Why Stuff Can Orbit, Part 12: How Fast Is An Orbit?

Previously:

And the supplemental reading:

On to the next piece of looking for stable, closed orbits of a central force. We start from a circular orbit of something around the sun or the mounting point or whatever. The center. I would have saved myself so much foggy writing if I had just decided to make this a sun-and-planet problem. But I had wanted to write the general problem. In this the force attracting the something towards the center has a strength that’s some constant times the distance to the center raised to a power. This is easy to describe in symbols. It’s cluttered to describe in words. This is why symbols are so nice.

The perturbed orbit, the one I want to see close up, looks like an oscillation around that circle. The fact it is a perturbation, a small nudge away from the equilibrium, means how big the perturbation is will oscillate in time. How far the planet (whatever) is from the center will make a sine wave in time. Whether it closes depends on what it does in space.

Part of what it does in space is easy. I just said what the distance from the planet to the center does. But to say where the planet is we need to know how far it is from the center and what angle it makes with respect to some reference direction. That’s a little harder. We also need to know where it is in the third dimension, but that’s so easy. An orbit like this is always in one plane, so we picked that plane to be that of our paper or whiteboard or tablet or whatever we’re using to sketch this out. That’s so easy to answer we don’t even count it as solved.

The angle, though. Here, I mean the angle made by looking at the planet, the center, and some reference direction. This angle can be any real number, although a lot of those angles are going to point to the same direction in space. We’re coming at this from a mathematical view, or a physics view. Or a mathematical physics view. It means we measure this angle as radians instead of degrees. That is, a right angle is $\frac{\pi}{2}$, not 90 degrees, thank you. A full circle is $2\pi$ and not 360 degrees. We aren’t doing this to be difficult. There are good reasons to use radians. They make the mathematics simpler. What else could matter?

We use $\theta$ as the symbol for this angle. It’s a popular choice. $\theta$ is going to change in time. We’ll want to know how fast it changes over time. This concept we call the angular velocity. For this there are a bunch of different possible notations. The one that I snuck in here two essays ago was ω.

We came at the physics of this orbiting planet from a weird direction. Well, I came at it, and you followed along, and thank you for that. But I never did something like set the planet at a particular distance from the center of the universe and give it a set speed so it would have a circular enough orbit. I set up that we should have some potential energy. That energy implies a central force. It attracts things to the center of the universe. And that there should be some angular momentum that the planet has in its movement. And from that, that there would be some circular orbit. That circular orbit is one with just the right radius and just the right change in angle over time.

From the potential energy and the angular momentum we can work out the radius of the circular orbit. Suppose your potential energy obeys a rule like $V(r) = Cr^n$ for some number ‘C’ and some power, another number, ‘n’. Suppose your planet has the mass ‘m’. Then you’ll get a circular orbit when the planet’s a distance ‘a’ from the center, if $a^{n + 2} = \frac{L^2}{n C m}$. And it turns out we can also work out the angular velocity of this circular orbit. It’s all implicit in the amount of angular momentum that the planet has. This is part of why a mathematical physicist looks for concepts like angular momentum. They’re easy to work with, and they yield all sorts of interesting information, given the chance.

I first introduced angular momentum as this number that was how much of something that our something had. It’s got physical meaning, though, reflecting how much … uh … our something would like to keep rotating around the way it has. And this can be written as a formula. The angular momentum ‘L’ is equal to the moment of inertia ‘I’ times the angular velocity ‘ω’. ‘L’ and ‘ω’ are really vectors, and ‘I’ is really a tensor. But we don’t have to worry about this because this kind of problem is easy. We can pretend these are all real numbers and nothing more.

The moment of inertia depends on how the mass of the thing rotating is distributed in space. And it depends on how far the mass is from whatever axis it’s rotating around. For real bodies this can be challenging to work out. It’s almost always a multidimensional integral, haunting students in Calculus III. For a mass in a central force problem, though, it’s easy once again. Please tell me you’re not surprised. If it weren’t easy I’d have some more supplemental reading pieces here first.

For a planet of mass ‘m’ that’s a distance ‘r’ from the axis of rotation, the moment of inertia ‘I’ is equal to ‘mr2‘. I’m fibbing. Slightly. This is for a point mass, that is, something that doesn’t occupy volume. We always look at point masses in this sort of physics. At least when we start. It’s easier, for one thing. And it’s not far off. The Earth’s orbit has a radius just under 150,000,000 kilometers. The difference between the Earth’s actual radius of just over 6,000 kilometers and a point-mass radius of 0 kilometers is a minor correction.

So since we know $L = I\omega$, and we know $I = mr^2$, we have $L = mr^2\omega$ and from this:

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

We know that ‘r’ changes in time. It oscillates from a maximum to a minimum value like any decent sine wave. So ‘r2‘ is going to oscillate too, like a … sine-squared wave. And then dividing the constant ‘L’ by something oscillating like a sine-squared wave … this implies ω changes in time. So it does. In a possibly complicated and annoying way. So it does. I don’t want to deal with that. So I don’t.

Instead, I am going to summon the great powers of approximation. This perturbed orbit is a tiny change from a circular orbit with radius ‘a’. Tiny. The difference between the actual radius ‘r’ and the circular-orbit radius ‘a’ should be small enough we don’t notice it at first glance. So therefore:

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

And this is going to be close enough. You may protest: what if it isn’t? Why can’t the perturbation be so big that ‘a’ is a lousy approximation to ‘r’? To this I say: if the perturbation is that big it’s not a perturbation anymore. It might be an interesting problem. But it’s a different problem from what I’m doing here. It needs different techniques. The Earth’s orbit is different from Halley’s Comet’s orbit in ways we can’t ignore. I hope this answers your complaint. Maybe it doesn’t. I’m on your side there. A lot of mathematical physics, and of analysis, is about making approximations. We need to find perturbations big enough to give interesting results. But not so big they need harder mathematics than you can do. It’s a strange art. I’m not sure I know how to describe how to do it. What I know I’ve learned from doing a lot of problems. You start to learn what kinds of approaches usually pan out.

But what we’re relying on is the same trick we use in analysis. We suppose there is some error margin in the orbit’s radius and angle that’s tolerable. Then if the perturbation means we’d fall outside that error margin, we just look instead at a smaller perturbation. If there is no perturbation small enough to stay within our error margin then the orbit isn’t stable. And we already know it is. Here, we’re looking for closed orbits. People could in good faith argue about whether some particular observed orbit is a small enough perturbation from the circular equilibrium. But they can’t argue about whether there exist some small enough perturbations.

Let me suppose that you’re all right with my answer about big perturbations. There’s at least one more good objection to have here. It’s this: where is the central force? The mass of the planet (or whatever) is there. The angular momentum is there. The equilibrium orbit is there. But where’s the force? Where’s the potential energy we started with? Shouldn’t that appear somewhere in the description of how fast this planet moves around the center?

It should. And it is there, in an implicit form. We get the radius of the circular, equilibrium orbit, ‘a’, from knowing the potential energy. But we’ll do well to tease it out more explicitly. I hope to get there next time.

Author: Joseph Nebus

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

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