Orbital Stability

The goal of this post — let me show you with a picture!

Why……… without leaving its orbit, can Earth just keep going around in a stable circular orbit??????

Why is it so hard to even imagine a tiny nudge knocking it off track and sending it flying off into the void?!!

That’s what we’re going to figure out.

OK so — let’s write down Earth’s equation of motion again!!!

(It’s exactly~~ the same as last time)

$$m\overrightarrow {\ddot {r}} = f(r)\hat {r}$$

Here $f(r)$ is gravity, but let’s keep it as $f(r)$ so we stay general.

$$\overrightarrow {\ddot {r}} = \left( \ddot {r} - r\dot {\theta}^{2} \right) \hat {r} + \left( r\ddot {\theta} + 2\dot {r}\dot {\theta} \right) \hat {\theta}\\ \text{so}\\ \begin{cases}{m\ddot {r} - mr\dot {\theta}^{2} = f(r)}\\{m\left( r\ddot {\theta} + 2\dot {r}\dot {\theta} \right) = 0}\end{cases}$$

The second one — same move as before:

$$0 = m\left( r\ddot {\theta} + 2\dot {r}\dot {\theta} \right) = \frac {d}{dt}\left( mr^{2}\dot {\theta} \right) =\frac {dL}{dt} \cdots \text{conservation of angular momentum}$$

Now let’s hit the first one:

$$m\left( \ddot {r} - r\dot {\theta}^{2} \right) = f(r)\\ m\left( \ddot {r} - r\frac {l^{2}}{r^{2}} \right) = f(r)\\ m\ddot {r} = f(r) + \frac {l^{2}}{r^{3}}$$

There’s our radial equation of motion.

Here’s where we’re going with this: I’m going to show you that a spring constant $k$ is hiding inside that equation of motion for $r$.

And once we dig it out and confirm it’s really there, think about what this whole equation starts to mean…

So what we actually want isn’t the equation of motion at exactly distance $r$ from the Sun!

What we want is the equation of motion juuust a tiny bit off from there.

$$r \cong a$$

— and with that in mind, we Taylor expand around $r=a$.

Still with me?

$$f(r) \cong f(a) + f'(a)(r-a) + \frac {1}{2!}f''(a)(r-a)^{2} + \cdots$$

And this one, well — same thing:

$$r^{-3} \cong \frac {1}{a^{3}} + \left( -3a^{-4} \right) (r-a) + \cdots \\ \frac {ml^{2}}{r^{3}} \cong \frac {ml^{2}}{a^{3}} + ml^{2}\left( -3a^{-4} \right) (r-a) + \cdots$$

So

$$m\ddot {r} \cong \left[ f(a) + f'(a)(r-a) + \cdots \right] + \left[ \frac {ml^{2}}{a^{3}} + ml^{2}\left( -3a^{-4} \right) (r-a) + \cdots \right] \\ \cong f(a) + f'(a)r - af(a) + \frac {ml^{2}}{a^{3}} - \frac {-3ml^{2}}{a^{4}}r + \frac {-3ml^{2}}{a^{3}}$$

It comes out like this, and — reason I color-coded those bits — all the constant terms are about to vanish!!

Let’s pretend we don’t know whether Earth is on an elliptical orbit, a circular orbit, or getting yeeted into deep space.

(Obviously, from observation, “yeeted into deep space” is not happening even if you beat the case to death — so no need to even humor that one.)

So I ran through the possibilities, and when I tried “circular orbit,” all those constant terms drop out. Check it out.

Circular orbit means $\ddot r = 0$:

$$\ddot {r} = 0 \text{ when}\\ \text{so,} \quad m\ddot {r} = f(r) + \frac {l^{2}}{r^{3}} \text{ this}\\ 0 = f(r) + \frac {l^{2}}{r^{3}} \text{ becomes}\\ \therefore \quad f(a) = -\frac {l^{2}}{a^{3}}$$

So

$$m\ddot {r} \cong f'(a)r - \frac {-3ml^{2}}{a^{4}}r \cong f'(a)r - \frac {3}{a}\frac {ml^{2}}{a^{2}}r\\ \text{hidden } f(a) \text{ found!}\\ \cong f'(a)r + \frac {3}{a}f(a)r\\ m\ddot {r} - \left[ f'(a) + \frac {3}{a}f(a) \right] r = 0$$

Boom — there’s the hidden spring constant.

Let’s write it like this:

$$m\ddot {x} + kx = 0$$

The spring equation,,, and it oscillates when $k>0$, right?!??!?!?!

(Has to be $k>0$ — because then $F=-kx$, meaning the force pulls back against the motion!!)

So in our case, we need

$$f'(a) + \frac {3}{a}f(a) < 0$$$$f(r) = -cr^{n} \text{ — if we let it be that} \\ \frac {3}{a}f(a) = -3ca^{n-1}\\ f'(a) = -nca^{n-1}$$

Then

$$f'(a) + \frac {3}{a}f(a) < 0 \\ -3ca^{n-1} - nca^{n-1} < 0 \\ -3-n<0 \\ -3$$

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