The Orbit Equation via Energy Methods [Classical Mechanics I Studied #10]

OK so last time, starting from conservation of angular momentum in spherical coordinates, we scored a goal all the way to the orbital equation!!!

This time — same destination, different road. We’re deriving the orbital equation from energy.

But before we dive in, I wanna talk about one thing.

Potential Energy — why is it negative???

You’ve probably heard the hand-wavy version:

“We set V = 0 at infinity (r = ∞), so everywhere closer than infinity has to be negative.”

Fine. But why is setting it to 0 at infinity the right thing to do? Let’s feel it out.

Suppose at some spot V = -1 and K = 1. What’s K when the thing reaches V = 0?

K = 0. (Because K + V is conserved. 1 + (-1) = 0 = 0 + 0.)

OK now flip it. At V = -1, say K = 0. What happens?

Nothing. It doesn’t move. Stuck.

What if at V = -1, K = 2? Then at V = 0, K = 1… meaning it still has juice left. It keeps going. It escapes.

So zooming out — if the conserved quantity K + V is

0,

the thing is not bound. It’ll fly off to infinity with leftover kinetic energy.

And if it’s

< 0,

it’s bound to that source over there.

The Earth doesn’t escape the Sun (as far as any of us have experienced, anyway), so K + V < 0~~~

That’s what I wanted to flag before we start.

Now, seriously — orbital equation from energy

V is easy:

$$V\quad =\quad -\int {F\cdot dr}\quad (\because \quad F\quad =\quad -\nabla V)\\ V\quad =\quad -\frac {GMm}{r}$$

And K is $\frac{1}{2}mv^2$. Writing $v^2$ in spherical coordinates,

$$\overrightarrow {v}\quad =\quad \dot {r}\hat {r}\quad +\quad r\dot {\theta}\hat {\theta}\\ \overrightarrow {v}\cdot \overrightarrow {v}\quad =\quad \dot {r}^{2}\quad +\quad r^{2}\dot {\theta}^{2}$$$$\text{And casually plugging in angular momentum conservation,}\\ r^{2}\dot {\theta}\quad =\quad l$$$$\left| v \right|^{2}\quad =\quad \dot {r}^{2}\quad +\quad \frac {l^{2}}{r^{2}}\\ \therefore \quad K\quad =\quad \frac {1}{2}m\left( \dot {r}^{2}\quad +\quad \frac {l^{2}}{r^{2}} \right)$$

Add ’em:

$$E\quad =\quad K\quad +\quad V\quad =\quad \frac {1}{2}m\left( \dot {r}^{2}\quad +\quad \frac {l^{2}}{r^{2}} \right) \quad -\quad \frac {GMm}{r}\quad =\quad \frac {1}{2}m\dot {r}^{2}\quad +\quad \frac {1}{2}m\frac {l^{2}}{r^{2}}-\quad \frac {k}{r}$$

Now we do the $r = 1/u$ substitution again. (We did this last time, so I’m just gonna yeet the results in!!!!!)

$$\dot {r}\quad =\quad -l\frac {du}{d\theta}\\ \ddot {r}\quad =\quad -l^{2}u^{2}\frac {d^{2}u}{d\theta^{2}}$$

Plugging those in:

$$\frac {1}{2}m\dot {r}^{2}\quad +\quad \frac {1}{2}m\frac {l^{2}}{r^{2}}-\quad \frac {k}{r}\quad =\quad E\\ \frac {1}{2}m\left( -l\frac {du}{d\theta} \right)^{2}\quad +\quad \frac {1}{2}ml^{2}u^{2}\quad -ku\quad =\quad E\\ ml^{2}\left( \frac {du}{d\theta} \right)^{2}\quad +\quad ml^{2}u^{2}\quad =\quad 2ku\quad +\quad 2E\\ \left( \frac {du}{d\theta} \right)^{2}\quad =\quad \frac {2ku}{ml^{2}}\quad +\quad \frac {2E}{ml^{2}}\quad -\quad u^{2} \\ \left( \frac {du}{d\theta} \right) \quad =\quad \sqrt {\frac {2ku}{ml^{2}}\quad +\quad \frac {2E}{ml^{2}}\quad -\quad u^{2}}$$$$\frac {1}{\sqrt {\frac {2ku}{ml^{2}}\quad +\quad \frac {2E}{ml^{2}}\quad -\quad u^{2}}}du\quad =\quad d\theta$$

Oh hell… an integral from the depths has appeared……

This is the kinda thing you look up in an integral table lol. Let’s not fight about where the formula comes from — we’re not math majors.

$$\int {\frac {du}{\sqrt {au^{2}+bu+c}}}\quad =\quad \frac {1}{\sqrt {-a}}\cos^{-1}\left( -\frac {b+2au}{\sqrt {b^{2}-4ac}} \right)$$

That’s the formula, apparently.

Alright, shove it in!!!

$$\int {\frac {1}{\sqrt {\frac {2ku}{ml^{2}}\quad +\quad \frac {2E}{ml^{2}}\quad -\quad u^{2}}}du}\quad \\ =\quad \frac {1}{\sqrt {1}}\cos^{-1}\left( -\frac {\frac {2k}{ml^{2}}+2(-1)u}{\sqrt {\left( \frac {2k}{ml^{2}} \right)^{2}-4(-1)\left( \frac {2E}{ml^{2}} \right)}} \right) \quad =\quad \theta \quad -\quad \theta_{0}$$

So:

$$\cos(\theta -\theta_{0})\quad =\quad -\frac {\frac {2k}{ml^{2}}-2u}{\sqrt {\left( \frac {2k}{ml^{2}} \right)^{2}-4(-1)\left( \frac {2E}{ml^{2}} \right)}} \\ \sqrt {\left( \frac {2k}{ml^{2}} \right)^{2}+4\left( \frac {2E}{ml^{2}} \right)}\cdot \cos(\theta -\theta_{0})\quad =\quad -\frac {2k}{ml^{2}}+2u$$$$2u\quad =\quad \sqrt {\left( \frac {2k}{ml^{2}} \right)^{2}+4\left( \frac {2E}{ml^{2}} \right)}\cdot \cos(\theta -\theta_{0})\quad +\quad \frac {2k}{ml^{2}}$$$$\frac {1}{r}\quad =\quad \sqrt {\left( \frac {k}{ml^{2}} \right)^{2}+\left( \frac {2E}{ml^{2}} \right)}\cdot \cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}$$$$r\quad =\quad \frac {1}{\sqrt {\left( \frac {k}{ml^{2}} \right)^{2}+\left( \frac {2E}{ml^{2}} \right)}\cdot \cos(\theta -\theta_{0})\quad +\quad \frac {k}{ml^{2}}}\\ \quad =\frac {ml^{2}/k}{1\quad +\quad \sqrt {1+\frac {2ml^{2}E}{k^{2}}}\cdot \cos(\theta -\theta_{0})}$$

Boom — we’ve got $r(\theta)$. And just staring at it~~~~, we can read off the eccentricity:

$$\epsilon \quad =\quad \sqrt {1+\frac {2ml^{2}E}{k^{2}}}$$

Eccentricity vs energy (ε vs E)

$$\epsilon^{2}\quad =\quad 1+\frac {2ml^{2}E}{k^{2}}\quad =\quad 1\quad +\quad \frac {2E}{k}\cdot \frac {ml^{2}}{k}\quad =\quad 1\quad +\quad \frac {2E}{k}\cdot \alpha \quad \left( \alpha \quad =\frac {ml^{2}}{k}\quad (\text{numerator}) \right) \\ \epsilon^{2}\quad =1\quad +\quad \frac {2E}{k}\cdot a(1-\epsilon^{2})$$$$\epsilon^{2}\quad -\quad 1\quad =\quad \frac {2E}{k}\cdot a(1-\epsilon^{2})\\ -1\quad =\quad \frac {2E}{k}\cdot a\\ \frac {2E}{k}\quad =\quad -\frac {1}{a}\\ E\quad =\quad -\frac {k}{2a}$$

Huh……… heh that just means E < 0.

Well, duh. Of course. heh

Now let’s look at it the other way:

$$\epsilon \quad =\quad \sqrt {1+\frac {2ml^{2}E}{k^{2}}}\\ \text{red (positive)}\\ \text{blue (negative)}\\ \quad =\quad \sqrt {1+\text{negative}}\quad <\quad 1$$

Eccentricity comes out less than 1!!!!!

0 < ε < 1 ☞ e-l-l-i-p-t-i-c-a-l o-r-b-i-t

To wrap this up: E = K + V. Depending on whether this is < 0, > 0, or = 0, the eccentricity shifts, and:

ε < 1 → ellipse or circle (bound state)
ε > 1 → hyperbolic orbit
ε = 1 → parabolic orbit!!

Coming at it from energy, the link between E and the shape of the orbit snaps into focus way more cleanly~~~!!!

The thing we glossed over last time

Let’s revisit something we kinda waved past before.

$$E\quad =\quad K+V\quad =\quad \frac {1}{2}m\left( \dot {r}^{2}+\frac {l^{2}}{r^{2}} \right) -\frac {k}{r}\quad =\quad \frac {m\dot {r}^{2}}{2}+\frac {ml^{2}}{2r^{2}}-\frac {k}{r}$$

The thing to feel here: the total energy E is a function of r only!!!!!!

(Notice θ is nowhere to be found!!!!)

And up to now we’ve been going on and on about orbital motion happening at some r!!!

If it’s circular motion, r stays constant, so

$$\dot {r}=0$$

the whole time~

And if it’s elliptical, r grows to a max, shrinks to a min, grows, shrinks, over and over, right?

What this reminds me of, honestly —

it looks like something bouncing on a spring. K + V = E stays constant, with energy sloshing back and forth as

$$K\xleftrightarrow _{\quad}^{\quad}V$$

in a nice periodic rhythm — feels like that!!! heh heh heh

So let’s try to look at it that way here too.

$$E\quad =\quad \frac {m\dot {r}^{2}}{2}+\frac {ml^{2}}{2r^{2}}+\quad V(r)\\ \text{But — this weird extra term showed up again.}\\ \text{So let's just lump it in together:}\\ \frac {ml^{2}}{2r^{2}}+\quad V(r)\quad =\quad U(r)\quad \text{— call the whole thing that.}$$$$U(r)\text{ isn't}\quad \text{a}\quad \text{real}\quad Potential\text{, but}\quad \\ E\quad =\quad \frac {m\dot {r}^{2}}{2}\quad +\quad U(r)\quad \text{— writing it like this,}$$$$K\xleftrightarrow _{\quad}^{\quad}V$$

it lets us view the whole thing as energy sloshing back and forth, just like before!!

U(r) is called the Effective Potential Energy — “effective potential” for short~~~

(This effective-potential idea is gonna show up again when we do the hydrogen atom in quantum mechanics. So tuck it away somewhere, heh heh heh.)

We’re dealing with celestial bodies right now, and yet the exact same machinery pops up for atoms in the microscopic world…… the universe is…… truly……… sigh…… heh.

Finding where $\dot{r} = 0$

OK, so with

$$E\quad =\quad \frac {m\dot {r}^{2}}{2}\quad +\quad U(r)$$

at the point where U(r) hits its maximum — at that spot,

$$\dot {r}$$

= 0, right!!!

Where would that be?

Since $\dot{r} = 0$ means we’re looking for where r was growing and starts to shrink, or where it was shrinking and starts to grow.

Predicting first: “maxima and minima” suggests one r that’s biggest and one that’s smallest — so those should land on the right-side and left-side endpoints of the major axis, right?

Alright, let’s do it!

$$0\quad +\quad U(r)\quad =\quad E\\ \frac {ml^{2}}{2r^{2}}-\frac {k}{r}\quad =\quad E\\ ml^{2}\quad -\quad 2rk\quad =\quad 2r^{2}E\\ 2Er^{2}\quad +\quad 2kr\quad -ml^{2}\quad =\quad 0$$

You remember the quadratic formula, right~~~

$$r\quad =\quad \frac {2k\quad \pm \sqrt {4k^{2}+8Eml^{2}}}{4E}\quad =\quad \frac {-k\quad \pm \sqrt {k^{2}+2Eml^{2}}}{2E}$$

The roots here! Depending on whether the discriminant is 0, (+), or (-), we get a repeated root, two real roots, or no real roots (two imaginary ones)~~

Let’s stare at the discriminant.

$$k^{2}+2Eml^{2}\quad =\quad 0\\ E\quad =\quad -\frac {k^{2}}{2ml^{2}}$$

So this E is the turning point. If E is bigger than this, two real roots. Smaller, no real roots. Exactly this, one real root.

Now — the “root” here is a root in r, so:

1 value of r satisfies it? → circular orbit, right??

2 values of r? → elliptical orbit.

No real r? → “open orbit” — parabolic or hyperbolic.

We just came at it from pure energy reasoning, and we got the exact same conclusion as when we came at it from the orbital equation before!

This is why I can’t quit physics (sobbing) lmao.

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