Kepler & Newton

I must confess that I haven't taken a single physics course while at university. In fact, the last time I took physics was in high school and this introduced me to the subject at the elementary level, but no further. Since PDEs are ubiquitous across all fields of science, I can't help but think that this must have been huge mistake on my part. Alas, the freedoms of the undergraduate. Still, the connections between mathematics and physics continue to fascinate me to the point that I persist in reading and writing about it as you have seen through my two previous posts. Even though I keep telling myself that maybe I should embark on writing a post more finely tuned towards pure mathematics, I can't help but to indulge once more on these two fields and elaborate on yet another fruitful relationship.

In 1609, Johannes Kepler published his Astronomia nova which contained his first two laws of planetary motion.

All planets move in ellipses with the Sun at one focus.

Planets sweep out equal areas in equal times.

He established these laws with the use of Tycho Brahe's observational data of the orbit of Mars. In fact, Kepler had actually set out to show that Mars traveled in an ovoid (egg-shaped) path around the Sun since he thought an elliptical trajectory was just too simple for other astronomers to overlook. As you would expect, his measurements came close to Brahe's data but was still off by a small margin. It was in 1605 that Kepler returned to the idea of the elliptical orbit and...eureka!

As for his second law, he also used the observational data on the orbit of Mars and constructed a formula in which a planet's rate of motion is inversely proportional to its distance from the Sun. To verify this formula, however, would have required extensive and tedious work and so Kepler reformulated the above via his second law.

Now enter a most beloved mathematician and physicist, Sir Isaac Newton. In 1687, Newton presented to the world his Philosophiae Naturalis Principia Mathematica which contained not only the very beginnings of calculus, but also two crucial laws of motion. One was his second fundamental law that

The net force on an object is equal to the rate of change of its linear momentum.

The second being his universal law of gravitation which states that

Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically speaking, we have that his second law of motion gives \[ \vec{F}=\frac{d\vec{p}}{dt}, \] where $\vec{p}=m\vec{v}$ with $m$ being the mass and $\vec{v}$ the velocity. His universal law of gravitation is written as \[ \vec{F}=-\frac{GMm}{r^2}\vec{u}, \ \ r=\|\vec{r}\|, \ \ \vec{u}=\frac{\vec{r}}{\|\vec{r}\|}, \] where $G=6.6720 \times 10^{-11} Nm^2/kg^2$ is Newton's gravitational constant, $M$ and $m$ are the two point masses and $r$ is the distance between them. The theme for the today's classroom is to show that we can derive Kepler's first two observational laws on planetary motion from the previous two laws of motion given to us by Newton. We begin by showing that planets travel in elliptical paths around the Sun. First, we note that \begin{align*} &\frac{d}{dt} (\vec{x} \cdot \vec{y})=\frac{d\vec{x}}{dt} \cdot \vec{y}+\vec{x}\cdot \frac{d\vec{y}}{dt} \\ &\frac{d}{dt} (\vec{x} \times \vec{y})=\frac{d\vec{x}}{dt} \times \vec{y}+\vec{x}\times \frac{d\vec{y}}{dt}. \end{align*} Now let's assume that our planet is revolving around the Sun and that $\vec{r}(t)$ is tracing out the planet's trajectory. We will now show that the planet's motion is planar and the Sun lies within this plane of motion. To do this, we will show that $\vec{r} \times \vec{v}$ is constant ($\vec{v}=\vec{r}'(t)$). So we have that \[ \frac{d}{dt} (\vec{r} \times \vec{v})=\frac{d\vec{r}}{dt} \times \vec{v}+\vec{r}\times \frac{d\vec{v}}{dt}=\vec{v} \times \vec{v}+\vec{r}\times\vec{a}=\vec{r} \times \vec{a}, \] where $\vec{a}=\vec{r}''(t)$. Taking into consideration Newton's two above laws, letting $M$ be the mass of the Sun and $m$ the mass of our planet we obtain \[ m\vec{a}=-\frac{GMm}{r^2}\vec{u}. \] Dividing both sides out by $m$, we have \[ \vec{a}=-\frac{GM}{r^3}\vec{r}. \] Thus, we have that \[ \frac{d}{dt} (\vec{r} \times \vec{v})=\vec{r} \times \vec{a}=\vec{r} \times \left(-\frac{GM}{r^3}\vec{r}\right)=0. \] This gives us that $\vec{r} \times \vec{v}=\vec{c}$, where $\vec{c}$ is a constant vector. Since this is true for all time $t$, we have that the motion of our planet is planar. Now let this planar motion be in the $xy$-plane. We will also find another expression for $\vec{c}$. Noting $\vec{r}=r\vec{u}$, we have that \[ \vec{v}=\frac{d}{dt}(r\vec{u})=r\frac{d\vec{u}}{dt}+\frac{dr}{dt}\vec{u}. \] From the above, we have \[ \vec{c}=\vec{r} \times \vec{v}=(r\vec{u}) \times \left(r\frac{d\vec{u}}{dt}+\frac{dr}{dt}\vec{u}\right)=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right)+r\frac{dr}{dt}(\vec{u}\times\vec{u})=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right). \] Taking note that the magnitude of $\vec{r}(t)$ is the polar coordinate $r$ of the planet's position, we now set out to find a polar coordinate description of the planet's trajectory. We have the following expression for $\vec{a} \times \vec{c}$. \begin{align*} \vec{a} \times \vec{c} &= \left(-\frac{GM}{r^2}\vec{u}\right) \times r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right) \\ &=-GM\left[ \vec{u} \times \left(\vec{u}\times\frac{d\vec{u}}{dt}\right)\right] \\ &=GM\left[\left(\vec{u}\times\frac{d\vec{u}}{dt}\right) \times \vec{u}\right] \\ &=GM\left[(\vec{u} \cdot \vec{u})\frac{d\vec{u}}{dt}-\left(\vec{u}\cdot \frac{d\vec{u}}{dt}\right)\vec{u}\right] \\ &=GM\frac{d\vec{u}}{dt} \\ &=\frac{d}{dt}(GM\vec{u}). \end{align*} On the other hand, we have \begin{align*} \vec{a} \times \vec{c} &= \frac{d\vec{v}}{dt} \times \vec{c} \\ &=\frac{d\vec{v}}{dt} \times \vec{c} + \vec{v} \times \frac{d\vec{c}}{dt} \\ &=\frac{d}{dt}(\vec{v} \times \vec{c}). \end{align*} So we have that \[ \vec{a}\times\vec{c}=\frac{d}{dt}(GM\vec{u})=\frac{d}{dt}(\vec{v} \times \vec{c}). \] Integrating with respect to $t$, we find \[ \vec{v} \times \vec{c} = GM\vec{u} + \vec{d} \] for some arbitrary constant vector $\vec{d}$. Since both $\vec{v} \times \vec{c}$ and $\vec{u}$ lie in the same plane, then so must $\vec{d}$. Furthermore, we have that \[ \vec{u} \cdot \vec{d}=\|\vec{u}\|\|\vec{d}\|\cos \theta = d\cos \theta, \] where $d$ is the magnitude of $\vec{d}$ and $\theta$ is the angle between vectors $\vec{u}$ and $\vec{d}$. Letting $c=\|\vec{c}\|$, we have that \begin{align*} c^2 &= \vec{c} \cdot \vec{c} \\ &=(\vec{r} \times \vec{v}) \cdot \vec{c} \\ &=\vec{r} \cdot (\vec{v} \times \vec{c}) \\ &=r\vec{u} \cdot (GM\vec{u}+\vec{d}). \end{align*} Hence, \[ c^2=GMr+rd\cos \theta. \] Solving for $r$, we obtain \[ r=\frac{c^2}{GM+d\cos \theta}. \] Dividing out by $GM$ on both numerator and denominator and letting $p=c^2/GM$ and $e=d/GM$, we have \[ r=\frac{p}{1+e\cos \theta}. \] Multiplying by $1+e\cos\theta$ and solving for $r$ again yields \[ r=p-er\cos\theta. \] Since $x=r\cos\theta$ and $r^2=x^2+y^2$, we have \[ x^2+y^2=p^2-2pex+e^2x^2. \] This gives us \[ (1-e^2)x^2+2pex+y^2=p^2, \] which describes an ellipse for $0<|e|<1$, a parabola for $e=\pm 1$ or a hyperbola for $|e|>1$. Now the parabolic and hyperbolic trajectories are possible for comets or other deep space objects of the same type. Planets, however, exhibit a closed orbit around the Sun which forces us to conclude that our orbit must be elliptical. Cleaning up the above by completing the square and assuming $0<|e|<1$, we obtain \[ \frac{(x+pe/(1-e^2))^2}{p^2/(1-e^2)^2}+\frac{y^2}{p^2/(1-e^2)}=1. \] Thus, we have an ellipse centered at $(-pe/(1-e^2)^2,0)$ with semimajor axis of length $a=p/(1-e^2)$ and semiminor axis of length $b=p/\sqrt{1-e^2}$. The foci of the ellipse are at distance \[ \sqrt{a^2-b^2} = \frac{p|e|}{1-e^2} \] from the center and we see that one of them must be the location of the Sun.

Proving the Kepler's second law, will be much easier than the first. Fix a point $P_0$ on the planet's orbit. Then we have that the area swept out between $P_0$ and a moving point $P$ is given by the integral \[ A(\theta)=\int_{\theta_0}^\theta\frac12 r^2 d\varphi. \] To show that planets sweep out equal area in equal time, all we need to show is that the time derivative of $A$ is constant. By chain rule, we have that \[ \frac{dA}{dt}=\frac{dA}{d\theta}\frac{d\theta}{dt}. \] First, we calculate \[ \frac{dA}{d\theta} = \frac12 \left[r(\theta)\right]^2 \] via the Fundamental Theorem of Calculus. So we have that \[ \frac{dA}{dt}=\frac12 r^2\frac{d\theta}{dt}. \] Now let's recall that $\vec{u}=\vec{r}/r$ and $\vec{r}=(r\cos\theta,r\sin \theta)$. Thus, \begin{align*} &\vec{u}=(\cos\theta,\sin\theta) \\ &\frac{d\vec{u}}{dt}=\left(-\sin\theta\frac{d\theta}{dt},\cos\theta\frac{d\theta}{dt}\right). \end{align*} By a calculation of a cross product, we have that \[ \vec{c}=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right)=\left(0,0,r^2\frac{d\theta}{dt}\right). \] So we have that the magnitude of $\vec{c}$ is \[ \|\vec{c}\|=c=r^2\frac{d\theta}{dt}, \] which implies that \[ \frac{dA}{dt}=\frac12 c. \] This concludes our proof of both Kepler's first and second laws of planetary motion. $\Box$

So there you have it, a mathematical derivation of Kepler's laws from Newton's. Until next time, I leave you with a quote from Newton that should be the watchword of all mathematicians.

If I have seen further it is by standing on the shoulders of giants.