# 克卜勒定律

## 克卜勒定律

### 克卜勒第二定律

$S_{AB}=S_{CD}=S_{EK}$

### 克卜勒第三定律

$\frac{a^3}{\tau^2}=K$

## 數學推導：由牛頓萬有引力定律導出克卜勒定律

### 克卜勒第二定律推導

$\boldsymbol{F}= - G \frac{mM}{r^2}\ \hat{\boldsymbol{r}}$

$\boldsymbol{F}=m\ddot{\boldsymbol{r}}$

$\ddot{\boldsymbol{r}}= - G \frac{M}{r^2}\ \hat{\boldsymbol{r}}$(1)

$\dot{\boldsymbol{r}}=\dot{r}\hat{\boldsymbol{r}} +r\dot{\theta}\hat{\boldsymbol{\theta}}$
$\ddot{\boldsymbol{r}} = \left(\ddot{r}\hat{\boldsymbol{r}} +\dot{r}\frac{\mathrm{d}\hat{\boldsymbol{r}}}{\mathrm{d}t}\right) + \left(\dot{r}\dot{\theta}\hat{\boldsymbol{\theta}} + r\ddot{\theta}\hat{\boldsymbol{\theta}} + r\dot{\theta}\frac{\mathrm{d}\hat{\boldsymbol{\theta}}}{\mathrm{d}t}\right) = (\ddot{r} - r\dot{\theta}^2) \hat{\boldsymbol{r}} + (r\ddot{\theta} + 2\dot{r}\dot{\theta}) \hat{\boldsymbol{\theta}}$(2)

$\frac{\mathrm{d}\hat{\boldsymbol{r}}}{\mathrm{d}t} = \dot{\theta} \hat{\boldsymbol{\theta}}$
$\frac{\mathrm{d}\hat{\boldsymbol{\theta}}}{\mathrm{d}t} = - \dot{\theta} \hat{\boldsymbol{r}}$

$(\ddot{r} - r\dot{\theta}^2) \hat{\boldsymbol{r}} + (r\ddot{\theta} + 2\dot{r} \dot{\theta}) \hat{\boldsymbol{\theta}}= - \frac{GM}{r^2}\hat{\boldsymbol{r}}$

$\ddot{r} - r\dot{\theta}^2 = - \frac{GM}{r^2}$(3)
$r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0$(4)

$\dot{\ell}=mr (2\dot{r}\dot{\theta}+r\ddot{\theta})=0$

$\Delta A=\int_{t_1}^{t_2}\frac{1}{2}\cdot r\cdot r\dot \theta\cdot \mathrm{d}t=\int_{t_1}^{t_2}\frac{\ell}{2m}\mathrm{d}t=\frac{\ell}{2m}\cdot(t_2-t_1)$

### 克卜勒第一定律推導

$\dot{\theta} =\frac{\ell}{mr^2}=\frac{\ell u^2}{m}$

$\frac{\mathrm{d}}{\mathrm{d}t}=\dot{\theta}\frac{\mathrm{d}}{\mathrm{d}\theta}=\frac{\ell u^2}{m} \frac{\mathrm{d}}{\mathrm{d}\theta}$

$\dot{r}=\frac{\ell u^2}{m} \frac{\mathrm{d}}{\mathrm{d}\theta}\frac{1}{u} = - \frac{\ell u^2}{m}\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}= - \frac{\ell }{m}\frac{\mathrm{d}u}{\mathrm{d}\theta}$

$\ddot{r} =\frac{\ell u^2}{m} \frac{\mathrm{d}\dot{r}}{\mathrm{d}\theta}=\frac{\ell u^2}{m} \frac{\mathrm{d}}{\mathrm{d}\theta} ( - \frac{\ell}{m}\frac{\mathrm{d}u}{\mathrm{d}\theta})= - \frac{\ell^2 u^2}{m^2}\frac{\mathrm{d}^2 u}{\mathrm{d}\theta^2}$

$- \frac{\ell^2 u^2}{m^2}\frac{\mathrm{d}^2 u}{\mathrm{d}\theta^2} - \frac{\ell^2 u^3}{m^2} = - GMu^2$

$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = \frac{GMm^2}{\ell^2}$

$u=\frac{GMm^2}{\ell^2}$

$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = 0$

$u=C\ \cos(\theta - \theta_0)$

$u=\frac{GMm^2}{\ell^2}+C\ \cos(\theta-\theta_0)$

$\frac{1}{r}=\frac{GMm^2}{\ell^2}(1+ e\ \cos{\theta})$

### 克卜勒第三定律推導

$\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{\ell}{2m}$

$\tau=\frac{2m\pi ab}{\ell}$(5)

$a=(r_A+r_B) / 2$(6)
$b=\sqrt{r_A\ r_B}$(7)

$E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2 - G\frac{mM}{r}$

$\dot{r}=0$

$E=\frac{1}{2}mr^2\dot{\theta}^2 - G\frac{mM}{r}= \frac{\ell^2}{2mr^2} - G\frac{mM}{r}$

$r^2+\frac{GmM}{E}r - \frac{\ell^2}{2mE}=0$

$r_A=\left( - \frac{GmM}{E} - \sqrt{\left(\frac{GmM}{E}\right)^2+\frac{2\ell^2}{mE}}\right) / 2$
$r_B=\left( - \frac{GmM}{E} + \sqrt{\left(\frac{GmM}{E}\right)^2+\frac{2\ell^2}{mE}}\right) / 2$

$a= - \frac{GmM}{2E}$
$b=\frac{\ell}{\sqrt{ - 2mE}}=\frac{\ell}{m}\frac{\sqrt{a}}{\sqrt{GM}}$

$\tau=\frac{2\pi a^{3/2}}{\sqrt{GM}}$

Q.E.D.