Jak Filipp Mari Binet tomonidan olingan **Binet tenglamasi** tekislik qutbli koordinatalarda orbital harakatning shakli berilgan markaziy kuch shaklini beradi. Tenglamadan maʼlum bir kuch qonuni uchun orbita shaklini olish uchun ham foydalanish mumkin, lekin bu odatda ikkinchi tartibli chiziqli boʻlmagan oddiy differentsial tenglamaning echimini oʻz ichiga oladi. Quvvat markazi atrofida aylanma harakatda yagona yechim mumkin emas.

Tenglama

Orbita shakli koʻpincha nisbiy masofa nuqtai nazaridan qulay tarzda tavsiflanadi {\displaystyle r} burchak funktsiyasi sifatida {\displaystyle \theta }. Binet tenglamasi uchun orbital shakli oʻrniga qisqaroq tasvirlangan {\displaystyle u=1/r} funktsiyasi sifatida {\displaystyle \theta }. Muayyan burchak momentumini quyidagicha aniqlang {\displaystyle h=L/m} qayerda {\displaystyle L} burchak impulsi va {\displaystyle m} massa hisoblanadi. Keyingi boʻlimda olingan Binet tenglamasi funksiya boʻyicha kuchni beradi {\displaystyle u(\theta )} :



F
(

u

−
1

)
=
−
m

h

2

u

2

(



d

2

u

d

θ

2


+
u

)

.

{\displaystyle F(u^{-1})=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right).}

{\displaystyle F(u^{-1})=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right).}

Chiqarish
---------

Nyutonning ikkinchi qonuni sof markaziy kuch uchun



F
(
r
)
=
m

(

r
¨

−
r

θ
˙

2

)

.

{\displaystyle F(r)=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right).}

{\displaystyle F(r)=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right).}
Burchak momentining saqlanishi shuni taqozo etadi


r

2


θ
˙

=
h
=

constant

.

{\displaystyle r^{2}{\dot {\theta }}=h={\text{constant}}.}

{\displaystyle r^{2}{\dot {\theta }}=h={\text{constant}}.}
ning hosilalari {\displaystyle r} vaqtga nisbatan hosila sifatida qayta yozilishi mumkin {\displaystyle u=1/r} burchakka nisbatan:





d

u

d

θ

=

d

d

t

(

1
r

)


d

t

d

θ

=
−

r
˙

r

2


θ
˙



=
−

r
˙

h




d

2

u

d

θ

2


=
−

1
h



d

r
˙



d

t


d

t

d

θ

=
−

r
¨

h

θ
˙



=
−

r
¨

h

2

u

2



{\displaystyle {\begin{aligned}&{\frac {\mathrm {d} u}{\mathrm {d} \theta }}={\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {1}{r}}\right){\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\dot {r}}{r^{2}{\dot {\theta }}}}=-{\frac {\dot {r}}{h}}\\&{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-{\frac {1}{h}}{\frac {\mathrm {d} {\dot {r}}}{\mathrm {d} t}}{\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\ddot {r}}{h{\dot {\theta }}}}=-{\frac {\ddot {r}}{h^{2}u^{2}}}\end{aligned}}}

{\displaystyle {\begin{aligned}&{\frac {\mathrm {d} u}{\mathrm {d} \theta }}={\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {1}{r}}\right){\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\dot {r}}{r^{2}{\dot {\theta }}}}=-{\frac {\dot {r}}{h}}\\&{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-{\frac {1}{h}}{\frac {\mathrm {d} {\dot {r}}}{\mathrm {d} t}}{\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\ddot {r}}{h{\dot {\theta }}}}=-{\frac {\ddot {r}}{h^{2}u^{2}}}\end{aligned}}}
Yuqoridagilarning barchasini birlashtirib, biz quyidagiga erishamiz



F
=
m

(

r
¨

−
r

θ
˙

2

)

=
−
m

(

h

2

u

2


d

2

u

d

θ

2


+

h

2

u

3

)

=
−
m

h

2

u

2

(



d

2

u

d

θ

2


+
u

)

{\displaystyle F=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right)=-m\left(h^{2}u^{2}{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+h^{2}u^{3}\right)=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}

{\displaystyle F=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right)=-m\left(h^{2}u^{2}{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+h^{2}u^{3}\right)=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}
Umumiy yechim 



θ
=

∫

r

0

r



d

r

r

2


2
m

L

2

(
E
−
V
)
−

1

r

2



+

θ

0

{\displaystyle \theta =\int \_{r\_{0}}^{r}{\frac {\mathrm {d} r}{r^{2}{\sqrt {{\frac {2m}{L^{2}}}(E-V)-{\frac {1}{r^{2}}}}}}}+\theta \_{0}}

{\displaystyle \theta =\int _{r_{0}}^{r}{\frac {\mathrm {d} r}{r^{2}{\sqrt {{\frac {2m}{L^{2}}}(E-V)-{\frac {1}{r^{2}}}}}}}+\theta _{0}}
bu yerda {\displaystyle (r\_{0},\theta \_{0})} zarrachaning dastlabki koordinatasidir.

Misollar
--------

### Kepler muammosi

#### Klassik

Teskari kvadrat qonunining orbitasini hisoblashning anʼanaviy Kepler muammosi differensial tenglamaning yechimi sifatida Binet tenglamasidan oʻqilishi mumkin.



−
k

u

2

=
−
m

h

2

u

2

(



d

2

u

d

θ

2


+
u

)

{\displaystyle -ku^{2}=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}

{\displaystyle -ku^{2}=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}



d

2

u

d

θ

2


+
u
=

k

m

h

2


≡

constant

>
0.

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {k}{mh^{2}}}\equiv {\text{constant}}>0.}

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {k}{mh^{2}}}\equiv {\text{constant}}>0.}
Agar burchak {\displaystyle \theta } periapsisdan oʻlchanadi, keyin (oʻzaro) qutb koordinatalarida ifodalangan orbita uchun umumiy yechim boʻladi.



l
u
=
1
+
ε
cos
⁡
θ
.

{\displaystyle lu=1+\varepsilon \cos \theta .}

{\displaystyle lu=1+\varepsilon \cos \theta .}
Yuqoridagi qutbli tenglama konus kesimlarini tasvirlaydi, bilan {\displaystyle l} yarim latus rektum (ga teng {\displaystyle h^{2}/\mu =h^{2}m/k}) va {\displaystyle \varepsilon } orbital ekssentriklik .

#### Relyativistik

Shvartsshild koordinatalari uchun olingan relyativistik tenglama 



d

2

u

d

θ

2


+
u
=

r

s

c

2

2

h

2


+

3

r

s

2

u

2

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r\_{s}c^{2}}{2h^{2}}}+{\frac {3r\_{s}}{2}}u^{2}}

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}}
bu yerda {\displaystyle c} yorugʻlik tezligi va {\displaystyle r\_{s}} Shvartsshild radiusidir. Va Reissner-Nordström metrikasi uchun biz olamiz



d

2

u

d

θ

2


+
u
=

r

s

c

2

2

h

2


+

3

r

s

2

u

2

−

G

Q

2

4
π

ε

0

c

4



(

c

2

h

2

u
+
2

u

3

)

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r\_{s}c^{2}}{2h^{2}}}+{\frac {3r\_{s}}{2}}u^{2}-{\frac {GQ^{2}}{4\pi \varepsilon \_{0}c^{4}}}\left({\frac {c^{2}}{h^{2}}}u+2u^{3}\right)}

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}-{\frac {GQ^{2}}{4\pi \varepsilon _{0}c^{4}}}\left({\frac {c^{2}}{h^{2}}}u+2u^{3}\right)}
bu yerda {\displaystyle Q} elektr zaryadidir va {\displaystyle \varepsilon \_{0}} vakuum oʻtkazuvchanligi hisoblanadi.

### Teskari Kepler muammosi

Teskari Kepler muammosini koʻrib chiqing. Qanday kuch qonuni ellips fokusi atrofida aylana boʻlmagan elliptik orbita (yoki umuman olganda aylana boʻlmagan konus kesimi) hosil qiladi?

Ellips uchun yuqoridagi qutb tenglamasini ikki marta differensiallash



l



d

2

u

d

θ

2


=
−
ε
cos
⁡
θ
.

{\displaystyle l\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-\varepsilon \cos \theta .}

{\displaystyle l\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-\varepsilon \cos \theta .}
Shuning uchun kuch qonuni



F
=
−
m

h

2

u

2

(

−
ε
cos
⁡
θ

l

+

1
+
ε
cos
⁡
θ

l

)

=
−

m

h

2

u

2

l

=
−

m

h

2

l

r

2


,

{\displaystyle F=-mh^{2}u^{2}\left({\frac {-\varepsilon \cos \theta }{l}}+{\frac {1+\varepsilon \cos \theta }{l}}\right)=-{\frac {mh^{2}u^{2}}{l}}=-{\frac {mh^{2}}{lr^{2}}},}

{\displaystyle F=-mh^{2}u^{2}\left({\frac {-\varepsilon \cos \theta }{l}}+{\frac {1+\varepsilon \cos \theta }{l}}\right)=-{\frac {mh^{2}u^{2}}{l}}=-{\frac {mh^{2}}{lr^{2}}},}
Bu kutilayotgan teskari kvadrat qonuni. Orbitalning mos kelishi {\displaystyle h^{2}/l=\mu } kabi jismoniy qadriyatlarga {\displaystyle GM} yoki {\displaystyle k\_{e}q\_{1}q\_{2}/m} Nyutonning universal tortishish qonunini yoki Kulon qonunini takrorlaydi.

Shvartsshild koordinatalari uchun samarali kuch 



F
=
−
G
M
m

u

2

(

1
+
3

(

h
u

c

)

2

)

=
−

G
M
m

r

2


(

1
+
3

(

h

r
c

)

2

)

.

{\displaystyle F=-GMmu^{2}\left(1+3\left({\frac {hu}{c}}\right)^{2}\right)=-{\frac {GMm}{r^{2}}}\left(1+3\left({\frac {h}{rc}}\right)^{2}\right).}

{\displaystyle F=-GMmu^{2}\left(1+3\left({\frac {hu}{c}}\right)^{2}\right)=-{\frac {GMm}{r^{2}}}\left(1+3\left({\frac {h}{rc}}\right)^{2}\right).}
bu yerda ikkinchi aʼzo periapsisning burchak siljishi kabi toʻrt kutupli taʼsirlarga mos keladigan teskari kvarts kuchdir (Uni kechiktirilgan potentsiallar orqali ham olish mumkin ).

Parametrlangan post-Nyuton formalizmida biz erishamiz



F
=
−

G
M
m

r

2


(

1
+
(
2
+
2
γ
−
β
)

(

h

r
c

)

2

)

.

{\displaystyle F=-{\frac {GMm}{r^{2}}}\left(1+(2+2\gamma -\beta )\left({\frac {h}{rc}}\right)^{2}\right).}

{\displaystyle F=-{\frac {GMm}{r^{2}}}\left(1+(2+2\gamma -\beta )\left({\frac {h}{rc}}\right)^{2}\right).}

  

bu yerda {\displaystyle \gamma =\beta =1} umumiy nisbiylik nazariyasi uchun va {\displaystyle \gamma =\beta =0} klassik holatda.

### Kotes spirallari

Teskari kub kuch qonuni shaklga ega



F
(
r
)
=
−

k

r

3

.

{\displaystyle F(r)=-{\frac {k}{r^{3}}}.}

{\displaystyle F(r)=-{\frac {k}{r^{3}}}.}

  

Teskari kub qonunining orbitalarining shakllari Cotes spirallari deb nomlanadi. Binet tenglamasi orbitalar tenglamaning yechimi boʻlishi kerakligini koʻrsatadi



d

2

u

d

θ

2


+
u
=

k
u

m

h

2


=
C
u
.

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {ku}{mh^{2}}}=Cu.}

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {ku}{mh^{2}}}=Cu.}
Differensial tenglama Kepler muammosining turli konus kesimlariga oʻxshab uch xil yechimga ega. Qachon {\displaystyle C<1}, yechim epispiral, shu jumladan toʻgʻri chiziqning patologik holati qachon {\displaystyle C=0}. Qachon {\displaystyle C=1}, yechim giperbolik spiraldir. Qachon {\displaystyle C>1} yechim Puinsot spiralidir .

### Eksadan tashqari aylanma harakat

Binet tenglamasi kuch markazi atrofida aylana harakati uchun yagona kuch qonunini bera olmasa ham, tenglama aylananing markazi va kuch markazi mos kelmasa, kuch qonunini berishi mumkin. Masalan, toʻgʻridan-toʻgʻri kuch markazidan oʻtadigan dumaloq orbitani koʻrib chiqing. Bunday dumaloq diametrli orbita uchun (oʻzaro) qutbli tenglama {\displaystyle D} hisoblanadi



D

u
(
θ
)
=
sec
⁡
θ
.

{\displaystyle D\,u(\theta )=\sec \theta .}

{\displaystyle D\,u(\theta )=\sec \theta .}
Farqlash {\displaystyle u} ikki marta va Pifagor kimligidan foydalanish imkonini beradi



D



d

2

u

d

θ

2


=
sec
⁡
θ

tan

2

⁡
θ
+

sec

3

⁡
θ
=
sec
⁡
θ
(

sec

2

⁡
θ
−
1
)
+

sec

3

⁡
θ
=
2

D

3

u

3

−
D

u
.

{\displaystyle D\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=\sec \theta \tan ^{2}\theta +\sec ^{3}\theta =\sec \theta (\sec ^{2}\theta -1)+\sec ^{3}\theta =2D^{3}u^{3}-D\,u.}

{\displaystyle D\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=\sec \theta \tan ^{2}\theta +\sec ^{3}\theta =\sec \theta (\sec ^{2}\theta -1)+\sec ^{3}\theta =2D^{3}u^{3}-D\,u.}
Kuch qonuni shunday



F
=
−
m

h

2

u

2

(

2

D

2

u

3

−
u
+
u

)

=
−
2
m

h

2

D

2

u

5

=
−

2
m

h

2

D

2

r

5

.

{\displaystyle F=-mh^{2}u^{2}\left(2D^{2}u^{3}-u+u\right)=-2mh^{2}D^{2}u^{5}=-{\frac {2mh^{2}D^{2}}{r^{5}}}.}

{\displaystyle F=-mh^{2}u^{2}\left(2D^{2}u^{3}-u+u\right)=-2mh^{2}D^{2}u^{5}=-{\frac {2mh^{2}D^{2}}{r^{5}}}.}
Eʼtibor bering, umumiy teskari masalani yechish, yaʼni jozibali orbitalarni qurish {\displaystyle 1/r^{5}} kuch qonuni, ancha qiyinroq muammodir, chunki u echishga tengdir



d

2

u

d

θ

2


+
u
=
C

u

3

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u=Cu^{3}}

{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u=Cu^{3}}
ikkinchi tartibli chiziqli boʻlmagan differentsial tenglama.

Manbalar
--------

1. ↑  Goldstein, Herbert*. Classical mechanics*. Reading, Mass.: Addison-Wesley Pub. Co, 1980. ISBN 0-201-02918-9. OCLC 5675073.
2. ↑  „Archived copy". 19-iyun 2010-yilda asl nusxadan arxivlandi. Qaraldi: 15-noyabr 2010-yil.
3. ↑   — The first-order orbital equation
4. ↑  Behera, Harihar; Naik, P. C (2003). "A flat space-time relativistic explanation for the perihelion advance of Mercury". arXiv:astro-ph/0306611.

Adabiyotlar
-----------

* Bohr-Sommerfeld quantization -Relativistic orbit
* Klassik markaziy kuch masalasi
* Umumiy nisbiylik nazariyasi
* Nisbiylik nazariyasida ikki jism masalasi
* Bertrand teoremasi