After last week’s dark matter post in which I mentioned that the outer planets are orbiting more slowly than the inner ones due to Kepler’s third law, Jim commented

In my head Neptune was going super-fast but over a gigantic distance which explained the longer time.

BIG MISTAKE.

Kepler’s laws are a good topic of discussion because while they do have relatively complicated mathematical expressions to describe them, they are also easily and concisely summed up in the form of simple words. Kepler’s laws are as follows:

1) The orbit of every planet is an ellipse, with the Sun at one of the foci.

2) A hypothetical line tethering a planet to the Sun will sweep out equal areas during equal periods of time.

3) The square of the period of the orbit is proportional to the cube of the semi-major axis of its orbit.

Sounds easy enough, but there’s still some snags in there; for example, what is a semi-major axis? Why is the line covering equal areas such a big deal? And if we go around the Sun why isn’t it at the centre of the ellipse rather than “at one of the foci”?  These are excellent questions, little Jimmy, and answering them goes a long way towards explaining why the solar system works the way it does.

Starting with the first law, the orbit of every planet (and everything else that goes around the Sun) being an ellipse has some profound implications for certain characteristics of those orbits such as the orbital velocity and where exactly the Sun is going to be in relation to the orbiting body. An ellipse looks like this:

The semi-major axis of an ellipse is the longest line that can be drawn from the ellipse’s edge towards its centre. An object found at the point where the major axis intersects with the ellipse will be at the furthest distance from the centre possible. The semi-minor axis is the reverse; it’s the shortest line that can be drawn from the edge of the ellipse towards its centre. The focii of the ellipse are two points located on the major axis which are equidistant from the centre. Exactly what that distance is is calculated by subtracting the square of the semi-minor axis from the square of the semi-major axis, and then taking a square root of the total.

(Everyone got that?)

For bodies orbiting the Sun, it is always located at one of the focii of the elliptical path they follow. If you have a body orbiting in a circular (or practically circular) orbit, then the length of the semi-major axis will be identical to the length of the semi-minor axis, and the distance of the focii from the centre of a circular orbit will turn out to be zero — which is why we go around the Sun in a nice, orderly fashion instead of screaming in for a close approach on a highly elliptical path like a comet. Turning an orbit into a circle also greatly simplifies laws two and three; since a body on a circular orbit will always be at a fixed distance from the Sun it will always have a fixed velocity, and you can calculate its orbital period by cubing its distance from the Sun at any point, since that distance is always the same.

However, while we can calculate the dynamics of circular (or near-circular¹) orbits using laws intended for ellipses, not all orbits are circular. Comets and Kuiper belt objects in particular tend to follow highly elliptical orbits; comets will famously make close approaches to the Sun during which they display a prominent tail due to surface outgassing blasting off ice and other particulate matter before retreating to the outer solar system and beyond, and Kepler’s laws explain their behaviour too.

This is the orbit of a hypothetical comet. As you can see it closely resembles the diagram of the ellipse with the Sun located at one of the focii, so that’s law one taken care of. Law two states that if we draw a hypothetical line between the Sun and the comet as it orbits around it, that line will sweep out equal areas in equal times. If you’re having difficulty visualising how this would work there is a neat animation on Wikipedia that illustrates it;  the blue shaded zone always has a constant area, and each slice of the ellipse in the red zone also has the same area and represents the distance an orbiting body will travel over a fixed period of time at that point in its orbit. The equal area thing basically means that if the area swept out by the imaginary line is a function of the distance of the comet from the Sun multiplied by the comet’s velocity, and if that area is always constant over a given period of time, then in order to keep that area constant as the comet’s distance from the Sun increases and decreases over time the velocity of the comet must increase and decrease too. When it is far away from the Sun the comet’s velocity will be small. When it is at its point of closest approach the comet’s velocity will be at its fastest.  This is a geometric representation of the comet converting gravitational potential energy into kinetic energy — and thus additional velocity — as it falls in towards the sun, and then losing that kinetic energy again as it moves outwards against the pull of the Sun’s gravity. The second law therefore tells how how the the velocity of an orbiting body will change as its distance from the Sun changes, which is crucial for plotting exactly where a comet or an asteroid is going to end up at a given point in the future. When astronomers find an Earth-threatening asteroid, Kepler’s second law is one of the things they’ll use to figure out if it’s going to hit us or not.

Finally there’s Kepler’s third law, which is the most interesting of the lot when you apply it to circular orbits. “The square of the period of the orbit is proportional to the cube of the semi-major axis of its orbit” doesn’t exactly roll off the tongue, but what this boils down to is that if the orbital period is determined by the distance of the orbiting body from the Sun, and if the orbital circumference — the distance that body covers over the orbital period — is also determined by the distance of the orbiting body from the Sun, as per 2πr, and if the orbital velocity is constant, as it must be for a circular orbit according to Kepler’s second law, then what you end up with is:

The velocity of a body in a circular orbit is a single, constant number which is also determined by the distance from that body to the Sun. In other words, if you’re X million kilometres away from the Sun there is only one velocity you can travel at to achieve a circular orbit; travelling faster than this will cause your orbit to shift outwards into an elliptical shape (or maybe even escape the solar system entirely), and travelling slower than this will cause you to fall inwards towards the Sun. This is best contextualised using the hoary old analogy of the cannon on a mountain firing a cannonball around the Earth. If it’s travelling too fast it’ll escape the Earth’s gravity and fly away from the Earth; too slow and it’ll succumb the the Earth’s gravity and crash into the surface. It needs to be travelling at a velocity that is just right in order to achieve a stable orbit.

This means that for each planet orbiting the Sun there is only one velocity it can possibly be moving at, determined by how far away from the Sun it is. This velocity decreases the further away from the Sun you go, with the orbital velocity of Mercury being 47.87 km/s compared to the orbital velocity of Neptune, a mere 5.42 km/s. Being able to calculate typical orbital velocities from just how far away from the Sun a planet or an asteroid is is very useful for several branches of solar system science; I’m most familiar with it as a way of estimating impact velocities — and consequently how destructive they’re going to be — as while these will vary depending on the direction of travel of both the impacting body and the body being impacted (head-on collisions are obviously more energetic) it’s perfectly possible to get upper and lower bounds from models based on Kepler’s second law. (For your information the typical impact speed of an asteroid hitting the Earth is likely to be between 10 and 70 kilometres per second, which is fast enough for even a small chunk of rock to really mess up somebody’s day. If you’re dealing with stuff in the Kuiper belt the range of impact speeds is much smaller — around 1 to 9 kilometres per second — since everything out there moves so slowly.) So again, Kepler’s laws aren’t just for orbital dynamics; they’re also useful for figuring out how we’re all going to die in some horrifying apocalyptic nightmare scenario!

The post When Science Posts Go Wrong. appeared first on The Scientific Gamer.

And now, the thing that indirectly led to last week’s post on Race Into Space: the Kerbal Space Program.

KSP has been in development for a while now. I first played it this time last year, when it was a free alpha and had a grand total of two rocket engines, two couplers and one capsule. Even then its potential was clear, because even though that version was the most basic bare-bones version of the concept, the concept just happens to be making a pseudo-accurate simulation of building and launching your own space rockets. It’s fantastic. No, really.

These days you have to pay $15 for the latest alpha version (although the old one is still free). You totally should, though; a year of development has added two moons you can land on, an improved UI that indicates atmospheric pressure and your projected orbital — or escape – trajectory, and a whole host of new rocket parts including RCS thrusters, jet and ramjet engines, wings, control surfaces, landing struts, fuel lines… it’s a very long list and the game isn’t even close to being finished yet, with a slowly-expanding BARIS-style space centre that remains mostly non-functional at the moment while the developers nail down the sandbox element of the game.

In Kerbal Space Program you are the omnipotent designer, builder, launcher and pilot of the Kerbal race’s Heath Robinson-esque attempts to escape the gravity well of their home planet, Kerbin.  You start your spaceship design with a command capsule containing three suicidal Kerbal astronauts. As your first rocket you might decide to keep things simple by adding just a fuel tank and an engine of some kind underneath the capsule; the builder is fairly intuitive with the modular rocket parts automatically snapping to pre-determined connection points, so this wouldn’t take more than thirty seconds or so. Then you rush it out to the launch pad to see what happens.

Now, I can’t speak for what will happen to this simple rocket design, but I can tell you what would probably happen. You’d ignite the engines, increase the throttle, lift off – and then nosedive into the ground because you forgot to include any control surfaces or the automatic SAS stabilisation system to keep the rocket stable as it ascends through the atmosphere. Or if that doesn’t happen, you burn your engines too hard and they overheat and explode. Or if that doesn’t happen, you run out of fuel well before you get anywhere near orbit, the rocket starts dropping back towards Kerbin, and you suddenly realise that not only is there no way to separate the command capsule from the main body of the rocket, but that even if there was the three Kerbal astronauts would be doomed anyway because you didn’t include that most basic piece of spacecraft equipment: a parachute,

There’s a lot of ways you can fail in KSP, and seeing just what kind of unanticipated disaster will befall your spacecraft next is half of KSP’s fun, especially when the failures are so spectacular. It can honestly take a bit of work just get a spacecraft design off the launchpad when they have a penchant for falling to bits if not structured correctly. Rocket engines that run on liquid fuel need some way for the fuel to get to them, either directly or through a fuel crossfeed. Somebody new to the game might be tempted to attach some solid fuel rocket boosters onto their spaceship, but while these have the great virtue of being simple they also  have the slight drawback of essentially being an unstable lump of explosive material stuck to the side of your rocket. Even if you manage to get it off the ground you have a hell of a task ahead of you just controlling your heading, orientation and fuel burns to get into orbit. KSP is a game where succeeding in your chosen goal for the first time – suborbital, orbital, moon landing, whatever — does not come easily. However, behind the succession of comedy failures lurks a surprisingly deep iterative learning process which mimics the way real space programs are developed.

(Except those tend to have less dead astronauts, obviously.)

KSP is a game that – for now, at least – fosters experimentation. At the moment it consists of a basic sandbox mode where all rocket parts are free and you can have as many as you want on your spaceship, while pile-driving your three astronauts into the bottom of a burning crater carries no penalties besides having to take that particular rocket design back to the drawing board. You are free to try as many launches as you want in order to make it work. And slowly, gradually, you’ll start to figure out what you should and shouldn’t be doing. It becomes apparent that more engines do not necessarily equate to more thrust when your burn all six of them at once and your rocket only gets about a foot off the launch pad, and that experimenting with explosive decouplers that separate a rocket into discrete stages is far more efficient. How many stages do you want, though? How many boosters per stage? Do you really need that tri-coupler there? Fine-tuning this, as well as the fuel/payload ratio, is what occupies you during the first couple of dozen launches. Eventually, though, you’re going to refine your design to the point where it’s capable of escaping Kerbin’s gravity well, and it’s at this point that you encounter a whole new challenge: getting into orbit.

This is not as simple as you might think. Orbit is often described as “falling without hitting the ground” and hey, falling’s pretty easy, right? It’s just a matter of giving yourself enough sideways velocity so that you fall towards the planet you’re trying to orbit at the same rate as the ground falls away from you due to its curved surface. Surely, then, orbit must be a matter of going up high enough and then turning your spacecraft ninety degrees to the side and making a full burn of all remaining engines.

Uh, no. Spacecraft kinematics are actually really difficult to get your head around when you’re trying to manage throttle and keep the thing pointed in the right direction without exploding like a giant firework. I did manage to make a successful orbit using this very stupid method – even got the astronauts back to Kerbin safely afterwards – but it looked like this:

A highly, highly elliptical orbit with the periapsis about 100km above the surface of Kerbin and the apoapsis out beyond the orbit of the sodding moon. It took over a day to complete one orbit. My problem was I went up to a height I thought would be sufficient to fall from without hitting the ground and then made my sideways burn, but I forgot that a spaceship isn’t a car and that it still had a vertical velocity of over 2km s^-1. The sideways burn didn’t help either because it was still pointed slightly up, hence the ridiculous orbital altitude.

Clearly I have some way to go before I can even think about attempting a moon landing. Here, again, I have some inkling that it’ll be far trickier than I think because if you point a spaceship directly at the moon and blast away for 48 hours then by the time you get there the moon will have moved on in its own orbit and you’ll have missed. A successful moon landing requires you to get yourself into an orbit that crosses the moon’s orbital path, and then for the two orbiting bodies – spacecraft and moon – to occupy the same area of space at roughly the same time. The patched conics system that makes up KSP’s map makes this a little kinder on the prospective lunar explorer (although it has been done without by a crazy person, which incidentally demonstrates how much easier it is to get out of a moon’s gravity well compared to a planet’s gravity well) but it’s still a challenge that is going to result in lots of failures, and hence lots of hapless Kerbal astronauts floating helplessly around the solar system inside their steel coffins.

KSP is something that everyone should try, I think. That alpha I played last year is still free, after all, and you’ll get some idea of what it’s like to try to launch a rocket into orbit. The full (well, fullest) version adds spaceplanes to the mix, giving you a whole new type of craft to ram into the ground at 5000 km h^-1, and the next release seems to be adding in EVAs for the little Kerbals giving them a physical presence in the game beyond the three perpetually-terrified portraits in the lower right-hand corner of the screen. It says something about Kerbal Space Program that I’m currently trying to figure out some way to tie it into genuine science education; it’d be excellent for demonstrating why staged rockets are superior, along with a lot of other basic concepts of rocketry. It’s nowhere near a 100% accurate simulation, but then I don’t think I’d want it to be. KSP successfully walks the fine line between player enjoyment and technical verisimilitude without ever falling off of it, and that, I think, is its greatest accomplishment. Kerbal Space Program  makes rockets fun.

The post In Praise Of: Kerbal Space Program. appeared first on The Scientific Gamer.

Jiiiiim asks

So, put a grapefruit and an egg in an area of space which has no other gravitational influences. Would they orbit each other and what would that orbit look like?

It rather depends on the initial parameters of the system, but assuming no other information my answer is going to be absolutely not.

If we have a completely closed system of the type you describe – a pocket universe with nothing in it except this grapefruit and this egg – and stick the grapefruit and the egg in it so that they are stationary with respect to one another (as we saw a couple of weeks ago, “stationary” is a rather relative term), then this is what will happen.

The grapefruit and the egg will exert a gravitational force on each other and begin accelerating towards one another, with the acceleration of each being inversely proportional to its mass (i.e. the heavier grapefruit will accelerate more slowly).  As they close the distance separating them r gets smaller and a₁ and a₂ get larger; in other words the rate of acceleration on the grapefruit and the egg increases as they get closer to one another. And eventually the egg will smash into the grapefruit, creating some sort of delicious grapefruit cake. This is what eventually happens to all two-body pairs in isolated systems of the kind you describe.

(If you’re wondering why the universe isn’t just one big glomp of matter after 13.7 billion years of mutual gravitational attraction, it’s because gravitational force acting over intergalactic distances is extremely tenuous and the universe is expanding more quickly than the acceleration this tiny force provides.)

If it’s an orbit you’re wanting, you need to give the egg some sort of lateral velocity relative to the grapefruit when you’re setting up your little pocket universe. Stuff in orbit is still falling towards the thing it’s orbiting around, it just also happens to be moving sideways at a fast enough speed that the surface of the Earth (or whatever) curves away from it at the same rate as it falls towards the Earth. In other words, orbit is falling in such a way that you never hit the ground.

There are some fairly standardised methods for working out the orbital velocity you’d need to orbit something at such-and-such a distance. While an eccentric (or oval-shaped) orbit would probably be possible in the grapefruit-egg scenario the maths for that is more complicated than I’m willing to get into right now, so we’ll stick with the basic circular variety.

So now all we have to do is find out the typical masses of a grapefruit and an egg. ENTER WOLFRAM ALPHA.

God that thing freaks me out. Anyway, it says the mass of a grapefruit is 140g and the mass of an egg is 35g, so if we assume a fairly modest orbital distance of 100m, we get an orbital velocity of 8.6 × 10^-7 ms^-1, which is pretty bloody tiny (it’s literally slower than a snail crawls). Because the masses we’re dealing with here are so small it would only take a very small nudge to put the egg in orbit around the grapefruit (or around the barycentre of the grapefruit-egg system). I can’t make any promises for the long term stability of this orbit since I haven’t really factored in the motion of the grapefruit here at all, so it might deteriorate after a few dozen orbits and we end up with our delicious grapefruit cake again, but it is possible.

The post Grapefruit + Egg = ??? appeared first on The Scientific Gamer.

Barycentre, barycentre. It’s the kind of word that sounds like it should be easy to pun, but really isn’t.

In preparation for the post on Lagrange points coming later on I should start by defining the concept of a gravitational barycentre (or centre of mass). For a single roughly-spherical object like a planet we can treat this as a single point mass located at the very centre of the sphere – that is, if we took the planet away and replaced it with a single infinitesimally small object with the same mass located at the planet’s centre of mass, this tiny object would (broadly) be gravitationally indistinguishable from the planet. If we add in another spherical object – say a moon – and attempt to calculate the centre of mass of the planet-moon system, things get a little trickier. The planet exerts a gravitational force on the moon and the moon exerts a gravitational force on the planet, meaning that the gravitational centre of mass of the combined system will be located somewhere in between.

This is actually kind of a bad picture for illustrating the concept because m₁ is the heavier mass, and yet it’s rendered as smaller than m₂. Never mind. Pretend m₁ is made of something really dense, like neutronium. The centre of mass of a two body system is defined by factoring each body’s mass by its distance from some arbitrary reference point – in this case x₁ and x₂ – and dividing by the system’s total mass. Since the reference point is entirely arbitrary we can cheat and say it’s located at the centre of mass of m₁; this means that x₁ is zero, x₂ is now the distance d between the two centres of mass and the equation reduces to

 In other words the centre of mass of the two body system is located a distance X_cm from the centre of mass of m₁along a line connecting the centres of mass of m₁ and m₂. This is the gravitational barycentre of the system, and it’s very important when considering orbital mechanics because in a two-body system like a moon and a planet, or a planet and the sun, both bodies will orbit the gravitational barycentre. If you have two masses of comparable size like a binary star system the barycentre will lie roughly in the middle of the two bodies, leading to very obvious mutual orbital behaviour like this:

(This is actually wikipedia’s example for Pluto + Charon, but I don’t like the one for the binary star system because it implies the stars have exactly the same mass, which would be unlikely.)

If one body is very much larger than the other, however, then X_cm will be smaller than the radius of the larger body, which is another way of saying that the gravitational barycentre will be located inside the larger mass. The existence of the barycentre is much less obvious in this case since the centre of mass of the combined system is so close to the centre of mass of one of the bodies. For example, in the case of the Earth-Sun system the barycentre is located

or about 450km away from the Sun’s centre of mass. Given that the Sun has a radius of 696,000km this does not produce much appreciable movement on the part of the Sun! I mention this so that the following gif doesn’t give you the wrong idea.

This is a very exaggerated portrayal of the sort of motion we’re talking about here; in actuality for most planets the motion on the part of the Sun will be tiny, and even a really big planet like Jupiter can only shift the barycentre of their mutual orbits to roughly the surface of the Sun, meaning that it’s less of a mutual orbit than it is a slight “wobble” on the part of the Sun. Nevertheless this motion can be detected, and as previously mentioned it’s one of the methods we use to detect exoplanets orbiting other stars. Hopefully by this point it should be obvious why we’ve mostly only found really big exoplanets, since they’re the ones which produce the biggest wobble in the motion of their parent star.

Now, remember, this applies for all two-body systems. The Earth and the Moon orbit a mutual gravitational barycentre. The Sun and Pluto orbit a mutual gravitational barycentre. Even tiny satellites like the Martian moons Phobos and Deimos will cause Mars to shift slightly in position, orbiting a mutual gravitational barycentre. This is an important factor to take into consideration when plotting celestial trajectories, and it also gives rise to some interesting side-effects which I’ll tackle on Thursday.

The post Barry’s Guide To Barycentres. appeared first on The Scientific Gamer.

Warning: post contains moderately difficult algebra along with hints of calculus. But don’t worry, it’s not that bad.

Rockets are kind of sucky for getting into space. Right now, though, they’re all we’ve got.

Rockets work by the classic principle of Newton’s Third Law: every action has an equal and opposite reaction. Push against a wall, and the wall also pushes against you with the same force. Friction between your feet and the ground stops you from moving backwards in response to this force, but if there wasn’t any friction – if you were standing on a big sheet of ice or something – then if you thumped the wall hard enough you would start to slide away from it. In a true Newtonian environment like space, even a simple action such as throwing a wrench away from you can nevertheless produce an appreciable impulse on you in the opposite direction to which you threw it, and that’s where you’d end up drifting.

This is basically how rockets work, except instead of a wrench they’re using thousands of kilograms of rocket fuel shot out of the back of the rocket at several kilometres per second. This creates an equal and opposite force which pushes the rocket upwards – thrust, in other words. When they’re considering how much fuel to put in a rocket, rocket scientists don’t think “Well, we want the rocket to reach such and such an altitude so we need this much fuel,” because that would be needlessly complex. Instead they think in terms of something called delta-V, or ΔV, which is the change in the velocity of the spacecraft that will be caused by burning X amount of rocket fuel. If your rocket’s delta-V is roughly equal to the Earth’s escape velocity, then congratulations! You’re on your way out of Earth orbit.

The amount of fuel you need to accelerate a given payload to a certain delta-V is dependent on how heavy that payload is. A bigger payload means more fuel. And as mentioned in the post on satellite orbits, it’s not enough just to add more fuel to the rocket, because now you’ve just added a whole bunch of extra kilograms that also needed to be lifted into orbit. So you need more fuel to lift the fuel, and then more fuel to lift the fuel to lift the fuel, and the whole thing would get dreadfully complicated if it weren’t for something called the Tsiolkovsky rocket equation. Because I’m trying to re-teach myself some basic physics I’m going to derive it from first principles. Don’t worry if you can’t or don’t want to follow what I’m doing, just skip to the end.

WARNING. WARNING. AWFUL CALCULUS STARTS HERE. SKIP FORWARD IF YOU’RE ALLERGIC TO MATHS.

Start with the principle of conservation of momentum.  Momentum is mass times velocity. Consider the state of the rocket at two different times: t, which for our purposes is the rocket sitting on the launch pad just before blast off, and t + dt, which is a time dt after that when the rocket has burned off all of its fuel.

(Note: dt is part of the Leibniz notation of calculus. The d doesn’t stand for a specific quantity, but instead basically means “change in”. So dt is the change in time, dm is the change in mass, and so on. They can be combined, too, so dm/dt would be the rate of change in mass over time. For a real world example, dV/dt is the rate of change in velocity over time, which is equivalent to acceleration a. Here, t + dt is similar to saying something like “D-Day +17”, where the +17 is equivalent to + dt.)

The rocket will start with a mass M(t), and will eject a quantity of exhaust gas as it burns its fuel which will have a total mass dm (because the mass of the rocket will change by this much). Due to conversation of momentum, the momentum of the payload after the fuel burn  — which for our purposes we will treat as everything on the rocket that isn’t the fuel (i.e. M(t) – dm) – will be equal and opposite to the momentum of the exhaust gas dm. We can model these separately to find out their respective momentums at the times t and t + dt, but to do this we need to make some assumptions about the velocity of the payload and the fuel.

At time t, the fuel is unburnt and flying along with the payload, so it and the payload have the same velocity V(t). At time t + dt, the fuel will have been converted into exhaust gas travelling at a constant velocity U (since for the purposes of simplicity we’ll say a single fuel type will burn at a constant rate). The payload will be travelling at the same velocity V(t) plus some positive change in velocity dV provided by burning the fuel, making its total velocity V(t) + dV. The fuel will be travelling in the opposite direction with velocity U, but it had a velocity of V(t) to start with, so its true velocity with respect to the payload will be V(t) – U. (The U is negative because the fuel is travelling in the opposite direction to V(t)).

Right, now we can get on with the interesting stuff. The momentum of the payload and the fuel can be expressed as:

To find out the change in momentum over time dt, we have to subtract the intial momentum at time t from the final momentum at time t + dt.

Because we’re treating the rocket as a closed system with no external forces, conservation of momentum means the sum of the change in momentum of the rocket and the payload will be zero. In other words, adding together the terms we’ve derived for ΔP_payload and ΔP_fuel should equal zero.

We’re trying to find the rocket’s change in velocity – its delta-V, or dV – so we rearrange this in terms of dV.

And now comes the part where I’ll probably lose even the more dedicated amongst you. The mass of the burned fuel dm will be equivalent to the change in mass of the rocket M(t), or -dM(t), so we can substitute that in for dm.

Then we do a wonderful, magical thing called integration. Say you have a line on a graph which is described by the equation y = x. Integrating that equation will give you the area underneath the line y = x. This is bloody complicated and there’s a whole bunch of different rules for doing it. For our simple example, the integral of y = x will be x²/2. Don’t understand? Don’t worry, I don’t either. The one thing you should grasp about integrals is that since most lines described by set equations will be technically infinite, we need to identify the bit we actually want to find the area under and integrate between upper and lower limits that define that part of the line.

We integrate the left hand side of the equation with respect to V between the limits V_o and V_final, the initial velocity of the rocket and the final velocity of the rocket respectively. This gives

Simple, right? Unfortunately we then have to integrate the right hand side with respect to M between the limits of M_payload and M_o, the mass of the payload and the original mass of the payload plus the fuel.

That’s substantially gnarlier. Fortunately there’s a neat way to simplify logarithms by smooshing them together, so

Since the final velocity minus the original velocity is the change in velocity, or delta-V, then

IF YOU ARE SKIPPING ALL THE HORRIBLE CALCULUS START READING AGAIN HERE.

It’s a lot of effort to go through for such a little thing, but the final product of all that work is Tsiolkovsky’s rocket equation. From it you can easily see that the delta-V you’ll get out of a given rocket will depend on the velocity of the exhaust U and the logarithm of the ratio of the total mass of the rocket divided by the payload mass. However, since I said rocket scientists usually think in terms of “How much fuel do I need to reach a given delta-V?” rather than “What delta-V do I want to reach today?”, a more useful form of the equation is

Everything on the right side of the equation is stuff the rocket scientist should know in advance – the exhaust velocity U, the desired delta-V and the mass of the payload. And if you subtract the mass of the payload from the total mass of the rocket M_o, what you have is the mass of the fuel. This can therefore be used to quickly and easily calculate what amount of fuel you’ll need to boost a given payload to a given delta-V. Just for the lols we can make a graph of how the fuel-to-payload ratio changes for that given delta-V depending on what sort of fuel you use. For a desired delta-V of 9.4 km s^-1 (the lower limit required for low Earth orbit)

Obviously there’s a law of diminishing returns in effect here, but we can see from the graph that in order for your rocket to even be feasible you need a fuel with an exhaust velocity of at least 4-5 km s^-1. Happily there are fuel mixtures which exist which are capable of this, and some go even higher, but there’s a limit to how big you can make U. Remember the Nedelin catastrophe I referenced last week? That neatly demonstrates that high-power rocket fuels are very, very volatile and prone to exploding, and so you don’t really want a U that’s too big. It’s essentially fixed at 4-6 km s-1. Assuming that you’re not willing to reduce the mass of your payload at all, is there any way we can further increase the efficiency of a rocket?

Well, I wouldn’t have asked the question if there wasn’t an answer. There is. It’s called rocket staging. The idea is that once you have burned X amount of fuel, whatever you were using to keep that X amount of fuel in is useless dead weight. Lifting it along with the rest of the spacecraft is inefficient; it’s better to just cut it loose entirely if you can. Rocket staging is basically a matter of perching a small rocket on top of a larger, more beefy rocket. The beefy rocket fires first and uses all of its fuel, and then a series of explosive bolts cut it loose from the small rocket which then starts burning its engines. We can work out how efficient this is by using the rocket equation for each successive stage.

Assume that your rocket requires 1% of storage mass for every 8% of fuel mass. For a one stage-rocket with a payload that is 1% of the total mass, this allows for a fuel mass of 88%. Your achievable delta-V will therefore be

Compare this with two rockets stacked on top of each other, using the same fuel/storage ratios. The first rocket has a fuel mass of 80% and a storage mass of 10%, with the second rocket making up the last 10% of the mass. The first stage will boost the second stage to a delta-V of

So this first stage has less delta-V than our big one-stage rocket. However, once it’s finished burning it’s cut loose and the second rocket fires as a standalone entity with

i.e. exactly the same as the first rocket since the fuel/storage mass ratio is identical. This is added to the delta-V from the first stage to give a total delta-V of 3.22 U – half as much again as a one-stage rocket carrying the same payload with the same total mass. Depending on how far you want to go and how complex you want to make your rocket you can add even more stages; the Saturn V which sent the Apollo astronauts to the Moon was a five-stage monster.

So! Hopefully this has at least been useful for crystallising in your minds the reason why we use multi-stage rockets. It’s certainly been helpful for me; I learned this stuff back in 2004 so it was good to spend an afternoon deriving the Tsiolkovsky rocket equation from first principles. Good. Yes.

*eye twitches*

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Edit: I forgot the best orbit of all, the Hohmann transfer orbit. This has now been amended.

Yes yes I’m twenty minutes late. Sue me. It occured to me while thinking of ideas for future science articles that it might be useful to talk about satellite orbits for a little bit so that I have some explanations on hand for why LEO sucks and why geostationary orbit is hard. There’s no ultimate conclusion here, just background information on the most common types of satellite orbit and their advantages/disadvantages. Enjoy.

Low Earth Orbit – The most boring kind of orbit. Anything orbiting below an altitude of 2000 km is in a low Earth orbit because 2000 km isn’t very high up – it’s not even high enough for the Earth’s gravity to be substantially diminished in any way, so the only reason you see astronauts larking about in what appears to be zero-G on the ISS is that they’re technically freefalling around the Earth. Nevertheless LEO is a very, very attractive orbit to put satellites in because it is comparatively a very easy and cheap orbit to reach. The main problem rocket engineers face in boosting stuff to higher orbits is that they need more fuel to do so. Then they need to carry that extra fuel up to where the rocket was originally going to do, so they need more fuel to do that. And to get the extra extra fuel up to where… anyway, it’s much less of a headache for them if you just park your satellite a few hundred kilometres above the surface of the Earth, which is a perfectly cromulent orbit for 90% of satellite applications.

 It’s not without its drawbacks, however; the major one being that if you don’t put your satellite very high up it’s still going to be partially within the Earth’s atmosphere, and thus subject to atmospheric drag as gas molecules bounce off the surface of the satellite and reduce its orbital velocity. If we just left satellites in Low Earth Orbit they’d eventually drop out of the sky and plummet back down to Earth because losing orbital velocity means losing orbital altitude. So satellites in Low Earth Orbit have to do a lot of what’s called station keeping – that is, orbital adjustments to counteract atmospheric drag and other effects which alter the satellite’s orbit in order to keep the satellite on station where it should be. For example, this is a graph of the orbital altitude of the ISS since it was launched; the altitude decreases over time because of drag, and every so often it gets boosted back up to a higher orbit by one of the supply vehicles they regularly send up to it.

Polar Orbit – A special variant of the Low Earth Orbit that’s been specifically shifted to have an orbit that’s inclined ninety degrees to the equator – in other words, it orbits from pole to pole. This is useful because it means that as the Earth rotates underneath the satellite, the satellite flies over a different part of the Earth every time it makes an orbit. If you orbit your satellite at a low enough altitude – say, 600–1000 km – then it will have both a very short orbital period (often 90 minutes or less) and be close enough to the Earth’s surface to get some very, very nice pictures of all the pretty scenery down there. The Polar Orbit is therefore of especial interest to people who like to look at stuff on Earth – i.e., Earth imaging scientists and spy agencies.

Geostationary Orbit – The higher an orbit you have, the longer it takes for you to go around the Earth. Since they aren’t very high, satellites in LEO orbit the Earth very very quickly. Sometimes this is desirable, as in the Polar Orbit example above. Most of the time, however, it is not. Quite aside from everything else, if you have a satellite zipping around the Earth at an orbital velocity of several kilometres per second it makes communicating with the bloody thing an absolute nightmare. Even if you know where it is and you have an antenna that can track it as it moves across the sky, eventually the satellite is going to disappear below the horizon and you’re going to be out of contact with it until it comes back around the other side of the Earth. This is not ideal, since if you want to be in communication with the satellite 100% of the time you have to build a daisy chain of satellite ground stations dotted around the planet so that it’s never out of your line of sight.

The solution to this problem is the geostationary orbit. If, as you go higher, the orbit of the satellite takes longer, then it follows that eventually you’re going to find an orbital altitude where the orbital period of the satellite is equal to the rotational period of the Earth – in other words, the satellite goes around the Earth once a day. This means that it orbits at the exact same rate as the Earth rotates, and so if it’s got an inclination of zero it constantly remains above the same point on the Earth’s equator. (Hence geostationary, or geosynchronous – it’s stationary with respect to the Earth.) This is a good orbit for communications satellites since once they’re moved into position they’ll never be out of contact with the country that launched them. It’s also good for other nation-specific applications, like weather monitoring. The drawback is that you need to go very, very high up in order to get an orbital period of 24 hours. 36,000 km up, in fact. That sort of thing requires a lot of fuel.

Molniya Orbit – Unfortunately the fact that you need to orbit stuff around the equator to achieve a geostationary orbit can be a bit difficult if you’re a country that happens to have most of its land mass located at high inclinations – say, for example, if you’re Russia. Since your line of sight to a geostationary satellite has a high angle of incidence, you’re going to need a lot more power to communicate with it. Additionally getting something to a geostationary orbit from the highly-inclined Russian launch sites is also very tricky. The Russian solution is the Molniya orbit. Unlike the other orbits described here a Molniya orbit is highly elliptical – that is, its perigee (point of closest approach to the Earth) is a thousand kilometres or so, but its apogee (point at which it is furthest away from the Earth) is 40,000 km up on the opposite side of the planet to the perigee. If you arrange matters so that the perigee is above the bit of the planet you want to look at/talk to, you can get almost all of the benefits of a geostationary satellite out of it since it’ll be spending 80% of its time on one side of the planet and while it’s doing this it’ll be in line of sight of your country of choice. Unfortunately while the Molniya is a very neat idea it’s not without its drawbacks; namely that you need directional antennae to track it as it moves through the sky as well as multiple satellites in a constellation to provide 24-hour coverage while one or more of your satellite constellation is out of contact on the other side of the planet.

Hohmann Transfer Orbit – Okay so these orbits are all very nice, but what if we want to orbit something around a planet/moon that is not the Earth? How do we even get there? Contrary to what Hollywood would have you believe, spacecraft that go to other planets don’t just point themselves at the planet and then blast off because this would take a shedload of fuel and also you have to account for the gravity of everything else in the Solar System at some point. Instead they enter something called a Hohmann transfer orbit. This is a special type of orbit because the spacecraft will only travel along it for as long as it takes to get to where it’s going.

The idea is this. You’ve got your spacecraft going round the Earth in what’s called a parking orbit. The objective is to get to a destination orbit that is different — geostationary orbit, say, for the most simple example, or an orbit around the Moon or Mars. Instead of trying to get there directly by burning fuel, the idea of the Hohmann transfer orbit is to let gravity do the work for you. The spacecraft just makes one brief burn to get itself onto the Hohmann orbit, which is an elliptical orbit going around the Earth or the Sun or whatever which just happens to intersect with both the parking orbit and the destination orbit. You sit back and wait for a couple of months, and then when you get to where you’re going you burn the engines again to shift the satellite to the destination orbit. It’s a very simple and elegant way of getting around the Solar System, and the only drawback is that it’s introduced the charming concept of the launch window — that is, you have to wait until the starting point (Earth) and the destination are in the right positions relative to each other for a Hohmann orbit to exist. Miss that window, and you have to wait for the next one. But then, as you should have learned by now, you don’t get anything for free where space travel is concerned.

Mlur. Sorry you had to sit through that. Hopefully there will be some payoff coming very, very soon.

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RRRrrrrrssssssssssnnnnnncccccccceeee.

I’m sorry, I was doing an impression of a resonating object there, mainly because I feel that starting with a description of what resonance is would be pointless. Everyone knows what “to resonate” means, and as far as physical principles goes all schoolchildren get taught about Galileo Galilei and oscillating pendulums when they’re doing their GCSEs. Basically if you have something undergoing a persistent, periodic motion, then it will transfer and store energy with more efficiency at certain specific frequencies resulting in a greater amplitude of motion. Most people understand this in the context of pendulums, as above – where, if you release a pendulum and let it settle for a few seconds, it will eventually swing with a period 

where l is the length of the pendulum and g is the pendulum’s acceleration due to gravity.

In a constant gravitational field – say, at the surface of the Earth where human beings tend to congregate – g will be a constant 9.81 ms^-1, making the period T entirely dependent on the length of the pendulum l. The frequency is just the number of oscillations per second, and so when the pendulum swings at a period T we can find its frequency 1/T. This is the pendulum’s resonant frequency; the frequency of oscillation at which it will produce the largest oscillation for the smallest energy outlay. Since it loses only a tiny amount of energy with each swing (and that only because the point where the pendulum is attached to whatever it’s swinging from will not be frictionless, and if it’s swinging in an atmosphere you’ve also got the question of drag to consider) the pendulum will swing with that period for a long, long time – hence why pendulums “prefer” to swing at this period T determined by their length: it’s the most efficient way they can swing.

So resonance is motion where an object oscillates/vibrates in a way which ensures the maximum possible motion for the minimum possible energy outlay. Objects can have many different resonant frequencies (which are an integer multiple of the fundamental frequency) and resonance is crucial to the real world operation of many things – opera singers breaking wine glasses by vibrating their vocal chords at the glass’s resonant frequency; tuning forks which, when struck, will quickly settle down to resonating at a certain set pitch governed by their resonant frequency; and string instruments like pianos which rely on hammers hitting steel strings of different lengths held under tension which then vibrate at their resonant frequency, producing a single tone depending on the length of the particular string. It can also lead to some amusing mishaps when applied to large-scale engineering projects, like this one:

Now, there are differing views on whether Tacoma Narrows really was an example of resonance. My undergraduate physics textbook would have me believe that it was, while Font Of All Knowledge Wikipedia says that it was not. However, since there are recorded examples of bridges which have collapsed due to resonance from, e.g., army formations marching over them in lockstep, and that as a result of this resonance is one of the things engineers have to account for when building bridges these days, it’s as good an example as any to illustrate that minor, repetitive perturbations of an object can have a very powerful effect on that object if they happen to match one of its resonant frequencies thanks to the efficient transfer and storage of the object’s energy. This holds true even when those minor, repetitive perturbations are one planet gravitationally interacting with another planet as it orbits around the sun.

Here is an example of orbital resonance:

This is the 4:2:1 resonance between Ganymede, Europa and Io. (You can think of those numbers as integer numbers of some fundamental frequency that governs orbits, as with mechanical resonance.)  As you can see, Europa orbits Jupiter twice for every orbit Ganymede makes, while Io orbits four times. This is called a mean motion orbital resonance because the ratios of the orbits of the moons are simple integers. Every time two of the moons are on the same side of the planet their mutual gravitational pull is greatly increased since they are closer together, with potentially catastrophic consequences.

Now, you may be thinking that it is something of a massive coincidence that Ganymede, Europa and Io just happen to have orbits that interlock so perfectly with one another, but it’s the same sort of massive coincidence that has led to you sitting here today reading the words that I have written. In other words, the Galilean moons did not just pop into existence in those positions and with those orbital resonances; they instead evolved over time until they reached their current stable, self-correcting orbits. We think of the Solar System as being immutable and unchanging, but that’s only over human timescales. Over million- and billion-year timescales the orbits of the planets and moons themselves change; this is either because they were not stable in the first place, or else because they are exchanging angular momentum with other, smaller bodies such as asteroids and comets (this will be important next week).  For example, there’s a laser reflector on the Moon that the Apollo astronauts left behind that scientists use to very, very accurately measure how far the Moon is from the Earth. What they found was that the Moon is gradually slipping away from the Earth at the rate of 4 cm per year – this may not sound like much, but over millions of years it all adds up¹. The orbits of the planets have similarly evolved and continue to evolve; as the Sun converts mass into radiative energy its gravitational hold on the planets will slacken and diminish, leading to them migrating further outwards in their orbits². And the same sort of thing is true of Io, Ganymede and Europa, which started closer in to Jupiter and migrated outwards until they reached their current arrangement, at which point their positions became self-perpetuating thanks to the orbital resonance.

The 4:2:1 Galilean resonance is an example of a stable resonance that resists any change in the orbits of its component moons, but not all orbital resonances are like this. Many orbital resonances are destabilising resonances, where the repeated gravitational perturbation of having two bodies in close proximity during their orbits moves one of them out of its stable orbit and ejects it from the Solar System. The best way to think about this is by imagining the gravitational force between the two bodies as an impossibly long rope strung between them. Usually the rope is just taut and under tension, but whenever the two bodies pass close by to each other the big one gives an almighty tug on the rope, yanking the smaller one towards it. Over thousands of orbits these repeated tugs on the gravitational rope can drag the smaller body off of its orbital path and from that point on it’s bye-bye, Solar System.

You’ve probably seen the effects of resonance, even though you might not have known it at the time. The most famous example is the rings of Saturn, the structure of which is a product of both stabilising and destabilising resonances. The gaps in the rings are caused by a number of different things, and one of these things is a destabilising resonance with certain of Saturn’s moons that are orbiting in phase with that portion of the rings³. On the other hand some of the rings exist in orbital regions where they should have dissipated long ago, and this is thanks to stabilising resonances with – again — large moons such as Titan. Other not so obvious examples of resonance in the Solar System include the Kirkwood gaps in the asteroid belt, which are governed by mean motion resonances with Jupiter. And finally there’s Pluto, which exists in a 2:3 resonance with Neptune.

Now, normally Pluto could never exist where it does. It has an orbital path with a high eccentricity which crosses that of Neptune (again, this should have been a big hint that Pluto Isn’t A Planet). If it’s going that close to the orbital path of one of the gas giants it should have been booted out of the Solar System a long time ago. However, Pluto survives thanks to its 2:3 resonance; any time it gets close enough to Neptune that it starts to become a hazard, Neptune gives it a gravitational kick that speeds it up and shifts it to a wider orbit further away from Neptune. Conversely, any time Pluto looks like it might be getting far enough away from Neptune that it will escape this resonance, Neptune gives it another gravitational kick from the opposite direction that slows it down and corrects its orbit. So while the orbit of Pluto is constantly shifting over time thanks to interactions with Neptune, these interactions and orbital shifts cancel each other out leading to Pluto existing in this stable 2:3 resonance.

(This is illustrated with a pretty animated gif on this website if you’re having trouble visualising exactly what is going on. The orbital paths are plotted with respect to Neptune which is stationary from the gif’s point of view, so Neptune is the blue dot which never moves. The orbit of Pluto is in pink, with the Neptune-crossing part of its orbit represented as the loop, or “twist”. As you can see, any time this dangerous part of its orbit looks like it might get too close to Neptune – which would be fatal for Pluto – the gravitational interactions of the 2:3 resonance “corrects” Pluto’s orbit and shunts it back in the opposite direction.)

And that’s orbital resonance. Nearly all the examples of resonance we see in the Solar System today are stabilising resonances; destabilising resonances are generally indicated by the absence of stuff where stuff should be as per the Kirkwood gaps. With Jupiter and the other gas giants being as large as they are their orbital resonances have had a very significant effect on the way the Solar System has formed and evolved over time, and particularly so in the case of Pluto and the other TNOs that form the Kuiper Belt, the Scattered Disc and the Oort Cloud. Which was kind of the point in me explaining resonance in the first place: I cannot adequately describe those areas without explaining why they are structured the way they are, and I cannot do that without talking about resonance. HOORAY.

1. The reason for this is tides, but not in the way you might think. Every time the Moon causes a high tide it creates a slight bulge in the ocean. This bulge is situated a little bit ahead of the Earth-Moon axis thanks to the rotation of the Earth, causing an uneven gravitational force that drags the Moon forward very, very slightly. The upshot of all this is that the Earth is losing rotational momentum to the Moon, which is converting it to orbital momentum and moving further away from the Earth. Wikipedia tells me that this interaction would in theory go on for another fifty billion years until the Earth and Moon become spin-locked in – what else – a resonance, but that the Moon will never get that far because the Sun is due to increase in size and luminosity and vaporise the Earth’s oceans 2.1 billion years from now. So that’s nice, I guess.

2. This, incidentally, is why nobody has any goddamn idea what is going to happen to the Earth when the Sun gets big and strong and turns into a red giant.

3. Other gaps are caused by moons orbiting inside the rings themselves, while the remainder are unexplained at this point in time.

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