物理

❴栖岸计划❵Binet's formula

$ \text{天体运动} $

$ \text{Binet formula} \quad \frac{d^2 u}{d\theta^2} + u = -\frac{m}{l^2} \frac{d}{du} V\left(\frac{1}{u}\right) $

$ \mathcal{E} = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) + V $

$ \dot{r} = \sqrt{ \frac{2}{m} \left( E - V - \frac{L^2}{2mr^2} \right) } $

$ dt = \frac{dr}{ \sqrt{ \frac{2}{m} \left( E - V - \frac{L^2}{2mr^2} \right) } } $

$ L = m r^2 \dot{\theta} $

$ dt = \frac{m r^2}{L} d\theta $

$ \frac{m r^2}{L} d\theta = \frac{dr}{ \sqrt{ \frac{2}{m} \left( E - V - \frac{L^2}{2mr^2} \right) } } $

$ d\theta = \frac{dr}{ \frac{m r^2}{L} \sqrt{ \frac{2}{m} \left( E - V - \frac{L^2}{2mr^2} \right) } } $

$ dt = \frac{dr}{ \sqrt{ \frac{2mE}{L^2} r^4 + \frac{2mV}{L^2} r^4 + r^2}} $

$ \text{设 } u = \frac{1}{r} \quad \text{平动化力 } V(r) = \mathcal{E} - \frac{k}{r} $

$ d\theta = \frac{ du \times r^2 }{ r^2 \sqrt{ \frac{2mE}{L^2} + \frac{2mV}{L^2} + \frac{1}{r^2}}} $

$ = \frac{ du }{ \sqrt{ \frac{2mE}{L^2} + \frac{2mV}{L^2} +u^2}}$

$ = \frac{ du }{ \sqrt{ \frac{2mE}{L^2} + \frac{2mk}{L^2} u + u^2 } } $

$ V(r) = -\frac{k}{r} = -k u $

$ \int \frac{dx}{\sqrt{a + \beta x + \gamma x^2}} = \frac{1}{\sqrt{-\gamma}} \arccos\left( \frac{-\beta + 2\gamma x}{\sqrt{q}} \right) $

$ q = \beta^2 - 4\alpha\gamma $

$ \alpha = \frac{2mE}{L^2}, \quad \beta = \frac{2mk}{L^2}, \quad \gamma = 1 $

$ \theta = \theta' - \arccos \frac{ \frac{L^2 u}{mk} - 1 }{ \sqrt{ 1 + \frac{2E L^2}{mk^2} } } $

$ u = \frac{mk}{L^2} \left[ 1 + \sqrt{ 1 + \frac{2E L^2}{mk^2} } \cos(\theta - \theta') \right] $

$ \text{极坐标下 } \frac{1}{r} = C [1 + e \cos(\theta - \theta')] $

$C=\frac{mk}{L^2}=a(1-e^2)$

$ e = \sqrt{ 1 + \frac{2E L^2}{mk^2} } $ > 1双曲线

$ e = 1 \quad \text{抛物线} $

 e < 1 椭圆

$ e = 0 \quad \text{圆} $

$ dt = \frac{dr}{ \sqrt{ \frac{2}{m} \left( E - V - \frac{L^2}{2mr^2} \right) } } $

$ t = \sqrt{ \frac{m}{2} } \int_{r_0}^{r} \frac{dr}{ \sqrt{ \frac{k}{r} + E - \frac{L^2}{2mr^2} } } $

$ \text{令 } u = \frac{1}{r} $

$ t = \sqrt{ \frac{m}{2} } \int_{r_0}^{r} \frac{ du \times r^2 }{ \sqrt{ k u + E - \frac{L^2}{2m} u^2 } } $

$ \text{进上式} $

$ dt = \frac{m r^2}{L} d\theta $

$ \frac{1}{r} = 1 + e \cos(\theta - \theta') $

$ dt = \frac{m}{L} \frac{d\theta}{ [1 + e \cos(\theta - \theta')]^2 } \times \frac{L^4}{m^2 k^2} $

$ t = \frac{L^3}{m k^2} \int_{\theta_0}^{\theta} \frac{d\theta}{ [1 + e \cos(\theta - \theta')] } $

$ e = 1 \quad \text{抛物线} $

$ t = \frac{L^3}{m k^2} \int_{\theta_0}^{\theta} \frac{d\theta}{ [1 + \cos(\theta - \theta')]^2 } $

$ 1 + \cos\theta = 2 \cos^2 \frac{\theta}{2} $

$ \cos^2\theta = \frac{1}{2} (1 + \cos 2\theta) $

$ \theta' = 0 \text{时} $

$ t = \frac{L^3}{m k^2} \int_{\theta_0}^{\theta} \sec^4\theta \, d\theta $

$ x = \tan\theta \quad \cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}} $

$ \sec^4\theta = (1 + x^2)^2 $

$ t = \frac{L^3}{m k^2} \int_0^{\tan\theta} (1 + x^2)^2 dx \times \cos^2\theta $

$ = \frac{L^3}{m k^2} \int_0^{x} 1 + x^2 dx $

$ = \frac{L^3}{m k^2} \left[ x + \frac{1}{3} x^3 \right] \Bigg|_0^{x = \tan\theta} $

$ = \frac{L^3}{m k^2} \left[ \tan\frac{\theta}{2} + \frac{1}{3} \tan^3\frac{\theta}{2} \right] $

e < 1 

$ \text{引入变量 } \psi \quad r = a (1 - e \cos\psi) $

$ t = \sqrt{ \frac{m}{2} } \int_{r_0}^{r} \frac{dr}{ \sqrt{ \frac{k}{r} + E - \frac{L^2}{2mr^2} } } $

$ \text{由 } ep = a (1 - e^2) = \frac{L^2}{G M m} = \frac{L^2}{mk} $

$ a (1 - e^2) = \frac{L^2}{G M m^2} = \frac{L^2}{mk} $

$ \text{先算上下限 } r $

$ t = \sqrt{ \frac{m}{2} } \int_{r_0}^{r} \frac{r dr}{ \sqrt{ G M m r + E r^2 - \frac{L^2}{2m} } } $

$ t = \sqrt{ \frac{m}{2 \cdot G M m} } \int_{r_0}^{r} \frac{r dr}{ \sqrt{ r + \frac{E r^2}{G M m} - \frac{L^2}{2 G M m^2} } } $

$ t = \sqrt{ \frac{m}{2k} } \int_{r_0}^{r} \frac{r dr}{ \sqrt{ r + \frac{E r^2}{k} - \frac{a(1-e^2)}{2} } } $

$ e = \frac{c}{a} = \frac{ \sqrt{b^2 + a^2} }{a} $

$ \text{由上述推导中，在轨道运动中} $

$ e = \sqrt{ 1 + \frac{2E L^2}{mk^2} } $

$ \frac{b^2 + a^2}{a^2} = 1 + \frac{2E L^2}{mk^2} $

$ \frac{b^2}{a^2} = \frac{2E L^2}{mk^2} $

$ 2E p = \frac{b^2}{a^2} $

$ 2E a (1 - e^2) = \frac{b^2}{a^2} k $

$ 1 - e^2 = \frac{a^2 - a^2 - b^2}{a^2} = -\frac{b^2}{a^2} $

$ -2E a = k $

$ E = -\frac{k}{2a} $

$ t = \sqrt{ \frac{m}{2k} } \int_{r_0}^{r} \frac{r dr}{ \sqrt{ r - \frac{r^2}{2a} - \frac{a(1-e^2)}{2} } } $

$ \text{代入 } r = a (1 - e \cos\psi) $

$ t =\sqrt{ \frac{m}{2k} }\int_0^{ψ}\frac{ a(1 - e \cosψ) d[a(1 - e \cosψ)] }{ \sqrt{ a(1 - e \cosψ) - \frac{a^2(1 - e \cosψ)^2}{2a} - \frac{a(1 - e^2)}{2} } } $

$ = \sqrt{ \frac{m a^3}{k} } \int_0^{\psi} (1 - e \cos\psi) d\psi $

$ \text{注：} \psi = \frac{ \text{area} }{ \frac{1}{2} a^2 } \Rightarrow \psi = \frac{ \text{area} }{ \frac{1}{2} a^2 } $

$ L = 2\pi a^2 \sqrt{ \frac{m}{k} } $

$ \text{Laplace-Lunge-Lenz 矢量} $

$ \vec{p} \times \vec{L} = f(\vec{r}) \frac{m}{r} \left[ \vec{r} \times (\vec{r} \times \dot{\vec{r}}) \right] $

$ = f(\vec{r}) \frac{m}{r} \left[ \vec{r} \cdot \dot{\vec{r}} - \frac{ \vec{r} \cdot \vec{r} }{ r^2 } \dot{\vec{r}} \right] $

$ \vec{r} \cdot \dot{\vec{r}} = r \cdot \dot{r} $

$ \dot{\vec{p}} \times \vec{L} = -f(\vec{r}) \frac{m}{r} \left[ r^2 ( \ddot{\vec{r}} - \frac{ \dot{\vec{r}} \cdot \dot{\vec{r}} }{ r^2 } \vec{r} ) \right] $

$ \frac{d}{dt} ( \vec{p} \times \vec{L} ) = -f(\vec{r}) \frac{m}{r} \left( \ddot{\vec{r}} - \frac{ \dot{\vec{r}} \cdot \dot{\vec{r}} }{ r^2 } \vec{r} \right) $

$ f(\vec{r}) = \frac{-k}{r^2} $

$ \frac{d}{dt} ( \vec{p} \times \vec{L} ) = m k \frac{d}{dt} \left( \frac{ \vec{r} }{ r } \right) $

$ \frac{d}{dt} \left[ \vec{p} \times \vec{L} - m k \frac{ \vec{r} }{ r } \right] = 0 $

$ \vec{A} = \vec{p} \times \vec{L} - m k \frac{ \vec{r} }{ r } = 0 $

$ \vec{A} \times \vec{L} = 0 $

$ \vec{A} \cdot \vec{r} = A r \cos\theta = \vec{r} \cdot \left[ \vec{p} \times \vec{L} - m k \frac{ \vec{r} }{ r } \right] $

$ \vec{r} \cdot ( \vec{p} \times \vec{L} ) = L \cdot ( \vec{r} \times \vec{p} ) = L^2 $

$ \vec{r} \cdot \left( \frac{ m k \vec{r} }{ r } \right) = m k r $

$ A r \cos\theta = L^2 - m k r $

$ r = \frac{m k}{L^2} \left( 1 + \frac{A}{m k} \cos\theta \right) $