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  • Joseph Nebus 6:00 pm on Wednesday, 1 February, 2017 Permalink | Reply
    Tags: cables, , orbits, podcasts   

    Mathematics Stuff To Read Or Listen To 


    I concede January was a month around here that could be characterized as “lazy”. Not that I particularly skimped on the Reading the Comics posts. But they’re relatively easy to do: the comics tell me what to write about, and I could do a couple paragraphs on most anything, apparently.

    While I get a couple things planned out for the coming month, though, here’s some reading for other people.

    The above links to a paper in the Proceedings of the National Academy of Sciences. It’s about something I’ve mentioned when talking about knot before. And it’s about something everyone with computer cables or, like the tweet suggests, holiday lights finds. The things coil up. Spontaneous knotting of an agitated string by Dorian M Raymer and Douglas E Smith examines when these knots are likely to form, and how likely they are. It’s not a paper for the lay audience, but there are a bunch of fine pictures. The paper doesn’t talk about Christmas lights, no matter what the tweet does, but the mathematics carries over to this.

    MathsByAGirl, meanwhile, had a post midmonth listing a couple of mathematics podcasts. I’m familiar with one of them, BBC Radio 4’s A Brief History of Mathematics, which was a set of ten-to-twenty-minute sketches of historically important mathematics figures. I’ll trust MathsByAGirl’s taste on other podcasts. I’d spent most of this month finishing off a couple of audio books (David Hackett Fischer’s Washington’s Crossing which I started listening to while I was in Trenton for a week, because that’s the sort of thing I think is funny, and Robert Louis Stevenson’s Doctor Jekyll and Mister Hyde And Other Stories) and so fell behind on podcasts. But now there’s some more stuff to listen forward to.

    And then I’ll wrap up with this from KeplerLounge. It looks to be the start of some essays about something outside the scope of my Why Stuff Can Orbit series. (Which I figure to resume soon.) We start off talking about orbits as if planets were “point masses”. Which is what the name suggests: a mass that fills up a single point, with no volume, no shape, no features. This makes the mathematics easier. The mathematics is just as easy if the planets are perfect spheres, whether hollow or solid. But real planets are not perfect spheres. They’re a tiny bit blobby. And they’re a little lumpy as well. We can ignore that if we’re doing rough estimates of how orbits work. But if we want to get them right we can’t ignore that anymore. And this essay describes some of how we go about dealing with that.

     
  • Joseph Nebus 6:00 pm on Friday, 28 October, 2016 Permalink | Reply
    Tags: , , , , , , orbits,   

    Why Stuff Can Orbit, Part 7: ALL the Circles 


    Previously:

    And some supplemental reading:


    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.

     
  • Joseph Nebus 6:00 pm on Friday, 21 October, 2016 Permalink | Reply
    Tags: , , , , , , , orbits,   

    Why Stuff Can Orbit, Part 6: Circles and Where To Find Them 


    Previously:

    And some supplemental reading:


    So now we can work out orbits. At least orbits for a central force problem. Those are ones where a particle — it’s easy to think of it as a planet — is pulled towards the center of the universe. How strong that pull is depends on some constants. But it only changes as the distance the planet is from the center changes.

    What we’d like to know is whether there are circular orbits. By “we” I mean “mathematical physicists”. And I’m including you in that “we”. If you’re reading this far you’re at least interested in knowing how mathematical physicists think about stuff like this.

    It’s easiest describing when these circular orbits exist if we start with the potential energy. That’s a function named ‘V’. We write it as ‘V(r)’ to show it’s an energy that changes as ‘r’ changes. By ‘r’ we mean the distance from the center of the universe. We’d use ‘d’ for that except we’re so used to thinking of distance from the center as ‘radius’. So ‘r’ seems more compelling. Sorry.

    Besides the potential energy we need to know the angular momentum of the planet (or whatever it is) moving around the center. The amount of angular momentum is a number we call ‘L’. It might be positive, it might be negative. Also we need the planet’s mass, which we call ‘m’. The angular momentum and mass let us write a function called the effective potential energy, ‘Veff(r)’.

    And we’ll need to take derivatives of ‘Veff(r)’. Fortunately that “How Differential Calculus Works” essay explains all the symbol-manipulation we need to get started. That part is calculus, but the easy part. We can just follow the rules already there. So here’s what we do:

    • The planet (or whatever) can have a circular orbit around the center at any radius which makes the equation \frac{dV_{eff}}{dr} = 0 true.
    • The circular orbit will be stable if the radius of its orbit makes the second derivative of the effective potential, \frac{d^2V_{eff}}{dr^2} , some number greater than zero.

    We’re interested in stable orbits because usually unstable orbits are boring. They might exist but any little perturbation breaks them down. The mathematician, ordinarily, sees this as a useless solution except in how it describes different kinds of orbits. The physicist might point out that sometimes it can take a long time, possibly millions of years, before the perturbation becomes big enough to stand out. Indeed, it’s an open question whether our solar system is stable. While it seems to have gone millions of years without any planet changing its orbit very much we haven’t got the evidence to say it’s impossible that, say, Saturn will be kicked out of the solar system anytime soon. Or worse, that Earth might be. “Soon” here means geologically soon, like, in the next million years.

    (If it takes so long for the instability to matter then the mathematician might allow that as “metastable”. There are a lot of interesting metastable systems. But right now, I don’t care.)

    I realize now I didn’t explain the notation for the second derivative before. It looks funny because that’s just the best we can work out. In that fraction \frac{d^2V_{eff}}{dr^2} the ‘d’ isn’t a number so we can’t cancel it out. And the superscript ‘2’ doesn’t mean squaring, at least not the way we square numbers. There’s a functional analysis essay in there somewhere. Again I’m sorry about this but there’s a lot of things mathematicians want to write out and sometimes we can’t find a way that avoids all confusion. Roll with it.

    So that explains the whole thing clearly and easily and now nobody could be confused and yeah I know. If my Classical Mechanics professor left it at that we’d have open rebellion. Let’s do an example.

    There are two and a half good examples. That is, they’re central force problems with answers we know. One is gravitation: we have a planet orbiting a star that’s at the origin. Another is springs: we have a mass that’s connected by a spring to the origin. And the half is electric: put a positive electric charge at the center and have a negative charge orbit that. The electric case is only half a problem because it’s the same as the gravitation problem except for what the constants involved are. Electric charges attract each other crazy way stronger than gravitational masses do. But that doesn’t change the work we do.

    This is a lie. Electric charges accelerating, and just orbiting counts as accelerating, cause electromagnetic effects to happen. They give off light. That’s important, but it’s also complicated. I’m not going to deal with that.

    I’m going to do the gravitation problem. After all, we know the answer! By Kepler’s something law, something something radius cubed something G M … something … squared … After all, we can look up the answer!

    The potential energy for a planet orbiting a sun looks like this:

    V(r) = - G M m \frac{1}{r}

    Here ‘G’ is a constant, called the Gravitational Constant. It’s how strong gravity in the universe is. It’s not very strong. ‘M’ is the mass of the sun. ‘m’ is the mass of the planet. To make sense ‘M’ should be a lot bigger than ‘m’. ‘r’ is how far the planet is from the sun. And yes, that’s one-over-r, not one-over-r-squared. This is the potential energy of the planet being at a given distance from the sun. One-over-r-squared gives us how strong the force attracting the planet towards the sun is. Different thing. Related thing, but different thing. Just listing all these quantities one after the other means ‘multiply them together’, because mathematicians multiply things together a lot and get bored writing multiplication symbols all the time.

    Now for the effective potential we need to toss in the angular momentum. That’s ‘L’. The effective potential energy will be:

    V_{eff}(r) = - G M m \frac{1}{r} + \frac{L^2}{2 m r^2}

    I’m going to rewrite this in a way that means the same thing, but that makes it easier to take derivatives. At least easier to me. You’re on your own. But here’s what looks easier to me:

    V_{eff}(r) = - G M m r^{-1} + \frac{L^2}{2 m} r^{-2}

    I like this because it makes every term here look like “some constant number times r to a power”. That’s easy to take the derivative of. Check back on that “How Differential Calculus Works” essay. The first derivative of this ‘Veff(r)’, taken with respect to ‘r’, looks like this:

    \frac{dV_{eff}}{dr} = -(-1) G M m r^{-2} -2\frac{L^2}{2m} r^{-3}

    We can tidy that up a little bit: -(-1) is another way of writing 1. The second term has two times something divided by 2. We don’t need to be that complicated. In fact, when I worked out my notes I went directly to this simpler form, because I wasn’t going to be thrown by that. I imagine I’ve got people reading along here who are watching these equations warily, if at all. They’re ready to bolt at the first sign of something terrible-looking. There’s nothing terrible-looking coming up. All we’re doing from this point on is really arithmetic. It’s multiplying or adding or otherwise moving around numbers to make the equation prettier. It happens we only know those numbers by cryptic names like ‘G’ or ‘L’ or ‘M’. You can go ahead and pretend they’re ‘4’ or ‘5’ or ‘7’ if you like. You know how to do the steps coming up.

    So! We allegedly can have a circular orbit when this first derivative is equal to zero. What values of ‘r’ make true this equation?

    G M m r^{-2} - \frac{L^2}{m} r^{-3} = 0

    Not so helpful there. What we want is to have something like ‘r = (mathematics stuff here)’. We have to do some high school algebra moving-stuff-around to get that. So one thing we can do to get closer is add the quantity \frac{L^2}{m} r^{-3} to both sides of this equation. This gets us:

    G M m r^{-2} = \frac{L^2}{m} r^{-3}

    Things are getting better. Now multiply both sides by the same number. Which number? r3. That’s because ‘r-3‘ times ‘r3‘ is going to equal 1, while ‘r-2‘ times ‘r3‘ will equal ‘r1‘, which normal people call ‘r’. I kid; normal people don’t think of such a thing at all, much less call it anything. But if they did, they’d call it ‘r’. We’ve got:

    G M m r = \frac{L^2}{m}

    And now we’re getting there! Divide both sides by whatever number ‘G M’ is, as long as it isn’t zero. And then we have our circular orbit! It’s at the radius

    r = \frac{L^2}{G M m^2}

    Very good. I’d even say pretty. It’s got all those capital letters and one little lowercase. Something squared in the numerator and the denominator. Aesthetically pleasant. Stinks a little that it doesn’t look like anything we remember from Kepler’s Laws once we’ve looked them up. We can fix that, though.

    The key is the angular momentum ‘L’ there. I haven’t said anything about how that number relates to anything. It’s just been some constant of the universe. In a sense that’s fair enough. Angular momentum is conserved, exactly the same way energy is conserved, or the way linear momentum is conserved. Why not just let it be whatever number it happens to be?

    (A note for people who skipped earlier essays: Angular momentum is not a number. It’s really a three-dimensional vector. But in a central force problem with just one planet moving around we aren’t doing any harm by pretending it’s just a number. We set it up so that the angular momentum is pointing directly out of, or directly into, the sheet of paper we pretend the planet’s orbiting in. Since we know the direction before we even start work, all we have to car about is the size. That’s the number I’m talking about.)

    The angular momentum of a thing is its moment of inertia times its angular velocity. I’m glad to have cleared that up for you. The moment of inertia of a thing describes how easy it is to start it spinning, or stop it spinning, or change its spin. It’s a lot like inertia. What it is depends on the mass of the thing spinning, and how that mass is distributed, and what it’s spinning around. It’s the first part of physics that makes the student really have to know volume integrals.

    We don’t have to know volume integrals. A single point mass spinning at a constant speed at a constant distance from the origin is the easy angular momentum to figure out. A mass ‘m’ at a fixed distance ‘r’ from the center of rotation moving at constant speed ‘v’ has an angular momentum of ‘m’ times ‘r’ times ‘v’.

    So great; we’ve turned ‘L’ which we didn’t know into ‘m r v’, where we know ‘m’ and ‘r’ but don’t know ‘v’. We’re making progress, I promise. The planet’s tracing out a circle in some amount of time. It’s a circle with radius ‘r’. So it traces out a circle with perimeter ‘2 π r’. And it takes some amount of time to do that. Call that time ‘T’. So its speed will be the distance travelled divided by the time it takes to travel. That’s \frac{2 \pi r}{T} . Again we’ve changed one unknown number ‘L’ for another unknown number ‘T’. But at least ‘T’ is an easy familiar thing: it’s how long the orbit takes.

    Let me show you how this helps. Start off with what ‘L’ is:

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

    Now let’s put that into the equation I got eight paragraphs ago:

    r = \frac{L^2}{G M m^2}

    Remember that one? Now put what I just said ‘L’ was, in where ‘L’ shows up in that equation.

    r = \frac{\left(2\pi m \frac{r^2}{T}\right)^2}{G M m^2}

    I agree, this looks like a mess and possibly a disaster. It’s not so bad. Do some cleaning up on that numerator.

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

    That’s looking a lot better, isn’t it? We even have something we can divide out: the mass of the planet is just about to disappear. This sounds bizarre, but remember Kepler’s laws: the mass of the planet never figures into things. We may be on the right path yet.

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

    OK. Now I’m going to multiply both sides by ‘T2‘ because that’ll get that out of the denominator. And I’ll divide both sides by ‘r’ so that I only have the radius of the circular orbit on one side of the equation. Here’s what we’ve got now:

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

    And hey! That looks really familiar. A circular orbit’s radius cubed is some multiple of the square of the orbit’s time. Yes. This looks right. At least it looks reasonable. Someone else can check if it’s right. I like the look of it.

    So this is the process you’d use to start understanding orbits for your own arbitrary potential energy. You can find the equivalent of Kepler’s Third Law, the one connecting orbit times and orbit radiuses. And it isn’t really hard. You need to know enough calculus to differentiate one function, and then you need to be willing to do a pile of arithmetic on letters. It’s not actually hard. Next time I hope to talk about more and different … um …

    I’d like to talk about the different … oh, dear. Yes. You’re going to ask about that, aren’t you?

    Ugh. All right. I’ll do it.

    How do we know this is a stable orbit? Well, it just is. If it weren’t the Earth wouldn’t have a Moon after all this. Heck, the Sun wouldn’t have an Earth. At least it wouldn’t have a Jupiter. If the solar system is unstable, Jupiter is probably the most stable part. But that isn’t convincing. I’ll do this right, though, and show what the second derivative tells us. It tells us this is too a stable orbit.

    So. The thing we have to do is find the second derivative of the effective potential. This we do by taking the derivative of the first derivative. Then we have to evaluate this second derivative and see what value it has for the radius of our circular orbit. If that’s a positive number, then the orbit’s stable. If that’s a negative number, then the orbit’s not stable. This isn’t hard to do, but it isn’t going to look pretty.

    First the pretty part, though. Here’s the first derivative of the effective potential:

    \frac{dV_{eff}}{dr} = G M m r^{-2} - \frac{L^2}{m} r^{-3}

    OK. So the derivative of this with respect to ‘r’ isn’t hard to evaluate again. This is again a function with a bunch of terms that are all a constant times r to a power. That’s the easiest sort of thing to differentiate that isn’t just something that never changes.

    \frac{d^2 V_{eff}}{dr^2} = -2 G M m r^{-3} - (-3)\frac{L^2}{m} r^{-4}

    Now the messy part. We need to work out what that line above is when our planet’s in our circular orbit. That circular orbit happens when r = \frac{L^2}{G M m^2} . So we have to substitute that mess in for ‘r’ wherever it appears in that above equation and you’re going to love this. Are you ready? It’s:

    -2 G M m \left(\frac{L^2}{G M m^2}\right)^{-3} + 3\frac{L^2}{m}\left(\frac{L^2}{G M m^2}\right)^{-4}

    This will get a bit easier promptly. That’s because something raised to a negative power is the same as its reciprocal raised to the positive of that power. So that terrible, terrible expression is the same as this terrible, terrible expression:

    -2 G M m \left(\frac{G M m^2}{L^2}\right)^3 + 3 \frac{L^2}{m}\left(\frac{G M m^2}{L^2}\right)^4

    Yes, yes, I know. Only thing to do is start hacking through all this because I promise it’s going to get better. Putting all those third- and fourth-powers into their parentheses turns this mess into:

    -2 G M m \frac{G^3 M^3 m^6}{L^6} + 3 \frac{L^2}{m} \frac{G^4 M^4 m^8}{L^8}

    Yes, my gut reaction when I see multiple things raised to the eighth power is to say I don’t want any part of this either. Hold on another line, though. Things are going to start cancelling out and getting shorter. Group all those things-to-powers together:

    -2 \frac{G^4 M^4 m^7}{L^6} + 3 \frac{G^4 M^4 m^7}{L^6}

    Oh. Well, now this is different. The second derivative of the effective potential, at this point, is the number

    \frac{G^4 M^4 m^7}{L^6}

    And I admit I don’t know what number that is. But here’s what I do know: ‘G’ is a positive number. ‘M’ is a positive number. ‘m’ is a positive number. ‘L’ might be positive or might be negative, but ‘L6‘ is a positive number either way. So this is a bunch of positive numbers multiplied and divided together.

    So this second derivative what ever it is must be a positive number. And so this circular orbit is stable. Give the planet a little nudge and that’s all right. It’ll stay near its orbit. I’m sorry to put you through that but some people raised the, honestly, fair question.

    So this is the process you’d use to start understanding orbits for your own arbitrary potential energy. You can find the equivalent of Kepler’s Third Law, the one connecting orbit times and orbit radiuses. And it isn’t really hard. You need to know enough calculus to differentiate one function, and then you need to be willing to do a pile of arithmetic on letters. It’s not actually hard. Next time I hope to talk about the other kinds of central forces that you might get. We only solved one problem here. We can solve way more than that.

     
    • howardat58 6:18 pm on Friday, 21 October, 2016 Permalink | Reply

      I love the chatty approach.

      Like

      • Joseph Nebus 5:03 am on Saturday, 22 October, 2016 Permalink | Reply

        Thank you. I realized doing Theorem Thursdays over the summer that it was hard to avoid that voice, and then that it was fun writing in it. So eventually I do learn, sometimes.

        Like

  • Joseph Nebus 6:00 pm on Wednesday, 28 September, 2016 Permalink | Reply
    Tags: , , , , , , , orbits   

    Why Stuff Can Orbit, Part 5: Why Physics Doesn’t Work And What To Do About It 


    Less way previously:


    My title’s hyperbole, to the extent it isn’t clickbait. Of course physics works. By “work” I mean “model the physical world in useful ways”. If it didn’t work then we would call it “pure” mathematics instead. Mathematicians would study it for its beauty. Physicists would be left to fend for themselves. “Useful” I’ll say means “gives us something interesting to know”. “Interesting” I’ll say if you want to ask what that means then I think you’re stalling.

    But what I mean is that Newtonian physics, the physics learned in high school, doesn’t work. Well, it works, in that if you set up a problem right and calculate right you get answers that are right. It’s just not efficient, for a lot of interesting problems. Don’t ask me about interesting again. I’ll just say the central-force problems from this series are interesting.

    Newtonian, high school type, physics works fine. It shines when you have only a few things to keep track of. In this central force problem we have one object, a planet-or-something, that moves. And only one force, one that attracts the planet to or repels the planet from the center, the Origin. This is where we’d put the sun, in a planet-and-sun system. So that seems all right as far as things go.

    It’s less good, though, if there’s constraints. If it’s not possible for the particle to move in any old direction, say. That doesn’t turn up here; we can imagine a planet heading in any direction relative to the sun. But it’s also less good if there’s a symmetry in what we’re studying. And in this case there is. The strength of the central force only changes based on how far the planet is from the origin. The direction only changes based on what direction the planet is relative to the origin. It’s a bit daft to bother with x’s and y’s and maybe even z’s when all we care about is the distance from the origin. That’s a number we’ve called ‘r’.

    So this brings us to Lagrangian mechanics. This was developed in the 18th century by Joseph-Louis Lagrange. He’s another of those 18th century mathematicians-and-physicists with his name all over everything. Lagrangian mechanics are really, really good when there’s a couple variables that describe both what we’d like to observe about the system and its energy. That’s exactly what we have with central forces. Give me a central force, one that’s pointing directly toward or away from the origin, and that grows or shrinks as the radius changes. I can give you a potential energy function, V(r), that matches that force. Give me an angular momentum L for the planet to have, and I can give you an effective potential energy function, Veff(r). And that effective potential energy lets us describe how the coordinates change in time.

    The method looks roundabout. It depends on two things. One is the coordinate you’re interested in, in this case, r. The other is how fast that coordinate changes in time. This we have a couple of ways of denoting. When working stuff out on paper that’s often done by putting a little dot above the letter. If you’re typing, dots-above-the-symbol are hard. So we mark it as a prime instead: r’. This works well until the web browser or the word processor assumes we want smart quotes and we already had the r’ in quote marks. At that point all hope of meaning is lost and we return to communicating by beating rocks with sticks. We live in an imperfect world.

    What we get out of this is a setup that tells us how fast r’, how fast the coordinate we’re interested in changes in time, itself changes in time. If the coordinate we’re interested in is the ordinary old position of something, then this describes the rate of change of the velocity. In ordinary English we call that the acceleration. What makes this worthwhile is that the coordinate doesn’t have to be the position. It also doesn’t have to be all the information we need to describe the position. For the central force problem r here is just how far the planet is from the center. That tells us something about its position, but not everything. We don’t care about anything except how far the planet is from the center, not yet. So it’s fine we have a setup that doesn’t tell us about the stuff we don’t care about.

    How fast r’ changes in time will be proportional to how fast the effective potential energy, Veff(r), changes with its coordinate. I so want to write “changes with position”, since these coordinates are usually the position. But they can be proxies for the position, or things only loosely related to the position. For an example that isn’t a central force, think about a spinning top. It spins, it wobbles, it might even dance across the table because don’t they all do that? The coordinates that most sensibly describe how it moves are about its rotation, though. What axes is it rotating around? How do those change in time? Those don’t have anything particular to do with where the top is. That’s all right. The mathematics works just fine.

    A circular orbit is one where the radius doesn’t change in time. (I’ll look at non-circular orbits later on.) That is, the radius is not increasing and is not decreasing. If it isn’t getting bigger and it isn’t getting smaller, then it’s got to be staying the same. Not all higher mathematics is tricky. The radius of the orbit is the thing I’ve been calling r all this time. So this means that r’, how fast r is changing with time, has to be zero. Now a slightly tricky part.

    How fast is r’, the rate at which r changes, changing? Well, r’ never changes. It’s always the same value. Anytime something is always the same value the rate of its change is zero. This sounds tricky. The tricky part is that it isn’t tricky. It’s coincidental that r’ is zero and the rate of change of r’ is zero, though. If r’ were any fixed, never-changing number, then the rate of change of r’ would be zero. It happens that we’re interested in times when r’ is zero.

    So we’ll find circular orbits where the change in the effective potential energy, as r changes, is zero. There’s an easy-to-understand intuitive idea of where to find these points. Look at a plot of Veff and imagine this is a smooth track or the cross-section of a bowl or the landscaping of a hill. Imagine dropping a ball or a marble or a bearing or something small enough to roll in it. Where does it roll to a stop? That’s where the change is zero.

    It’s too much bother to make a bowl or landscape a hill or whatnot for every problem we’re interested in. We might do it anyway. Mathematicians used to, to study problems that were too complicated to do by useful estimates. These were “analog computers”. They were big in the days before digital computers made it no big deal to simulate even complicated systems. We still need “analog computers” or models sometimes. That’s usually for problems that involve chaotic stuff like turbulent fluids. We call this stuff “wind tunnels” and the like. It’s all a matter of solving equations by building stuff.

    We’re not working with problems that complicated. There isn’t the sort of chaos lurking in this problem that drives us to real-world stuff. We can find these equilibriums by working just with symbols instead.

     
  • Joseph Nebus 6:00 pm on Thursday, 8 September, 2016 Permalink | Reply
    Tags: , , , , , , orbits, , ,   

    Why Stuff Can Orbit, Part 4: On The L 


    Less way previously:


    We were chatting about central forces. In these a small object — a satellite, a planet, a weight on a spring — is attracted to the center of the universe, called the origin. We’ve been studying this by looking at potential energy, a function that in this case depends only on how far the object is from the origin. But to find circular orbits, we can’t just look at the potential energy. We have to modify this potential energy to account for angular momentum. This essay I mean to discuss that angular momentum some.

    Let me talk first about the potential energy. Mathematical physicists usually write this as a function named U or V. I’m using V. That’s what my professor used teaching this, back when I was an undergraduate several hundred thousand years ago. A central force, by definition, changes only with how far you are from the center. I’ve put the center at the origin, because I am not a madman. This lets me write the potential energy as V = V(r).

    V(r) could, in principle, be anything. In practice, though, I am going to want it to be r raised to a power. That is, V(r) is equal to C rn. The ‘C’ here is a constant. It’s a scaling constant. The bigger a number it is the stronger the central force. The closer the number is to zero the weaker the force is. In standard units, gravity has a constant incredibly close to zero. This makes orbits very big things, which generally works out well for planets. In the mathematics of masses on springs, the constant is closer to middling little numbers like 1.

    The ‘n’ here is a deceiver. It’s a constant number, yes, and it can be anything we want. But the use of ‘n’ as a symbol has connotations. Usually when a mathematician or a physicist writes ‘n’ it’s because she needs a whole number. Usually a positive whole number. Sometimes it’s negative. But we have a legitimate central force if ‘n’ is any real number: 2, -1, one-half, the square root of π, any of that is good. If you just write ‘n’ without explanation, the reader will probably think “integers”, possibly “counting numbers”. So it’s worth making explicit when this isn’t so. It’s bad form to surprise the reader with what kind of number you’re even talking about.

    (Some number of essays on we’ll find out that the only values ‘n’ can have that are worth anything are -1, 2, and 7. And 7 isn’t all that good. But we aren’t supposed to know that yet.)

    C rn isn’t the only kind of central force that could exist. Any function rule would do. But it’s enough. If we wanted a more complicated rule we could just add two, or three, or more potential energies together. This would give us V(r) = C_1 r^{n_1} + C_2 r^{n_2} , with C1 and C2 two possibly different numbers, and n1 and n2 two definitely different numbers. (If n1 and n2 were the same number then we should just add C1 and C2 together and stop using a more complicated expression than we need.) Remember that Newton’s Law of Motion about the sum of multiple forces being something vector something something direction? When we look at forces as potential energy functions, that law turns into just adding potential energies together. They’re well-behaved that way.

    And if we can add these r-to-a-power potential energies together then we’ve got everything we need. Why? Polynomials. We can approximate most any potential energy that would actually happen with a big enough polynomial. Or at least a polynomial-like function. These r-to-a-power forces are a basis set for all the potential energies we’re likely to care about. Understand how to work with one and you understand how to work with them all.

    Well, one exception. The logarithmic potential, V(r) = C log(r), is really interesting. And it has real-world applicability. It describes how strongly two vortices, two whirlpools, attract each other. You can write the logarithm as a polynomial. But logarithms are pretty well-behaved functions. You might be better off just doing that as a special case.

    Still, at least to start with, we’ll stick with V(r) = C rn and you know what I mean by all those letters now. So I’m free to talk about angular momentum.

    You’ve probably heard of momentum. It’s got something to do with movement, only sports teams and political campaigns are always gaining or losing it somehow. When we talk of that we’re talking of linear momentum. It describes how much mass is moving how fast in what direction. So it’s a vector, in three-dimensional space. Or two-dimensional space if you’re making the calculations easier. To find what the vector is, we make a list of every object that’s moving. We take its velocity — how fast it’s moving and in what direction — and multiply that by its mass. Mass is a single number, a scalar, and we’re always allowed to multiply a vector by a scalar. This gets us another vector. Once we’ve done that for everything that’s moving, we add all those product vectors together. We can always add vectors together. And this gives us a grand total vector, the linear momentum of the system.

    And that’s conserved. If one part of the system starts moving slower it’s because other parts are moving faster, and vice-versa. In the real world momentum seems to evaporate. That’s because some of the stuff moving faster turns out to be air objects bumped into, or particles of the floor that get dragged along by friction, or other stuff we don’t care about. That momentum can seem to evaporate is what makes its use in talking about ports teams or political campaigns make sense. It also annoys people who want you to know they understand science words better than you. So please consider this my authorization to use “gaining” and “losing” momentum in this sense. Ignore complainers. They’re the people who complain the word “decimate” gets used to mean “destroy way more than ten percent of something”, even though that’s the least bad mutation of an English word’s meaning in three centuries.

    Angular momentum is also a vector. It’s also conserved. We can calculate what that vector is by the same sort of process, that of calculating something on each object that’s spinning and adding it all up. In real applications it can seem to evaporate. But that’s also because the angular momentum is going into particles of air. Or it rubs off grease on the axle. Or it does other stuff we wish we didn’t have to deal with.

    The calculation is a little harder to deal with. There’s three parts to a spinning thing. There’s the thing, and there’s how far it is from the axis it’s spinning around, and there’s how fast it’s spinning. So you need to know how fast it’s travelling in the direction perpendicular to the shortest line between the thing and the axis it’s spinning around. Its angular momentum is going to be as big as the mass times the distance from the axis times the perpendicular speed. It’s going to be pointing in whichever axis direction makes its movement counterclockwise. (Because that’s how physicists started working this out and it would be too much bother to change now.)

    You might ask: wait, what about stuff like a wheel that’s spinning around its center? Or a ball being spun? That can’t be an angular momentum of zero? How do we work that out? The answer is: calculus. Also, we don’t need that. This central force problem I’ve framed so that we barely even need algebra for it.

    See, we only have a single object that’s moving. That’s the planet or satellite or weight or whatever it is. It’s got some mass, the value of which we call ‘m’ because why make it any harder on ourselves. And it’s spinning around the origin. We’ve been using ‘r’ to mean the number describing how far it is from the origin. That’s the distance to the axis it’s spinning around. Its velocity — well, we don’t have any symbols to describe what that is yet. But you can imagine working that out. Or you trust that I have some clever mathematical-physics tool ready to introduce to work it out. I have, kind of. I’m going to ignore it altogether. For now.

    The symbol we use for the total angular momentum in a system is \vec{L} . The little arrow above the symbol is one way to denote “this is a vector”. It’s a good scheme, what with arrows making people think of vectors and it being easy to write on a whiteboard. In books, sometimes, we make do just by putting the letter in boldface, L, which is easier for old-fashioned word processors to do. If we’re sure that the reader isn’t going to forget that L is this vector then we might stop highlighting the fact altogether. That’s even less work to do.

    It’s going to be less work yet. Central force problems like this mean the object can move only in a two-dimensional plane. (If it didn’t, it wouldn’t conserve angular momentum: the direction of \vec{L} would have to change. Sounds like magic, but trust me.) The angular momentum’s direction has to be perpendicular to that plane. If the object is spinning around on a sheet of paper, the angular momentum is pointing straight outward from the sheet of paper. It’s pointing toward you if the object is moving counterclockwise. It’s pointing away from you if the object is moving clockwise. What direction it’s pointing is locked in.

    All we need to know is how big this angular momentum vector is, and whether it’s positive or negative. So we just care about this number. We can call it ‘L’, no arrow, no boldface, no nothing. It’s just a number, the same as is the mass ‘m’ or distance from the origin ‘r’ or any of our other variables.

    If ‘L’ is zero, this means there’s no total angular momentum. This means the object can be moving directly out from the origin, or directly in. This is the only way that something can crash into the center. So if setting L to be zero doesn’t allow that then we know we did something wrong, somewhere. If ‘L’ isn’t zero, then the object can’t crash into the center. If it did we’d be losing angular momentum. The object’s mass times its distance from the center times its perpendicular speed would have to be some non-zero number, even when the distance was zero. We know better than to look for that.

    You maybe wonder why we use ‘L’ of all letters for the angular momentum. I do. I don’t know. I haven’t found any sources that say why this letter. Linear momentum, which we represent with \vec{p} , I know. Or, well, I know the story every physicist says about it. p is the designated letter for linear momentum because we used to use the word “impetus”, as in “impulse”, to mean what we mean by momentum these days. And “p” is the first letter in “impetus” that isn’t needed for some more urgent purpose. (“m” is too good a fit for mass. “i” has to work both as an index and as that number which, squared, gives us -1. And for that matter, “e” we need for that exponentials stuff, and “t” is too good a fit for time.) That said, while everybody, everybody, repeats this, I don’t know the source. Perhaps it is true. I can imagine, say, Euler or Lagrange in their writing settling on “p” for momentum and everybody copying them. I just haven’t seen a primary citation showing this is so.

    (I don’t mean to sound too unnecessarily suspicious. But just because everyone agrees on the impetus-thus-p story doesn’t mean it’s so. I mean, every Star Trek fan or space historian will tell you that the first space shuttle would have been named Constitution until the Trekkies wrote in and got it renamed Enterprise. But the actual primary documentation that the shuttle would have been named Constitution is weak to nonexistent. I’ve come to the conclusion NASA had no plan in mind to name space shuttles until the Trekkies wrote in and got one named. I’ve done less poking around the impetus-thus-p story, in that I’ve really done none, but I do want it on record that I would like more proof.)

    Anyway, “p” for momentum is well-established. So I would guess that when mathematical physicists needed a symbol for angular momentum they looked for letters close to “p”. When you get into more advanced corners of physics “q” gets called on to be position a lot. (Momentum and position, it turns out, are nearly-identical-twins mathematically. So making their symbols p and q offers aesthetic charm. Also great danger if you make one little slip with the pen.) “r” is called on for “radius” a lot. Looking on, “t” is going to be time.

    On the other side of the alphabet, well, “o” is just inviting danger. “n” we need to count stuff. “m” is mass or we’re crazy. “l” might have just been the nearest we could get to “p” without intruding on a more urgently-needed symbol. (“s” we use a lot for parameters like length of an arc that work kind of like time but aren’t time.) And then shift to the capital letter, I expect, because a lowercase l looks like a “1”, to everybody’s certain doom.

    The modified potential energy, then, is going to include the angular momentum L. At least, the amount of angular momentum. It’s also going to include the mass of the object moving, and the radius r that says how far the object is from the center. It will be:

    V_{eff}(r) = V(r) + \frac{L^2}{2 m r^2}

    V(r) was the original potential, whatever that was. The modifying term, with this square of the angular momentum and all that, I kind of hope you’ll just accept on my word. The L2 means that whether the angular momentum is positive or negative, the potential will grow very large as the radius gets small. If it didn’t, there might not be orbits at all. And if the angular momentum is zero, then the effective potential is the same original potential that let stuff crash into the center.

    For the sort of r-to-a-power potentials I’ve been looking at, I get an effective potential of:

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

    where n might be an integer. I’m going to pretend a while longer that it might not be, though. C is certainly some number, maybe positive, maybe negative.

    If you pick some values for C, n, L, and m you can sketch this out. If you just want a feel for how this Veff looks it doesn’t much matter what values you pick. Changing values just changes the scale, that is, where a circular orbit might happen. It doesn’t change whether it happens. Picking some arbitrary numbers is a good way to get a feel for how this sort of problem works. It’s good practice.

    Sketching will convince you there are energy minimums, where we can get circular orbits. It won’t say where to find them without some trial-and-error or building a model of this energy and seeing where a ball bearing dropped into it rolls to a stop. We can do this more efficiently.

     
  • Joseph Nebus 6:00 pm on Thursday, 25 August, 2016 Permalink | Reply
    Tags: , , , , , , orbits,   

    Why Stuff Can Orbit, Part 3: It Turns Out Spinning Matters 


    Way previously:

    Before the big distractions of Theorem Thursdays and competitive pinball events and all that I was writing up the mathematics of orbits. Last time I’d got to establishing that there can’t be such a thing as an orbit. This seems to disagree with what a lot of people say we can observe. So I want to resolve that problem. Yes, I’m aware I’m posting this on a Thursday, which I said I wasn’t going to do because it’s too hard on me to write. I don’t know how it worked out like that.

    Let me get folks who didn’t read the previous stuff up to speed. I’m using as model two things orbiting each other. I’m going to call it a sun and a planet because it’s way too confusing not to give things names. But they don’t have to be a sun and a planet. They can be a planet and moon. They can be a proton and an electron if you want to pretend quantum mechanics isn’t a thing. They can be a wood joist and a block of rubber connected to it by a spring. That’s a legitimate central force. They can even be stuff with completely made-up names representing made-up forces. So far I’m supposing the things are attracted or repelled by a force with a strength that depends on how far they are from each other but on nothing else.

    Also I’m supposing there are only two things in the universe. This is because the mathematics of two things with this kind of force is easy to do. An undergraduate mathematics or physics major can do it. The mathematics of three things is too complicated to do. I suppose somewhere around two-and-a-third things the mathematics hard enough you need an expert but the expert can do it.

    Mathematicians and physicists will call this sort of problem a “central force” problem. We can make it easier by supposing the sun is at the center of the universe, or at least our coordinate system. So we don’t have to worry about it moving. It’s just there at the center, the “origin”, and it’s only the planet that moves.

    Forces are tedious things to deal with. They’re vectors. In this context that makes them bundles of three quantities each related to the other two. We can avoid a lot of hassle by looking at potential energy instead. Potential energy is a scalar, a single number. Numbers are nice and easy. Calculus tells us how to go from potential energy to forces, in case we need the forces. It also tells us how to go from forces to potential energy, so we can do the easier problem instead. So we do.

    To write about potential energy mathematical physicists use exactly the letter you would guess they’d use if every other letter were unavailable for some reason: V. Or U, if they prefer. I’ll stick with V. Right now I don’t want to say anything about what rule determines the values of V. I just want to allow that its value changes as the planet’s distance from the star — the radius ‘r’ of its orbit — changes. So we make that clear by writing the potential energy is V = V(r). (The potential energy might change with the mass of the planet or sun, or the strength of gravity in the universe, or whatever. But we’re going to pretend those don’t change, not for the problem we’re doing, so we don’t have to write them out.)

    If you draw V(r) versus r you can discover right away circular orbits. They’re ones that are local maximums or local minimums of V(r). Physical intuition will help us here. Imagine the graph of the potential energy as if it were a smooth bowl. Drop a marble into it. Where would the marble come to rest? That’s a local minimum. The radius of that minimum is a circular orbit. (Oh, a local maximum, where the marble is at the top of a hill and doesn’t fall to either side, could be a circular orbit. But it isn’t going to be stable. The marble will roll one way or another given the slightest chance.)

    The potential energy for a force like gravity or electric attraction looks like the distance, r, raised to a power. And then multiplied by some number, which is where we hide gravitational constants and masses and all that stuff. Generally, it looks like V(r) = C rn where C is some number and n is some other number. For gravity and electricity that number is -1. For two particles connected by a spring that number n is +2. Could be anything.

    The trouble is if you draw these curves you realize that a marble dropped in would never come to a stop. It would roll down to the center, the planet falling into the sun. Or it would roll away forever, the planet racing into deep space. Either way it doesn’t orbit or do anything near orbiting. This seems wrong.

    It’s not, though. Suppose the force is repelling, that is, the potential energy gets to be smaller and smaller numbers as the distance increases. Then the two things do race away from each other. Physics students are asked to imagine two positive charges let loose next to each other. Physics students understand they’ll go racing away from each other, even though we don’t see stuff in the real world that does that very often. We suppose the students understand, though. These days I guess you can make an animation of it and people will accept that as if it’s proof of anything.

    Suppose the force is attracting. Imagine just dropping a planet out somewhere by a sun. Set it carefully just in place and let it go and get out of the way before happens. This is what we do in physics and mathematics classes, so that’s the kind of fun stuff you skipped if you majored in something else. But then we go on to make calculations about it. But that’ll orbit, right? It won’t just drop down into the sun and get melted or something?

    Not so, the way I worded it. If we set the planet into space so it was holding still, not moving at all, then it will fall. Plummet, really. The planet’s attracted to the sun, and it moves in that direction, and it’s just going to keep moving that way. If it were as far from the center as the Earth is from the Sun it’ll take its time, yes, but it’ll fall into the sun and not do anything remotely like orbiting. And yet there’s still orbits. What’s wrong?

    What’s wrong is a planet isn’t just sitting still there waiting to fall into the sun. Duh, you say. But why isn’t it just sitting still? That’s because it’s moving. Might be moving in any direction. We can divide that movement up into two pieces. One is the radial movement, how fast it’s moving towards or away from the center, that is, along the radius between sun and planet. If it’s a circular orbit this speed is zero; the planet isn’t moving any closer or farther away. If this speed isn’t zero it might affect how fast the planet falls into the sun, but it won’t affect the fact of whether it does or not. No more than how fast you toss a ball up inside a room changes whether it’ll eventually hit the floor. </p.

    It’s the other part, the transverse velocity, that matters. This is the speed the thing is moving perpendicular to the radius. It’s possible that this is exactly zero and then the planet does drop into the sun. It’s probably not. And what that means is that the planet-and-sun system has an angular momentum. Angular momentum is like regular old momentum, only for spinning. And as with regular momentum, the total is conserved. It won’t change over time. When I was growing up this was always illustrated by thinking of ice skaters doing a spin. They pull their arms in, they spin faster. They put their arms out, they spin slower.

    (Ice skaters eventually slow down, yes. That’s for the same reasons they slow down if they skate in a straight line even though regular old momentum, called “linear momentum” if you want to be perfectly clear, is also conserved. It’s because they have to get on to the rest of their routine.)

    The same thing has to happen with planets orbiting a sun. If the planet moves closer to the sun, it speeds up; if it moves farther away, it slows down. To fall into the exact center while conserving angular momentum demands the planet get infinitely fast. This they don’t typically do.

    There was a tipoff to this. It’s from knowing the potential energy V(r) only depends on the distance between sun and planet. If you imagine taking the system and rotating it all by any angle, you wouldn’t get any change in the forces or the way things move. It would just change the values of the coordinates you used to describe this. Mathematical physicists describe this as being “invariant”, which means what you’d imagine, under a “continuous symmetry”, which means a change that isn’t … you know, discontinuous. Rotating thing as if they were on a pivot, that is, instead of (like) reflecting them through a mirror.

    And invariance under a continuous symmetry like this leads to a conservation law. This is known from Noether’s Theorem. You can find explained quite well on every pop-mathematics and pop-physics blog ever. It’s a great subject for pop-mathematics/physics writing. The idea, that the geometry of a problem tells us something about its physics and vice-versa, is important. It’s a heady thought without being so exotic as to seem counter-intuitive. And its discoverer was Dr Amalie Emmy Noether. She’s an early-20th-century demonstration of the first-class work that one can expect women to do when they’re not driven out of mathematics. You see why the topic is so near irresistible.

    So we have to respect the conservation of angular momentum. This might sound like we have to give up on treating circular orbits as one-variable problems. We don’t have to just yet. We will, eventually, want to look at not just how far the planet is from the origin but also in what direction it is. We don’t need to do that yet. We have a brilliant hack.

    We can represent the conservation of angular momentum as a slight repulsive force. It’s not very big if the angular momentum is small. It’s not going to be a very big force unless the planet gets close to the origin, that is, until r gets close to zero. But it does grow large and acts as if the planet is being pushed away. We consider that a pseudoforce. It appears because our choice of coordinates would otherwise miss some important physics. And that’s fine. It’s not wrong any more than, say, a hacksaw is the wrong tool to cut through PVC pipe just because you also need a vise.

    This pseudoforce can be paired with a pseduo-potential energy. One of the great things about the potential-energy view of physics is that adding two forces together is as easy as adding their potential energies together. We call the sum of the original potential energy and the angular-momentum-created pseudopotential the “effective potential energy”. Far from the origin, for large radiuses r, this will be almost identical to the original potential energy. Close to the origin, this will be a function that rises up steeply. And as a result there can suddenly be a local minimum. There can be a circular orbit.

    Spring potential, which is a parabola growing with the distance r from the origin. And the effective potential, which grows to a vertical asymptote where the radius is zero.

    Figure 1. The potential energy of a spring — the red line — and the effective potential energy — the blue line — when the angular momentum is added as a pseudoforce. Without angular momentum in consideration the only equilibrium is at the origin. With angular momentum there’s some circular orbit, somewhere. Don’t pay attention to the numbers on the axes. They don’t mean anything.


    Gravitational potential, with a vertical asymptote at the radius equalling zero going down to negative infinitely great numbers and a horizontal asymptote at the radius going off to infinity. And the effective potential, with the vertical asymptote at radius of zero going to positive infinitely great numbers, forcing there to be some minimum: a circular orbit.

    Figure 2. The potential energy of a gravitational attraction — the red line — and the effective potential energy — the blue line — when the angular momentum is added as a pseudoforce. Without angular momentum in consideration there’s no equilibrium. The thing, a planet, falls into the center, the sun. With angular momentum there’s some circular orbit. As before the values of the numbers don’t matter and you should just ignore them.

    The location of the minimum — the radius of the circular orbit — will depend on the original potential, of course. It’ll also depend on the angular momentum. The smaller the angular momentum the closer to the origin will be the circular orbit. If the angular momentum is zero we have the original potential and the planet dropping into the center again. If the angular momentum is large enough there might not even be a minimum anymore. That matches systems where the planet has escape velocity and can go plunging off into deep space. And we can see this by looking at the plot of the effective velocity even before we calculate things.

    Gravitational potential, with a vertical asymptote at the radius equalling zero going down to negative infinitely great numbers and a horizontal asymptote at the radius going off to infinity. And then the effective potential for a very large angular momentum. There's a vertical asymptote at radius of zero going to positive infinitely great numbers, and so large that there isn't any local minimum except at an infinitely large radius.

    Figure 3. Gravitational potential energy — the red line — and the effective potential energy — the blue line — when angular momentum is considered. In this case the angular momentum is so large, that is, the planet is moving so fast, that there are no orbits. The planet’s reached escape velocity and can go infinitely far away from the sun.

    This only goes so far as demonstrating a circular orbit should exist. Or giving some conditions for which a circular orbit wouldn’t. We might want to know something more, like where that circular orbit is. Or if it’s possible for there to be an elliptic orbit. Or other shapes. I imagine it’s possible to work this out with careful enough drawings. But at some point it gets easier to just calculate things. We’ll get to that point soon.

     
  • Joseph Nebus 3:00 pm on Friday, 27 May, 2016 Permalink | Reply
    Tags: , , , , , orbits   

    Why Stuff Can Orbit, Part 2: Why Stuff Can’t Orbit 


    Previously:

    As I threatened last week, I want to talk some about central forces. They’re forces in which one particle attracts another with a force that depends only on how far apart the two are. Last week’s essay described some of the assumptions behind the model.

    Mostly, we can study two particles interacting as if it were one particle hovering around the origin. The origin is some central reference point. If we’re looking for a circular orbit then we only have to worry about one variable. This would be ‘r’, the radius of the orbit: how far the planet is from the sun it orbits.

    Now, central forces can follow any rule you like. Not in reality, of course. In reality there’s two central forces you ever see. One is gravity (electric attraction or repulsion follows the same rule) and the other is springs. But we can imagine there being others. At the end of this string of essays I hope to show why there’s special things about these gravity and spring-type forces. And by imagining there’s others we can learn something about why we only actually see these.

    So now I’m going to stop talking about forces. I’ll talk about potential energy instead. There’s several reasons for this, but they all come back to this one: energy is easier to deal with. Energy is a scalar, a single number. A force is a vector, which for this kind of physics-based problem is an array of numbers. We have less to deal with if we stick to energy. If we need forces later on we can get them from the energy. We’ll need calculus to do that, but it won’t be the hard parts of calculus.

    The potential energy will be some function. As a central force it’ll depend only on the distance, r, that a particle is from the origin. It’s convenient to have a name for this. So I will use a common name: V(r). V is a common symbol to use for potential energy. U is another. The (r) emphasizes that this is some function which depends on r. V(r) doesn’t commit us to any particular function, not at this point.

    You might ask: why is the potential energy represented with V, or with U? And I don’t really know. Sometimes we’ll use PE to mean potential energy, which is as clear a shorthand name as we could hope for. But a name that’s two letters like that tends to be viewed with suspicion when we have to do calculus work on it. The label looks like the product of P and E, and derivatives of products get tricky. So it’s a less popular label if you know you’re going take the derivative of the potential energy anytime soon. EP can also get used, and the subscript means it doesn’t look like the product of any two things. Still, at least in my experience, U and V are most often used.

    As I say, I don’t know just why it should be them. It might just be that the letters were available when someone wrote a really good physics textbook. If we want to assume there must be some reason behind this letter choice I have seen a plausible guess. Potential energy is used to produce work. Work is W. So potential energy should be a letter close to W. That suggests U and V, both letters that are part of the letter W. (Listen to the name of ‘W’, and remember that until fairly late in the game U and V weren’t clearly distinguished as letters.) But I do not know of manuscript evidence suggesting that’s what anyone every thought. It is at best a maybe useful mnemonic.

    Here’s an advantage that using potential energy will give us: we can postpone using calculus a little. Not for quantitative results. Not for ones that describe exactly where something should orbit. But it’s good for qualitative results. We can answer questions like “is there a circular orbit” and “are there maybe several plausible orbits” just by looking at a picture.

    That picture is a plot of the values of V(r) against r. And that can be anything. I mean it. Take your preferred drawing medium and draw any wiggly curve you like. It can’t loop back or cross itself or something like that, but it can be as smooth or as squiggly as you like. That’s your central-force potential energy V(r).

    Are there any circular orbits for this potential? Calculus gives us the answer, but we don’t need that. For a potential like our V(r), that depend on one variable, we can just look. (We could also do this for a potential that depends on two variables.) Take your V(r). Imagine it’s the sides of a perfectly smooth bowl or track or something. Now imagine dropping a marble or a ball bearing or something nice and smooth on it. Does the marble come to a rest anywhere? That’s your equilibrium. That’s where a circular orbit can happen.

    A generic wiggly shape with a bunch of peaks and troughs.

    Figure 1. A generic yet complicated V(r). Spoiler: I didn’t draw this myself because I figured using Octave was easier than using ArtRage on my iPad.

    We’re using some real-world intuition to skip doing analysis. That’s all right in this case. Newtonian mechanics say that a particle’s momentum changes in the direction of a force felt. If a particle doesn’t change its mass, then that means it accelerates where the force, uh, forces it. And this sort of imaginary bowl or track matches up the potential energy we want to study with a constrained gravitational potential energy.

    My generic V(r) was a ridiculous function. This sort of thing doesn’t happen in the real world. But they might have. Wiggly functions like that were explored in the 19th century by physicists trying to explain chemistry. They hoped complicated potentials would explain why gases expanded when they warmed and contracted when they cooled. The project failed. Atoms follow quantum-mechanics laws that match only vaguely match Newtonian mechanics like this. But just because functions like these don’t happen doesn’t mean we can’t learn something from them.

    We can’t study every possible V(r). Not at once. Not without more advanced mathematics than I want to use right now. What I’d like to do instead is look at one family of V(r) functions. There will be infinitely many different functions here, but they’ll all resemble each other in important ways. If you’ll allow me to introduce two new numbers we can describe them all with a single equation. The new numbers I’ll name C and n. They’re both constants, at least for this problem. They’re some numbers and maybe at some point I’ll care which ones they are, but it doesn’t matter. If you want to pretend that C is another way to write “eight”, go ahead. n … well, you can pretend that’s just another way to write some promising number like “two” for now. I’ll say when I want to be more specific about it.

    The potential energy I want to look at has a form we call a power law, because it’s all about raising a variable to a power. And we only have the one variable, r. So the potential energy looks like this:

    V(r) = C r^n

    There are some values of n that it will turn out are meaningful. If n is equal to 2, then this is the potential energy for two particles connected by a spring. You might complain there are very few things in the world connected to other things by springs. True enough, but a lot of things act as if they were springs. This includes most anything that’s near but pushed away from a stable equilibrium. It’s a potential worth studying.

    If n is equal to -1, then this is the potential energy for two particles attracting each other by gravity or by electric charges. And here there’s an important little point. If the force is attractive, like gravity or like two particles having opposite electric charges, then we need C to be a negative number. If the force is repulsive, like two particles having the same electric charge, then we need C to be a positive number.

    Although n equalling two, and n equalling negative one, are special cases they aren’t the only ones we can imagine. n may be any number, positive or negative. It could be zero, too, but in that case the potential is a flat line and there’s nothing happening there. That’s known as a “free particle”. It’s just something that moves around with no impetus to speed up or slow down or change direction or anything.

    So let me sketch the potentials for positive n, first for a positive C and second for a negative C. Don’t worry about the numbers on either the x- or the y-axes here; they don’t matter. The shape is all we care about right now.

    The curve starts at zero and rises ever upwards as the radius r increases.

    Figure 2. V(r) = C rn for a positive C and a positive n.


    The curve starts at zero and drops ever downwards as the radius r increases.

    Figure 3. V(r) = C rn for a negative C and a positive n.

    Now let me sketch the potentials for a negative n, first for a positive C and second for a negative C.

    The curve starts way high up and keeps dropping, but levelling out, as the radius r increases

    Figure 4. V(r) = C rn for a positive C and a negative n.


    The curve starts way down low and rises, but levelling out, as the radius r increases.

    Figure 5. V(r) = C rn for a negative C and a negative n.

    And now we can look for equilibriums, for circular orbits. If we have a positive n and a positive C, then — well, do the marble-in-a-bowl test. Start from anywhere; the marble rolls down to the origin where it smashes and stops. The only circular orbit is at a radius r of zero.

    With a positive n and a negative C, start from anywhere except a radius r of exactly zero and the marble rolls off to the right, without ever stopping. The only circular orbit is at a radius r of zero.

    With a negative n and a positive C, the marble slides down a hill that gets more shallow but that never levels out. It rolls off getting ever farther from the origin. There’s no circular orbits.

    With a negative n and a negative C, start from anywhere and the marble rolls off to the left. The marble will plummet down that ever-steeper hill. The only circular orbit is at a radius r of zero.

    So for all these cases, with a potential V(r) = C rn, the only possible “orbits” have both particles zero distance apart. Otherwise the orbiting particle smashes right down into the center or races away never to be seen again. Clearly something has gone wrong with this little project.

    If you’ve spotted what’s gone wrong please don’t say what it is right away. I’d like people to ponder it a little before coming back to this next week. That will come, I expect, shortly after the first Theorem Thursday post. If you have any requests for that project, please get them in, the sooner the better.

     
  • Joseph Nebus 3:00 pm on Friday, 20 May, 2016 Permalink | Reply
    Tags: , , orbits, polar coordinates   

    Why Stuff Can Orbit, Part 1: Laying Some Groundwork 


    My recent talking about central forces got me going. There’s interesting stuff about what central forces allow things to orbit one another, and what forces allow for closed orbits. And I feel like trying out a bit of real mathematics, the kind that physics majors do as undergraduates, around here. I should get something for the student loans I’m still paying off and I’ll accept “showing off on my meager little blog here” as something.

    Central forces are, uh, forces. Pairs of particles attract each other. The strength of the attraction depends on how far apart they are. The direction of the attraction is exactly towards the other in the pair. So it works like gravity or electric attraction. It might follow a different rule, although I know I’m going to casually refer to things as “gravity” or “gravitational” because that’s just too familiar a reference. I’m formally talking about a problem in classical mechanics, but the ideas and approaches come from orbital mechanics. The language of orbital mechanics comes along with it.

    And it is too possible that the force would point some other way. Electric charges in a magnetic field feel a force perpendicular to the magnet. And we can represent vortices, things that swirl around the way cyclones do, as particles pushing each other in perpendicular directions. We’re not going to deal with those.

    The easiest kind of orbit to find is a circular one, made by a single pair of particles. I so want to describe that, but if I do, I’m just going to make things more confusing. It’s an orbit that’s a circle. And we’re sticking to a single pair of particles because it turns out it’s easy to describe the central-force movement of two particles. And it’s kind of impossible to describe the central-force movement three particles. So, let’s stick to two.

    When we start thinking about what we need to describe the system it’s easy to despair. We need the x, y, and z coordinates for two particles. Plus there’s the mass of both particles. Plus there’s some gravitational constant, however strong the force itself is. That’s at least nine things to keep track of.

    We don’t need all that. Physics helps us. Ever hear of the Conservation of Angular Momentum? It’s that thing that makes an ice skater twirling around speed up by pulling in his arms and slow down by reaching them out again. In an argument I’m not dealing with here, the Conservation of Angular Momentum tells us the two particles are going to keep to a single plane. They can move together or apart, but they’ll trace out paths in a two-dimensional slice of space. We can, without loss of generality, suppose it to be the horizontal plane. That is, that the z-coordinate for both planets starts as zero and stays there. So we’re down to seven things to keep track of.

    We can simplify some other stuff. For example, suppose we have one really big mass and one really small one: a sun and a planet, or a planet and a satellite. The sun isn’t going to move very much; the planet hasn’t got enough gravity to matter. We can pretend the sun doesn’t move. We’ll make a little error, but it’ll be small enough we don’t have to care. So we’re down to five things to keep track of.

    And we’ll do better. The strength of the attractive force isn’t going to change because we don’t need a universe that complicated. The mass of the sun and the planet? Well, that could change, if we wanted to work out how rockets behave. We don’t. So their masses are not going to change. So that’s three things whose value we might not have, but which aren’t going to change. We’ll give those numbers labels that will be letters, but there’s nothing to keep track of. They don’t change. We only have to worry about the x- and y-coordinates of the planet.

    But we don’t even have to do that, not really. The force between the sun and the planet depends on how far apart they are. This almost begs us to use polar coordinates instead of Cartesian coordinates. In polar coordinates we identify a point by two things. First is how far it is from the origin. Second is what angle the line from the origin to that point makes with some reference line. And if we’re looking for a circular orbit, then we don’t care what the angle is. It’s going to start at some arbitrary value and increase (or decrease) steadily in time. We don’t have to keep track of it. The only thing that changes that we have to keep track of is the distance between the sun and the planet. Since this is a distance, we naturally call this ‘r’. Well, it’s the radius of the circle traced out by the planet. That’s why it makes sense.

    So we have one thing that changes, r. And we have a couple things whose value we don’t know, but which aren’t going to change during the problem. This is getting manageable. (Later on, when we’ll want to allow for elliptic or other funny-shaped orbits, we’ll need an angle θ. But by then we’ll be so comfortable with one variable we’ll be looking to get the thrill of the challenge back.)

    When I pick this up again I mean to introduce all the kinds of central forces that we might possibly look at. And then how right away we can see there’s no such thing as an orbit. Should be fun.

     
    • davekingsbury 2:47 pm on Saturday, 21 May, 2016 Permalink | Reply

      No such thing as an orbit … food for thought, indeed!

      Like

      • Joseph Nebus 3:31 am on Thursday, 2 June, 2016 Permalink | Reply

        Well, I do that by leaving out a piece that’s important. But it’s one that’s easy to leave out if you’re just skimming through a physics book instead of doing all the pieces along it.

        Like

    • davekingsbury 9:30 am on Thursday, 2 June, 2016 Permalink | Reply

      I’m sure the insight is true, though. If gravity affects everything there can be no fixed points in the dance.

      Like

      • Joseph Nebus 7:45 pm on Saturday, 4 June, 2016 Permalink | Reply

        Well, now, “fixed” is a funny word in this context. But there are orbits that keep to a fixed distance, after all. And if all we’re looking at is how far apart two things are, then that circular, fixed-distance orbit looks on my little plots here like a particle staying still.

        I just haven’t got to explaining how they can happen given that, at least to start with, it looks like pairs of particles should crash together or race apart from one another.

        Liked by 1 person

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