第二法則はいずれの導出においてもその全部が第一法則の前半の導出と同一である。
第一法則
導出1
中心天体 C \text{C} C M M M A \text{A} A m m m A \text{A} A B \text{B} B G G G
m d 2 r d t 2 = − G M m ∣ r ∣ 2 r ∣ r ∣ m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} = -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t 2 d 2 r = − G ∣ r ∣ 2 M m ∣ r ∣ r ここで角運動量 L \boldsymbol{L} L
L = m ( r × d r d t ) \boldsymbol{L} = m\left(\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right) L = m ( r × d t d r ) 両辺を時間 t t t 2 2 2 d d t ( a ( t ) × b ( t ) ) = d a ( t ) d t × b ( t ) + a ( t ) × d b ( t ) d t \dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{a}(t) \times \boldsymbol{b}(t)) = \dfrac{\mathrm{d}\boldsymbol{a}(t)}{\mathrm{d}t} \times \boldsymbol{b}(t) + \boldsymbol{a}(t) \times \dfrac{\mathrm{d}\boldsymbol{b}(t)}{\mathrm{d}t} d t d ( a ( t ) × b ( t )) = d t d a ( t ) × b ( t ) + a ( t ) × d t d b ( t )
d L d t = m ( d r d t × d r d t + r × d 2 r d t 2 ) \dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} = m \left( \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} + \boldsymbol{r} \times \dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} \right) d t d L = m ( d t d r × d t d r + r × d t 2 d 2 r ) したがって右辺第 1 1 1 r × r = 0 \boldsymbol{r} \times \boldsymbol{r} = \boldsymbol{0} r × r = 0 2 2 2 m d 2 r d t 2 = − G M m ∣ r ∣ 2 r ∣ r ∣ m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} = -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t 2 d 2 r = − G ∣ r ∣ 2 M m ∣ r ∣ r
d L d t = r × ( m d 2 r d t 2 ) = r × ( − G M m ∣ r ∣ 2 r ∣ r ∣ ) = − G M m ∣ r ∣ 3 ( r × r ) = 0 \begin{aligned}
\dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} &= \boldsymbol{r} \times \left( m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} \right) \\
&= \boldsymbol{r} \times \left( -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} \right) \\
&= -G\dfrac{Mm}{|\boldsymbol{r}|^3} (\boldsymbol{r} \times \boldsymbol{r}) \\
&= \boldsymbol{0}
\end{aligned} d t d L = r × ( m d t 2 d 2 r ) = r × ( − G ∣ r ∣ 2 M m ∣ r ∣ r ) = − G ∣ r ∣ 3 M m ( r × r ) = 0 ここで，v = d r d t \boldsymbol{v}=\dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} v = d t d r d v d t × L \dfrac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} \times \boldsymbol{L} d t d v × L 2 2 2 d d t ( a ( t ) × b ( t ) ) = d a ( t ) d t × b ( t ) + a ( t ) × d b ( t ) d t \dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{a}(t) \times \boldsymbol{b}(t)) = \dfrac{\mathrm{d}\boldsymbol{a}(t)}{\mathrm{d}t} \times \boldsymbol{b}(t) + \boldsymbol{a}(t) \times \dfrac{\mathrm{d}\boldsymbol{b}(t)}{\mathrm{d}t} d t d ( a ( t ) × b ( t )) = d t d a ( t ) × b ( t ) + a ( t ) × d t d b ( t ) d L d t = 0 \dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} = \boldsymbol{0} d t d L = 0
d d t ( v × L ) = d v d t × L + v × d L d t = d v d t × L \begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{v} \times \boldsymbol{L}) &= \dfrac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} \times \boldsymbol{L} + \boldsymbol{v} \times \dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t}\\
&= \dfrac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} \times \boldsymbol{L}
\end{aligned} d t d ( v × L ) = d t d v × L + v × d t d L = d t d v × L 運動方程式 m d v d t = − G M m ∣ r ∣ 2 r ∣ r ∣ m \dfrac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} = -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t d v = − G ∣ r ∣ 2 M m ∣ r ∣ r L = m ( r × v ) \boldsymbol{L} = m(\boldsymbol{r} \times \boldsymbol{v}) L = m ( r × v )
d d t ( v × L ) = ( − G M r 3 r ) × ( m ( r × v ) ) = − G M m r 3 ( r × ( r × v ) ) \begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{v} \times \boldsymbol{L})
&= \left( -\dfrac{GM}{r^3} \boldsymbol{r} \right) \times \Big(m (\boldsymbol{r} \times \boldsymbol{v})\Big) \\
& = -\dfrac{GMm}{r^3} \Big( \boldsymbol{r} \times (\boldsymbol{r} \times \boldsymbol{v}) \Big)
\end{aligned} d t d ( v × L ) = ( − r 3 GM r ) × ( m ( r × v ) ) = − r 3 GM m ( r × ( r × v ) ) ベクトル三重積を展開する数学公式 a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c \boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c}) = (\boldsymbol{a} \cdot \boldsymbol{c})\boldsymbol{b} - (\boldsymbol{a} \cdot \boldsymbol{b})\boldsymbol{c} a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
− G M m r 3 ( r × ( r × v ) ) = − G M m r 3 ( ( r ⋅ v ) r − ( r ⋅ r ) v ) = − G M m r 3 ( ( r ⋅ v ) r − r 2 v ) \begin{aligned}
-\dfrac{GMm}{r^3} \Big( \boldsymbol{r} \times (\boldsymbol{r} \times \boldsymbol{v}) \Big) &=
-\dfrac{GMm}{r^3} \Big( (\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{r} - (\boldsymbol{r} \cdot \boldsymbol{r})\boldsymbol{v} \Big)\\
&=-\dfrac{GMm}{r^3} \Big( (\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{r} - r^2 \boldsymbol{v} \Big)
\end{aligned} − r 3 GM m ( r × ( r × v ) ) = − r 3 GM m ( ( r ⋅ v ) r − ( r ⋅ r ) v ) = − r 3 GM m ( ( r ⋅ v ) r − r 2 v ) ここで，r = ∣ r ∣ r = |\boldsymbol{r}| r = ∣ r ∣ r ⋅ r = r 2 \boldsymbol{r} \cdot \boldsymbol{r} = r^2 r ⋅ r = r 2
2 r ⋅ v = 2 r d r d t r ⋅ v = r d r d t \begin{aligned}
2\boldsymbol{r} \cdot \boldsymbol{v} &= 2r \dfrac{\mathrm{d}r}{\mathrm{d}t}\\
\boldsymbol{r} \cdot \boldsymbol{v} &= r \dfrac{\mathrm{d}r}{\mathrm{d}t}
\end{aligned} 2 r ⋅ v r ⋅ v = 2 r d t d r = r d t d r ∣ r ∣ ≠ 0 |\boldsymbol{r}| \neq \boldsymbol{0} ∣ r ∣ = 0 d d x ( f ( x ) g ( x ) ) = d f ( x ) d x g ( x ) − f ( x ) d g ( x ) d x { g ( x ) } 2 \dfrac{\mathrm{d}}{\mathrm{d}x} \left( \dfrac{f(x)}{g(x)} \right) = \dfrac{\dfrac{\mathrm{d}f(x)}{\mathrm{d}x} g(x) - f(x) \dfrac{\mathrm{d}g(x)}{\mathrm{d}x}}{\{g(x)\}^2} d x d ( g ( x ) f ( x ) ) = { g ( x ) } 2 d x d f ( x ) g ( x ) − f ( x ) d x d g ( x )
d d t ( r r ) = v r − r d r d t r 2 d d t ( r r ) = − 1 r 3 ( ( r ⋅ v ) r − r 2 v ) ( r ⋅ v ) r − r 2 v = − r 3 d d t ( r r ) \begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\boldsymbol{r}}{r}\right) &= \dfrac{\boldsymbol{v}r - \boldsymbol{r}\dfrac{\mathrm{d}r}{\mathrm{d}t}}{r^2}\\
\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\boldsymbol{r}}{r}\right) &= -\dfrac{1}{r^3} \Big( (\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{r} - r^2 \boldsymbol{v} \Big)\\
(\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{r} - r^2 \boldsymbol{v} &= -r^3\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\boldsymbol{r}}{r}\right)
\end{aligned} d t d ( r r ) d t d ( r r ) ( r ⋅ v ) r − r 2 v = r 2 v r − r d t d r = − r 3 1 ( ( r ⋅ v ) r − r 2 v ) = − r 3 d t d ( r r ) したがって
d d t ( v × L ) = − G M m r 3 ( ( r ⋅ v ) r − r 2 v ) d d t ( v × L ) = G M m d d t ( r r ) \begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{v} \times \boldsymbol{L}) &= -\dfrac{GMm}{r^3} \Big( (\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{r} - r^2 \boldsymbol{v} \Big)\\
\dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{v} \times \boldsymbol{L}) &= GMm \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\boldsymbol{r}}{r}\right)
\end{aligned} d t d ( v × L ) d t d ( v × L ) = − r 3 GM m ( ( r ⋅ v ) r − r 2 v ) = GM m d t d ( r r ) 両辺を時間 t t t G M m K GMm\boldsymbol{K} GM m K
v × L = G M m r r + G M m K v × L = G M m ( r r + K ) \begin{aligned}
\boldsymbol{v} \times \boldsymbol{L} &= GMm \dfrac{\boldsymbol{r}}{r} + GMm\boldsymbol{K}\\
\boldsymbol{v} \times \boldsymbol{L} &= GMm \left( \dfrac{\boldsymbol{r}}{r} + \boldsymbol{K} \right)
\end{aligned} v × L v × L = GM m r r + GM m K = GM m ( r r + K ) ここで，次のように両辺と位置ベクトル r \boldsymbol{r} r
r ⋅ ( v × L ) = ( G M m ( r r + K ) ) ⋅ r \boldsymbol{r} \cdot (\boldsymbol{v} \times \boldsymbol{L}) = \bigg(GMm \left(\dfrac{\boldsymbol{r}}{r} + \boldsymbol{K}\right)\bigg)\cdot \boldsymbol{r} r ⋅ ( v × L ) = ( GM m ( r r + K ) ) ⋅ r 左辺につきスカラー三重積の巡回性に関する数学公式 a ⋅ ( b × c ) = ( a × b ) ⋅ c \boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} a ⋅ ( b × c ) = ( a × b ) ⋅ c ( k a ) ⋅ b = k ( a ⋅ b ) (k\boldsymbol{a})\cdot\boldsymbol{b}=k(\boldsymbol{a}\cdot\boldsymbol{b}) ( k a ) ⋅ b = k ( a ⋅ b ) a ⋅ ( b + c ) = a ⋅ b + a ⋅ c \boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c}) = \boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c} a ⋅ ( b + c ) = a ⋅ b + a ⋅ c
( r × v ) ⋅ L = G M m ( r r + K ) ⋅ r ( L m ) ⋅ L = G M m ( r ⋅ r r + r ⋅ K ) ∣ L ∣ 2 m = G M m ( ∣ r ∣ + r ⋅ K ) ∣ r ∣ + r ⋅ K = ∣ L ∣ 2 G M m 2 \begin{aligned}
(\boldsymbol{r} \times \boldsymbol{v}) \cdot \boldsymbol{L} &= GMm \left(\dfrac{\boldsymbol{r}}{r} + \boldsymbol{K}\right)\cdot \boldsymbol{r}\\
\left(\dfrac{\boldsymbol{L}}{m}\right) \cdot \boldsymbol{L} &= GMm \left(\dfrac{\boldsymbol{r} \cdot \boldsymbol{r}}{r} + \boldsymbol{r} \cdot \boldsymbol{K}\right)\\
\dfrac{|\boldsymbol{L}|^2}{m} &= GMm \left(|\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{K}\right)\\
|\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{K} &= \dfrac{|\boldsymbol{L}|^2}{GMm^2}
\end{aligned} ( r × v ) ⋅ L ( m L ) ⋅ L m ∣ L ∣ 2 ∣ r ∣ + r ⋅ K = GM m ( r r + K ) ⋅ r = GM m ( r r ⋅ r + r ⋅ K ) = GM m ( ∣ r ∣ + r ⋅ K ) = GM m 2 ∣ L ∣ 2 右辺の定数を p = ∣ L ∣ 2 G M m 2 p = \dfrac{|\boldsymbol{L}|^2}{GMm^2} p = GM m 2 ∣ L ∣ 2 G > 0 G>0 G > 0 M > 0 M>0 M > 0 p > 0 p>0 p > 0
∣ r ∣ + r ⋅ K = p |\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{K} = p ∣ r ∣ + r ⋅ K = p ここで定数ベクトル K \boldsymbol{K} K e \boldsymbol{e} e
r = ∣ r ∣ r=|\boldsymbol{r}| r = ∣ r ∣ K = ∣ K ∣ K=|\boldsymbol{K}| K = ∣ K ∣ r \boldsymbol{r} r K \boldsymbol{K} K θ \theta θ a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ \boldsymbol{a} \cdot \boldsymbol{b} = \vert{}\boldsymbol{a}\vert{} \vert{}\boldsymbol{b}\vert{} \cos\theta a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ
r + r K cos θ = p r ( 1 + K cos θ ) = p r = p 1 + K cos θ \begin{aligned}
r + r K \cos\theta &= p\\
r (1 + K \cos\theta) &= p \\
r &= \dfrac{p}{1 + K \cos\theta}
\end{aligned} r + r K cos θ r ( 1 + K cos θ ) r = p = p = 1 + K cos θ p ［1 1 1 K \boldsymbol{K} K
軌道上で原点にある中心天体からの距離 r r r p > 0 p>0 p > 0 K K K r r r 1 + K cos θ 1 + K \cos\theta 1 + K cos θ − 1 ≤ cos θ ≤ 1 -1 \le \cos\theta \le 1 − 1 ≤ cos θ ≤ 1 cos θ = 1 \cos\theta = 1 cos θ = 1 θ = 0 \theta = 0 θ = 0 θ = 0 \theta = 0 θ = 0 r \boldsymbol{r} r K \boldsymbol{K} K e \boldsymbol{e} e K \boldsymbol{K} K e \boldsymbol{e} e
［2 2 2 K \boldsymbol{K} K
近日点距離を r p r_\text{p} r p r a r_\text{a} r a θ = 0 \theta=0 θ = 0 r r r 1 + K cos θ 1 + K \cos\theta 1 + K cos θ θ = π \theta=\pi θ = π
{ r p = p 1 + K cos 0 = p 1 + K r a = p 1 + K cos π = p 1 − K \left\{\;
\begin{aligned}
r_\text{p} &= \dfrac{p}{1 + K \cos 0} = \dfrac{p}{1 + K} \\[8pt]
r_\text{a} &= \dfrac{p}{1 + K \cos \pi} = \dfrac{p}{1 - K}
\end{aligned}
\right. ⎩ ⎨ ⎧ r p r a = 1 + K cos 0 p = 1 + K p = 1 + K cos π p = 1 − K p 軌道を楕円とすれば長軸の長さ 2 a 2a 2 a 2 2 2
2 a = r p + r a = p 1 + K + p 1 − K = 2 p 1 − K 2 p = a ( 1 − K 2 ) \begin{aligned}
2a &= r_\text{p} + r_\text{a} \\[5pt]
&= \dfrac{p}{1 + K} + \dfrac{p}{1 - K} \\[8pt]
&= \dfrac{2p}{1 - K^2} \\[8pt]
p &= a(1 - K^2)
\end{aligned} 2 a p = r p + r a = 1 + K p + 1 − K p = 1 − K 2 2 p = a ( 1 − K 2 ) これを r p = p 1 + K r_\text{p} = \dfrac{p}{1 + K} r p = 1 + K p
r p = p 1 + K = a ( 1 − K 2 ) 1 + K = a − a K \begin{aligned}
r_\text{p} &= \dfrac{p}{1 + K}\\[8pt]
&= \dfrac{a(1 - K^2)}{1 + K}\\[8pt]
&= a - aK
\end{aligned} r p = 1 + K p = 1 + K a ( 1 − K 2 ) = a − a K ここで中心から焦点（原点）までの距離を c c c r p r_\text{p} r p a a a
r p = a − c r_\text{p} = a - c r p = a − c と比較して
c = a K K = c a c=aK\\
K=\dfrac{c}{a} c = a K K = a c これは中心から焦点までの距離 c c c a a a c a \dfrac{c}{a} a c e e e
以上［1 1 1 2 2 2 e \boldsymbol{e} e
K = e \boldsymbol{K} = \boldsymbol{e} K = e これらより，∣ r ∣ + r ⋅ K = p |\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{K} = p ∣ r ∣ + r ⋅ K = p p = a ( 1 − K 2 ) p = a(1 - K^2) p = a ( 1 − K 2 ) K = ∣ K ∣ K=|\boldsymbol{K}| K = ∣ K ∣ K = e \boldsymbol{K} = \boldsymbol{e} K = e
∣ r ∣ + r ⋅ e = a ( 1 − ∣ e ∣ 2 ) |\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{e} = a(1 - |\boldsymbol{e}|^2) ∣ r ∣ + r ⋅ e = a ( 1 − ∣ e ∣ 2 ) 導出2
中心天体 C \text{C} C M M M A \text{A} A m m m A \text{A} A
m d 2 r d t = − G M m r 2 r ∣ r ∣ m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t}=-G\dfrac{Mm}{\boldsymbol{r}^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t d 2 r = − G r 2 M m ∣ r ∣ r ここで，天体 A \text{A} A
時刻 t = 0 t=0 t = 0 A \text{A} A r 0 \boldsymbol{r}_0 r 0 v 0 \boldsymbol{v}_0 v 0
［1 1 1 r 0 \boldsymbol{r}_0 r 0 v 0 \boldsymbol{v}_0 v 0
力は常にその直線方向を向くため，運動は自明に直線運動し，同一平面上を運動する。
［2 2 2 r 0 \boldsymbol{r}_0 r 0 v 0 \boldsymbol{v}_0 v 0
ベクトル r 0 \boldsymbol{r}_0 r 0 v 0 \boldsymbol{v}_0 v 0 x y xy x y z z z t = 0 t=0 t = 0 A \text{A} A z z z t t t
z ( 0 ) = 0 z(0) = 0 z ( 0 ) = 0 かつ速度ベクトル v 0 \boldsymbol{v}_0 v 0 t = 0 t=0 t = 0 z z z d z d t \dfrac{\mathrm{d}z}{\mathrm{d}t} d t d z 0 0 0
d z d t ∣ t = 0 = 0 \left.\dfrac{\mathrm{d}z}{\mathrm{d}t}\right|_{t=0} = 0 d t d z t = 0 = 0 時刻 t t t r \boldsymbol{r} r t t t r ( t ) = x ( t ) 2 + y ( t ) 2 + z ( t ) 2 \displaystyle r(t) = \displaystyle\sqrt{x(t)^2 + y(t)^2 + z(t)^2} r ( t ) = x ( t ) 2 + y ( t ) 2 + z ( t ) 2 z z z
m d 2 z d t 2 = − ( G M m r ( t ) 3 ) z ( t ) d 2 z d t 2 = − ( G M r ( t ) 3 ) z ( t ) \begin{aligned}
m\dfrac{\mathrm{d}^2z}{\mathrm{d}t^2} &= -\left( \dfrac{GMm}{r(t)^3} \right) z(t)\\
\dfrac{\mathrm{d}^2z}{\mathrm{d}t^2} &= -\left( \dfrac{GM}{r(t)^3} \right) z(t)
\end{aligned} m d t 2 d 2 z d t 2 d 2 z = − ( r ( t ) 3 GM m ) z ( t ) = − ( r ( t ) 3 GM ) z ( t ) f ( t ) = G M r ( t ) 3 f(t) = \dfrac{GM}{r(t)^3} f ( t ) = r ( t ) 3 GM
d 2 z d t 2 + f ( t ) z = 0 \dfrac{\mathrm{d}^2z}{\mathrm{d}t^2} + f(t) z = 0 d t 2 d 2 z + f ( t ) z = 0 ここで z ( t ) = 0 z(t) = 0 z ( t ) = 0 1 1 1
z ( 0 ) = 0 z(0) = 0 z ( 0 ) = 0 および 2 2 2
d z d t ∣ t = 0 = 0 \left.\dfrac{\mathrm{d}z}{\mathrm{d}t}\right|_{t=0} = 0 d t d z t = 0 = 0 をそれぞれ満たし，次の通り明らかに方程式を満たす。
d 2 d t 2 ( 0 ) + f ( t ) ⋅ 0 = 0 \dfrac{\mathrm{d}^2}{\mathrm{d}t^2}(0) + f(t)\cdot 0 = 0 d t 2 d 2 ( 0 ) + f ( t ) ⋅ 0 = 0 天体 A \text{A} A F \boldsymbol{F} F r > 0 r>0 r > 0 A \text{A} A r > 0 r>0 r > 0 1 1 1 z ( t ) z(t) z ( t )
ゆえに任意の時刻 t t t A \text{A} A
z ( t ) = 0 z(t) = 0 z ( t ) = 0 したがって，天体 A \text{A} A r \boldsymbol{r} r O \mathrm{O} O ( x , y ) (x, y) ( x , y ) ( r , θ ) (r, \theta) ( r , θ )
{ x = r cos θ y = r sin θ \begin{cases}
x = r\cos\theta \\
y = r\sin\theta
\end{cases} { x = r cos θ y = r sin θ 天体 A \text{A} A r \boldsymbol{r} r
r = ( x y ) = ( r cos θ r sin θ ) \begin{aligned}
\boldsymbol{r} &= \begin{pmatrix} x \\ y \end{pmatrix}\\
&= \begin{pmatrix} r\cos\theta \\ r\sin\theta \end{pmatrix}
\end{aligned} r = ( x y ) = ( r cos θ r sin θ ) 天体 A \text{A} A v \boldsymbol{v} v x x x y y y v x v_x v x v y v_y v y d d x ( f ( x ) g ( x ) ) = d f ( x ) d x g ( x ) + f ( x ) d g ( x ) d x \dfrac{\mathrm{d}}{\mathrm{d}x} \left( f(x)g(x) \right) = \dfrac{\mathrm{d}f(x)}{\mathrm{d}x} g(x) + f(x) \dfrac{\mathrm{d}g(x)}{\mathrm{d}x} d x d ( f ( x ) g ( x ) ) = d x d f ( x ) g ( x ) + f ( x ) d x d g ( x ) d y d x = d y d u d u d x \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \dfrac{\mathrm{d}u}{\mathrm{d}x} d x d y = d u d y d x d u
v = d r d t = ( v x v y ) = ( d r cos θ d t d r sin θ d t ) = ( d r d t cos θ − r sin θ d θ d t d r d t sin θ + r cos θ d θ d t ) \begin{aligned}
\boldsymbol{v} &= \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\\
&= \begin{pmatrix}
v_x \\
v_y
\end{pmatrix}\\
&= \begin{pmatrix}
\dfrac{\mathrm{d}r\cos\theta}{\mathrm{d}t} \\[8pt]
\dfrac{\mathrm{d}r\sin\theta}{\mathrm{d}t}
\end{pmatrix}\\
&= \begin{pmatrix}
\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos\theta - r\sin\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t} \\[8pt]
\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin\theta + r\cos\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}
\end{pmatrix}
\end{aligned} v = d t d r = ( v x v y ) = d t d r cos θ d t d r sin θ = d t d r cos θ − r sin θ d t d θ d t d r sin θ + r cos θ d t d θ ここでまず，速さの 2 2 2 v 2 = ∣ v ∣ 2 v^2=|\boldsymbol{v}|^2 v 2 = ∣ v ∣ 2 cos 2 θ + sin 2 θ = 1 \cos^2\theta + \sin^2\theta = 1 cos 2 θ + sin 2 θ = 1
v 2 = ( d r d t cos θ − r sin θ d θ d t ) 2 + ( d r d t sin θ + r cos θ d θ d t ) 2 = ( d r d t ) 2 ( cos 2 θ + sin 2 θ ) + r 2 ( d θ d t ) 2 ( sin 2 θ + cos 2 θ ) = ( d r d t ) 2 + r 2 ( d θ d t ) 2 \begin{aligned}
v^2 &= \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos\theta - r\sin\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 + \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin\theta + r\cos\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 \\[8pt]
&= \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 (\cos^2\theta + \sin^2\theta) + r^2\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 (\sin^2\theta + \cos^2\theta) \\[8pt]
&= \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 + r^2 \left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2
\end{aligned} v 2 = ( d t d r cos θ − r sin θ d t d θ ) 2 + ( d t d r sin θ + r cos θ d t d θ ) 2 = ( d t d r ) 2 ( cos 2 θ + sin 2 θ ) + r 2 ( d t d θ ) 2 ( sin 2 θ + cos 2 θ ) = ( d t d r ) 2 + r 2 ( d t d θ ) 2 また，原点まわりの角運動量ベクトル L \boldsymbol{L} L
L = m ( r × d r d t ) \boldsymbol{L} = m\left(\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right) L = m ( r × d t d r ) r × d r d t \boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} r × d t d r v = d r d t = ( v x v y 0 ) \displaystyle\boldsymbol{v} = \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} = \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} v = d t d r = v x v y 0
r × v = ( y ⋅ 0 − 0 ⋅ v y 0 ⋅ v x − x ⋅ 0 x v y − y v x ) = ( 0 0 x v y − y v x ) \begin{aligned}
\boldsymbol{r} \times \boldsymbol{v}
&= \begin{pmatrix}
y \cdot 0 - 0 \cdot v_y \\
0 \cdot v_x - x \cdot 0 \\
x v_y - y v_x
\end{pmatrix}\\
&= \begin{pmatrix}
0 \\
0 \\
x v_y - y v_x
\end{pmatrix}
\end{aligned} r × v = y ⋅ 0 − 0 ⋅ v y 0 ⋅ v x − x ⋅ 0 x v y − y v x = 0 0 x v y − y v x ここで，z z z r = ( r cos θ r sin θ ) \boldsymbol{r} = \begin{pmatrix} r\cos\theta \\ r\sin\theta \end{pmatrix} r = ( r cos θ r sin θ ) v = ( d r d t cos θ − r sin θ d θ d t d r d t sin θ + r cos θ d θ d t ) \boldsymbol{v} = \begin{pmatrix}
\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos\theta - r\sin\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t} \\[8pt]
\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin\theta + r\cos\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}
\end{pmatrix} v = d t d r cos θ − r sin θ d t d θ d t d r sin θ + r cos θ d t d θ cos 2 θ + sin 2 θ = 1 \cos^2\theta + \sin^2\theta = 1 cos 2 θ + sin 2 θ = 1
x v y − y v x = ( r cos θ ) ( d r d t sin θ + r cos θ d θ d t ) − ( r sin θ ) ( d r d t cos θ − r sin θ d θ d t ) = ( r d r d t cos θ sin θ + r 2 cos 2 θ d θ d t ) − ( r d r d t sin θ cos θ − r 2 sin 2 θ d θ d t ) = r 2 ( cos 2 θ + sin 2 θ ) d θ d t = r 2 d θ d t \begin{aligned}
x v_y - y v_x &= (r\cos\theta)\left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin\theta + r\cos\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right) - (r\sin\theta)\left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos\theta - r\sin\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right) \\[8pt]
&= \left(r\dfrac{\mathrm{d}r}{\mathrm{d}t}\cos\theta\sin\theta + r^2\cos^2\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right) - \left(r\dfrac{\mathrm{d}r}{\mathrm{d}t}\sin\theta\cos\theta - r^2\sin^2\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right) \\[8pt]
&= r^2 (\cos^2\theta + \sin^2\theta) \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \\[8pt]
&= r^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t}
\end{aligned} x v y − y v x = ( r cos θ ) ( d t d r sin θ + r cos θ d t d θ ) − ( r sin θ ) ( d t d r cos θ − r sin θ d t d θ ) = ( r d t d r cos θ sin θ + r 2 cos 2 θ d t d θ ) − ( r d t d r sin θ cos θ − r 2 sin 2 θ d t d θ ) = r 2 ( cos 2 θ + sin 2 θ ) d t d θ = r 2 d t d θ したがって角運動量ベクトル L \boldsymbol{L} L L L L
L = m r 2 d θ d t d θ d t = L m r 2 \begin{aligned}
L &= m r^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
\dfrac{\mathrm{d}\theta}{\mathrm{d}t} &= \dfrac{L}{mr^2}
\end{aligned} L d t d θ = m r 2 d t d θ = m r 2 L 万有引力は保存力だから，運動エネルギーと万有引力による位置エネルギーの和すなわち力学的エネルギー E E E
E = 1 2 m v 2 − G M m r E = \dfrac{1}{2}m v^2 - G\dfrac{Mm}{r} E = 2 1 m v 2 − G r M m これに速度の式を代入すると
E = 1 2 m { ( d r d t ) 2 + r 2 ( d θ d t ) 2 } − G M m r E = \dfrac{1}{2}m \left\{ \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 + r^2 \left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 \right\} - \dfrac{GMm}{r} E = 2 1 m { ( d t d r ) 2 + r 2 ( d t d θ ) 2 } − r GM m d θ d t = L m r 2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \dfrac{L}{mr^2} d t d θ = m r 2 L
E = 1 2 m ( d r d t ) 2 + 1 2 m r 2 ( L m r 2 ) 2 − G M m r E = 1 2 m ( d r d t ) 2 + L 2 2 m r 2 − G M m r 1 2 m ( d r d t ) 2 = E + G M m r − L 2 2 m r 2 ( d r d t ) 2 = 2 E m + 2 G M r − L 2 m 2 r 2 \begin{aligned}
E &= \dfrac{1}{2}m \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 + \dfrac{1}{2}m r^2 \left(\dfrac{L}{mr^2}\right)^2 - \dfrac{GMm}{r} \\[8pt]
E &= \dfrac{1}{2}m \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 + \dfrac{L^2}{2mr^2} - \dfrac{GMm}{r}\\[8pt]
\dfrac{1}{2}m \left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 &= E + \dfrac{GMm}{r} - \dfrac{L^2}{2mr^2} \\[8pt]
\left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2 &= \dfrac{2E}{m} + \dfrac{2GM}{r} - \dfrac{L^2}{m^2 r^2}
\end{aligned} E E 2 1 m ( d t d r ) 2 ( d t d r ) 2 = 2 1 m ( d t d r ) 2 + 2 1 m r 2 ( m r 2 L ) 2 − r GM m = 2 1 m ( d t d r ) 2 + 2 m r 2 L 2 − r GM m = E + r GM m − 2 m r 2 L 2 = m 2 E + r 2 GM − m 2 r 2 L 2 合成関数の微分に関する数学公式 d y d x = d y d u ⋅ d u d x = d y d u d u d x \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \cdot \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{\dfrac{\mathrm{d}y}{\mathrm{d}u} }{\dfrac{\mathrm{d}u}{\mathrm{d}x}} d x d y = d u d y ⋅ d x d u = d x d u d u d y
( d r d θ ) 2 = ( d r d t ) 2 ( d θ d t ) 2 = 2 E m + 2 G M r − L 2 m 2 r 2 ( L m r 2 ) 2 = ( m r 2 L ) 2 ( 2 E m + 2 G M r − L 2 m 2 r 2 ) = r 4 L 2 ( 2 m E + 2 G M m 2 r − L 2 r 2 ) \begin{aligned}
\left(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2 &= \dfrac{\left(\dfrac{\mathrm{d}r}{\mathrm{d}t}\right)^2}{\left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2} = \dfrac{\dfrac{2E}{m} + \dfrac{2GM}{r} - \dfrac{L^2}{m^2 r^2}}{\left(\dfrac{L}{mr^2}\right)^2} \\[10pt]
&= \left(\dfrac{mr^2}{L}\right)^2 \left( \dfrac{2E}{m} + \dfrac{2GM}{r} - \dfrac{L^2}{m^2 r^2} \right) \\[10pt]
&= \dfrac{r^4}{L^2} \left( 2mE + \dfrac{2GMm^2}{r} - \dfrac{L^2}{r^2} \right)
\end{aligned} ( d θ d r ) 2 = ( d t d θ ) 2 ( d t d r ) 2 = ( m r 2 L ) 2 m 2 E + r 2 GM − m 2 r 2 L 2 = ( L m r 2 ) 2 ( m 2 E + r 2 GM − m 2 r 2 L 2 ) = L 2 r 4 ( 2 m E + r 2 GM m 2 − r 2 L 2 ) したがって d r d θ \dfrac{\mathrm{d}r}{\mathrm{d}\theta} d θ d r
d r d θ = ± r 2 L 2 m E + 2 G M m 2 r − L 2 r 2 d θ = ± L r 2 d r 2 m E + 2 G M m 2 r − L 2 r 2 \begin{aligned}
\dfrac{\mathrm{d}r}{\mathrm{d}\theta} &= \pm \dfrac{r^2}{L} \displaystyle\sqrt{2mE + \dfrac{2GMm^2}{r} - \dfrac{L^2}{r^2}}\\
\mathrm{d}\theta &= \pm \dfrac{\dfrac{L}{r^2} \, \mathrm{d}r}{\displaystyle\sqrt{2mE + \dfrac{2GMm^2}{r} - \dfrac{L^2}{r^2}}}
\end{aligned} d θ d r d θ = ± L r 2 2 m E + r 2 GM m 2 − r 2 L 2 = ± 2 m E + r 2 GM m 2 − r 2 L 2 r 2 L d r ここで u = 1 r u = \dfrac{1}{r} u = r 1 d u = − 1 r 2 d r \mathrm{d}u = -\dfrac{1}{r^2} \, \mathrm{d}r d u = − r 2 1 d r 1 r 2 d r = − d u \dfrac{1}{r^2} \, \mathrm{d}r = -\mathrm{d}u r 2 1 d r = − d u
d θ = ∓ L d u 2 m E + 2 G M m 2 u − L 2 u 2 \mathrm{d}\theta = \mp \dfrac{L \, \mathrm{d}u}{\displaystyle\sqrt{2mE + 2GMm^2 u - L^2 u^2}} d θ = ∓ 2 m E + 2 GM m 2 u − L 2 u 2 L d u 根号の中身について，u u u 2 2 2
− L 2 u 2 + 2 G M m 2 u + 2 m E = − L 2 ( u 2 − 2 G M m 2 L 2 u ) + 2 m E = − L 2 ( u − G M m 2 L 2 ) 2 + G 2 M 2 m 4 L 2 + 2 m E \begin{aligned}
-L^2 u^2 + 2GMm^2 u + 2mE &= -L^2 \left( u^2 - \dfrac{2GMm^2}{L^2} u \right) + 2mE \\[8pt]
&= -L^2 \left( u - \dfrac{GMm^2}{L^2} \right)^2 + \dfrac{G^2 M^2 m^4}{L^2} + 2mE
\end{aligned} − L 2 u 2 + 2 GM m 2 u + 2 m E = − L 2 ( u 2 − L 2 2 GM m 2 u ) + 2 m E = − L 2 ( u − L 2 GM m 2 ) 2 + L 2 G 2 M 2 m 4 + 2 m E ここで，定数 p = L 2 G M m 2 p = \dfrac{L^2}{GMm^2} p = GM m 2 L 2
G M m 2 L 2 = 1 p \dfrac{GMm^2}{L^2} = \dfrac{1}{p} L 2 GM m 2 = p 1 また，定数 K = 1 + 2 E L 2 G 2 M 2 m 3 K = \displaystyle\sqrt{1 + \dfrac{2E L^2}{G^2 M^2 m^3}} K = 1 + G 2 M 2 m 3 2 E L 2
G 2 M 2 m 4 L 2 + 2 m E = G 2 M 2 m 4 L 2 ( 1 + 2 E L 2 G 2 M 2 m 3 ) = ( G M m 2 L ) 2 K 2 = ( L K p ) 2 \dfrac{G^2 M^2 m^4}{L^2} + 2mE = \dfrac{G^2 M^2 m^4}{L^2} \left( 1 + \dfrac{2E L^2}{G^2 M^2 m^3} \right) = \left(\dfrac{GMm^2}{L}\right)^2 K^2 = \left(\dfrac{L K}{p}\right)^2 L 2 G 2 M 2 m 4 + 2 m E = L 2 G 2 M 2 m 4 ( 1 + G 2 M 2 m 3 2 E L 2 ) = ( L GM m 2 ) 2 K 2 = ( p L K ) 2 したがって根号の中身は
− L 2 ( u − 1 p ) 2 + ( L e p ) 2 = L 2 { ( K p ) 2 − ( u − 1 p ) 2 } -L^2 \left( u - \dfrac{1}{p} \right)^2 + \left(\dfrac{Le}{p}\right)^2 = L^2 \left\{ \left(\dfrac{K}{p}\right)^2 - \left(u - \dfrac{1}{p}\right)^2 \right\} − L 2 ( u − p 1 ) 2 + ( p L e ) 2 = L 2 { ( p K ) 2 − ( u − p 1 ) 2 } また，u − 1 p = K p cos ϕ ( 0 ≤ ϕ ≤ π ) u - \dfrac{1}{p} = \dfrac{K}{p} \cos\phi \quad (0 \le \phi \le \pi) u − p 1 = p K cos ϕ ( 0 ≤ ϕ ≤ π ) d u = − K p sin ϕ d ϕ \mathrm{d}u = -\dfrac{K}{p}\sin\phi \, \mathrm{d}\phi d u = − p K sin ϕ d ϕ
( K p ) 2 − ( u − 1 p ) 2 = K p sin ϕ \displaystyle\sqrt{\left(\dfrac{K}{p}\right)^2 - \left(u - \dfrac{1}{p}\right)^2} = \dfrac{K}{p}\sin\phi ( p K ) 2 − ( u − p 1 ) 2 = p K sin ϕ よって d θ \mathrm{d}\theta d θ
d θ = ∓ d u ( K p ) 2 − ( u − 1 p ) 2 = ∓ − K p sin ϕ d ϕ K p sin ϕ = ± d ϕ \begin{aligned}
\mathrm{d}\theta &= \mp \dfrac{\mathrm{d}u}{\displaystyle\sqrt{\left(\dfrac{K}{p}\right)^2 - \left(u - \dfrac{1}{p}\right)^2}}\\
&= \mp \dfrac{-\dfrac{K}{p}\sin\phi \, \mathrm{d}\phi}{\dfrac{K}{p}\sin\phi}\\
&= \pm \mathrm{d}\phi
\end{aligned} d θ = ∓ ( p K ) 2 − ( u − p 1 ) 2 d u = ∓ p K sin ϕ − p K sin ϕ d ϕ = ± d ϕ 両辺を積分すると，積分定数を θ 0 \theta_0 θ 0
∫ d θ = ∫ ± d ϕ ϕ = ± ( θ − θ 0 ) \begin{aligned}
\int \mathrm{d}\theta &= \int \pm \mathrm{d}\phi\\
\phi &= \pm (\theta - \theta_0)
\end{aligned} ∫ d θ ϕ = ∫ ± d ϕ = ± ( θ − θ 0 ) 数学公式 cos ( − x ) = cos x \cos (-x) = \cos x cos ( − x ) = cos x cos ϕ = cos ( θ − θ 0 ) \cos\phi = \cos(\theta - \theta_0) cos ϕ = cos ( θ − θ 0 )
u − 1 p = K p cos ( θ − θ 0 ) u = 1 + K cos ( θ − θ 0 ) p \begin{aligned}
u - \dfrac{1}{p} &= \dfrac{K}{p} \cos(\theta - \theta_0)\\
u &= \dfrac{1 + K\cos(\theta - \theta_0)}{p}
\end{aligned} u − p 1 u = p K cos ( θ − θ 0 ) = p 1 + K cos ( θ − θ 0 ) u = 1 r u = \dfrac{1}{r} u = r 1
r = p 1 + K cos ( θ − θ 0 ) r = \dfrac{p}{1 + K\cos(\theta - \theta_0)} r = 1 + K cos ( θ − θ 0 ) p 中心天体からの距離 r r r θ = 0 \theta = 0 θ = 0 θ 0 = 0 \theta_0 = 0 θ 0 = 0
r = p 1 + K cos θ \begin{aligned}
r &= \dfrac{p}{1 + K\cos\theta}
\end{aligned} r = 1 + K cos θ p ここで定数 K K K e \boldsymbol{e} e e e e
近日点距離を r p r_\text{p} r p r a r_\text{a} r a θ = 0 \theta=0 θ = 0 r r r 1 + K cos θ 1 + K \cos\theta 1 + K cos θ θ = π \theta=\pi θ = π
{ r p = p 1 + K cos 0 = p 1 + K r a = p 1 + K cos π = p 1 − K \left\{\;
\begin{aligned}
r_\text{p} &= \dfrac{p}{1 + K \cos 0} = \dfrac{p}{1 + K} \\[8pt]
r_\text{a} &= \dfrac{p}{1 + K \cos \pi} = \dfrac{p}{1 - K}
\end{aligned}
\right. ⎩ ⎨ ⎧ r p r a = 1 + K cos 0 p = 1 + K p = 1 + K cos π p = 1 − K p 軌道を楕円とすれば長軸の長さ 2 a 2a 2 a 2 2 2
2 a = r p + r a = p 1 + K + p 1 − K = 2 p 1 − K 2 p = a ( 1 − K 2 ) \begin{aligned}
2a &= r_\text{p} + r_\text{a} \\[5pt]
&= \dfrac{p}{1 + K} + \dfrac{p}{1 - K} \\[8pt]
&= \dfrac{2p}{1 - K^2} \\[8pt]
p &= a(1 - K^2)
\end{aligned} 2 a p = r p + r a = 1 + K p + 1 − K p = 1 − K 2 2 p = a ( 1 − K 2 ) これを r p = p 1 + K r_\text{p} = \dfrac{p}{1 + K} r p = 1 + K p
r p = p 1 + K = a ( 1 − K 2 ) 1 + K = a − a K \begin{aligned}
r_\text{p} &= \dfrac{p}{1 + K}\\[8pt]
&= \dfrac{a(1 - K^2)}{1 + K}\\[8pt]
&= a - aK
\end{aligned} r p = 1 + K p = 1 + K a ( 1 − K 2 ) = a − a K ここで中心から焦点（原点）までの距離を c c c r p r_\text{p} r p a a a
r p = a − c r_\text{p} = a - c r p = a − c と比較して
c = a K K = c a c=aK\\
K=\dfrac{c}{a} c = a K K = a c これは中心から焦点までの距離 c c c a a a c a \dfrac{c}{a} a c e e e
K = e K=e K = e ここで，r ⋅ e = r e cos θ \boldsymbol{r}\cdot\boldsymbol{e} = re\cos\theta r ⋅ e = r e cos θ p = a ( 1 − K 2 ) p = a(1 - K^2) p = a ( 1 − K 2 ) e = ∣ e ∣ e=|\boldsymbol{e}| e = ∣ e ∣ r = ∣ r ∣ r=|\boldsymbol{r}| r = ∣ r ∣
r = p 1 + K cos θ r + r K cos θ = a ( 1 − K 2 ) r + r e cos θ = a ( 1 − e 2 ) ∣ r ∣ + r ⋅ e = a ( 1 − ∣ e ∣ 2 ) \begin{aligned}
r &= \dfrac{p}{1 + K\cos\theta}\\
r + rK\cos\theta &= a(1 - K^2)\\
r + re\cos\theta &= a(1 - e^2)\\
|\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{e} &= a(1 - |\boldsymbol{e}|^2)
\end{aligned} r r + r K cos θ r + r e cos θ ∣ r ∣ + r ⋅ e = 1 + K cos θ p = a ( 1 − K 2 ) = a ( 1 − e 2 ) = a ( 1 − ∣ e ∣ 2 ) 第二法則
中心天体 C \text{C} C M M M A \text{A} A m m m A \text{A} A B \text{B} B G G G
m d 2 r d t 2 = − G M m ∣ r ∣ 2 r ∣ r ∣ m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} = -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t 2 d 2 r = − G ∣ r ∣ 2 M m ∣ r ∣ r ここで原点まわりの角運動量 L \boldsymbol{L} L
L = m ( r × d r d t ) \boldsymbol{L} = m\left(\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right) L = m ( r × d t d r ) 両辺を時間 t t t 2 2 2 d d t ( a ( t ) × b ( t ) ) = d a ( t ) d t × b ( t ) + a ( t ) × d b ( t ) d t \dfrac{\mathrm{d}}{\mathrm{d}t} (\boldsymbol{a}(t) \times \boldsymbol{b}(t)) = \dfrac{\mathrm{d}\boldsymbol{a}(t)}{\mathrm{d}t} \times \boldsymbol{b}(t) + \boldsymbol{a}(t) \times \dfrac{\mathrm{d}\boldsymbol{b}(t)}{\mathrm{d}t} d t d ( a ( t ) × b ( t )) = d t d a ( t ) × b ( t ) + a ( t ) × d t d b ( t )
d L d t = m ( d r d t × d r d t + r × d 2 r d t 2 ) \dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} = m \left( \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} + \boldsymbol{r} \times \dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} \right) d t d L = m ( d t d r × d t d r + r × d t 2 d 2 r ) したがって右辺第 1 1 1 r × r = 0 \boldsymbol{r} \times \boldsymbol{r} = \boldsymbol{0} r × r = 0 2 2 2 m d 2 r d t 2 = − G M m ∣ r ∣ 2 r ∣ r ∣ m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} = -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} m d t 2 d 2 r = − G ∣ r ∣ 2 M m ∣ r ∣ r
d L d t = r × ( m d 2 r d t 2 ) = r × ( − G M m ∣ r ∣ 2 r ∣ r ∣ ) = − G M m ∣ r ∣ 3 ( r × r ) = 0 \begin{aligned}
\dfrac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} &= \boldsymbol{r} \times \left( m\dfrac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2} \right) \\
&= \boldsymbol{r} \times \left( -G\dfrac{Mm}{|\boldsymbol{r}|^2}\dfrac{\boldsymbol{r}}{|\boldsymbol{r}|} \right) \\
&= -G\dfrac{Mm}{|\boldsymbol{r}|^3} (\boldsymbol{r} \times \boldsymbol{r}) \\
&= \boldsymbol{0}
\end{aligned} d t d L = r × ( m d t 2 d 2 r ) = r × ( − G ∣ r ∣ 2 M m ∣ r ∣ r ) = − G ∣ r ∣ 3 M m ( r × r ) = 0 したがって，天体 A \text{A} A L \boldsymbol{L} L L L L
L = m ∣ r × d r d t ∣ ∣ r × d r d t ∣ = L m \begin{aligned}
L &= m\left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right|\\
\left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right| &= \dfrac{L}{m}
\end{aligned} L r × d t d r = m r × d t d r = m L ここで動径方向の単位ベクトルを e r \boldsymbol{e}_r e r θ \theta θ e θ \boldsymbol{e}_\theta e θ r = ∣ r ∣ r = |\boldsymbol{r}| r = ∣ r ∣
r = r e r \boldsymbol{r} = r \boldsymbol{e}_r r = r e r 動径方向の単位ベクトル e r \boldsymbol{e}_r e r θ \theta θ 1 1 1
e r = ( cos θ sin θ ) \boldsymbol{e}_r = \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix} e r = ( cos θ sin θ ) 円周方向の単位ベクトル e θ \boldsymbol{e}_\theta e θ e r \boldsymbol{e}_r e r θ \theta θ 90 ∘ 90^\circ 9 0 ∘ 1 1 1
e θ = ( cos ( θ + 90 ∘ ) sin ( θ + 90 ∘ ) ) = ( − sin θ cos θ ) \boldsymbol{e}_\theta = \begin{pmatrix} \cos(\theta + 90^\circ) \\ \sin(\theta + 90^\circ) \end{pmatrix} = \begin{pmatrix} -\sin\theta \\ \cos\theta \end{pmatrix} e θ = ( cos ( θ + 9 0 ∘ ) sin ( θ + 9 0 ∘ ) ) = ( − sin θ cos θ ) 角度 θ \theta θ t t t d y d x = d y d u ⋅ d u d x = d y d u d u d x \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \cdot \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{\dfrac{\mathrm{d}y}{\mathrm{d}u} }{\dfrac{\mathrm{d}u}{\mathrm{d}x}} d x d y = d u d y ⋅ d x d u = d x d u d u d y
d e r d t = d θ d t d e r d θ = d θ d t ( d d θ cos θ d d θ sin θ ) = d θ d t ( − sin θ cos θ ) = d θ d t e θ \begin{aligned}
\dfrac{\mathrm{d}\boldsymbol{e}_r}{\mathrm{d}t}
&= \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \dfrac{\mathrm{d}\boldsymbol{e}_r}{\mathrm{d}\theta}\\
&= \dfrac{\mathrm{d}\theta}{\mathrm{d}t}
\begin{pmatrix}
\dfrac{\mathrm{d}}{\mathrm{d}\theta}\cos\theta \\[6pt]
\dfrac{\mathrm{d}}{\mathrm{d}\theta}\sin\theta
\end{pmatrix} \\[10pt]
&= \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \begin{pmatrix} -\sin\theta \\ \cos\theta \end{pmatrix}\\
&= \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \boldsymbol{e}_\theta
\end{aligned} d t d e r = d t d θ d θ d e r = d t d θ d θ d cos θ d θ d sin θ = d t d θ ( − sin θ cos θ ) = d t d θ e θ 天体 A \text{A} A d r d t \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} d t d r d d x ( f ( x ) g ( x ) ) = d f ( x ) d x g ( x ) + f ( x ) d g ( x ) d x \dfrac{\mathrm{d}}{\mathrm{d}x} \left( f(x)g(x) \right) = \dfrac{\mathrm{d}f(x)}{\mathrm{d}x} g(x) + f(x) \dfrac{\mathrm{d}g(x)}{\mathrm{d}x} d x d ( f ( x ) g ( x ) ) = d x d f ( x ) g ( x ) + f ( x ) d x d g ( x )
d r d t = r d e r d t = d r d t e r + r d e r d t = d r d t e r + r d θ d t e θ \begin{aligned}
\dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} &= r\dfrac{\mathrm{d}\boldsymbol{e}_r}{\mathrm{d}t}\\
&= \dfrac{\mathrm{d}r}{\mathrm{d}t} \boldsymbol{e}_r + r \dfrac{\mathrm{d}\boldsymbol{e}_r}{\mathrm{d}t} \\
&= \dfrac{\mathrm{d}r}{\mathrm{d}t} \boldsymbol{e}_r + r \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \boldsymbol{e}_\theta
\end{aligned} d t d r = r d t d e r = d t d r e r + r d t d e r = d t d r e r + r d t d θ e θ ここで位置ベクトル r \boldsymbol{r} r d r d t \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} d t d r
r × d r d t = ( r e r ) × ( d r d t e r + r d θ d t e θ ) = r d r d t ( e r × e r ) + r 2 d θ d t ( e r × e θ ) \begin{aligned}
\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} &= (r \boldsymbol{e}_r) \times \left( \dfrac{\mathrm{d}r}{\mathrm{d}t} \boldsymbol{e}_r + r \dfrac{\mathrm{d}\theta}{\mathrm{d}t} \boldsymbol{e}_\theta \right) \\[6pt]
&= r \dfrac{\mathrm{d}r}{\mathrm{d}t} (\boldsymbol{e}_r \times \boldsymbol{e}_r) + r^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} (\boldsymbol{e}_r \times \boldsymbol{e}_\theta)
\end{aligned} r × d t d r = ( r e r ) × ( d t d r e r + r d t d θ e θ ) = r d t d r ( e r × e r ) + r 2 d t d θ ( e r × e θ ) e r × e r = 0 \boldsymbol{e}_r \times \boldsymbol{e}_r = \boldsymbol{0} e r × e r = 0 e r × e θ \boldsymbol{e}_r \times \boldsymbol{e}_\theta e r × e θ
r × d r d t = r 2 d θ d t ( e r × e θ ) \boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} = r^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} (\boldsymbol{e}_r \times \boldsymbol{e}_\theta) r × d t d r = r 2 d t d θ ( e r × e θ ) 両辺の大きさをとって整理すると
∣ r × d r d t ∣ = r 2 d θ d t = ∣ r ∣ 2 d θ d t 1 2 ∣ r ∣ 2 d θ d t = 1 2 ∣ r × d r d t ∣ \begin{aligned}
\left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right| = r^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = |\boldsymbol{r}|^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
\dfrac{1}{2} |\boldsymbol{r}|^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \dfrac{1}{2} \left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right|
\end{aligned} r × d t d r = r 2 d t d θ = ∣ r ∣ 2 d t d θ 2 1 ∣ r ∣ 2 d t d θ = 2 1 r × d t d r 右辺について ∣ r × d r d t ∣ = L m \left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right| = \dfrac{L}{m} r × d t d r = m L
1 2 ∣ r ∣ 2 d θ d t = 1 2 ∣ r × d r d t ∣ = L 2 m = const. \begin{aligned}
\dfrac{1}{2} |\boldsymbol{r}|^2 \dfrac{\mathrm{d}\theta}{\mathrm{d}t} &= \dfrac{1}{2} \left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right| \\
&= \dfrac{L}{2m} \\
&= \text{const.}
\end{aligned} 2 1 ∣ r ∣ 2 d t d θ = 2 1 r × d t d r = 2 m L = const. 第三法則
第一法則より，原点に位置する中心天体 C \text{C} C A \text{A} A r \boldsymbol{r} r e \boldsymbol{e} e a a a
∣ r ∣ + r ⋅ e = a ( 1 − ∣ e ∣ 2 ) |\boldsymbol{r}| + \boldsymbol{r} \cdot \boldsymbol{e} = a(1 - |\boldsymbol{e}|^2) ∣ r ∣ + r ⋅ e = a ( 1 − ∣ e ∣ 2 ) 近日点方向からの回転角を θ \theta θ e \boldsymbol{e} e e = ∣ e ∣ e = |\boldsymbol{e}| e = ∣ e ∣ A \text{A} A r \boldsymbol{r} r r = ∣ r ∣ r = |\boldsymbol{r}| r = ∣ r ∣ r ⋅ e = r e cos θ \boldsymbol{r} \cdot \boldsymbol{e} = r e \cos\theta r ⋅ e = r e cos θ C \text{C} C A \text{A} A r r r
r = a ( 1 − e 2 ) 1 + e cos θ r = \dfrac{a(1 - e^2)}{1 + e \cos\theta} r = 1 + e cos θ a ( 1 − e 2 ) 天体 A \text{A} A C \text{C} C θ = 0 \theta = 0 θ = 0 θ = π \theta = \pi θ = π r p , r a r_{\text{p}}, r_\text{a} r p , r a
{ r p = a ( 1 − e ) r a = a ( 1 + e ) \left\{
\begin{aligned}
r_{\text{p}} &= a(1 - e)\\
r_\text{a} &= a(1 + e)
\end{aligned}
\right. { r p r a = a ( 1 − e ) = a ( 1 + e ) ここで近日点および遠日点における天体 A \text{A} A d r d t \dfrac{\mathrm{d}r}{\mathrm{d}t} d t d r d y d x = d y d u d u d x \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \dfrac{\mathrm{d}u}{\mathrm{d}x} d x d y = d u d y d x d u d d x ( f ( x ) g ( x ) ) = d f ( x ) d x g ( x ) − f ( x ) d g ( x ) d x { g ( x ) } 2 \dfrac{\mathrm{d}}{\mathrm{d}x} \left( \dfrac{f(x)}{g(x)} \right) = \dfrac{\dfrac{\mathrm{d}f(x)}{\mathrm{d}x} g(x) - f(x) \dfrac{\mathrm{d}g(x)}{\mathrm{d}x}}{\{g(x)\}^2} d x d ( g ( x ) f ( x ) ) = { g ( x ) } 2 d x d f ( x ) g ( x ) − f ( x ) d x d g ( x )
d r d t = d r d θ ⋅ d θ d t = d d θ a ( 1 − e 2 ) 1 + e cos θ ⋅ d θ d t = a ( 1 − e 2 ) ⋅ e sin θ ( 1 + e cos θ ) 2 d θ d t \begin{aligned}
\dfrac{\mathrm{d}r}{\mathrm{d}t} &= \dfrac{\mathrm{d}r}{\mathrm{d}\theta} \cdot \dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
&= \dfrac{\mathrm{d}}{\mathrm{d}\theta}\dfrac{a(1 - e^2)}{1 + e \cos\theta} \cdot \dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
&= \dfrac{a(1 - e^2) \cdot e\sin\theta}{(1 + e\cos\theta)^2}\dfrac{\mathrm{d}\theta}{\mathrm{d}t}
\end{aligned} d t d r = d θ d r ⋅ d t d θ = d θ d 1 + e cos θ a ( 1 − e 2 ) ⋅ d t d θ = ( 1 + e cos θ ) 2 a ( 1 − e 2 ) ⋅ e sin θ d t d θ 近日点（θ = 0 \theta = 0 θ = 0 θ = π \theta = \pi θ = π
{ sin 0 = 0 sin π = 0 \left\{
\begin{aligned}
\sin 0 &= 0\\
\sin \pi &= 0
\end{aligned}
\right. { sin 0 sin π = 0 = 0 だからいずれの場合でも
d r d θ = 0 \dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 0 d θ d r = 0 よって動径方向の速度 d r d t \dfrac{\mathrm{d}r}{\mathrm{d}t} d t d r
d r d t = a ( 1 − e 2 ) ⋅ e sin θ ( 1 + e cos θ ) 2 d θ d t = a ( 1 − e 2 ) ⋅ e ⋅ 0 ( 1 + e cos θ ) 2 d θ d t = 0 \begin{aligned}
\dfrac{\mathrm{d}r}{\mathrm{d}t} &= \dfrac{a(1 - e^2) \cdot e\sin\theta}{(1 + e\cos\theta)^2}\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
&= \dfrac{a(1 - e^2) \cdot e \cdot 0}{(1 + e\cos\theta)^2}\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\\
&=0
\end{aligned} d t d r = ( 1 + e cos θ ) 2 a ( 1 − e 2 ) ⋅ e sin θ d t d θ = ( 1 + e cos θ ) 2 a ( 1 − e 2 ) ⋅ e ⋅ 0 d t d θ = 0 動径方向の速度が 0 0 0 A \text{A} A A \text{A} A r p , r a {\boldsymbol{r}_\text{p}}, {\boldsymbol{r}_\text{a}} r p , r a v p , v a {\boldsymbol{v}_\text{p}}, {\boldsymbol{v}_\text{a}} v p , v a v p , v a {v_\text{p}}, {v_\text{a}} v p , v a
{ d r p d t = v p = ( 0 v p ) d r a d t = v a = ( 0 v a ) \left\{
\begin{aligned}
\dfrac{\mathrm{d}\boldsymbol{r}_\text{p}}{\mathrm{d}t} &= \boldsymbol{v}_\text{p}
=
\begin{pmatrix}
0\\
v_\text{p}
\end{pmatrix}\\[10pt]
\dfrac{\mathrm{d}\boldsymbol{r}_\text{a}}{\mathrm{d}t} &= \boldsymbol{v}_\text{a}
=
\begin{pmatrix}
0\\
v_\text{a}
\end{pmatrix}
\end{aligned}
\right. ⎩ ⎨ ⎧ d t d r p d t d r a = v p = ( 0 v p ) = v a = ( 0 v a ) ケプラーの第二法則より，天体 A \text{A} A r \boldsymbol{r} r d r d t \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} d t d r
1 2 ∣ r × d r d t ∣ = const. \dfrac{1}{2} \left|\boldsymbol{r} \times \dfrac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}\right| = \text{const.} 2 1 r × d t d r = const. 左辺を C C C
r p × d r p d t = ( r p 0 0 ) × ( 0 v p 0 ) = ( 0 ⋅ 0 − 0 ⋅ v p 0 ⋅ 0 − r p ⋅ 0 r p ⋅ v p − 0 ⋅ 0 ) = ( 0 0 r p v p ) \begin{aligned}
\boldsymbol{r}_\text{p} \times \dfrac{\mathrm{d}\boldsymbol{r}_\text{p}}{\mathrm{d}t}
&= \begin{pmatrix} r_\text{p} \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ v_\text{p} \\ 0 \end{pmatrix} \\
&= \begin{pmatrix}
0 \cdot 0 - 0 \cdot v_\text{p} \\
0 \cdot 0 - r_\text{p} \cdot 0 \\
r_\text{p} \cdot v_\text{p} - 0 \cdot 0
\end{pmatrix} \\
&= \begin{pmatrix} 0 \\ 0 \\ r_\text{p} v_\text{p} \end{pmatrix}
\end{aligned} r p × d t d r p = r p 0 0 × 0 v p 0 = 0 ⋅ 0 − 0 ⋅ v p 0 ⋅ 0 − r p ⋅ 0 r p ⋅ v p − 0 ⋅ 0 = 0 0 r p v p 同様に遠日点においても
r a × d r a d t = ( 0 0 r a v a ) \begin{aligned}
\boldsymbol{r}_\text{a} \times \dfrac{\mathrm{d}\boldsymbol{r}_\text{a}}{\mathrm{d}t} = \begin{pmatrix} 0 \\ 0 \\ r_\text{a} v_\text{a} \end{pmatrix}
\end{aligned} r a × d t d r a = 0 0 r a v a であるから
1 2 ∣ r p × d r p d t ∣ = 1 2 ∣ r a × d r a d t ∣ = C 1 2 r p v p = 1 2 r a v a = C \dfrac{1}{2} \left| \boldsymbol{r}_\text{p} \times \dfrac{\mathrm{d}\boldsymbol{r}_\text{p}}{\mathrm{d}t} \right| =\dfrac{1}{2} \left| \boldsymbol{r}_\text{a} \times \dfrac{\mathrm{d}\boldsymbol{r}_\text{a}}{\mathrm{d}t} \right| = C\\
\dfrac{1}{2} r_\text{p} v_\text{p} =\dfrac{1}{2} r_\text{a} v_\text{a} = C 2 1 r p × d t d r p = 2 1 r a × d t d r a = C 2 1 r p v p = 2 1 r a v a = C したがって，速さ v p , v a {v_\text{p}}, {v_\text{a}} v p , v a r p = a ( 1 − e ) r_{\text{p}} = a(1 - e) r p = a ( 1 − e ) r a = a ( 1 + e ) r_\text{a} = a(1 + e) r a = a ( 1 + e )
{ v p = 2 C r p = 2 C a ( 1 − e ) v a = 2 C r a = 2 C a ( 1 + e ) \left\{
\begin{aligned}
{v_\text{p}} &= \dfrac{2C}{r_\text{p}} = \dfrac{2C}{a(1 - e)}\\[10pt]
{v_\text{a}} &= \dfrac{2C}{r_\text{a}} = \dfrac{2C}{a(1 + e)}
\end{aligned}
\right. ⎩ ⎨ ⎧ v p v a = r p 2 C = a ( 1 − e ) 2 C = r a 2 C = a ( 1 + e ) 2 C 次に，中心天体 C \text{C} C M M M A \text{A} A m m m G G G A \text{A} A C \text{C} C
1 2 m v p 2 − G M m r p = 1 2 m v a 2 − G M m r a 1 2 ( v p 2 − v a 2 ) = G M ( 1 r p − 1 r a ) \begin{aligned}
\dfrac{1}{2} m {v_\text{p}}^2 - G\dfrac{Mm}{r_{\text{p}}} = \dfrac{1}{2} m {v_\text{a}}^2 - G\dfrac{Mm}{r_\text{a}}\\
\dfrac{1}{2}({v_\text{p}}^2 - {v_\text{a}}^2) = GM \left( \dfrac{1}{r_{\text{p}}} - \dfrac{1}{r_\text{a}} \right)
\end{aligned} 2 1 m v p 2 − G r p M m = 2 1 m v a 2 − G r a M m 2 1 ( v p 2 − v a 2 ) = GM ( r p 1 − r a 1 ) v p = 2 C a ( 1 − e ) {v_\text{p}} = \dfrac{2C}{a(1 - e)} v p = a ( 1 − e ) 2 C v a = 2 C a ( 1 + e ) {v_\text{a}} = \dfrac{2C}{a(1 + e)} v a = a ( 1 + e ) 2 C r p = a ( 1 − e ) r_{\text{p}} = a(1 - e) r p = a ( 1 − e ) r a = a ( 1 + e ) r_\text{a} = a(1 + e) r a = a ( 1 + e )
1 2 { 4 C 2 a 2 ( 1 − e ) 2 − 4 C 2 a 2 ( 1 + e ) 2 } = G M { 1 a ( 1 − e ) − 1 a ( 1 + e ) } 2 C 2 a 2 ( 1 + e ) 2 − ( 1 − e ) 2 ( 1 − e 2 ) 2 = G M a ( 1 + e ) − ( 1 − e ) 1 − e 2 8 C 2 e a 2 ( 1 − e 2 ) 2 = 2 G M e a ( 1 − e 2 ) \begin{aligned}
\dfrac{1}{2} \left\{ \dfrac{4C^2}{a^2(1 - e)^2} - \dfrac{4C^2}{a^2(1 + e)^2} \right\} &= GM \left\{ \dfrac{1}{a(1 - e)} - \dfrac{1}{a(1 + e)} \right\}\\
\dfrac{2C^2}{a^2} \dfrac{(1 + e)^2 - (1 - e)^2}{\left(1 - e^2\right)^2}&=\dfrac{GM}{a} \dfrac{(1 + e) - (1 - e)}{1 - e^2} \\
\dfrac{8C^2 e}{a^2\left(1 - e^2\right)^2}&=\dfrac{2GMe}{a(1 - e^2)}
\end{aligned} 2 1 { a 2 ( 1 − e ) 2 4 C 2 − a 2 ( 1 + e ) 2 4 C 2 } a 2 2 C 2 ( 1 − e 2 ) 2 ( 1 + e ) 2 − ( 1 − e ) 2 a 2 ( 1 − e 2 ) 2 8 C 2 e = GM { a ( 1 − e ) 1 − a ( 1 + e ) 1 } = a GM 1 − e 2 ( 1 + e ) − ( 1 − e ) = a ( 1 − e 2 ) 2 GM e ［1 1 1 e ≠ 0 e \neq 0 e = 0
両辺を 2 e a ( 1 − e 2 ) \dfrac{2e}{a(1 - e^2)} a ( 1 − e 2 ) 2 e C > 0 C>0 C > 0
C 2 = G M 4 a ( 1 − e 2 ) C = 1 2 G M a ( 1 − e 2 ) \begin{aligned}
C^2 &= \dfrac{GM}{4} a(1 - e^2)\\
C &= \dfrac{1}{2}\displaystyle\sqrt{GMa(1 - e^2)}
\end{aligned} C 2 C = 4 GM a ( 1 − e 2 ) = 2 1 GM a ( 1 − e 2 ) ［2 2 2 e = 0 e = 0 e = 0
e = 0 e=0 e = 0 0 0 0 A \text{A} A A \text{A} A m m m a a a v v v C \text{C} C G G G
m v 2 a = G M m a 2 v = G M a \begin{aligned}
m \dfrac{v^2}{a} &= G \dfrac{Mm}{a^2}\\
v &= \displaystyle\sqrt{\dfrac{GM}{a}}
\end{aligned} m a v 2 v = G a 2 M m = a GM 第二法則より
C = 1 2 r v = 1 2 a G M a = 1 2 G M a \begin{aligned}
C &= \dfrac{1}{2} r v\\
&= \dfrac{1}{2} a \displaystyle\sqrt{\dfrac{GM}{a}} \\
&= \dfrac{1}{2} \displaystyle\sqrt{GMa}
\end{aligned} C = 2 1 r v = 2 1 a a GM = 2 1 GM a これはe ≠ 0 e \neq 0 e = 0 C = 1 2 G M a ( 1 − e 2 ) C = \dfrac{1}{2}\displaystyle\sqrt{GMa(1 - e^2)} C = 2 1 GM a ( 1 − e 2 ) e = 0 e = 0 e = 0
次に楕円軌道全体の面積 S S S O \mathrm{O} O θ \theta θ r ( θ ) \boldsymbol{r}(\theta) r ( θ ) θ + Δ θ \theta + \Delta\theta θ + Δ θ r ( θ + Δ θ ) \boldsymbol{r}(\theta + \Delta\theta) r ( θ + Δ θ ) Δ θ \Delta\theta Δ θ 2 2 2 r ( θ ) \boldsymbol{r}(\theta) r ( θ ) r ( θ + Δ θ ) \boldsymbol{r}(\theta + \Delta\theta) r ( θ + Δ θ ) Δ S \Delta S Δ S S = 1 2 a h S=\dfrac{1}{2}ah S = 2 1 ah
Δ S = 1 2 ∣ r ( θ ) ∣ ∣ r ( θ + Δ θ ) ∣ sin ( Δ θ ) = 1 2 r ( θ ) r ( θ + Δ θ ) sin ( Δ θ ) \Delta S = \dfrac{1}{2} |\boldsymbol{r}(\theta)| |\boldsymbol{r}(\theta + \Delta\theta)| \sin(\Delta\theta) = \dfrac{1}{2} r(\theta) \, r(\theta + \Delta\theta) \sin(\Delta\theta) Δ S = 2 1 ∣ r ( θ ) ∣∣ r ( θ + Δ θ ) ∣ sin ( Δ θ ) = 2 1 r ( θ ) r ( θ + Δ θ ) sin ( Δ θ ) 両辺を 0 0 0 Δ θ \Delta\theta Δ θ Δ θ → 0 \Delta\theta \to 0 Δ θ → 0 lim Δ θ → 0 sin ( Δ θ ) Δ θ = 1 \displaystyle\lim_{\Delta\theta \to 0} \dfrac{\sin(\Delta\theta)}{\Delta\theta} = 1 Δ θ → 0 lim Δ θ sin ( Δ θ ) = 1
d S d θ = lim Δ θ → 0 Δ S Δ θ = 1 2 lim Δ θ → 0 { r ( θ ) r ( θ + Δ θ ) ⋅ sin ( Δ θ ) Δ θ } = 1 2 r ( θ ) ⋅ r ( θ ) ⋅ 1 = 1 2 r ( θ ) 2 \begin{aligned}
\dfrac{\mathrm{d}S}{\mathrm{d}\theta} &= \lim_{\Delta\theta \to 0} \dfrac{\Delta S}{\Delta\theta}\\
&= \dfrac{1}{2} \lim_{\Delta\theta \to 0} \left\{ r(\theta) \, r(\theta + \Delta\theta) \cdot \dfrac{\sin(\Delta\theta)}{\Delta\theta} \right\} \\[6pt]
&= \dfrac{1}{2} r(\theta) \cdot r(\theta) \cdot 1 = \dfrac{1}{2} r(\theta)^2
\end{aligned} d θ d S = Δ θ → 0 lim Δ θ Δ S = 2 1 Δ θ → 0 lim { r ( θ ) r ( θ + Δ θ ) ⋅ Δ θ sin ( Δ θ ) } = 2 1 r ( θ ) ⋅ r ( θ ) ⋅ 1 = 2 1 r ( θ ) 2 変数を分離して両辺を積分して
∫ 0 S d S = ∫ 0 2 π 1 2 r ( θ ) 2 d θ S = ∫ 0 2 π 1 2 r ( θ ) 2 d θ \begin{aligned}
\int_{0}^{S}\mathrm{d}S &= \int_{0}^{2\pi}\dfrac{1}{2} r(\theta)^2 \mathrm{d}\theta\\
S &= \int_{0}^{2\pi}\dfrac{1}{2} r(\theta)^2 \mathrm{d}\theta
\end{aligned} ∫ 0 S d S S = ∫ 0 2 π 2 1 r ( θ ) 2 d θ = ∫ 0 2 π 2 1 r ( θ ) 2 d θ 距離 r ( θ ) = a ( 1 − e 2 ) 1 + e cos θ r(\theta) = \dfrac{a(1 - e^2)}{1 + e\cos\theta} r ( θ ) = 1 + e cos θ a ( 1 − e 2 ) r ( θ ) r(\theta) r ( θ )
S = a 2 ( 1 − e 2 ) 2 ∫ 0 2 π d θ 2 ( 1 + e cos θ ) 2 = a 2 ( 1 − e 2 ) 2 ∫ 0 π d θ ( 1 + e cos θ ) 2 \begin{aligned}
S &= a^2 \left(1 - e^2\right)^2 \int_0^{2\pi} \dfrac{\mathrm{d}\theta}{2(1 + e\cos\theta)^2}\\
&= a^2 \left(1 - e^2\right)^2 \int_0^{\pi} \dfrac{\mathrm{d}\theta}{(1 + e\cos\theta)^2}
\end{aligned} S = a 2 ( 1 − e 2 ) 2 ∫ 0 2 π 2 ( 1 + e cos θ ) 2 d θ = a 2 ( 1 − e 2 ) 2 ∫ 0 π ( 1 + e cos θ ) 2 d θ 積分部分につき次のように置換して求めることとする。新たに変数 E E E cos θ = cos E − e 1 − e cos E \cos\theta = \dfrac{\cos E - e}{1 - e\cos E} cos θ = 1 − e cos E cos E − e θ \theta θ 0 0 0 π \pi π E E E 0 0 0 π \pi π
1 + e cos θ = 1 + e ( cos E − e 1 − e cos E ) = 1 − e 2 1 − e cos E \begin{aligned}
1 + e\cos\theta
&= 1 + e \left( \dfrac{\cos E - e}{1 - e\cos E} \right) \\
&= \dfrac{1 - e^2}{1 - e\cos E}
\end{aligned} 1 + e cos θ = 1 + e ( 1 − e cos E cos E − e ) = 1 − e cos E 1 − e 2 sin 2 θ = 1 − cos 2 θ \sin^2\theta = 1 - \cos^2\theta sin 2 θ = 1 − cos 2 θ
sin θ = 1 − e 2 sin E 1 − e cos E \sin\theta = \dfrac{\displaystyle\sqrt{1 - e^2} \sin E}{1 - e\cos E} sin θ = 1 − e cos E 1 − e 2 sin E cos θ = cos E − e 1 − e cos E \cos\theta = \dfrac{\cos E - e}{1 - e\cos E} cos θ = 1 − e cos E cos E − e
d ( cos θ ) = − sin θ d θ \mathrm{d}(\cos\theta) = -\sin\theta \,\mathrm{d}\theta d ( cos θ ) = − sin θ d θ 右辺の微分について商の微分に関する数学公式 d ( f g ) = g d f − f d g g 2 \mathrm{d}\left( \dfrac{f}{g} \right) = \dfrac{g\,\mathrm{d}f - f\,\mathrm{d}g}{g^2} d ( g f ) = g 2 g d f − f d g
d ( cos E − e 1 − e cos E ) = ( − sin E ) ( 1 − e cos E ) d E − ( cos E − e ) ( e sin E ) d E ( 1 − e cos E ) 2 = ( − sin E + e sin E cos E − e sin E cos E + e 2 sin E ) d E ( 1 − e cos E ) 2 = − ( 1 − e 2 ) sin E ( 1 − e cos E ) 2 d E sin θ d θ = ( 1 − e 2 ) sin E ( 1 − e cos E ) 2 d E \begin{aligned}
\mathrm{d}\left(\dfrac{\cos E - e}{1 - e\cos E}\right) &= \dfrac{(-\sin E)(1 - e\cos E)\,\mathrm{d}E - (\cos E - e)(e\sin E)\,\mathrm{d}E}{(1 - e\cos E)^2} \\
&= \dfrac{\Big( -\sin E + e\sin E\cos E - e\sin E\cos E + e^2\sin E \Big)\,\mathrm{d}E}{(1 - e\cos E)^2} \\
&= -\dfrac{(1 - e^2)\sin E}{(1 - e\cos E)^2} \,\mathrm{d}E\\
\sin\theta \,\mathrm{d}\theta &= \dfrac{(1 - e^2)\sin E}{(1 - e\cos E)^2} \,\mathrm{d}E
\end{aligned} d ( 1 − e cos E cos E − e ) sin θ d θ = ( 1 − e cos E ) 2 ( − sin E ) ( 1 − e cos E ) d E − ( cos E − e ) ( e sin E ) d E = ( 1 − e cos E ) 2 ( − sin E + e sin E cos E − e sin E cos E + e 2 sin E ) d E = − ( 1 − e cos E ) 2 ( 1 − e 2 ) sin E d E = ( 1 − e cos E ) 2 ( 1 − e 2 ) sin E d E ここで左辺を E E E sin 2 θ = 1 − cos 2 θ \sin^2\theta = 1 - \cos^2\theta sin 2 θ = 1 − cos 2 θ cos θ = cos E − e 1 − e cos E \cos\theta = \dfrac{\cos E - e}{1 - e\cos E} cos θ = 1 − e cos E cos E − e
sin 2 θ = 1 − ( cos E − e 1 − e cos E ) 2 = ( 1 − e cos E ) 2 − ( cos E − e ) 2 ( 1 − e cos E ) 2 = ( 1 − 2 e cos E + e 2 cos 2 E ) − ( cos 2 E − 2 e cos E + e 2 ) ( 1 − e cos E ) 2 = 1 − e 2 − cos 2 E + e 2 cos 2 E ( 1 − e cos E ) 2 = ( 1 − e 2 ) − cos 2 E ( 1 − e 2 ) ( 1 − e cos E ) 2 = ( 1 − e 2 ) ( 1 − cos 2 E ) ( 1 − e cos E ) 2 = ( 1 − e 2 ) sin 2 E ( 1 − e cos E ) 2 \begin{aligned}
\sin^2\theta &= 1 - \left(\dfrac{\cos E - e}{1 - e\cos E}\right)^2 \\
&= \dfrac{(1 - e\cos E)^2 - (\cos E - e)^2}{(1 - e\cos E)^2}\\
&= \dfrac{(1 - 2e\cos E + e^2\cos^2 E) - (\cos^2 E - 2e\cos E + e^2)}{(1 - e\cos E)^2}\\
&= \dfrac{1 - e^2 - \cos^2 E + e^2\cos^2 E}{(1 - e\cos E)^2}\\
&= \dfrac{(1 - e^2) - \cos^2 E(1 - e^2)}{(1 - e\cos E)^2}\\
&= \dfrac{(1 - e^2)(1 - \cos^2 E)}{(1 - e\cos E)^2}\\
&= \dfrac{(1 - e^2)\sin^2 E}{(1 - e\cos E)^2}\\
\end{aligned} sin 2 θ = 1 − ( 1 − e cos E cos E − e ) 2 = ( 1 − e cos E ) 2 ( 1 − e cos E ) 2 − ( cos E − e ) 2 = ( 1 − e cos E ) 2 ( 1 − 2 e cos E + e 2 cos 2 E ) − ( cos 2 E − 2 e cos E + e 2 ) = ( 1 − e cos E ) 2 1 − e 2 − cos 2 E + e 2 cos 2 E = ( 1 − e cos E ) 2 ( 1 − e 2 ) − cos 2 E ( 1 − e 2 ) = ( 1 − e cos E ) 2 ( 1 − e 2 ) ( 1 − cos 2 E ) = ( 1 − e cos E ) 2 ( 1 − e 2 ) sin 2 E 0 ≤ θ ≤ π 0 \le \theta \le \pi 0 ≤ θ ≤ π 0 ≤ E ≤ π 0 \le E \le \pi 0 ≤ E ≤ π sin θ ≥ 0 \sin\theta \ge 0 sin θ ≥ 0 sin E ≥ 0 \sin E \ge 0 sin E ≥ 0 0 < e < 1 0 < e < 1 0 < e < 1
sin θ = 1 − e 2 sin E 1 − e cos E \sin\theta = \dfrac{\displaystyle\sqrt{1 - e^2}\sin E}{1 - e\cos E} sin θ = 1 − e cos E 1 − e 2 sin E これを sin θ d θ = ( 1 − e 2 ) sin E ( 1 − e cos E ) 2 d E \sin\theta \,\mathrm{d}\theta = \dfrac{(1 - e^2)\sin E}{(1 - e\cos E)^2} \,\mathrm{d}E sin θ d θ = ( 1 − e cos E ) 2 ( 1 − e 2 ) sin E d E
( 1 − e 2 sin E 1 − e cos E ) d θ = ( 1 − e 2 ) sin E ( 1 − e cos E ) 2 d E d θ = 1 − e 2 1 − e 2 ⋅ 1 1 − e cos E d E d θ = 1 − e 2 1 − e cos E d E \begin{aligned}
\left(\dfrac{\displaystyle\sqrt{1 - e^2}\sin E}{1 - e\cos E}\right) \mathrm{d}\theta &= \dfrac{(1 - e^2)\sin E}{(1 - e\cos E)^2} \,\mathrm{d}E\\
\mathrm{d}\theta &= \dfrac{1 - e^2}{\displaystyle\sqrt{1 - e^2}} \cdot \dfrac{1}{1 - e\cos E} \,\mathrm{d}E\\
\mathrm{d}\theta &= \dfrac{\displaystyle\sqrt{1 - e^2}}{1 - e\cos E} \, \mathrm{d}E\\
\end{aligned} ( 1 − e cos E 1 − e 2 sin E ) d θ d θ d θ = ( 1 − e cos E ) 2 ( 1 − e 2 ) sin E d E = 1 − e 2 1 − e 2 ⋅ 1 − e cos E 1 d E = 1 − e cos E 1 − e 2 d E したがって軌道楕円全体の面積 S = a 2 ( 1 − e 2 ) 2 ∫ 0 π d θ ( 1 + e cos θ ) 2 \displaystyle S = a^2 \left(1 - e^2\right)^2 \int_0^{\pi} \dfrac{\mathrm{d}\theta}{(1 + e\cos\theta)^2} S = a 2 ( 1 − e 2 ) 2 ∫ 0 π ( 1 + e cos θ ) 2 d θ cos θ = cos E − e 1 − e cos E \cos\theta = \dfrac{\cos E - e}{1 - e\cos E} cos θ = 1 − e cos E cos E − e d θ = 1 − e 2 1 − e cos E d E \mathrm{d}\theta = \dfrac{\displaystyle\sqrt{1 - e^2}}{1 - e\cos E} \, \mathrm{d}E d θ = 1 − e cos E 1 − e 2 d E θ \theta θ 0 0 0 π \pi π E E E 0 0 0 π \pi π
S = a 2 ( 1 − e 2 ) 2 ∫ 0 π ( 1 − e cos E 1 − e 2 ) 2 ( 1 − e 2 1 − e cos E d E ) = a 2 1 − e 2 ∫ 0 π ( 1 − e cos E ) d E = a 2 1 − e 2 [ E − e sin E ] 0 π = π a 2 1 − e 2 \begin{aligned}
S &= a^2 \left(1 - e^2\right)^2 \int_0^\pi \left( \dfrac{1 - e\cos E}{1 - e^2} \right)^2 \left( \dfrac{\displaystyle\sqrt{1 - e^2}}{1 - e\cos E} \, \mathrm{d}E \right) \\[10pt]
&= a^2 \displaystyle\sqrt{1 - e^2} \int_0^\pi (1 - e\cos E) \, \mathrm{d}E\\
&= a^2 \displaystyle\sqrt{1 - e^2}\Big[ E - e\sin E \Big]_0^\pi\\
&= \pi a^2 \displaystyle\sqrt{1 - e^2}
\end{aligned} S = a 2 ( 1 − e 2 ) 2 ∫ 0 π ( 1 − e 2 1 − e cos E ) 2 ( 1 − e cos E 1 − e 2 d E ) = a 2 1 − e 2 ∫ 0 π ( 1 − e cos E ) d E = a 2 1 − e 2 [ E − e sin E ] 0 π = π a 2 1 − e 2 一方，第二法則より天体 A \text{A} A C C C T T T S S S
S = ∫ 0 T C d t = C T S = \int_0^T C \, \mathrm{d}t = CT S = ∫ 0 T C d t = C T したがって公転周期 T T T C = 1 2 G M a ( 1 − e 2 ) C = \dfrac{1}{2}\displaystyle\sqrt{GMa(1 - e^2)} C = 2 1 GM a ( 1 − e 2 ) S = π a 2 1 − e 2 S=\pi a^2 \displaystyle \displaystyle\sqrt{1 - e^2} S = π a 2 1 − e 2
T = S C = π a 2 1 − e 2 1 2 G M a ( 1 − e 2 ) = 2 π G M a 3 / 2 T 2 = 4 π 2 G M a 3 \begin{aligned}
T &= \dfrac{S}{C} \\[6pt]
&= \dfrac{\pi a^2 \displaystyle\sqrt{1 - e^2}}{\dfrac{1}{2}\displaystyle\sqrt{GMa(1 - e^2)}} \\[10pt]
&= \dfrac{2\pi}{\displaystyle\sqrt{GM}} a^{3/2}\\
T^2 &= \dfrac{4\pi^2}{GM} a^3
\end{aligned} T T 2 = C S = 2 1 GM a ( 1 − e 2 ) π a 2 1 − e 2 = GM 2 π a 3/2 = GM 4 π 2 a 3 両辺を a 3 a^3 a 3
T 2 a 3 = 4 π 2 G M = const. \dfrac{T^2}{a^3} = \dfrac{4\pi^2}{GM} = \text{const.} a 3 T 2 = GM 4 π 2 = const.