa protivolezhashei uglu storony na sinus tret'ei storony minus proizvedenie sinusa na kosinus etih zhe storon, umnozhennoe na kosinus ugla mezhdu nimi".

Ispol'zuya ciklicheskuyu perestanovku storon i uglov, mozhno poluchit' sleduyushie uravneniya:

$\displaystyle \sin a\cos C$ $\displaystyle =\cos c\sin b - \sin c\cos b \cos A,$    
$\displaystyle \sin b\cos A$ $\displaystyle =\cos a\sin c - \sin a\cos c \cos B,$    
$\displaystyle \sin b\cos C$ $\displaystyle =\cos c\sin a - \sin c\cos a \cos B,$ (41)
$\displaystyle \sin c\cos A$ $\displaystyle =\cos a\sin b - \sin a\cos b \cos C,$    
$\displaystyle \sin c\cos B$ $\displaystyle =\cos b\sin a - \sin b\cos a \cos C.$    

Formuly (2.40) i (2.41) izvestny kak formuly pyati elementov ili formuly podobiya.

Na osnove formul sinusov, kosinusov i formul podobiya mozhno poluchit' ryad drugih uravnenii, svyazyvayushih ugly i storony sfericheskogo treugol'nika. S vyvodom etih uravnenii mozhno oznakomit'sya v sootvetstvuyushih uchebnikah. Tak kak v astronomii naibolee chasto ispol'zuyutsya formuly sinusov, kosinusov i podobiya, na vyvode drugih uravnenii my ne budem ostanavlivat'sya.

Dlya usvoeniya osnovnyh formul sfericheskoi rassmotrim reshenie sleduyushih zadach.



Zadacha 1. Vychislit' kratchaishee rasstoyanie mezhdu tochkami $ A$ i $ B$ na poverhnosti Zemli, koordinaty kotoryh ravny $ \varphi_1,\lambda_1$ i $ \varphi_2,\lambda_2$, sootvetstvenno. Zemlyu schitat' sferoi radiusa $ R$.

Reshenie. Tak kak kratchaishim rasstoyaniem na sfere yavlyaetsya duga okruzhnosti bol'shogo kruga, ispol'zuem dlya resheniya zadachi formulu kosinusov. Rassmotrim sfericheskii treugol'nik s vershinami $ A,B,C$, v kotorom tochka $ C$ yavlyaetsya severnym polyusom. Togda duga $ \widehat{AC}$ ravna $ \pi/2-\varphi_1$, duga $ \widehat{BC}$ ravna $ \pi/2-\varphi_2$, duga $ \widehat{AB}$ -- $ \lambda_2-\lambda_1$ (budem schitat', chto $ \lambda_2\gt \lambda_1$). Esli $ \lambda_2=\lambda_1$, to, ochevidno, rasstoyanie v uglovoi mere ravno $ \vert\varphi_2-\varphi_1\vert$, v lineinoi mere $ R\vert\varphi_2-\varphi_1\vert$ ( $ \varphi_1,\varphi_2$ vyrazheny v radianah).

Soglasno (2.16) poluchim:

$\displaystyle \cos \widehat{AB}=\sin\varphi_1\sin\varphi_2+ \cos\varphi_1\cos\varphi_2\cos(\lambda_2-\lambda_1). $

V lineinoi mere rasstoyanie mezhdu dvumya tochkami ravno $ R \arccos(\cos \widehat{AB})$.

Pri vychisleniyah na nebesnoi sfere zaranee, kak pravilo, neizvestno, kakoi mozhet byt' dlina dugi. Poetomu, obyazatel'no odnovremenno s ispol'zovaniem formuly kosinusov neobhodimo nahodit' sinus dugi (ili ugla). V etom sluchae velichiny sinusa i kosinusa odnoznachno opredelyayut velichinu dugi (ugla).



Zadacha 2. Vychislit' koordinaty $ \varphi,\lambda$ samoi severnoi tochki dugi bol'shogo kruga, prohodyashego cherez tochki $ A,B$ na poverhnosti Zemli s koordinatami $ \varphi_1,\lambda_1$ i $ \varphi_2,\lambda_2$, sootvetstvenno.

Reshenie. Rassmotrim sfericheskii treugol'nik $ ABC$, v kotorom tochka $ C$ yavlyaetsya severnym polyusom. Opredelim severnyi polyus kak tochku, s kotoroi vrashenie proishodit protiv chasovoi strelki.

Oboznachim samuyu severnuyu tochku dugi bol'shogo kruga, prohodyashego cherez tochki $ A,B$, cherez $ D$. Provedem cherez tochku $ D$ ploskost', perpendikulyarnuyu osi, soedinyayushei polyusy. Takaya ploskost' peresechet sferu po okruzhnosti, nazyvaemoi parallel'yu. Ochevidno, chto duga $ \widehat{AB}$ bol'shogo kruga kasaetsya v t.$ D$ paralleli, i, sledovatel'no, meridian $ CD$ peresekaet dugu $ \widehat{AB}$ pod pryamym uglom. Znachit ugly $ \angle CDA$, $ \angle CDB$ ravny $ 90^\circ$, dugi $ \widehat{AC}$ i $ \widehat{BC}$ ravny $ \pi/2-\varphi_1, \pi/2-\varphi_2$, sootvetstvenno, i po formule sinusov dlya treugol'nika $ CAD$ poluchim:

$\displaystyle \frac{\sin \widehat{CD}}{\sin(\angle CAD)}=\frac{\sin(\pi/2-\varphi_1)}{\sin(\angle CDA)} $

ili

$\displaystyle \sin \widehat{CD}=\cos\varphi_1\sin(\angle CAD). $

Po formule sinusov dlya treugol'nika $ ABC$ naidem:

$\displaystyle \frac{\sin(\angle CAD)}{\sin(\pi/2-\varphi_2)} = \frac{\sin(\lambda_2-\lambda_1)}{\sin\widehat{AB}}. $

Tak kak dlina dugi $ \widehat{CD}$ ravna $ \pi/2-\varphi$ i $ \sin(\angle CAD)=\sin(\angle CAB)$, to poluchim:

$\displaystyle \cos\varphi = \cos\varphi_1\cos\varphi_2 \frac{\sin(\lambda_2-\lambda_1)}{\sin\widehat{AB}}.$ (2.42)

Sinus dugi $ \widehat{AB}$ legko vychislit', ispol'zuya reshenie predydushei zadachi. Kak izvestno, $ \sin\widehat{AB}=\pm\sqrt{1-\cos^2\widehat{AB}}$. Nuzhnyi znak vybiraetsya iz usloviya $ \cos\varphi\gt 0$, tak kak trebuetsya naiti samuyu severnuyu tochku dugi bol'shogo kruga.

Dolgotu tochki $ D$ naidem iz sleduyushih soobrazhenii. Ispol'zuyu formulu podobiya dlya pryamougol'nogo treugol'nika $ CAD$, poluchim:

$\displaystyle \sin\widehat{AD}\cos(\angle CDA) = \cos(\pi/2-\varphi_1)\sin(\pi/2-\varphi) - \sin(\pi/2-\varphi_1)\cos(\pi/2-\varphi) \cos(\lambda-\lambda_1) $

ili:

$\displaystyle \cos(\lambda-\lambda_1)=\tg\varphi_1\textrm{ctg}\varphi. $

Opredelenie dolgoty stanovitsya nevozmozhnym v dvuh sluchayah: esli tochki $ A$ i $ B$ lezhat na ekvatore (v etom sluchae $ \varphi_1= \varphi_2=0$; $ \sin\widehat{AB}=\sin(\lambda_2-\lambda_1)$; iz (2.42) sleduet, chto $ \varphi=0$; znachit $ \cos(\lambda-\lambda_1)=0\cdot \infty$) i esli tochki $ A$ i $ B$ lezhat na odnom meridiane (v etom sluchae $ \lambda_1=\lambda_2$ ili $ \lambda_1=\lambda_2+\pi$ i $ \varphi=\pi/2$, t.e. duga prohodit cherez polyus).

<< 2.3. Sfericheskaya sistema koordinat | Oglavlenie | 3. Astronomicheskie sistemy koordinat >>

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