> [!key-idea] > Kepler's Law's of Planetary Motion are: > 1) Bound motion of two body systems move in elliptical orbits. $r(\phi) = \frac{a \, (1 - \varepsilon^{2})}{1 + \varepsilon \cos \phi} \hspace{1.5cm} a (1 - \varepsilon^{2}) = \frac{L^{2}}{\mu k} \hspace{1.5cm} \varepsilon = \sqrt{1 + \frac{2 E L^{2}}{\mu^{2} k^{2}}}$ > 2) The radius vector of the orbiting body sweeps out equal areas during equal intervals of time. $\frac{\mathrm{d} A}{\mathrm{d} t} = \frac{L}{2 \mu} \hspace{1cm} \text{where} \hspace{1cm} L = \mu r^{2} \dot{\phi}$ > 3) Period-Radius Relationship: $P^{2} \propto a^{3}$ ## Parameters for Orbits *(https://en.wikipedia.org/wiki/Orbital_elements)* For bodies bound in an orbit, they are constrained to a mutual plane ($xy$-plane) with a shared angular momentum ($z$-direction). We can describe their motion through elliptical orbits and Euler Angles $(\Omega \, , \, i \, , \, \omega) \equiv (\alpha \, , \, \beta \, , \, \gamma)$ from a fixed observational plane ($XY$-plane). The following parameters with simplify these orbital systems such that they solvable. - Shape of the Orbit - $a \equiv$ semi-major axis - $\varepsilon \equiv$ eccentricity - $b \equiv$ semi-minor axis $=a\sqrt{1 - \varepsilon^{2}}$ - Orientation of the Orbit (Euler Angles) - $i \equiv$ inclination angle (see [[Mnemonics#Inclination Angle|mnemonic]]) - $\Omega \equiv$ longitude of ascending node - $\omega \equiv$ argument/longitude of periapse - Time Position - $T \equiv$ time of reference, usually periapse - $P \equiv$ period of orbit - $\phi(t) \equiv$ true anomaly - physical angle from periaspe of the object from center of mass - $\Psi(t) \equiv$ eccentric anomaly - handy variable that lets us get time dependence - $\omega t \equiv$ mean anomaly - convenient but fictional angle that varies linearly with time **General Elliptical Orbit** ![[ellipse.png|align:center]] **Euler Angles** ![[eulerAngles_orbit.png|aling:center|500]] > [!note] Another definition of Euler Angles... > ![[eulerAngles.png]] > > - $\alpha$ or $\phi$ (precession) represents a rotation around the _z_ axis, > - $\beta$ or $\theta$ (nutation) represents a rotation around the _N_ or x′ axis -- inclination > - $\gamma$ or $\psi$ (intrinsic rotation) represents a rotation around the _Z_ or z″ axis. ## Kepler's 1st Law Bound motion of two body systems move in elliptical orbits. $ r(\phi) = \frac{a \, (1 - \varepsilon^{2})}{1 + \varepsilon \cos \phi} $ > [!derivation] Proof of Statement (1) > > Using [[Center of Mass & Relative Coordinates]] with Lagrangian Mechanics... > $ \begin{align} > T &= \frac{1}{2} \mu \left( \dot{r}^{2} + r^{2} \dot{\phi}^{2} \right) \\ > U &= - \frac{G M \mu}{r} = - \frac{k}{r} \\ > \\ > \mathcal{L} &= T - U = \frac{1}{2} \mu \left( \dot{r}^{2} + r^{2} \dot{\phi}^{2} \right) + \frac{k}{r} > \end{align} > $ > ...we can find the equations of motion for $r$ and $\phi$ through the Euler-Lagrange equation. > > **Equation of Motion for $\phi(t)$ :** > > $ \begin{align} > \frac{\mathrm{d} }{\mathrm{d} t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \right) - \frac{\partial \mathcal{L}}{\partial \phi} &= 0 \\ > \frac{\mathrm{d} }{\mathrm{d} t} \left( \mu r^{2} \dot{\phi} \right) &= 0 > \hspace{1cm} \Rightarrow \hspace{1cm} > \frac{\mathrm{d} L}{\mathrm{d} t} = 0 > \hspace{1cm} \text{where} \hspace{1cm} > L \equiv \mu r^{2} \dot{\phi} > \end{align} > $ > This is a statement of the **Conservation of Angular Momentum ($L$)**. Re-writing our Lagrangian in terms of $L$... > $ > \mathcal{L} = \frac{1}{2} \mu \dot{r}^{2} + \frac{L^{2}}{2 \mu r^{2}} + \frac{k}{r} > $ > > **Equation of Motion for $r(t)$ :** > > $ \begin{align} > \frac{\mathrm{d} }{\mathrm{d} t} \left( \frac{\partial \mathcal{L}}{\partial \dot{r}} \right) - \frac{\partial \mathcal{L}}{\partial r} &= 0 \\ > \frac{\mathrm{d} }{\mathrm{d} t} \left( \mu \dot{r} \right) - \left( \mu r \dot{\phi}^{2} - \frac{k}{r^{2}} \right) &= 0 \\ > \frac{\mathrm{d} }{\mathrm{d} t} \left( \mu \dot{r} \right) - \mu r \dot{\phi}^{2} + \frac{k}{r^{2}} &= 0 \\ > \mu \ddot{r} - \frac{L^{2}}{\mu r^{3}} + \frac{k}{r^{2}} &= 0 > \end{align} > $ > $ > \left[ \hspace{1cm} \begin{align} > \dot{r} = \frac{\mathrm{d} r}{\mathrm{d} t} &= \frac{\mathrm{d} \phi}{\mathrm{d} t} \frac{\mathrm{d} r}{\mathrm{d} \phi} \hspace{1cm} \textcolor{gray}{\left\{ \dot{\phi} = \frac{\mathrm{d} \phi}{\mathrm{d} t} = \frac{L}{\mu r^{2}} \right\}} \\ > &= \frac{L}{\mu r^{2}} \frac{\mathrm{d} r}{\mathrm{d} \phi} \\ > \\ > \ddot{r} = \frac{\mathrm{d}^{2} r}{\mathrm{d} t^{2}} &= \frac{\mathrm{d} }{\mathrm{d} t} \left( \frac{\mathrm{d} r}{\mathrm{d} t} \right) \\ > &= \frac{\mathrm{d} \phi}{\mathrm{d} t} \frac{\mathrm{d} }{\mathrm{d} \phi} \left( \frac{\mathrm{d} \phi}{\mathrm{d} t} \frac{\mathrm{d} r}{\mathrm{d} \phi} \right) \\ > &= \frac{L^{2}}{\mu^{2} r^{2}} \frac{\mathrm{d} }{\mathrm{d} \phi} \left( \frac{1}{r^{2}} \frac{\mathrm{d} r}{\mathrm{d} \phi} \right) > \end{align} > \hspace{1cm} \right] > $ > > $ \begin{align} > \mu \ddot{r} - \frac{L^{2}}{\mu r^{3}} + \frac{k}{r^{2}} &= 0 \\ > \mu \left[ \frac{L^{2}}{\mu^{2} r^{2}} \frac{\mathrm{d} }{\mathrm{d} \phi} \left( \frac{1}{r^{2}} \frac{\mathrm{d} r}{\mathrm{d} \phi} \right) \right] - \frac{L^{2}}{\mu r^{3}} + \frac{k}{r^{2}} &= 0 \\ > \frac{\mathrm{d} }{\mathrm{d} \phi} \left( \frac{1}{r^{2}} \frac{\mathrm{d} r}{\mathrm{d} \phi} \right) - \frac{1}{r} + \frac{\mu k}{L^{2}} &= 0 \\ > \frac{\mathrm{d} }{\mathrm{d} \phi} \left( - \frac{\mathrm{d} (\tfrac{1}{r})}{\mathrm{d} \phi} \right) - \frac{1}{r} + \frac{\mu k}{L^{2}} &= 0 \\ > - \frac{\mathrm{d}^{2} W}{\mathrm{d} \phi^{2}} - W + C &= 0 \hspace{1cm} \text{where} \hspace{1cm} > W(r) = \frac{1}{r} \hspace{1cm} > C = \frac{\mu k}{L^{2}} \\ > \frac{\mathrm{d}^{2} W}{\mathrm{d} \phi^{2}} + W - C &= 0 \\ > \frac{\mathrm{d}^{2} Y}{\mathrm{d} \phi^{2}} + Y &= 0 \hspace{1cm} \text{where} \hspace{1cm} Y(r) = W(r) - C \\ > \end{align} > $ > This is the differential equation for the simple harmonic oscillator equation. After simplifying the solution form... > $ > Y(r) = B \cos (\phi - \phi_{0}) = \frac{1}{r} - \frac{\mu k}{L^{2}} > \hspace{1cm} \Rightarrow \hspace{1cm} \begin{aligned}[t] > \frac{1}{r} \left(\frac{L^{2}}{\mu k}\right) &= 1 + \left(\frac{L^{2} B}{\mu k}\right) \cos (\phi - \phi_{0}) > \end{aligned} > $ > ...we can correlate the equation structure to a conic section with the focus at the origin. > > ![[conicSections.png|align:center]] > > 1) If $\varepsilon = 0$, then we have bounded motion in a *circular orbit*. $\left( \text{radius} = r = a = \frac{L^{2}}{G M \mu^{2}} \right)$ > 2) If $0 < \varepsilon < 1$, then we have bounded elliptical motion. > 3) If $\varepsilon = 1$, then we have unbounded parabolic motion. > 4) If $\varepsilon > 1$, then we have unbounded hyperbolic motion. > > $ > \frac{\alpha}{r} = 1 - \varepsilon \cos (\phi - \phi_{0}) \hspace{0.5cm} \therefore \hspace{0.5cm} \varepsilon = \frac{L^{2} B}{\mu k} \hspace{0.5cm} , \hspace{0.5cm} \boxed{\alpha = a (1 - \varepsilon^{2}) = \frac{L^{2}}{\mu k} = \frac{L^{2}}{G M \mu^{2}}} > $ > $ > r(\phi) = \frac{L^{2} /\mu k}{1 + \left(\tfrac{L^{2} B}{\mu k}\right) \cos \phi} > \hspace{1cm} \Rightarrow \hspace{1cm} > \boxed{r(\phi) = \frac{a \, (1 - \varepsilon^{2})}{1 + \varepsilon \cos \phi}} > $ > [!note] > With the [[#Conservation of Energy]], we can express the *eccentricity ($\varepsilon$)* in terms of total energy ($E$). > $ > \begin{align} > a (1 - \varepsilon^{2}) = \frac{L^{2}}{\mu k} \hspace{1cm} \Rightarrow \hspace{1cm} E &= - \frac{\mu k}{2 a} = - \frac{\mu^{2} k^{2} (1 - \varepsilon^{2})}{2 L^{2}} \\ > & \\ > &\boxed{\varepsilon = \sqrt{1 + \frac{2 E L^{2}}{\mu^{2} k^{2}}}} > \end{align} > $ ## Kepler's 2nd Law The radius vector of the orbiting body sweeps out equal areas during equal intervals of time. $\frac{\mathrm{d} A}{\mathrm{d} t} = \frac{L}{2 \mu} \hspace{1cm} \text{where} \hspace{1cm} L = \mu r^{2} \dot{\phi}$ ![[kepler2a.png|300]] ![[kepler2b.png|300]] > [!derivation] Proof of Statement (2) > We look at a small period of time ($\mathrm{d} t$) and take the time derivative of the area of a the created triangle... > $ > \begin{align} > \mathrm{d} A &= \frac{1}{2} r \, (r \, \mathrm{d} \phi) + \cancelto{\rm \; 0, \; second \ order}{\frac{1}{2} (r \, \mathrm{d} \phi) \, \mathrm{d} r} \\ > \\ > \frac{\mathrm{d} A}{\mathrm{d} t} &= \frac{\mathrm{d} }{\mathrm{d} t} \left( \frac{1}{2} r^{2} \, \mathrm{d} \phi \right) = \frac{1}{2} r^{2} \dot{\phi} = \frac{L}{2 \mu} \; \boxed{= \text{constant}} > \end{align} > $ ## Kepler's 3rd Law The square of a planet's orbital period ($P$) is proportional to the cubed-length of the semi-major axis ($a$) of the orbit. $ \text{Period-Radius Relationship:} \hspace{1cm} P^{2} = \frac{4 \pi^{2} a^{3}}{G M_{\rm tot}} \hspace{1cm} \Rightarrow \hspace{1cm} P^{2} \propto a^{3} $ > [!derivation] Proof of Statement (3) We compare the area of an ellipse ($A = \pi a b$) to the rate of change in the area ($\frac{\mathrm{d} A}{\mathrm{d} t}$). > > $ > \begin{align} > \pi a b &= \int_{0}^{P} \left( \frac{\mathrm{d} A}{\mathrm{d} t} \right) \; \mathrm{d} t > \hspace{1.4cm} > \textcolor{gray}{\left[ b = a \sqrt{1 - \varepsilon^{2}} \right]} \\ > \pi a^{2} \sqrt{1 - \varepsilon^{2}} &= \int_{0}^{P} \left( \frac{L}{2 \mu} \right) \; \mathrm{d} t \\ > \pi a^{2}\sqrt{1 - \varepsilon^{2}} &= \left(\frac{L}{2 \mu}\right) \int_{0}^{P} \mathrm{d} t > \hspace{1.5cm} > \textcolor{gray}{\left[ a (1 - \varepsilon^{2}) = \frac{L^{2}}{k \mu} = \frac{L^{2}}{G M \mu^{2}} \right]} \\ > \\ > \pi a^{2}\sqrt{1 - \varepsilon^{2}} &= \frac{P}{2} \sqrt{G M a (1 - \varepsilon^{2})} \\ > \\ > P^{2} &= \frac{4 \pi^{2} a^{3}}{G M} \hspace{1cm} \Rightarrow \hspace{1cm} \boxed{ \;P^{2} \propto a^{3} \;} > \end{align} > $ ## Conservation of Energy In a bound Keplerian Orbit, energy is conserved. > [!derivation] Proof of Statement (E) > Instead of using Lagrangian Mechanics ($\mathcal{L} = T - U$), we will look at the total energy of the system ($E = T + U$) and its time derivative. > > $ > E = T + U = \frac{1}{2} \mu \dot{r}^{2} + \frac{L^{2}}{2 \mu r^{2}} - \frac{k}{r} \hspace{1cm} \text{where} \hspace{1cm} k = G M \mu > $ > > Taking the time-derivative, we can prove that energy is conserved. > $ > \frac{\mathrm{d} E}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t} \left( \frac{1}{2} \mu \dot{r}^{2} + \frac{L^{2}}{2 \mu r^{2}} - \frac{k}{r} \right) = \mu \dot{r} \ddot{r} - \frac{L^{2} \dot{r}}{\mu r^{3}} + \frac{k \dot{r}}{r^{2}} > $ > $ > \left[ \hspace{1cm} > \begin{align} > \text{Equation of Motion for $r(t)$:} \hspace{1cm} \ddot{r} = \frac{L^{2}}{\mu^{2} r^{3}} - \frac{k}{\mu r^{2}} > \end{align} > \hspace{1cm} \right] > $ > $ > \frac{\mathrm{d} E}{\mathrm{d} t} = \left( \frac{L^{2} \dot{r}}{\mu r^{3}} - \frac{k \dot{r}}{r^{2}} \right) - \frac{L^{2} \dot{r}}{\mu r^{3}} + \frac{k \dot{r}}{r^{2}} = 0 > \hspace{1cm} \Rightarrow \hspace{1cm} Conserved. > $ **Energy in Terms of the Semi-Major Axis:** With energy conserved throughout the orbit, we can write the energy in terms of of the semi-major axis ($a$) by choosing to specifically look at one position of the orbit. By convention, we take the initial position of $m_{2}$ to be at the pericenter. $ \begin{align} \phi_{0} \equiv \phi(t_{0} = 0) = 0 &\hspace{1cm} \Rightarrow \hspace{1cm} r(\phi_{0}) = \frac{a (1 - \varepsilon^{2})}{1 + \varepsilon} = a (1 - \varepsilon) \\ \\ E &= \frac{1}{2} \mu \dot{r}^{2} + \frac{L^{2}}{2 \mu r^{2}} - \frac{k}{r} \hspace{2.5cm} \textcolor{gray}{\left[ \; \dot{r}(\phi_{0}) = 0 \; \right]} \\ &= \frac{L^{2}}{2 \mu a^{2} (1 - \varepsilon)^{2}} - \frac{k}{a(1 - \varepsilon)} \hspace{1cm} \textcolor{gray}{\left[ \; a (1 - \varepsilon^{2}) = \frac{L^{2}}{\mu k} \; \right]} \\ &= \frac{k (1 - \varepsilon^{2})}{2 a (1 - \varepsilon)^{2}} - \frac{k}{a(1 - \varepsilon)} \\ &= \frac{k}{2a} \left( \frac{(1 - \varepsilon^{2})}{(1 - \varepsilon)^{2}} - \frac{2}{(1 - \varepsilon)} \right)\\ &= - \frac{k}{2a} \\ &= - \frac{G M \mu}{2a} \\ \end{align} $ $ \boxed{ \; E = - \frac{G M \mu}{2 a} = - \frac{G M_{1} M_{2}}{2 a} \; } $