# Why Stuff Can Orbit, Part 7: ALL the Circles

Previously:

Last time around I showed how to do a central-force problem for normal gravity. That’s one where a planet, or moon, or satellite, or whatever is drawn towards the center of space. It’s drawn by a potential energy that equals some constant times the inverse of the distance from the origin. That is, V(r) = C r-1. With a little bit of fussing around we could find out what distance from the center lets a circular orbit happen. And even Kepler’s Third Law, connecting how long an orbit takes to how big it must be.

There are two natural follow-up essays. One is to work out elliptical orbits. We know there are such things; all real planets and moons have them, and nearly all satellites do. The other is to work out circular orbits for another easy-to-understand example, like a mass on a spring. That’s something with a potential energy that looks like V(r) = C r2.

I want to do the elliptical orbits later on. The mass-on-a-spring I could do now. So could you, if you look follow last week’s essay and just change the numbers a little. But, you know, why bother working out one problem? Why not work out a lot of them? Why not work out every central-force problem, all at once?

Because we can’t. I mean, I can describe how to do that, but it isn’t going to save us much time. Like, the quadratic formula is great because it’ll give you the roots of a quadratic polynomial in one step. You don’t have to do anything but a little arithmetic. We can’t get a formula that easy if we try to solve for every possible potential energy.

But we can work out a lot of central-force potential energies all at once. That is, we can solve for a big set of similar problems, a “family” as we call them. The obvious family is potential energies that are powers of the planet’s distance from the center. That is, they’re potential energies that follow the rule

$V(r) = C r^n$

Here ‘C’ is some number. It might depend on the planet’s mass, or the sun’s mass. Doesn’t matter. All that’s important is that it not change over the course of the problem. So, ‘C’ for Constant. And ‘n’ is another constant number. Some numbers turn up a lot in useful problems. If ‘n’ is -1 then this can describe gravitational attraction. If ‘n’ is 2 then this can describe a mass on a spring. This ‘n’ can be any real number. That’s not an ideal choice of letter. ‘n’ usually designates a whole number. By using that letter I’m biasing people to think of numbers like ‘2’ at the expense of perfectly legitimate alternatives such as ‘2.1’. But now that I’ve made that explicit maybe we won’t make a casual mistake.

So what I want is to find where there are stable circular orbits for an arbitrary radius-to-a-power force. I don’t know what ‘C’ and ‘n’ are, but they’re some numbers. To find where a planet can have a circular orbit I need to suppose the planet has some mass, ‘m’. And that its orbit has some angular momentum, a number called ‘L’. From this we get the effective potential energy. That’s what the potential energy looks like when we remember that angular momentum has to be conserved.

$V_{eff}(r) = C r^n + \frac{L^2}{2m} r^{-2}$

To find where a circular orbit can be we have to take the first derivative of Veff with respect to ‘r’. The circular orbit can happen at a radius for which this first derivative equals zero. So we need to solve this:

$\frac{dV_{eff}}{dr} = n C r^{n-1} - 2\frac{L^2}{2m} r^{-3} = 0$

That derivative we know from the rules of how to take derivatives. And from this point on we have to do arithmetic. We want to get something which looks like ‘r = (some mathematics stuff here)’. Hopefully it’ll be something not too complicated. And hey, in the second term there, the one with L2 in it, we have a 2 in the numerator and a 2 in the denominator. So those cancel out and that’s simpler. That’s hopeful, isn’t it?

$n C r^{n-1} - \frac{L^2}{m}r^{-3} = 0$

OK. Add $\frac{L^2}{m}r^{-3}$ to both sides of the equation; we’re used to doing that. At least in high school algebra we are.

$n C r^{n-1} = \frac{L^2}{m}r^{-3}$

Not looking much better? Try multiplying both left and right sides by ‘r3‘. This gets rid of all the ‘r’ terms on the right-hand side of the equation.

$n C r^{n+2} = \frac{L^2}{m}$

Now we’re getting close to the ideal of ‘r = (some mathematics stuff)’. Divide both sides by the constant number ‘n times C’.

$r^{n+2} = \frac{L^2}{n C m}$

I know how much everybody likes taking (n+2)-nd roots of a quantity. I’m sure you occasionally just pick an object at random — your age, your telephone number, a potato, a wooden block — and find its (n+2)-nd root. I know. I’ll spoil some of the upcoming paragraphs to say that it’s going to be more useful knowing ‘rn + 2‘ than it is knowing ‘r’. But I’d like to have the radius of a circular orbit on the record. Here it is.

$r = \left(\frac{L^2}{n C m}\right)^{\frac{1}{n + 2}}$

Can we check that this is right? Well, we can at least check that things aren’t wrong. We can check against the example we already know. That’s the gravitational potential energy problem. For that one, ‘C’ is the number ‘G M m’. That’s the gravitational constant of the universe times the mass of the sun times the mass of the planet. And for gravitational potential energy, ‘n’ is equal to -1. This implies that, for a gravitational potential energy problem, we get a circular orbit when

$r_{grav} = \left(\frac{L^2}{n G M m^2}\right)^{\frac{1}{1}}$

I’m labelling it ‘rgrav‘ to point out it’s the radius of a circular orbit for gravitational problems. Might or might not need that in the future, but the label won’t hurt anything.

Go ahead and guess whether that agrees with last week’s work. I’m feeling confident.

OK, so, we know where a circular orbit might turn up for an arbitrary power function potential energy. Is it stable? We know from the third “Why Stuff Can Orbit” essay that it’s not a sure thing. We can have potential energies that don’t have any circular orbits. So it must be possible there are unstable orbits.

Whether our circular orbit is stable demands we do the same work we did last time. It will look a little harder to start, because there’s one more variable in it. What had been ‘-1’ last time is now an ‘n’, and stuff like ‘-2’ becomes ‘n-1’. Is that actually harder? Really?

So here’s the second derivative of the effective potential:

$\frac{d^2V_{eff}}{dr^2} = (n-1)nCr^{n - 2} + 3\frac{L^2}{m}r^{-4}$

My first impulse when I worked this out was to take the ‘r’ for a circular orbit, the thing worked out five paragraphs above, and plug it in to that expression. This is madness. Don’t do it. Or, you know, go ahead and start doing it and see how long it takes before you regret the errors of your ways.

The non-madness-inducing way to work out if this is a positive number? It involves noticing $r^{n-2}$ is the same number as $r^{n+2}\cdot r^{-4}$. So we have this bit of distribution-law magic:

$\frac{d^2V_{eff}}{dr^2} = (n-1)nCr^{n + 2}r^{-4} + 3\frac{L^2}{m}r^{-4}$

$\frac{d^2V_{eff}}{dr^2} = \left((n-1)nCr^{n + 2} + 3\frac{L^2}{m}\right) \cdot r^{-4}$

I’m sure we all agree that’s better, right? No, honestly, let me tell you why this is better. When will this expression be true?

$\left((n-1)nCr^{n + 2} + 3\frac{L^2}{m}\right) \cdot r^{-4} > 0$

That’s the product of two expressions. One of them is ‘r-4‘. ‘r’ is the radius of the planet’s orbit. That has to be a positive number. It’s how far the planet is from the origin. The number can’t be anything but positive. So we don’t have to worry about that.

SPOILER: I just palmed a card there. Did you see me palm a card there? Because I totally did. Watch for where that card turns up. It’ll be after this next bit.

So let’s look at the non-card-palmed part of this. We’re going to have a stable equilibrium when the other factor of that mess up above is positive. We need to know when this is true:

$(n-1)nCr^{n + 2} + 3\frac{L^2}{m} > 0$

OK. Well. We do know what ‘rn+2‘ is. Worked that out … uhm … twelve(?) paragraphs ago. I’ll say twelve and hope I don’t mess that up in editing. Anyway, what’s important is $r^{n+2} = \frac{L^2}{n C m}$. So we put that in where ‘rn+2‘ appeared in that above expression.

$(n-1)nC\frac{L^2}{n C m} + 3 \frac{L^2}{m} > 0$

This is going to simplify down some. Look at that first term, with an ‘n C’ in the numerator and again in the denominator. We’re going to be happier soon as we cancel those out.

$(n-1)\frac{L^2}{m} + 3\frac{L^2}{m} > 0$

And now we get to some fine distributive-law action, the kind everyone likes:

$\left( (n-1) + 3 \right)\frac{L^2}{m} > 0$

Well, we know $\frac{L^2}{m}$ has to be positive. The angular momentum ‘L’ might be positive or might be negative but its square is certainly positive. The mass ‘m’ has to be a positive number. So we’ll get a stable equilibrium whenever $(n - 1) + 3$ is greater than 0. That is, whenever $n > -2$. Done.

No we’re not done. That’s nonsense. We knew that going in. We saw that a couple essays ago. If your potential energy were something like, say, $V(r) = -2 r^3$ you wouldn’t have any orbits at all, never mind stable orbits. But 3 is certainly greater than -2. So what’s gone wrong here?

Let’s go back to that palmed card. Remember I mentioned how the radius of our circular orbit was a positive number. This has to be true, if there is a circular orbit. What if there isn’t one? Do we know there is a radius ‘r’ that the planet can orbit the origin? Here’s the formula giving us that circular orbit’s radius once again:

$r = \left(\frac{L^2}{n C m}\right)^{\frac{1}{n + 2}}$

Do we know that’s going to exist? … Well, sure. That’s going to be some meaningful number as long as we avoid obvious problems. Like, we can’t have the power ‘n’ be equal to zero, because dividing by zero is all sorts of bad. Also we can’t have the constant ‘C’ be zero, again because dividing by zero is bad.

Not a problem, though. If either ‘C’ or ‘n’ were zero, or if both were, then the original potential energy would be a constant number. V(r) would be equal to ‘C’ (if ‘n’ were zero), or ‘0’ (if ‘C’ were zero). It wouldn’t change with the radius ‘r’. This is a case called the ‘free particle’. There’s no force pushing the planet in one direction or another. So if the planet were not moving it would never start. If the planet were already moving, it would keep moving in the same direction in a straight line. No circular orbits.

Similarly if ‘n’ were equal to ‘-2’ there’d be problems because the power we raise that parenthetical expression to would be equal to one divided by zero, which is bad. Is there anything else that could be trouble there?

What if the thing inside parentheses is a negative number? I may not know what ‘n’ is. I don’t. We started off by supposing we didn’t know beyond that it was a number. But I do know that the n-th root of a negative number is going to be trouble. It might be negative. It might be complex-valued. But it won’t be a positive number. And we need a radius that’s a positive number. So that’s the palmed card. To have a circular orbit at all, positive or negative, we have to have:

$\frac{L^2}{n C m} > 0$

‘L’ is a regular old number, maybe positive, maybe negative. So ‘L2‘ is a positive number. And the mass ‘m’ is a positive number. We don’t know what ‘n’ and C’ are. But as long as their product is positive we’re good. The whole equation will be true. So ‘n’ and ‘C’ can both be negative numbers. We saw that with gravity: $V(r) = -\frac{GMm}{r}$. ‘G’ is the gravitational constant of the universe, a positive number. ‘M’ and ‘m’ are masses, also positive.

Or ‘n’ and ‘C’ can both be positive numbers. That turns up with spring problems: $V(r) = K r^2$, where ‘K’ is the ‘spring constant’. That’s some positive number again.

That time we found potential energies that didn’t have orbits? They were ones that had a positive ‘C’ and negative ‘n’, or a negative ‘C’ and positive ‘n’. The case we just worked out doesn’t have circular orbits. It’s nice to have that sorted out at least.

So what does it mean that we can’t have a stable orbit if ‘n’ is less than or equal to -2? Even if ‘C’ is negative? It turns out that if you have a negative ‘C’ and big negative ‘n’, like say -5, the potential energy drops way down to something infinitely large and negative at smaller and smaller radiuses. If you have a positive ‘C’, the potential energy goes way up at smaller and smaller radiuses. For large radiuses the potential drops to zero. But there’s never the little U-shaped hill in the middle, the way you get for gravity-like potentials or spring potentials or normal stuff like that. Yeah, who would have guessed?

What if we do have a stable orbit? How long does an orbit take? How does that relate to the radius of the orbit? We used this radius expression to work out Kepler’s Third Law for the gravity problem last week. We can do that again here.

Last week we worked out what the angular momentum ‘L’ had to be in terms of the radius of the orbit and the time it takes to complete one orbit. The radius of the orbit we called ‘r’. The time an orbit takes we call ‘T’. The formula for angular momentum doesn’t depend on what problem we’re doing. It just depends on the mass ‘m’ of what’s spinning around and how it’s spinning. So:

$L = 2\pi m \frac{r^2}{T}$

And from this we know what ‘L2‘ is.

$L^2 = 4\pi^2 m^2 \frac{r^4}{T^2}$

That’s convenient because we have an ‘L2‘ term in the formula for what the radius is. I’m going to stick with the formula we got for ‘rn+2‘ because that is so, so much easier to work with than ‘r’ by itself. So we go back to that starting point and then substitute what we know ‘L2‘ to be in there.

$r^{n + 2} = \frac{L^2}{n C m}$

This we rewrite as:

$r^{n + 2} = \frac{4 \pi^2 m^2}{n C m}\frac{r^4}{T^2}$

Some stuff starts cancelling out again. One ‘m’ in the numerator and one in the denominator. Small thing but it makes our lives a bit better. We can multiply the left side and the right side by T2. That’s more obviously an improvement. We can divide the left side and the right side by ‘rn + 2‘. And yes that is too an improvement. Watch all this:

$r^{n + 2} = \frac{4 \pi^2 m}{n C}\frac{r^4}{T^2}$

$T^2 \cdot r^{n + 2} = \frac{4 \pi^2 m}{n C}r^4$

$T^2 = \frac{4 \pi^2 m}{n C}r^{2 - n}$

And that last bit is the equivalent of Kepler’s Third Law for our arbitrary power-law style force.

Are we right? Hard to say offhand. We can check that we aren’t wrong, at least. We can check against the gravitational potential energy. For this ‘n’ is equal to -1. ‘C’ is equal to ‘-G M m’. Make those substitutions; what do we get?

$T^2 = \frac{4 \pi^2 m}{(-1) (-G M m)}r^{2 - (-1)}$

$T^2 = \frac{4 \pi^2}{G M}r^{3}$

Well, that is what we expected for this case. So the work looks good, this far. Comforting.

## 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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