Kepler's Laws: Ellipse Law, Equal-Area Law, and Harmonic Law (Part 2)

Quick pull from the last post, then we keep going.

We had the orbital equation:

$$u = A\cos(\theta - \theta_0) + \frac{k}{ml^2}$$$$\frac{1}{r} = A\cos(\theta - \theta_0) + \frac{k}{ml^2}$$$$\therefore r = \frac{1}{A\cos(\theta - \theta_0) + \frac{k}{ml^2}} = \frac{\left(\frac{ml^2}{k}\right)}{\left(\frac{ml^2}{k}\right)A\cos(\theta - \theta_0) + 1}$$

And this thing is supposedly an ellipse…..

But it’s in polar coordinates, so I honestly can’t feel it~~!!!

Why~~~~ is this an ellipse? Let me actually derive a Real real ellipse in polar coordinates!!! And then compare.

Then hopefully I hit the “ahhh~~~ that’s why that equation is an ellipse~~~~~~” moment!!! That’s the direction I’m going!!!

OK so — definition of an ellipse first: “the set of points where the sum of distances to the two foci is constant!!!!”

Let’s derive the locus equation straight from that definition!!

We literally did this in high school math class.

Only difference — we’re doing it in polar this time!!! heh heh heh

Draw an ellipse. Call the semi-major axis $a$.

And the focus $f$ sits at a distance $\varepsilon a$ from the origin.

$$\text{point } f = (\varepsilon a,\ 0)$$

Now pick a point where the sum of distances to the two foci is the same.

Call the distance from $f'$ as $r'$,

and the distance from $f$ as $r$.

And set $\theta$ like this.

Ta-daaaa~~~~

Variables defined.~~

By the ellipse definition:

$$r + r' = 2a$$

We can write $r'$ in terms of $\theta$ and $r$~!

How? Courtesy of Mr. Pythagoras. Pythago-goras. Pythagoras.

$$r'^2 = (r\sin\theta)^2 + (2\varepsilon a + r\cos\theta)^2$$

But!! We also have $r' = 2a - r$, so:

$$\sqrt{(r\sin\theta)^2 + (2\varepsilon a + r\cos\theta)^2} = 2a - r$$$$(r\sin\theta)^2 + (2\varepsilon a + r\cos\theta)^2 = (2a - r)^2$$

We should be able to get $r(\theta)$ out of this. Let’s grind it out.

$$4a^2 - 4ar + r^2 = r^2\sin^2\theta + 4\varepsilon^2 a^2 + 4\varepsilon ar\cos\theta + r^2\cos^2\theta$$$$4a^2 - 4\varepsilon^2 a^2 = 4ar + 4\varepsilon ar\cos\theta$$$$a - \varepsilon^2 a = r(1 + \varepsilon\cos\theta)$$$$\therefore r = \frac{a(1 - \varepsilon^2)}{1 + \varepsilon\cos\theta} = \frac{\text{constant}'}{1 + \text{constant}\cdot\cos\theta}$$

There it is — the ellipse, as $r(\theta)$!

One quick thing before we move on!

What if $\varepsilon = 0$???????

Then $r = a$. That’s a circle~!!!

$\varepsilon$ is called eccentricity!

• $\varepsilon > 1$ ☞ hyperbola
• $\varepsilon = 1$ ☞ parabola
• $0 < \varepsilon < 1$ ☞ ellipse
• $\varepsilon = 0$ ☞ circle

Anyway — the $r(\theta)$ we got by plugging universal gravity into the orbital equation (in a central force field) has the exact same shape as above!!!

It’s an ellipse!!!!!!!

$$r = \frac{\left(\frac{ml^2}{k}\right)}{\left(\frac{ml^2}{k}\right)A\cos(\theta - \theta_0) + 1} = \frac{\text{constant}'}{1 + \text{constant}\cdot\cos\theta}$$

Chotto matte kudasai!!!!!!! (hold up!!)

The constant playing the role of $\varepsilon$,

$$A\left(\frac{ml^2}{k}\right)$$

we’d need to confirm this actually sits between 0 and 1!!!

And there’s really no clean way to calculate $A$ from first principles… heh. You basically have to get it from observation… experimentally….

Let me show you~ heh heh

lol lol lol lol

Earth runs on a neeeeeeeearly circular orbit!!!! heh heh heh

Third one — the Harmonic Law

Kepler’s 3rd, the law of harmonies. It says: the square of a planet’s period $T$ is proportional to the cube of the semi-major axis $a$ of its elliptical orbit. Hence:

“Wooooo~~~ wow~~~ beautiful~~~~~~ the universe achieves such harmony~~~~~~~”

……………………………..my ass, what the hell is beautiful about it, not a damn thing. (sob)(sob)(sob)

Deriving it is easy, actually. This one also falls out of angular momentum conservation.

By the constant-areal-velocity law, the areal velocity is

$$\dot{A} = \frac{L}{2m}$$

Right.

Integrate over one period $T$ with respect to $dt$!!!

$$\int_0^T \dot{A}\, dt = \int_0^T \frac{L}{2m}\, dt = \frac{L}{2m}T = \pi a b \quad (\text{area of ellipse})$$

Now let me write the semi-minor axis $b$ in terms of $a$ ttttttt!!!

$$b = \sqrt{a^2 - \varepsilon^2 a^2} = a\sqrt{1 - \varepsilon^2}$$

So~

$$T = \frac{2m}{L}\pi a b = \frac{2m}{L}\pi a \left(a\sqrt{1 - \varepsilon^2}\right) = \frac{2m}{L}\pi a^2 \sqrt{1 - \varepsilon^2} = \frac{2}{l}\pi a^2 \sqrt{1 - \varepsilon^2}$$$$T^2 = \frac{4}{l^2}\pi^2 a^4 (1 - \varepsilon^2)$$

On top of that~ on top of this thing built purely from ellipse info,

let’s plug in the actual info — Earth and Sun~.

Grab the constants from the orbital equation and drop them in:

$$a(1 - \varepsilon^2) = \frac{ml^2}{k}, \qquad \varepsilon = A\frac{ml^2}{k}$$

Plugging in, heh heh heh.

$$T^2 = \frac{4}{l^2}\pi^2 a^3 \cdot a(1 - \varepsilon^2) = \frac{4}{l^2}\pi^2 a^3 \cdot \frac{ml^2}{k} = 4\pi^2 a^3 \cdot \frac{m}{GMm} = \frac{4\pi^2}{GM}a^3$$$$\therefore T^2 = \frac{4\pi^2}{GM}a^3$$

And there it is — the law of harmonies, derived~~!!

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Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.