Rockets suck. This is a thing that we have established here; they’re terrifically awful ways of getting into space that are only used because nobody has really come up with anything better. There’s all sorts of ideas for wacky drive systems once your spacecraft is actually in space – ion drives, solar sails, Bussard ramjets – but these all sidestep the real problem facing future space travel, which is that you have to get out of the Earth’s gravity well first. This is not easy; even though the Earth is pretty small for a planet it’s still the heaviest of the four terrestrials and has what is to us a very hefty gravitational pull.

In order to exit Earth’s gravity well, an object being launched on a ballistic trajectory from the Earth’s surface must attain escape velocity. Escape velocity is calculated by

where G is the gravitational constant, M is the mass of the Earth and R is the radius of the Earth. For “ballistic” you can read “in freefall” – i.e. an object moving without any other forces acting on it. Rockets do have external forces acting on them thanks to their propulsion systems, so a rocket doesn’t necessarily have to reach escape velocity of 11.2 kilometres per second in order to escape the Earth’s gravity so long as it’s under constant power. However, the amount of energy it will have to expend will be the same as this hypothetical ballistic object, and that can be worked out by using the equation for kinetic energy.

where m is the mass of the spacecraft. With Earth’s escape velocity being 11.2 km s^-1, we can rearrange this equation to find out how much energy per kilo of mass a launch vehicle will need to have to escape Earth’s gravitational pull, and this turns out to be about 63 megajoules per kilogram. I always get a little lost when trying to visualise what a joule is in everyday terms, but Wikipedia tells me that 63 megajoules is roughly the amount of energy you’d release if you detonated fifteen kilograms of TNT. So for a crude method of visualising this, go and look at a picture of a rocket. Imagine a pile of TNT fifteen times as big. Blowing up this pile of TNT would release the same amount of energy as would be needed to get that rocket out of the Earth’s gravity well.

(One significant caveat here: rockets going into orbit don’t need to expend this much energy. They just need to go fast enough that they never hit the ground.)

This amount of energy is fixed. There is no way of getting around it; if you want to go into space, you have to expend at least 63 MJ per kilo of spacecraft mass and this energy has to come from somewhere. For rockets, it comes from rocket fuel; unfortunately rocket fuel increases the mass of the rocket, which requires more rocket fuel, which increases the mass of the- actually I think I’ve done this already, haven’t I. Anyway, rockets suck, but we still use them because the alternative solutions are, to put it mildly, just a little bit far-fetched.

Solution 1: Build better rockets.

Ahaha. Ahahaha. Ahaha. No. Rocket science isn’t necessarily all that complicated – I can understand it, after all – and one of the things that struck me when I was studying space science in the second year of my undergraduate degree was that rockets are a dead end. Well, that’s a bit harsh. Better to say that they’ve been refined as much as they can be.

Two things affect the amount of thrust you can get out of a rocket: the expansion velocity of the exhaust gas you’re creating by burning the fuel, and the size of the rocket nozzle. Rapidly expanding rocket fuels tend to be rather unstable – again, see the Nedelin catastrophe – and so there’s an upper limit on just how volatile you want to make that. What about the rocket nozzle? How wide/narrow this is determines the pressure at which the exhaust gases exit the rocket. For reasons I won’t go into, rockets produce the maximum amount of thrust when the pressure of the exhaust gas equals the pressure of the ambient atmosphere through which the rocket is flying. Atmospheric pressure decreases as you ascend into space, so ideally you want a rocket nozzle that automatically adjusts in size with altitude to keep the rocket working as efficiently as possible. This has been done. After that, there’s no real way to further improve a chemical rocket. We’ve taken them just about as far as we’re going to.

Solution 2: Don’t launch them from Earth.

One of the attractive things about a Moon colony – aside from the whaling opportunities – is that the Moon has a gravitational pull less than one-sixth that of Earth’s. It has an escape velocity of 2.38 km s^-1, and consequently it would only take 2.83 MJ to get one kilogram of mass out of the Moon’s gravity – under a twentieth the energy you’d need to get the same kilo off of Earth! Probably the best way to illustrate this is the ascent module on the Apollo lunar landers; that little capsule was able to create enough thrust to return to orbit to rendezvous with the command module.

It would be far easier, therefore, if we could somehow launch rockets from the Moon instead. Unfortunately this idea merely replaces the problem of getting out of the Earth’s gravity well with the even larger problem of building the necessary industrial base on the Moon to build and launch rockets. I’m not saying it couldn’t be done, and if somebody did manage to do it there’s every chance we’d see the commonplace spaceflight depicted in movies like 2001 (if you were lucky enough not to live on Earth, anyway), but the level of commitment and resources it would require would be staggering.

Solution 3: Mass drivers.

This one is simple, and operates on the principle that while achieving 63 MJ per kilo is a tall order for a chemical rocket, if we could somehow use electricity instead it’d be far easier to generate the energy required. The idea is you have a long, long, long, long launch track, kind of like a very high-tech version of Japan’s bullet train network (see maglevs for further information), and you accelerate the thing you want to launch down this track using superconducting electromagnetism. The track slopes upwards towards the end, and so once it reaches the end of the line your launch vehicle is shot into the stratosphere. At this point some small rocket boosters would take over and move the launch vehicle to the desired orbital trajectory. Building this thing would be a bit more feasible than the Moon colony, but there’s just one tiny snag: room temperature superconductors capable of carrying the currents required haven’t been invented yet.

Solution 4: Project Orion.

In which the Orion spacecraft is supposed to fart out shaped nuclear explosions, the brunt of which is directed against an impact plate on the ass of the spacecraft which makes the nukes “push” the spacecraft along. Utterly, utterly mad.

Solution 5: A space elevator. Also unicorns.

A very old concept that’s rather in vogue in 4X games, for some reason. As mentioned here geostationary satellites orbit the earth with a period of one day – that is, they remain above the same point relative to the Earth’s surface throughout their orbit. In theory, if you had access to a material strong enough as well as the combined and focused resources of the largest superpowers on Earth over a half-century or more, you could lower a cable from geostationary orbit all the way down to the surface of the Earth and simply have vehicles ride up and down the cable just like an elevator car. This is attractive because the energy costs are similar to those of a mass driver launch with the added bonus that you get most of it back as elevator cars come back down from orbit. If we then wanted to launch stuff further out into the solar system we could take advantage of the fact that the space elevator would kind of act like a giant sling to any payload launched from the far end.

Sounds fun, right? Perhaps so, but space elevators have several minor niggles past the huge advances in space technology, robotic manufacturing, materials science and megastructure construction (I just made that last one up, but someone’s going to have to invent it before they can build the elevator) that would be required, not to mention the totalitarian world government that’d have to be in place to keep everyone pointed in the right direction long enough to finish the bloody thing.

• In order to make it work you need a big counterweight of some kind to produce the necessary centrifugal force to keep the elevator cable taut. These days space elevator designs merely pay out the cable a little further past geostationary orbit to provide a counterweight mass rather than the previous idea of moving an asteroid to GEO to act as counterweight, but even if you do this you still need the asteroid because a) you need to build the elevator from both ends at once and so you’re going to need a base of operations in space from which to do it, and b) an asteroid would provide necessary raw materials for cable construction. So first you need to move several million tonnes of rock into geostationary orbit. This is just a little bit tricky to do, and I imagine it might make people down on the surface a little bit nervous as well what with the potential consequences if something goes wrong.

• There is no material which can currently be mass-produced in the quantities required which has the necessary tensile strength to support thousands of kilometres worth of its own weight. People always point to carbon nanotubes as the catch-all solution, but while they have the theoretical capacity to support the quantity of mass required, it would require the cable to be structurally flawless over its entire 40,000 km-odd length. Even one flaw would introduce a weakness which could be potentially fatal to the whole shebang.

• Transmitting power to the elevator cars is also going to be a bit of a bugger, since you still need that 63 MJ/kg to get into orbit. A nuclear power source would do it, but that’d probably make passengers a bit uncomfortable. Solar panels would increase the weight of the car, while using the cable itself to transmit power is going to run into the very inefficiency problems the space elevator is supposed to avoid. The current favoured proposal is wireless energy transfer via lasers or something, which has the tiny drawback that it is literally space magic.

• Elevator cars would travel fairly slowly, with an ascent to GEO taking anywhere from 6-12 hours to a full day. This brings the Van Allen radiation belts into play; the cars would travel through them slowly enough that anyone inside would end up taking lethal doses of radiation. It would be necessary to shield the cars in some way, which again would add to the weight.

Personally I think the space elevator is the ultimate manifestation of Archimedes and his lever; something that is theoretically possible but practically stupid. We can’t even agree on the best way to fly three people up to LEO; we abandoned the moon after sending barely two dozen guys up there. We are nowhere near being ready to take on the incredible engineering challenges of building something like this, and so anyone who thinks the space elevator is going to be built in the next half-millenia needs their head examined.

Where do I think the likely future of spaceflight lies? Well, assuming the world doesn’t enter a period of terminal dystopian senescence in the next couple of decades I think the mass driver concept is probably the most feasible solution as long as somebody comes along to solve the superconductor problem. It’s the smallest engineering project, it has the very great advantage of the thing being built on Earth rather than in space or on another planet, and it has the fewest question marks over necessary technology advances. Plus, if aliens ever try to invade we can use it to shoot rocks at them. Welcome to Earth.

The post The Future Of Spaceflight. 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*

The post You Have Discovered Rocketry. appeared first on The Scientific Gamer.

How do atmospheres work? As it turns out, the same way as everything else; that is, the gas molecules making up a planet’s atmosphere are bound to it by the planet’s gravity. However, there are some key differences. Inside the atmosphere all the different gas molecules are constantly moving around, colliding with each other, exchanging energy and then moving off in different directions. This means there are a couple of special rules that apply thanks to the difference in behaviour between a gas molecule and the unwieldy mass of squishy cells and organs that makes up the average human being. The general behaviour of a gas is described by the equation

where P is the pressure of its gas, V is its volume, N is the number of molecules of gas enclosed in that volume, k is something called the Boltzmann constant and T is the temperature of the gas in Kelvin.

The Boltzmann constant is necessary because temperature is not a property that can be applied to an individual gas molecule. There is no such thing as a gas molecule with a temperature of 300K; instead, the molecule will be moving at such-and-such a speed which gives it a certain kinetic energy, which will be different from the speed and energy of an adjacent molecule in the same volume. This speed is constantly changing as the molecule collides with other molecules and loses or gains energy, making it impossible to get discrete speed/energy measurements for a single gas molecule. The only way we can deal with a gas in any meaningful sense is by measuring the average kinetic energy of all the molecules inside it; this is expressed as its temperature. So in order to convert temperature — a measure designed to describe the general behaviour of a whole bunch of gas molecules — to the energy of a single molecule of the gas, we need to chuck in the Boltzmann constant k. The general idea is that the quantity kT will be on the order of (that is, somewhere close to, but probably not the same as) the energy of a given gas molecule in the volume V.

Understanding the difference between temperature and the kinetic energy of a single molecule is important for understanding how atmospheres work. All the gas molecules in a planet’s atmosphere are whizzing around at completely different speeds and kinetic energies even though the atmosphere itself has a certain temperature T. Most of the molecule velocities will cluster around the molecule speed described by that temperature T, but there will be many, many outliers which travel slower or faster. The range of molecule speeds and how they change with temperature is described by something called the Maxwell-Boltzmann distribution.

That graph may look a bit bewildering, but hopefully once I’ve explained what’s going on it won’t be all that complicated. The number on the vertical (or y) axis is the probability function, while the number on the horizontal x-axis is the speed of the molecule. What the graph is describing is what the probability is of a certain molecule in a gas travelling at a certain speed given a temperature a. If a single molecule has a 0.6 (or 60%) chance of travelling at a certain speed, then it follows that 60% of all the molecules in the gas will be travelling at that speed. The three different coloured plots on the graph show how this probability distribution changes at three different relative temperatures, a = 1, 2 and 5.

So from the graph we can see that for a low temperature of 1, all the gas molecules will be tightly clustered around a speed of 1-2, with none of them exceeding a speed of five³. Increasing the temperature to 2 changes the shape of the distribution; the average speed of a gas molecule is now 3-4 but the range of speeds at which the gas molecules as a whole travel is now much larger, as shown by the wider base and shallow peak of the distribution. Finally, for a high temperature of 5 the speeds of the gas molecules are much more evenly distributed, with a small peak at about 7 but with a long tail that stretches off all the way up to 15.

This long tail is the thing that interests us. Even if the temperature of an atmosphere is low, and the majority of the gas molecules in it are travelling at low speeds, some of the molecules will be travelling faster. A few of them will be travelling much faster – and if they’re travelling fast enough, they’ll reach what is called the planet’s escape velocity.

The escape velocity is a measure of how fast something has to be going in order to escape the planet’s gravity well permanently. You could strap yourself into a rocket and blast yourself into space, but if your rocket wasn’t powerful enough to propel you up to escape velocity before it ran out of fuel you’d eventually plummet back down to Earth if you didn’t manage to get into a stable orbit. The escape velocity changes from planet to planet as every planet has a different mass and therefore a different level of gravity, and furthermore since the gravitational force a planet exerts on an object diminishes the further away that object is from it, the escape velocity is greatest at the surface of a planet and diminishes as you progress upwards into space.

Now that we have all the background information down, we can finally start to look at why atmospheres are the way are. In reality atmospheres are not composed of a single homogenous gas but instead a whole variety of different elements. Even inside an atmosphere with a single uniform temperature the heavier gas molecules such as nitrogen and oxygen might have the same kinetic energy as a hydrogen molecule, but they’ll be travelling at lower speeds thanks to their greater mass. This causes the various gaseous elements inside an atmosphere to differentiate themselves according to weight; the heavier, slower stuff can’t get very far off the ground and is found close to the surface, while the lighter molecules can and slowly migrate to the higher portions of the atmosphere over time. If a molecule is light enough and the air around it is thin enough (so that it doesn’t get its direction of travel changed by a collision with another molecule) then the velocity of that molecule can easily exceed the planet’s escape velocity and escape into space – and this is true even if the temperature isn’t high enough for the average velocity of a gas molecule to exceed the escape velocity because the long tail of the Maxwell distribution ensures that some of the molecules will be moving fast enough, causing the planet to slowly (or not so slowly) bleed portions of its atmosphere away into space.

How likely a given molecule of gas is to remain part of the atmosphere can be calculated mathematically. My notes here don’t go into the specifics of how the Maxwell distribution and the interactions between molecules dictate this, but if a molecular constituent’s thermal velocity is near one-third the escape velocity, then about half of that molecule type will have escaped from the atmosphere within weeks. If the thermal velocity is one-fifth of the escape velocity, then the planet will lose half of that molecule type after a billion years. And if the thermal velocity is one-tenth of the escape velocity, then the planet will retain that molecular constituent indefinitely.

Therefore we run two parallel equations.

where G is the gravitational constant and k is the Boltzmann constant.

Everything in these equations that isn’t a number or a constant is a property that affects the likelihood of a planet losing a certain type of gas – the mass of the planet, the radius of the planet (since gravity diminishes the further away you go from the centre of mass), the temperature of the atmosphere and the mass of the gas molecule. If, at the end of the day, v_thermal is more than 0.1 v_escape, the planet will eventually lose that gas type. How much more v_thermal is is what dictates how quickly this happens.

From this the compositions of the atmospheres of the planets become much clearer. Earth has very little hydrogen or helium in its atmosphere because it’s not massive enough to hold on to them. Jupiter has no such problems (not to mention forming in an area where light materials were much more abundant) and so it’s still got most of its primordial hydrogen – this is why Jupiter and the other outer planets are referred to as gas giants. Mars is a relatively light planet and so much of its atmosphere has seeped away over time, but it still retains a fair amount of the heavier stuff such as carbon dioxide. Poor old Mercury gets hit with a double whammy: not only is it situated very close to the Sun — raising its surface temperature and thus its escape rate — but it’s also constantly being blasted by the solar wind which also serves to strip away atmosphere, as a result of which it doesn’t really have one any more.

So how is this all relevant to the terraforming argument? Well, if we run the equations above for an oxygen molecule in the Martian atmosphere we get a mean thermal velocity of 630 metres per second. Mars has an escape velocity of 5 kilometres per second. While Mars is losing oxygen molecules to thermal escape over time it’s doing it very slowly. This means that the fact that Mars has a crappy atmosphere in terms of potential human habitation has very little to do with how small it is, and is in fact heavily related to two other factors:

• Attrition by the solar wind; something we don’t have to worry about so much here on Earth thanks to the magnetosphere.
• Mars getting absolutely creamed by asteroid/comet impacts (along with every other terrestrial planet) during the Late Heavy Bombardment. Like, so much so that there’s still chunks of Mars dating from around about then floating around the Solar System which occasionally fall to Earth. This is because the impacts were so violent they threw up ejecta and debris from the surface so high and so fast that they reached escape velocity and were never seen again. From the point of view of Mars, anyway. If the Late Heavy Bombardment did that sort of thing to solid rock, imagine what it did to the atmosphere.

The first point is something that any prospective terraformers would still have to worry about, and given the lack of any magnetosphere on Mars the solar wind is likely to contribute far more to atmosphere loss than thermal escape. Barring some sort of cataclysmic event we don’t have to worry about the second ever happening again, at least over human timescales. As a result, while any Martian atmosphere we generate would dissipate relatively quickly, that “relatively” is relatively to the lifetime of the planets and the Solar System. In timescales relevant to humans even the most pessimistic estimates have a usable Martian atmosphere sticking around for 200,000 – 300,000 years, and it’s more likely that it’d last for a million plus. Are we really going to say terraforming the Martian surface isn’t worth it because the atmosphere will “only” last for 300,000 years?

1. It was a talk for kids aged 12-16 so I couldn’t say what I really thought of them.

2. The moral of this story is either that you should never use even mild language like “crazy lunatics” in case there are some crazy lunatics in your audience, or else that no matter what you do you should be prepared to go through life inadvertently offending an awful lot of people.

3. This graph uses dimensionless measures because the shape of it will be the same no matter what system of measurement you use to measure the temperature and speed of the gas.

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