Kepler's laws

Proving Kepler’s laws

2 body interactive system.png

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} $$

  1. Finding the total energy

    $$ \begin{aligned} \mu\ddot{\vec r}&=-\frac{\text dU}{\text dr}\hat r \\ \mu\,(\dot{\vec r}\cdot\ddot{\vec r})&=-\frac{\text dU}{\text dr}\underbrace{\hat r\cdot\dot{\vec r}}_{=\dot r} \\ \mu\,(\dot{\vec r}\cdot\ddot{\vec r})&=-\dot r\frac{\text dU}{\text dr}=-\frac{\text dU}{\text dt} \\ {\color{979A9B}\text{which implies that: }}E&=\frac 12 \mu|\dot{\vec r}|^2+U(\vec r)=\text{Constant} \end{aligned} $$

  2. 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

  3. Since motion is in a plane we can define everything in 2D polar coordinates

    $$ \begin{aligned} E&=\frac 12\mu\left( \dot r^2+r^2\dot\theta^2 \right)+U(r) \\ &=\frac 12\mu \dot r^2+\underbrace{\frac{L_\text{CoM}^2}{2\mu r^2}+U(r)}{U\text{eff}} \\ &=\frac 12 \mu \dot r^2+\underbrace{U_\text{eff}(r)}{\tiny{\begin{matrix} {\color{979A9B}\text{Potential for effective}} \\ {\color{979A9B}\text{1D motion}} \end{matrix}}} \end{aligned} \quad ; \quad \begin{aligned} \vec L\text{CoM}&=\mu\left[ \vec r\times( \dot r\hat r+r\dot\theta\hat\theta ) \right] \\ &=\mu r^2\dot\theta\hat e_z =\text{Constant} \end{aligned} $$

  4. 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

  5. To find $r(\theta)$ we will introduce $u=1/r$ with $\dot u=-\dot r/r^2=-\dot ru^2$ and $\dot \theta=L_\text{CoM}u^2/\mu$

    $$ \begin{aligned} &\rhd \;\left ( \frac{\text du}{\text d\theta} \right)^2=\left( \frac{\dot u}{\dot \theta} \right)^2-\left( \frac{\mu}{L_\text{CoM}} \right)^2\dot r^2=\frac{2\mu}{L_\text{CoM}^2}\left[ E-U\left ( \frac{1}{u}\right) \right]-u^2 \\ {\color{979A9B}\text{We define: }}&U=-\frac{\alpha}{r}=-\alpha u {\color{979A9B}\text{ where }}\alpha \left \{ \begin{aligned} &=-Gm_1m_2 &\text{ for gravity}\qquad\;\; \\ &=-\frac{q_1q_2}{4\pi\epsilon_0} &\;\;\text{ for electrostatics} \end{aligned} \right . \\ &\rhd \; \left ( \frac{\text du}{\text d\theta} \right )^2=\frac{2\mu}{L_\text{CoM}^2}(E+\alpha u)-u^2=\epsilon^2 u_0^2-(u-u_0)^2 \\ {\color{979A9B}\text{Where: }}&u_0=\frac{1}{r_0}=\frac{\alpha \mu}{L_\text{CoM}^2} \quad ; \quad \epsilon^2=1+2\frac{L_\text{CoM}^2E}{\alpha^2\mu} \quad ; \quad \bar{u}=\frac{u-u_0}{\epsilon u_0} \\ &\rhd\; \left ( \frac{\text d\bar u}{\text d\theta} \right )^2=1-\bar u^2 \\ {\color{979A9B}\text{Taking lim: }}&u=u_0 \;, \; r=r_0 \;, \;\theta=\frac{\pi}{2} \; {\color{979A9B}\text{we get: }}\bar u=\cos\theta {\color{979A9B}\text{ thus:}} \\ &\rhd\; u=u_0(1+\epsilon\cos\theta) \quad ; \quad \boxed{r=\frac{r_0}{1+\epsilon\cos\theta}} \end{aligned} $$

  6. 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

  7. 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

    Motion in the effective potential

    Consider a potential $U=-\alpha/r$

    🛩️ Let’s first look when $\alpha>0$

    $$ \frac{\text d U_\text{eff}}{\text dr}=-\frac{L^2}{\mu r^3}+\frac{\alpha}{r^2} $$

    $$ \frac{\text d U^2_\text{eff}}{\text dr^2}=\frac 1 {r^4}\left( \frac{3L^2}\mu-2\alpha r \right) $$

    $$ U_\text{eff}(r_0)=-\frac{\mu \alpha^2}{2L^2}=-\frac{\alpha}{2r_0}=\frac 12 U(r_0) $$

    $$ \lim_{r\to\text{small}}(U_\text{eff})\sim \frac{L^2}{2\mu r^2} \quad ; \quad \lim_{r\to0}=\infin \quad ; \quad \lim_{r\to\text{large}}\sim -\frac{\alpha}{r} \quad ; \quad \lim_{r\to\infin}=0^- $$

    The curve that we expect

    The curve that we expect

    🖥️ Properties

    1. When $E=-\mu\alpha^2/(2L^2), \epsilon=0$ the particle is at the minimum potential and in this case there is stable circular orbit with $r=r_0$. There will be small oscillations about the circular orbit with effective spring constant $k=L^2/(\mu r^4_0)$. Here the orbit is nearly circular
    2. When $-\mu \alpha^2/(2L^2)<E<0, 0<\epsilon<1$ the particle will be in close approach, it will roll around in the bottom of the potential but does not have sufficient energy to escape. Min and Max can be calculated by setting $\dot r=0$. Here the radius varies, so it is an elliptical orbit between an $r_\text{min}$ (pericenter) and $r_\text{max}$(apocenter)
    3. $E=0,\epsilon=1$ the particle comes from infinity, rolls down to $r_0$ , has no velocity at $r_0$ then goes back to infinity reaching it with zero velocity, this is called an unbound parabolic orbit (think of a gravity assist)
    4. $E>0, \epsilon>1$ same as 3. except with more energy, called an unbound hyperbolic orbit.

    ✈️ Now when $\alpha<0$

    🚧 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 $r=0$, $U_\text{eff}\to\infin$

    Here at $r=0$, $U_\text{eff}\to\infin$

    Stability of circular orbits

    $$ k=\frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0) $$

    🖥️ General properties

    $$ \frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0)>0 $$

    desmos-graph.png

    $$ \frac{\text d^2 U_\text{eff}}{\text dr^2}(r_0)<0 $$

    desmos-graph (1).png