править
В Кеплеровой задаче можно вычислить интеграл (1.4.28)
t
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r
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m
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E
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U
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M
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m
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{\displaystyle t=\int \displaystyle {\frac {dr}{\sqrt {{\frac {2}{m}}\left(-\vert E\vert -U\right)-{\frac {M^{2}}{m^{2}r^{2}}}}}}}
(1.4.49) (1.4.38) (1.4.49)
U
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r
t
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r
d
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r
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E
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{\displaystyle {\begin{aligned}&U=-{\frac {\alpha }{r}}\\&t={\sqrt {\frac {m}{2\vert E\vert }}}\int {\frac {rdr}{\displaystyle {\sqrt {-{r^{2}}+r{\frac {\alpha }{\vert E\vert }}-{\frac {M^{2}}{2m\vert E\vert }}}}}}\\\end{aligned}}}
(1.4.50) (1.4.42) (1.4.44) (1.4.50)
t
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{\displaystyle t={\sqrt {\frac {ma}{\alpha }}}\int {\frac {rdr}{\sqrt {a^{2}e^{2}-(r-a)^{2}}}}}
(1.4.51) (1.4.51)
r
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a
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e
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θ
r
d
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e
cos
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a
e
sin
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d
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t
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m
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a
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a
e
cos
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a
e
sin
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d
θ
a
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cos
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θ
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m
a
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1
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e
cos
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d
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t
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m
a
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α
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sin
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+
c
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t
{\displaystyle {\begin{aligned}&r-a=-ae\cos \theta \\&rdr=\left(a-ae\cos \theta \right)ae\sin \theta d\theta \\&{\begin{matrix}t=\displaystyle {\sqrt {\frac {ma}{\alpha }}}\int {\frac {\left(a-ae\cos \theta \right)ae\sin \theta d\theta }{ae{\sqrt {1-\cos ^{2}\theta }}}}={\sqrt {\frac {ma^{3}}{\alpha }}}\int {\left(1-e\cos \theta \right)d\theta }\\\Downarrow \\\displaystyle t={\sqrt {\frac {ma^{3}}{\alpha }}}\left(\theta -e\sin \theta \right)+const\\\end{matrix}}\\\end{aligned}}}
(1.4.52) (1.4.52)
r
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t
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sin
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{\displaystyle {\begin{aligned}&r=a\left(1-e\cos \theta \right)\\&t={\sqrt {\frac {ma^{3}}{\alpha }}}\left(\theta -e\sin \theta \right)\\\end{aligned}}}
(1.4.53) (1.4.53)
θ
{\displaystyle \theta }
(1.4.43)
x
=
r
cos
φ
y
=
r
sin
φ
p
=
r
+
e
r
cos
φ
}
⇒
e
x
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e
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−
r
⏟
a
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1
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e
cos
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=
a
e
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cos
θ
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e
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⇓
{
x
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a
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cos
θ
−
e
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y
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r
2
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x
2
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a
1
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e
2
sin
θ
{\displaystyle {\begin{matrix}\left.{\begin{aligned}&x=r\cos \varphi \\&y=r\sin \varphi \\&p=r+er\cos \varphi \\\end{aligned}}\right\}\Rightarrow ex=\underbrace {p} _{a\left(1-e^{2}\right)}-\underbrace {r} _{a\left(1-e\cos \theta \right)}=ae\left(\cos \theta -e\right)\\\Downarrow \\\left\{{\begin{aligned}&x=a\left(\cos \theta -e\right)\\&y={\sqrt {r^{2}-x^{2}}}=a{\sqrt {1-e^{2}}}\sin \theta \\\end{aligned}}\right.\\\end{matrix}}}
(1.4.54) (1.4.28)
r
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a
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e
⋅
ch
θ
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1
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;
t
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m
a
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e
⋅
sh
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x
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y
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sh
θ
{\displaystyle {\begin{aligned}&{\begin{matrix}r=a\left(e\cdot {\text{ch}}\theta -1\right);&t={\sqrt {\frac {ma^{3}}{\alpha }}}\left(e\cdot {\text{sh}}\theta -\theta \right)\\\end{matrix}}\\&{\begin{matrix}x=a\left(e-{\text{ch}}\theta \right);&y=a{\sqrt {e^{2}-1}}\cdot {\text{sh}}\\\end{matrix}}\theta \\\end{aligned}}}
(1.4.55)
