Kepler's Laws

Ellipse Properties

If an ellipse is centered at the origin, its equation is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ The longest line passing through the origin from with start point and end point lie on the ellipse is called the major axis. Similarly, the shortest line is called the minor axis. The two foci are located at a distance $$c = \sqrt{a^2 - b^2}$$ from the origin on both side on the major axis.

Integrate $y$ with respect to $x$ will yield the area $$\pi ab$$ The eccentricity is $$e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}$$ If one of the focus of the ellipse is on the origin, the equation in polar coordinate is $$r(\theta) = \frac{a(1-\epsilon^2)}{1+\epsilon\cos\theta}$$ where $$\begin{align} \epsilon>1 &\to \text{hyperbola}\\ \epsilon=1 &\to \text{parabola}\\ \epsilon<1 &\to \text{ellipse}\\ \epsilon=0 &\to \text{circle} \end{align}$$

Kepler's Laws

Through observation, Kepler concluded that

1. Each planet orbits the Sun in an ellipse with the sun on one of the foci
2. Let a line drawn from the Sun to the planet. The area swept by this line as the planet orbits is the same in anywhere in the locus if the time interval is the same
3. The square of the time intervals for the planet to complete one revolution is proportional to the cube of the semi-major axis of its ellipse locus

Kepler's Second Law Implication

By definition, the magnitude of angular momentum is given by, $$\begin{align} \vec{L} &= \vec{r}\times\vec{p} \\ &= \vec{r}\times(m\vec{v}) \\ L &= mr(r\omega) \\ &= mr^2\frac{d\theta}{dt} \end{align}$$ In a small fraction of an ellipse $$\begin{align} dA &= \frac{1}{2}r^2d\theta \\ \frac{dA}{dt} &= \frac{1}{2}r^2\frac{d\theta}{dt} \\ &= \frac{L}{2m} \\ A_{swept} &= \int^{t_f}_{t_i}\frac{L}{2m} dt \end{align}$$ Since the area swept only depends on time regardless the position, the Kepler's second law implies that the magnitude of the angular momentum, $L$, is constant. By the definition of angular momentum, when the magnitude of angular momentum is conserved, $$\frac{d\textbf{L}}{dt} = \textbf{r}\times\textbf{F} = 0,$$ that is, the net torque is zero. Conclusion: the gravity force is parallel to $\textbf{r}$, i.e. centripetal.

Kepler's Third Law Implication

Let $$l \equiv \frac{L}{m}$$ and $$u \equiv \frac{1}{r}$$ So, $$\begin{align} \dot{r} &= -\frac{1}{u^2}\dot{u} \\ &= -\frac{1}{u^2}\dot{\theta}\frac{du}{d\theta} \\ &= -l\frac{du}{d\theta} \\ \end{align}$$ $$\begin{align} \ddot{r} &= -l\frac{d}{dt}\frac{du}{d\theta} \\ &= -l\dot{\theta}\frac{d^2u}{d\theta^2} \\ &= -l^2u^2\frac{d^2u}{d\theta^2} \\ \end{align}$$ As we have known from the Kepler's second law that there is only centripetal force exerted on the planet. Let $f(r)$ be the centripetal force $$m(\ddot{r} - r\dot{\theta}^2) = f(r)$$ Replace $r$ with $u$ $$\frac{d^2u}{d\theta^2} + u = -\frac{1}{ml^2u^2}f(u^{-1})$$ Suppose $f(r)$ is proportional to $-\frac{1}{r^2}$ $$\begin{align} \frac{d^2u}{d\theta^2} + u &= -\frac{k}{ml^2} \\ u &= B \cos(\theta - \theta_0) + \frac{k}{ml^2} \\ r &= \frac{1}{b\cos(\theta-\theta_0) + \frac{k}{ml^2}} \\ &= \frac{ml^2/k}{1+(Bml^2/k)\cos\theta} \end{align}$$ which is in the form of ellipse equation. $\theta_0$ can be neglected because the planet orbits the Sun again and again such that initial condition is not important. From the ellipse equation $$\epsilon = \frac{Aml^2}{k}$$ and define $$\alpha \equiv \frac{ml^2}{k}$$ The locus becomes $$\bbox[5px,border:2px solid #666] { r(\theta) = \frac{\alpha}{1+\epsilon\cos\theta} }$$ From the ellipse equation, we know that $$\alpha = a(1-\epsilon^2)$$ and $$b = a\sqrt{1-\epsilon^2}$$ We go back to the area to get the period, $\tau$, of the orbit $$\begin{align} \dot{A} &= \frac{L}{2m} \\ A &= \int^{\tau}_0 \dot{A} dt \\ &= \frac{l}{2}\tau \\ \tau &= \frac{2A}{l} \\ &= \frac{2\pi ab}{l} \\ &= \frac{2\pi a^2\sqrt{1-\epsilon^2}}{l} \\ \tau^2 &= \frac{4\pi^2a^4}{l^2}(1-\epsilon^2) \\ &= \frac{4\pi^2a^4}{l^2}\frac{\alpha}{a} \\ &= 4\pi^2a^3\frac{\alpha}{l} \end{align}$$ Thus, $$\tau^2 \propto a^3$$ This fits the observation stated in Kepler's law.

Conclusion: the force exerted on the planet is always towards the Sun and proportional to the inverse of distance $$f(r) \propto -\frac{1}{r^2}$$ One may check this force is conservative.

Eccentricity & Energy

By $$r(\theta) = \frac{ml^2/k}{1+(Bml^2/k)\cos\theta}$$ At $r_{min}$, $\cos\theta$ is maximum, i.e. $$r(\theta) = \frac{ml^2/k}{1+(Bml^2/k)}$$ As the $f(r)$ is conservative, its potential energy is well-defined $$f(r) = -\int^r_{\infty}-\frac{k}{r^2} dr = -\frac{k}{r}$$ Since $f(r)$ is the only force exerted on the planet, the change in potential energy is equivalent of the change in kinetic energy $$E(r) = \frac{1}{2}mv^2 - \frac{k}{r} = \frac{1}{2}m\dot{r}^2 + \frac{ml^2}{r^2} - \frac{k}{r}$$ Since at $r_{min}$ the planet is not going anymore away from the sun, $\dot{r}$ should be zero $$\begin{align} E &= \frac{1}{2}\frac{ml^2}{r_{min}^2}-\frac{k}{r_{min}} \\ 0 &= \frac{1}{r_{min}^2}-\left(\frac{2k}{ml^2}\right)\frac{1}{r_{min}}-\frac{2E}{ml^2}\\ r_{min} &= \frac{ml^2/k}{1 + \sqrt{1 + \frac{2E}{k}\frac{ml^2}{k}}}\\ \end{align}$$ Thus, $$\bbox[5px,border:2px solid #666] { \epsilon = \sqrt{1 + \frac{2E}{k}\frac{ml^2}{k}} }$$

Speed

Previously, we know that $$\alpha = a(1-\epsilon^2) = \frac{ml^2}{k}$$ The total energy of the planet $$\begin{align} \epsilon &= \sqrt{1 + \frac{2E}{k}\frac{ml^2}{k}} \\ E &= \frac{(\epsilon^2 -1)k^2}{2ml^2} \\ &= -\frac{\alpha k}{a}\frac{1}{2\alpha} \\ &= -\frac{k}{2a} \\ \end{align}$$ Thus, the total energy of a planet orbiting in an ellipse can be expressed as $$\bbox[5px,border:2px solid #666] { E = -\frac{k}{2a} }$$ Since the total energy is equal to kinetic energy plus potential energy $$\begin{align} \frac{1}{2}mv^2 - \frac{k}{r} &= -\frac{k}{2a} \\ \frac{1}{2}mv^2 &= \frac{k}{r} - \frac{k}{2a} \\ mv^2 &= \frac{2k}{r} - \frac{k}{a} \\ v^2 &= \frac{k}{m}\left(\frac{2}{r} - \frac{1}{a}\right) \\ \end{align}$$

Two-body Central Force Motion

When the centre of mass of a two-body system is at distance away from both bodies that is not negligible, some modification is required. Consider an isolated system with two particles.

The centre of mass vector is $$\vec{R}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$$ Since there is no external force, the equation of motion of the centre of mass vector is $$\ddot{\vec{R}}=0.$$ Thus, the solution is $$\vec{R} = \vec{R}_0 + \vec{V}t,$$ where $\vec{R}_0$ and $\vec{V}$ are constant vector depending on choice of coordinates and initial conditions. One may choose an inertial frame such that $\vec{R}_0=0$ and $\vec{V}=0$. So, the centre of mass of the system is at rest.

The two particles are interacting under a central force $f(r)$. $$\begin{align} \vec{r} &= \vec{r}_1-\vec{r}_2\\ r &= |\vec{r}|\\ &= |\vec{r}_1-\vec{r}_2|. \end{align}$$ The equations of motion are $$\begin{align} m_1\ddot{\vec{r}}_1 &= f(r)\hat{r}\\ m_2\ddot{\vec{r}}_2 &= -f(r)\hat{r}.\\ \end{align}$$ Dividing the equations by $m_1$ and $m_2$ respectively and subtract, we have $$\begin{align} \ddot{\vec{r}}_1-\ddot{\vec{r}}_2&=\left(\frac{1}{m_1}+\frac{1}{m_2}\right)f(r)\hat{r}\\ \left(\frac{m_1m_2}{m_1+m_2}\right)(\ddot{\vec{r}}_1-\ddot{\vec{r}}_2)&=f(r)\hat{r} \end{align}$$ Define the reduced mass as $$\mu\equiv \frac{m_1m_2}{m_1+m_2}$$ and substitute $\vec{r}=\vec{r}_1-\vec{r}_2$, we have $$\mu\ddot{r}=f(r)\hat{r}.$$ As solved in above section, the solution is the same except with reduced mass. The gravitational force constant $k=Gm_1m_2$ is the same as the single particle motion. Therefore, $$\begin{align} \alpha &= \frac{\mu l^2}{Gm_1m_2}\\ \epsilon &= \sqrt{1+\frac{2E\mu l^2}{(Gm_1m_2)^2}}\\ r(\theta) &= \frac{\alpha}{1+\epsilon\cos\theta}, \end{align}$$ where the total energy $E$ is $$E=\frac{1}{2}\mu r^2+\frac{1}{2}\mu\frac{l^2}{r^2}-\frac{Gm_1m_2}{r}$$

Example

An isolated system of two point masses of mass $m$ and $2m$ that interact with each other with gravitational force. At $t=0$, mass $2m$ is momentarily at rest with respect to the inertial frame. What speed should the mass $m$ have at $t=0$ so that it will move in a circle relative to the heavier mass? What is the speed of the centre of mass?

Solution: Let $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of mass $m$ and $2m$ respectively. The correct equation of motion relative to the mass $2m$ is $$\mu\ddot{\vec{r}}=f(r)\hat{r},$$ where $\mu=\frac{Mm}{M+m}$ and $\vec{r}=\vec{r}_1-\vec{r}_2$. To have circular motion, $$|\vec{r}(t)|=r_0$$ and the radial acceleration is $$\mu a_r = \mu\frac{v^2}{r}=\frac{G(2m)(m)}{r^2}$$ $$\frac{(2m)m}{2m+m}\frac{v^2}{r_0}=\frac{2Gm^2}{r^2_0}$$ So, speed of mass $m$ is $$v=\sqrt{\frac{3Gm}{r_0}}.$$ The centre of mass of this system is $$\vec{R}=\frac{\vec{r}_1m+\vec{r}_2(2m)}{m+2m}.$$ Then, $$\dot{\vec{R}}=\frac{\dot{\vec{r}}_1m+\dot{\vec{r}}_2(2m)}{m+2m}$$ At $t=0$, $|\dot{\vec{r}}_1|=v$ and $|\dot{\vec{r}}_2|=0$, so $$|\dot{\vec{R}}(t=0)|=\frac{1}{3}v=\frac{1}{3}\sqrt{\frac{3Gm}{r_0}}$$ Since there is no external force, $\ddot{\vec{R}}=0$, $\dot{\vec{R}}$ is a constant. $$|\dot{\vec{R}}(t)|=\frac{1}{3}v=\frac{1}{3}\sqrt{\frac{3Gm}{r_0}}$$