Kepler's laws
- Planetary orbits around elliptical with the star at one of the two foci
- Lines between the planets and the star sweep out equal areas in equal intervals
- The orbit period of a planet squared is prop. to the cube of the semi-major axis
What we know:
$$ \begin{aligned} M&=m_1+m_2 \\ \mu&=\frac{m_1m_2}{m_1+m_2} \\ \vec R&=\frac{m_1\vec r_1+m_2\vec r_2}{M} \\ \vec r&=\vec r_1-\vec r_2 \\ \dot R&=0 \text{ (in CoM frame)} \\ \vec L_\text{CoM}&=\mu\,(\vec r\times\dot{\vec r}) \\ \mu \ddot{\vec r}&=\vec F=-\frac{\text dU}{\text dr}\,\hat r
\end{aligned} $$
Finding the total energy
$$ \begin{aligned} \mu\ddot{\vec r}&=-\frac{\text dU}{\text dr}\hat r \\ \mu\,(\dot{\vec r}\cdot\ddot{\vec r})&=-\frac{\text dU}{\text dr}\underbrace{\hat r\cdot\dot{\vec r}}_{=\dot r} \\ \mu\,(\dot{\vec r}\cdot\ddot{\vec r})&=-\dot r\frac{\text dU}{\text dr}=-\frac{\text dU}{\text dt} \\ {\color{979A9B}\text{which implies that: }}E&=\frac 12 \mu|\dot{\vec r}|^2+U(\vec r)=\text{Constant} \end{aligned} $$
Momentum in the CoM frame
$$ \begin{aligned} \vec L_\text{CoM}&=\mu\,(\vec r\times\dot{\vec r}) \\ \frac{\text d \vec L_\text{CoM}}{\text dt}&=\mu\,(\dot{\vec r}\times\dot{\vec r}+\vec r\times\ddot{\vec r})=-r\times\frac{\text dU}{\text dr}=0 \\ {\color{979A9B}\text{Which implies: }}\;1.\qquad\vec L_\text{CoM}&=\text{Constant} \\ 2.\;\;\; \vec L_\text{CoM}\cdot \dot{\vec r}&= \mu\left [\vec{r}\times(\dot{\vec r}\cdot \dot{\vec r})\right ] =0 {\color{979A9B}\text{ thus: }} \vec L_\text{CoM}\perp\dot{\vec r} \end{aligned} $$
💡 This proves part of Kepler’s First law as motion has to be on a 2D plane
Since motion is in a plane we can define everything in 2D polar coordinates
$$ \begin{aligned} E&=\frac 12\mu\left( \dot r^2+r^2\dot\theta^2 \right)+U(r) \\ &=\frac 12\mu \dot r^2+\underbrace{\frac{L_\text{CoM}^2}{2\mu r^2}+U(r)}{U\text{eff}} \\ &=\frac 12 \mu \dot r^2+\underbrace{U_\text{eff}(r)}{\tiny{\begin{matrix} {\color{979A9B}\text{Potential for effective}} \\ {\color{979A9B}\text{1D motion}} \end{matrix}}} \end{aligned} \quad ; \quad \begin{aligned} \vec L\text{CoM}&=\mu\left[ \vec r\times( \dot r\hat r+r\dot\theta\hat\theta ) \right] \\ &=\mu r^2\dot\theta\hat e_z =\text{Constant} \end{aligned} $$
Consider a segment $\text d\vec r$ and $\vec r+\text d\vec r$ with $\text d\theta$ the angle between them
$$ \begin{aligned} \text dA&=\frac 12 r^2\text d\theta \\ \frac {\text dA} {\text dt}&=\frac 12 r^2\dot\theta =\frac{L_\text{CoM}}{2\mu} \\ &=\text{Constant} \end{aligned} $$
💡 The conservation of momentum leads to the proving Kepler’s Second law
To find $r(\theta)$ we will introduce $u=1/r$ with $\dot u=-\dot r/r^2=-\dot ru^2$ and $\dot \theta=L_\text{CoM}u^2/\mu$
$$ \begin{aligned} &\rhd \;\left ( \frac{\text du}{\text d\theta} \right)^2=\left( \frac{\dot u}{\dot \theta} \right)^2-\left( \frac{\mu}{L_\text{CoM}} \right)^2\dot r^2=\frac{2\mu}{L_\text{CoM}^2}\left[ E-U\left ( \frac{1}{u}\right) \right]-u^2 \\ {\color{979A9B}\text{We define: }}&U=-\frac{\alpha}{r}=-\alpha u {\color{979A9B}\text{ where }}\alpha \left \{ \begin{aligned} &=-Gm_1m_2 &\text{ for gravity}\qquad\;\; \\ &=-\frac{q_1q_2}{4\pi\epsilon_0} &\;\;\text{ for electrostatics} \end{aligned} \right . \\ &\rhd \; \left ( \frac{\text du}{\text d\theta} \right )^2=\frac{2\mu}{L_\text{CoM}^2}(E+\alpha u)-u^2=\epsilon^2 u_0^2-(u-u_0)^2 \\ {\color{979A9B}\text{Where: }}&u_0=\frac{1}{r_0}=\frac{\alpha \mu}{L_\text{CoM}^2} \quad ; \quad \epsilon^2=1+2\frac{L_\text{CoM}^2E}{\alpha^2\mu} \quad ; \quad \bar{u}=\frac{u-u_0}{\epsilon u_0} \\ &\rhd\; \left ( \frac{\text d\bar u}{\text d\theta} \right )^2=1-\bar u^2 \\ {\color{979A9B}\text{Taking lim: }}&u=u_0 \;, \; r=r_0 \;, \;\theta=\frac{\pi}{2} \; {\color{979A9B}\text{we get: }}\bar u=\cos\theta {\color{979A9B}\text{ thus:}} \\ &\rhd\; u=u_0(1+\epsilon\cos\theta) \quad ; \quad \boxed{r=\frac{r_0}{1+\epsilon\cos\theta}} \end{aligned} $$
We can convert the answer to cartesian coordinate
$$ \begin{aligned} r^2=(r_0-\epsilon x)&\rightarrow x^2+y^2=r_0^2-2\epsilon r_0 x+\epsilon^2 x^2 \\ x^2+\frac{2\epsilon r_0}{1-\epsilon^2} x+\frac{y^2}{1-\epsilon^2}=\frac{r_0^2}{1-\epsilon^2} &\rightarrow \left ( x+ \frac{\epsilon r_0}{1-\epsilon^2} \right)^2+\frac{y^2}{1-\epsilon^2}=\frac{r_0^2}{(1-\epsilon^2)^2} \\ {\color{979A9B}\text{if }}0\le\epsilon <1 {\color{979A9B}\text{ then: }}\quad 1&=\frac{\left( x+\frac{\epsilon r_0}{1-\epsilon} \right)^2}{\left( \frac{r_0}{1-\epsilon^2}\right)^2}+\frac{y^2}{\left( \frac{r_0}{\sqrt{1-\epsilon^2}} \right)^2} \end{aligned} \\ \boxed{
\frac{\overbrace{X^2}^{\left( x+\frac{\epsilon r_0}{1-\epsilon} \right)^2}}{\underbrace{a^2}{\left( \frac{r_0}{1-\epsilon^2}\right)^2}}+\frac{\overbrace{Y^2}^{y^2}}{\underbrace{b^2}{\left( \frac{r_0}{\sqrt{1-\epsilon^2}} \right)^2}}=1
} $$
💡 We find an equation of an ellipse with semi-major axis $a$ and semi-minor axis $b$. We also get $b=\sqrt{1-\epsilon^2}\, a$ and $\epsilon^2=1-(b/a)^2$. $\epsilon$ is the eccentricity. Therefore orbits are elliptical, which proves Kepler’s First law
Finding the period
$$ 1-\epsilon^2=-2\frac{L_\text{CoM}^2E}{\alpha^2\mu}=-2\frac{E r_0}{\alpha} \quad ; \quad E<0 \quad ; \quad 0\le \epsilon_0<1 \\ {\color{979A9B}\text{ Implies: }} -\frac{\alpha^2\mu}{2L^2_\text{CoM}}=-\frac{\alpha}{2r_0}\le E<0 \\ {\color{979A9B}\text{Using K2: }}\quad\frac{\text dA}{\text dt}=\frac{L_\text{CoM}}{2\mu} \;{\color{979A9B}\text{ thus:}} \\ \frac{L_\text{CoM}T}{2\mu}=A=\pi ab=\pi a^2\sqrt{1-\epsilon^2} \\ \frac{L}{\mu}=\left( \frac{\alpha}{\mu} \right)^\frac{1}{2} \underbrace{\sqrt{r_0}}{r_0=L^2\text{CoM}/(\alpha \mu)}=\sqrt{\frac{\alpha}{\mu}}\sqrt{a}\sqrt{1-\epsilon^2} \\ \boxed{T=2\pi\frac{\mu}{L_\text{CoM}}a^2\sqrt{1-\epsilon^2}=2\pi a^\frac 32 \sqrt{\frac{\mu}{\alpha}}}
$$
💡 Hence, we find that Kepler’s First and Second law lead naturally to Kepler’s Third law
Consider a potential $U=-\alpha/r$
$$ \frac{\text d U_\text{eff}}{\text dr}=-\frac{L^2}{\mu r^3}+\frac{\alpha}{r^2} $$
$$ \frac{\text d U^2_\text{eff}}{\text dr^2}=\frac 1 {r^4}\left( \frac{3L^2}\mu-2\alpha r \right) $$
$$ U_\text{eff}(r_0)=-\frac{\mu \alpha^2}{2L^2}=-\frac{\alpha}{2r_0}=\frac 12 U(r_0) $$
$$ \lim_{r\to\text{small}}(U_\text{eff})\sim \frac{L^2}{2\mu r^2} \quad ; \quad \lim_{r\to0}=\infin \quad ; \quad \lim_{r\to\text{large}}\sim -\frac{\alpha}{r} \quad ; \quad \lim_{r\to\infin}=0^- $$
The curve that we expect
🚧 This is impossible in Newtonian gravity but is possible in electrostatics for the scattering of particles with the same sign for the charge. It is a classical model for Rutherford scattering of $\alpha$ particles off a nuclear.
$$ \lim_{r\to\infin}(U_\text{eff})=0^+ $$
Here at $r=0$, $U_\text{eff}\to\infin$
$$ k=\frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0) $$
$$ \frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0)>0 $$
$$ \frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0)<0 $$