Kepler's Laws: Ellipse Law, Equal-Area Law, and Harmonic Law (Part 1)
A breezy Newton sidebar on universal gravitation sets the stage for Kepler's laws โ including a quick sanity check on whether treating planets as point masses is actually legit, heh.
Before we get into Kepler’s laws, let me lay down some background on universal gravitation.
Quick Newton sidebar. The cartoon everyone uses for this dude is this one. (Warning: contains some language.)
Newton, deep in thought like this,
apparently went, “yo, if you fire an apple cannon like this, it won’t fall back down~~~” โ or so the story goes????
But you’ve probably also heard that Newton himself said he was just “a small boy standing on the shoulders of giants”????
What that means is: the motion of stuff on Earth (inertia included) โ Galileo
had alreeeeady basically nailed it before Newton even showed up.
And the motion of celestial bodies out there in space โ everyone knows this one โ
Kepler had already wrapped up beautifully in mathematical form. (Also pre-Newton.)
So what Newton actually did was take these two things everyone thought were totally different and, with universal gravitation
$$F\quad =\quad G\frac {Mm}{r^{2}}$$clee~~~anly merged them into one. heh heh heh heh
OK so now I’ve got to talk about “reductive thinking” for a second.
Because we’re about to head into Section 2, “Gravity between a uniform sphere and a particle”!!!
Let me unpack “reductive thinking” a bit:
“If you strip everything down to its bones, you’ll find it’s just a conclusion that follows scientific cause and effect. Everything has a scientific cause!!!!!”
And apparently this kind of reductive way of thinking was the philosophical hot take in Newton’s era?? lol
(Would the statement “Love is nothing but the mischief of hormones fizzing around in your brain!!!!!!!!!!!!!!” count as the same flavor of thinking???)
OK so what I’m getting at right now is โ
$$F\quad =\quad G\frac {Mm}{r^{2}}$$I want to actually verify this.
Here $r$ is the distance between the centers of the two bodies, and $M$ and $m$ are the total masses!!!
Can we really just write it like that!!
Here’s how Newton thought about it:
Take the Earth and the Moon, and ~
chop them up into absurdly tiny pieces like that, then sum up the universal gravitation between every tiny piece of one and every tiny piece of the other, and that total force
is what universal gravitation actually is โ that’s what he said.
But we just go “yeah $F$ is big-M little-m over $r$-squared” and roll right past it, right???
Is that even legal!!!! Can we really treat the whole thing as a point mass!!!!!!!!!!!!!!!!! We need to actually verify this, quickly.
It’s literally the same calculation as the electric-force one, so I’ll let you verify it on your own for now!!!! heh โ spoiler: the conclusion is ‘yep, treating it as a point mass is totally fine!!’
I’ll post it as an appendix later!
OK, on to Kepler’s laws.
What are Kepler’s laws โ the ones that, before Newton, already basically nailed celestial motion in mathematical form~~!!!!
Law 1. Ellipse Law (1609): Every planet’s orbit is an ellipse with the Sun at one focus!
Law 2. Equal Area Law (1609): The line connecting the Sun and a planet sweeps out equal areas in equal times as the planet orbits.
Law 3. Harmonic Law (1618): The square of a planet’s period (time to go once around the Sun) is proportional to the cube of the semi-major axis of its orbit.
Now, our job is to prove these using Newtonian mechanics!!!
Let’s start with Law 2!!!!!!
OK so we know Earth goes around the Sun, but apparently nobody really knew back then whether it was a circle, or an ellipse, or some slow outward spiral.
But this much we do know!!! The external force is zero!! heh
External force? Zero.
If the external force is 0, then the torque that drives rotational motion โ $r \times F$ โ is also 0.
And if torque is 0, angular momentum $L$ is constant!!!!
$$F=0\overrightarrow {\quad}r\times F=N=0\overrightarrow {\quad}N=\frac {d\overrightarrow {L}}{dt}\overrightarrow {\quad}\text{therefore}\quad \overrightarrow {L}\quad \text{is}\quad \text{constant}$$Now let’s write it like this:
$$L\text{ magnitude}\quad \text{constant}\quad =\quad \left| \overrightarrow {r}\times (m\overrightarrow {v}) \right| =m\left| r\hat {r}\times \left( \dot {r}\hat {r}+r\dot {\theta}\hat {\theta} \right) \right| \\ =\quad m\left| r^{2}\dot {\theta}(\hat {r}\times \hat {\theta}) \right| \quad =\quad mr^{2}\dot {\theta}\quad \text{constant} \\ l\equiv \frac {L}{m}=r^{2}\dot {\theta}\quad \text{this is}\quad \text{constant}$$Now let’s compute the areal velocity as the planet moves. First, the area itself!!!!
I’m going to compute this much area~~~
Pretty easy to see, right??
$$\therefore dA\quad =\quad \frac {1}{2}r^{2}d\theta$$Divide both sides by $dt$ and take the limit:
$$\dot {A}\quad =\quad \frac {1}{2}r^{2}\dot {\theta}$$WAIT!!!! There’s $r^2\dot{\theta}$!!!!
And we already said angular momentum is constant!!
So the conclusion is that the areal velocity is constant!!!!!~~~~~
And just like that, Kepler’s 2nd law is proven.
Now for Kepler’s 1st law โ the ellipse law.
But first, instead of writing universal gravitation as
$$F\quad =\quad G\frac {Mm}{r^{2}}$$let’s just write it as
$$F\quad =\quad f(r)\hat {r}$$for now.
We’re treating it as something that only depends on $r$~~~~~~
Then the equation of motion becomes
$$m\overrightarrow {\ddot {r}}\quad =\quad f(r)\hat {r}$$Like that. Now I’ll solve this equation of motion! heh
We already did $\ddot{r}$ back in Chapter 2, but let me write it out once more and keep moving!
$$\overrightarrow {r}\quad =\quad r\hat {r}\\ \overrightarrow {\dot {r}}\quad =\quad \overrightarrow {v}\quad =\quad \frac {d}{dt}\left( r\hat {r} \right) \quad =\quad \dot {r}\hat {r}\quad +\quad r\dot {\theta}\hat {\theta}\\ \overrightarrow {\ddot {r}}\quad =\quad \overrightarrow {a}\quad =\quad \frac {d}{dt}\left( \dot {r}\hat {r}\quad +\quad r\dot {\theta}\hat {\theta} \right) \quad =\quad \ddot {r}\hat {r}\quad +\dot {\quad r}\frac {d\hat {r}}{dt}\quad +\quad \dot {r}\dot {\theta}\hat {\theta}\quad +\quad r\ddot {\theta}\hat {\theta}\quad +\quad r\dot {\theta}\frac {d\hat {\theta}}{dt}\\ \quad =\quad \ddot {r}\hat {r}\quad +\dot {\quad r}\left( \dot {\theta}\hat {\theta} \right) \quad +\quad \dot {r}\dot {\theta}\hat {\theta}\quad +\quad r\ddot {\theta}\hat {\theta}\quad +\quad r\dot {\theta}\left( -\dot {\theta}\hat {r} \right) \\ \quad =\quad \left( \ddot {r}\quad -\quad r\dot {\theta}^{2} \right) \hat {r}\quad +\quad \left( r\ddot {\theta}\quad +\quad 2\dot {r}\dot {\theta} \right) \hat {\theta}$$OK OK OK OK OK. So the equation of motion above is
$$m\left( \ddot {r}\quad -\quad r\dot {\theta}^{2} \right) \hat {r}\quad +\quad m\left( r\ddot {\theta}\quad +\quad 2\dot {r}\dot {\theta} \right) \hat {\theta}\quad =\quad f(r)\hat {r}$$$$\therefore \quad \text{two equations pop out}\quad !\\ m\left( \ddot {r}\quad -\quad r\dot {\theta}^{2} \right) =\quad f(r)\\ m\left( r\ddot {\theta}\quad +\quad 2\dot {r}\dot {\theta} \right) \quad =\quad 0$$What does that bottom equation mean?
$$m\left( r\ddot {\theta}\quad +\quad 2\dot {r}\dot {\theta} \right) \quad =\quad 0\\ \frac {d}{dt}(mr^{2}\dot {\theta})\quad =\quad 0\quad \text{writing it this way is the same thing}!! \\ \text{ohh}\approx\ \text{universal gravitation already has conservation of angular momentum baked right into it}\approx ~~~\approx$$OK, so now let’s focus on
$$m\left( \ddot {r}\quad -\quad r\dot {\theta}^{2} \right) \quad =\quad f(r)$$this one.
What we’re going to do here is substitute
$$r=\frac {1}{u}$$I’ll explain why this substitution in a bit,
and we need to express $\ddot {r}$ in terms of the new $u$, which is kind of a headache…. ha; (annoyed face)
$$r\quad =\quad \frac {1}{u}\quad =\quad u^{-1}\\ \dot {r}\quad =\quad -u^{-2}\frac {du}{dt}\quad chain\text{-}rule\quad gogo\\ \quad =\quad -u^{-2}\frac {d\theta}{dt}\frac {du}{d\theta}\\ \quad =\quad -u^{-2}\dot {\theta}\frac {du}{d\theta}\quad \text{oh}!?\quad \text{angular momentum just popped out}!\quad \text{you're}\quad l\quad \text{now}\sim\\ \quad =\quad -l\frac {du}{d\theta}$$$$\dot {r}\quad =\quad -l\frac {du}{d\theta}$$What just happened is โ the information about time got hidden away on the right side. The $\dot{\theta}$, which was the thing carrying the time info,
teamed up with $r^2$, disguised itself as angular momentum, became a constant โ and just like that, time vanished… (cry)
Tiny spoiler for where we’re heading,,,, we’re heading toward the orbital equation!!!
OK, we did $\dot{r}$, now onto double-dot!!!!
$$\ddot {r}\quad =\quad \frac {d}{dt}\left( -l\frac {du}{d\theta} \right) \quad =\quad -l\frac {d}{dt}\frac {du}{d\theta}\quad \text{chain-rule round 2}\quad gogogogo\\ \quad =\quad -l\left( \frac {d\theta}{1}\frac {d}{dt}\frac {du}{d\theta}\frac {1}{d\theta} \right) \\ \quad =\quad -l\left( \frac {d\theta}{dt}\frac {d}{d\theta}\frac {du}{d\theta} \right) \\ \quad =\quad -l\left( \dot {\theta}\frac {d^{2}u}{d\theta^{2}} \right) \quad r^{2}\dot {\theta}\quad =\quad l\quad \text{therefore}\quad \dot {\theta}\quad =\quad \frac {l}{r^{2}}\quad =\quad lu^{2}\\ \quad =\quad -l\left( lu^{2}\frac {d^{2}u}{d\theta^{2}} \right) \\ \quad =\quad -l^{2}u^{2}\frac {d^{2}u}{d\theta^{2}}$$Done!!!! $\ddot{r}$ expressed in terms of $l$, $\theta$, and $u$!!!
Back to the equation of motion!!!
$$m\left( \ddot {r}\quad -\quad r\dot {\theta}^{2} \right) \quad =\quad f(r)$$$$m\left( -l^{2}u^{2}\frac {d^{2}u}{d\theta^{2}}\quad -\frac {1}{u}\left( lu^{2} \right)^{2} \right) \quad =\quad f(u^{-1})$$For $\dot{\theta}$, same move as before:
$$\quad r^{2}\dot {\theta}\quad =\quad l\quad \text{therefore}\quad \dot {\theta}\quad =\quad \frac {l}{r^{2}}\quad =\quad lu^{2}$$That’s what I used~~~~~
Keep going~~~~
$$-ml^{2}u^{2}\frac {d^{2}u}{d\theta^{2}}\quad -ml^{2}u^{3}\quad =\quad f(u^{-1})\\ -\frac {d^{2}u}{d\theta^{2}}\quad -u\quad =\quad \frac {1}{ml^{2}u^{2}}f(u^{-1})\\ \frac {d^{2}u}{d\theta^{2}}\quad +\quad u\quad =\quad -\frac {1}{ml^{2}u^{2}}f(u^{-1})\quad \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{orbital equation}$$This one is called the “orbital equation,” apparently.
It gives you a relation between $r$ and $\theta$.
But there’s a big fat premise lurking here!!!! $f(r)$ has to be a central force field (isotropic force โ direction is $\hat{r}$).
Once we’re in a central force field, we can pull the force out just by looking at $r(\theta)$!~ Let’s keep rolling!!
Should we solve one example and move on?????????? Ah, actually no.
OK so we’ve got this orbital equation โ got it.
But then โ why, of all things, doesn’t the Earth go in a circle,
and not even some random crumpled shape, but specifically an ellipse โ shouldn’t figuring that out come first!!!! heh
(Actually its orbit is almoooost a circle lol lol lol โ we’ll get to that later)
OK, now let me plug actual universal gravitation into $f(r)$~~~ (since we’re dealing with the real Earth now!)
$$f(r)\quad =\quad -G\frac {Mm}{r^{2}}\quad =\quad -\frac {k}{r^{2}}(\text{for convenience})\\ \frac {d^{2}u}{d\theta^{2}}\quad +\quad u\quad =\quad -\frac {1}{ml^{2}u^{2}}f(u^{-1})\quad =\frac {k}{ml^{2}} \\ \frac {d^{2}u}{d\theta^{2}}\quad +\quad u\quad =\quad \frac {k}{ml^{2}}$$Holy crap. What this is secretly telling you is: universal gravitation,
of all things, is inversely proportional to $r$ squared?!??!~~?~? โ you’re supposed to really feel that, apparently. heh heh heh
Our Creator cooked up universal gravitation as inverse-square, of all possible choices,,,
and made life hell for us physics students…… (expletive)….
Ah wait, but if it hadn’t been inverse-square for no particular reason,,,
would it have been even worse?????????? hehehe thanks thanks thanks thanks
OK, now let me actually solve the differential equation!!!!
$$\frac {d^{2}u}{d\theta^{2}}\quad +\quad u\quad =\quad \frac {k}{ml^{2}}$$Let’s crack this one with pure gut intuition!
$$\text{second derivative}\quad +\quad \text{itself}\quad =\quad \text{constant}$$Second derivative plus itself equals a constant??????
Let’s first look at what happens if the right side is 0:
$$\frac {d^{2}u}{d\theta^{2}}\quad +\quad u\quad =\quad 0$$If $u$ is sin or cos, the story works beautifully….??!
Trial solution, here we go:
$$if.\quad u\quad =\quad A\cos(\theta -\theta_{0})\\ \frac {d^{2}u}{d\theta^{2}}\quad =\quad -A\cos(\theta -\theta_{0})$$Boom! It works!!!
Then the general solution for the version with a constant on the right is
$$u\quad =\quad A\cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}$$and that satisfies the constant-on-the-right differential equation too????
Yep~~~!!~
So we’ve officially expressed $u$ in terms of $\theta$!!
We’ve got $u(\theta)$ satisfying the orbital equation. heh heh heh hehe
Now let’s flip it into $r(\theta)$:
$$u\quad =\quad A\cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}\\ \frac {1}{r}\quad =\quad A\cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}\\ \therefore \quad r\quad =\quad \frac {1}{A\cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}}\quad =\quad \frac {\left( \frac {ml^{2}}{k} \right)}{\left( \frac {ml^{2}}{k} \right) A\cos\left( \theta -\theta_{0} \right) \quad +\quad 1}$$And this thing is supposedly an ellipse…………………..
But it’s in polar coordinates, so it just doesn’t feel like one, right~~~!!!
To convince ourselves this equation really is an ellipse โ
let me actually derive a real ellipse in polar coordinates from scratch, compare the two,
and then go, “Ohhh~~~ yeah, that’s an ellipse all right~~~~~~~”!!!
First, the definition of an ellipse: “the set of points where the sum of the distances from the two foci is constant!!!”
Let’s use that definition to derive the locus equation of an ellipse!!
You actually did this back in high school math โ the only twist is we’re doing it in polar. heh
This post has gotten ridiculously long, so I’ll continue in the next one!!!!
Straight into part 2!!!! heh
Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.