# Why Stuff Can Orbit, Part 13: To Close A Loop

Previously:

Today’s is one of the occasional essays in the Why Stuff Can Orbit sequence that just has a lot of equations. I’ve tried not to write everything around equations because I know what they’re like to read. They’re pretty to look at and after about four of them you might as well replace them with a big grey box that reads “just let your eyes glaze over and move down to the words”. It’s even more glaze-y than that for non-mathematicians.

But we do need them. Equations are wonderfully compact, efficient ways to write about things that are true. Especially things that are only true if exacting conditions are met. They’re so good that I’ll often find myself checking a textbook for an explanation of something and looking only at the equations, letting my eyes glaze over the words. That’s a chilling thing to catch yourself doing. Especially when you’ve written some obscure textbooks and a slightly read mathematics blog.

What I had been looking at was a perturbed central-force orbit. We have something, generically called a planet, that orbits the center of the universe. It’s attracted to the center of the universe by some potential energy, which we describe as ‘U(r)’. It’s some number that changes with the distance ‘r’ the planet has from the center of the universe. It usually depends on other stuff too, like some kind of mass of the planet or some constants or stuff. The planet has some angular momentum, which we can call ‘L’ and pretend is a simple number. It’s in truth a complicated number, but we’ve set up the problem where we can ignore the complicated stuff. This angular momentum implies the potential energy allows for a circular orbit at some distance which we’ll call ‘a’ from the center of the universe.

From ‘U(r)’ and ‘L’ we can say whether this is a stable orbit. If it’s stable, a little perturbation, a nudging, from the circular orbit will stay small. If it’s unstable, a little perturbation will keep growing and never stop. If we perturb this circular orbit the planet will wobble back and forth around the circular orbit. Sometimes the radius will be a little smaller than ‘a’, and sometimes it’ll be a little larger than ‘a’. And now I want to see whether we get a stable closed orbit.

The orbit will be closed if the planet ever comes back to the same position and same momentum that it started with. ‘Started’ is a weird idea in this case. But it’s common vocabulary. By it we mean “whatever properties the thing had when we started paying attention to it”. Usually in a problem like this we suppose there’s some measure of time. It’s typically given the name ‘t’ because we don’t want to make this hard on ourselves. The start is some convenient reference time, often ‘t = 0’. That choice usually makes the equations look simplest.

The position of the planet we can describe with two variables. One is the distance from the center of the universe, ‘r’, which we know changes with time: ‘r(t)’. Another is the angle the planet makes with respect to some reference line. The angle we might call ‘θ’ and often do. This will also change in time, then, ‘θ(t)’. We can pick other variables to describe where something is. But they’re going to involve more algebra, more symbol work, than this choice does so who needs it?

Momentum, now, that’s another set of variables we need to worry about. But we don’t need to worry about them. This particular problem is set up so that if we know the position of the planet we also have the momentum. We won’t be able to get both ‘r(t)’ and ‘θ(t)’ back to their starting values without also getting the momentum there. So we don’t have to worry about that. This won’t always work, as see my future series, ‘Why Statistical Mechanics Works’.

So. We know, because it’s not that hard to work out, how long it takes for ‘r(t)’ to get back to its original, ‘r(0)’, value. It’ll take a time we worked out to be (first big equation here, although we found it a couple essays back):

$T_r = 2\pi\sqrt{ \frac{m}{ -F'(a) - \frac{3}{a} F(a) }}$

Here ‘m’ is the mass of the planet. And ‘F’ is a useful little auxiliary function. It’s the force that the planet feels when it’s a distance from the origin. It’s defined as $F(r) = -\frac{dU}{dr}$. It’s convenient to have around. It makes equations like this one simpler, for one. And it’s weird to think of a central force problem where we never, ever see forces. The peculiar thing is we define ‘F’ for every distance the planet might be from the center of the universe. But all we care about is its value at the equilibrium, circular orbit distance of ‘a’. We also care about its first derivative, also evaluated at the distance of ‘a’, which is that $F'(a)$ talk early on in that denominator.

So in the time between time ‘0’ and time ‘Tr‘ the perturbed radius will complete a full loop. It’ll reach its biggest value and its smallest value and get back to the original. (It is so much easier to suppose the perturbation starts at its biggest value at time ‘0’ that we often assume it has. It doesn’t have to be. But if we don’t have something forcing the choice of what time to call ‘0’ on us, why not pick one that’s convenient?) The question is whether ‘θ(t)’ completes a full loop in that time. If it does then we’ve gotten back to the starting position exactly and we have a closed orbit.

Thing is that the angle will never get back to its starting value. The angle ‘θ(t)’ is always increasing at a rate we call ‘ω’, the angular velocity. This number is constant, at least approximately. Last time we found out what this number was:

$\omega = \frac{L}{ma^2}$

So the angle, over time, is going to look like:

$\theta(t) = \frac{L}{ma^2} t$

And ‘θ(Tr)’ will never equal ‘θ(0)’ again, not unless ‘ω’ is zero. And if ‘ω’ is zero then the planet is racing away from the center of the universe never to be seen again. Or it’s plummeting into the center of the universe to be gobbled up by whatever resides there. In either case, not what we traditionally think of as orbits. Even if we allow these as orbits, these would be nudges too big to call perturbations.

So here’s the resolution. Angles are right awful pains in mathematical physics. This is because increasing an angle by 2π — or decreasing it by 2π — has no visible effect. In the language of the hew-mon, adding 360 degrees to a turn leaves you back where you started. A 45 degree angle is indistinguishable from a 405 degree angle, or a 765 degree angle, or a -315 degree angle, or so on. This makes for all sorts of irritating alternate cases to consider when you try solving for where one thing meets another. But it allows us to have closed orbits.

Because we can have a closed orbit, now, if the radius ‘r(t)’ completes a full oscillation in the time it takes ‘θ(t)’ to grow by 2π. Or to grow by π. Or to grow by ½π. Or a third of π. Or so on.

So. Last time we worked out that the angular velocity had to be this number:

$\omega = \frac{L}{ma^2}$

And that looked weird because the central force doesn’t seem to be there. It’s in there. It’s just implicit. We need to know what the central force is to work out what ‘a’ is. But we can make it explicit by using that auxiliary little function ‘F(r)’. In particular, at the circular orbit radius of ‘a’ we have that:

$F(a) = -\frac{L^2}{ma^3}$

I am going to use this to work out what ‘L’ has to be, in terms of ‘F’ and ‘m’ and ‘a’. First, multiply both sides of this equation by ‘ma3‘:

$F(a) \cdot ma^3 = -L^2$

And then both sides by -1:

$-ma^3 F(a) = L^2$

Take the square root — don’t worry, that it will turn out that ‘F(a)’ is a negative number so we’re not doing anything suspicious —

$\sqrt{-ma^3 F(a)} = L$

Now, take that ‘L’ we’ve got and put it back into the equation for angular velocity:

$\omega = \frac{L}{ma^2} = \frac{\sqrt{-ma^3 F(a)}}{ma^2}$

We might look stuck and at what seems like an even worse position. It’s not. When you do enough of these problems you get used to some tricks. For example, that ‘ma2‘ in the denominator we could move under the square root if we liked. This we know because $ma^2 = \sqrt{ \left(ma^2\right)^2 }$ at least as long as ‘ma2‘ is positive. It is.

So. We fall back on the trick of squaring and square-rooting the denominator and so generate this mess:

$\omega = \sqrt{\frac{-ma^3 F(a)}{\left(ma^2\right)^2}} \\ \omega = \sqrt{\frac{-ma^3 F(a)}{m^2 a^4}} \\ \omega = \sqrt{\frac{-F(a)}{ma}}$

That’s getting nice and simple. Let me go complicate matters. I’ll want to know the angle that the planet sweeps out as the radius goes from its largest to its smallest value. Or vice-versa. This time is going to be half of ‘Tr‘, the time it takes to do a complete oscillation. The oscillation might have started at time ‘t’ of zero, maybe not. But how long it takes will be the same. I’m going to call this angle ‘ψ’, because I’ve written “the angle that the planet sweeps out as the radius goes from its largest to its smallest value” enough times this essay. If ‘ψ’ is equal to π, or one-half π, or one-third π, or some other nice rational multiple of π we’ll get a closed orbit. If it isn’t, we won’t.

So. ‘ψ’ will be one-half times the oscillation time times that angular velocity. This is easy:

$\psi = \frac{1}{2} \cdot T_r \cdot \omega$

Put in the formulas we have for ‘Tr‘ and for ‘ω’. Now it’ll be complicated.

$\psi = \frac{1}{2} 2\pi \sqrt{\frac{m}{-F'(a) - \frac{3}{a} F(a)}} \sqrt{\frac{-F(a)}{ma}}$

Now we’ll make this a little simpler again. We have two square roots of fractions multiplied by each other. That’s the same as the square root of the two fractions multiplied by each other. So we can take numerator times numerator and denominator times denominator, all underneath the square root sign. See if I don’t. Oh yeah and one-half of two π is π but you saw that coming.

$\psi = \pi \sqrt{ \frac{-m F(a)}{-\left(F'(a) + \frac{3}{m}F(a)\right)\cdot ma} }$

OK, so there’s some minus signs in the numerator and denominator worth getting rid of. There’s an ‘m’ in the numerator and the denominator that we can divide out of both sides. There’s an ‘a’ in the denominator that can multiply into a term that has a denominator inside the denominator and you know this would be easier if I could use little cross-out symbols in WordPress LaTeX. If you’re not following all this, try writing it out by hand and seeing what makes sense to cancel out.

$\psi = \pi \sqrt{ \frac{F(a)}{aF'(a) + 3F(a)} }$

This is getting not too bad. Start from a potential energy ‘U(r)’. Use an angular momentum ‘L’ to figure out the circular orbit radius ‘a’. From the potential energy find the force ‘F(r)’. And then, based on what ‘F’ and the first derivative of ‘F’ happen to be, at the radius ‘a’, we can see whether a closed orbit can be there.

I’ve gotten to some pretty abstract territory here. Next time I hope to make things simpler again.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 6 thoughts on “Why Stuff Can Orbit, Part 13: To Close A Loop”

1. I am getting fascinated with this, at last!
Am I right in saying that a closed orbit can have many turns before it reaches the starting position? Rational numbers?
And did I spot the F'(a)/F(a) lurking in there somewhere?

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1. Thank you, and I’m glad you enjoy.

I’m not aware of any reason we can’t have many turns before the orbit closed. The only examples I can think of where this happens are Lissajous figures, from masses on springs connected in multiple dimensions, and those aren’t central forces the way I’ve set this frame up. But for peculiar enough powers ‘n’ there are what look like closed and complicated orbits to me.

And yeah, F'(a)/F(a) is lurking in there. I’m hoping to fit at least one, maybe two, more of this sequence in-between A To Z posts and that should make the F'(a)/F(a) explicit.

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