I spent this weekend fielding a lot of questions from teenagers about how long it would take to get to various destinations in the solar system, with the general assumption being that if we simply strapped a bigger engine to a spaceship we’d get there faster. This is broadly true but I didn’t have the time (and their brains were too small) for me to explain all the nuances of the situation to them, so I’m just going to explain it to you instead. Lucky you.

Happily the restrictions on getting somewhere fast revolve on three things I’ve talked about on here before: delta-v costs, the exponential fuel requirements of rocket staging, and Hohmann transfer orbits, which all combine into one unholy Megazord of making space travel as complicated as possible. We will start with the simplest scenario of sheer brute force: let us say we have a rocket sitting in Earth orbit with as much fuel as we need it to and we want to use it to take us to a faraway planet/moon/whatever. We figure out how quickly our rocket is going to be going and how long it’s going to take for us to get there, point the rocket at where we think the destination planet is going to be at that date, and start firing the engine. Even discounting gravitational interference from nearby orbiting bodies, doing interplanetary travel like this is going to present us with a couple of problems that should hopefully be fairly obvious if you’ve been paying attention.

• Accelerating directly against the pull of the Sun’s gravity will require a lot of fuel.
• Decelerating back down to the orbital velocity of the destination planet will also require a lot of fuel (remember, speeding up and slowing down are the same thing as far as delta-v is concerned.)

If we had infinite fuel this would not be a problem. Infinite fuel would allow us to get around the solar system very very quickly as long as spaceship passengers were willing to put up with the constant g-forces. Unfortunately we live in a world where fuel efficiency is the number one limitation on interplanetary spaceflight. It doesn’t matter how big your engine is or how fast it could potentially make you go if you don’t have the fuel to run it. Any fuel used for maneuvering in between the planets is effectively a very large payload that has to be lifted out of earth’s gravity well. This requires a lot of fuel. And that fuel requires a lot of fuel to lift, and that fuel requires yet more fuel to lift… and I think you get the idea. If you’re launching something from the earth’s surface the fuel requirements increase exponentially with the payload mass, so we want to keep our payload as small as possible. This means drastically, drastically limiting the amount of fuel on our interplanetary spaceship, and for most spacecraft it forces us to get creative when considering how to spend that fuel.

Going back to the delta-v chart I talked about last week you’ll remember that it featured “free” delta-v changes in the form of aerobraking. This is a maneuver that uses no fuel but which nevertheless changes the velocity of the spacecraft through interaction with an outside force. In general you want to put as many of these free maneuvers into your flight plan as you can in order to conserve as much fuel as possible, and aside from aerobraking there are two things you can do to get free delta-v: gravitational slingshotting and Hohmann transfer orbits.

Gravitational slingshotting is easy to sum up in one sentence but hard to describe in useful detail. As a spacecraft passes close by to a planet it passes around the rim of that planet’s gravity well, and as it does so it trades gravitational potential energy to the planet in exchange for kinetic energy (or velocity). The spacecraft speeds up as a result; however, its direction of travel also changes as its trajectory is altered by the planet’s gravity. Do it right and you can end up with something like this.

In this diagram the spacecraft is approaching the booster planet with velocity v. The planet is moving in the opposite direction with velocity U. This means the total velocity of the spacecraft relative to the planet is U + v. The key to gravitational slingshotting is this: after completing the slingshot maneuver the velocity of the spacecraft relative to the planet will still be U + v. However, its direction of travel has changed as it skimmed around the planet’s gravity well and it’s now moving in the same direction as the planet, which is still moving at velocity U. In order to still be moving at U + v relative to the planet, which itself is moving at velocity U, the spacecraft has to be moving at U + U + v, or 2U + v.

By doing this gravitational slingshot the spacecraft has increased its velocity by 2U for free (without burning any fuel), but the slingshot maneuver does have some strings attached; namely that you need a large gravity source somewhere near to the desired trajectory of your spacecraft in order to do it, and planets are curiously uncooperatively in this regard, rarely being in a good enough alignment to make multiple slingshot maneuvers possible without taking the spacecraft a very long way off course. For example, the Voyager missions took advantage of a fortuitous alignment of the outer planets to embark on its famous Grand Tour, doing a slingshot manuever as they encountered each gas giant, but such an alignment of the gas giants is not going to happen again until the 22^nd century. The requirement that the slingshot gravity source be moving relative to the destination of the spacecraft (because that 2U of extra speed is picked up as a result of the gravity source’s own motion through space) is also a bit of a kicker since it means we can’t use the Sun for slingshots.

Then there’s the Hohmann transfer orbit. This is a special kind of elliptical orbit used to transfer between two circular orbits at different altitudes. This picture should explain far better than I can:

The idea is that the elliptical orbit runs off at a tangent from each of the circular orbits. A spacecraft orbiting (for example) the Earth already has a certain amount of orbital velocity, so it only requires a limited burn of the engine  in order to bump the spacecraft’s velocity up just slightly and put it onto an elliptical trajectory. Some time passes – one of the drawbacks of the Hohmann method is that it’s relatively slow, relying on natural Newtonian motion through space rather than rockets or engines – and then, if you’ve planned things right, your spacecraft will eventually reach the destination circular orbit at the same time as the planet you happen to be trying to get to is passing through the area.  Because the orbital velocity of planets decreases the further away from the Sun they are the spacecraft then has to make a deceleration burn in order to bleed off some speed and enter a stable orbit around the destination, but after that the job is done. That’s all it takes in an ideal scenario: one brief engine burn at the start of a spacecraft’s journey, and another brief burn at the end of it. Gravity takes care of the rest.

As ever with space travel, though, you don’t get anything for free. As already mentioned, the Hohmann may be very economical in terms of fuel but it is also rather slow when compared to other more energetic methods of space travel.  Additionally you have to arrange things so that the destination planet shows up at the same time and place as your spacecraft does; given the alignments of the planets and the different speeds at which they orbit the Sun this means you have to launch the spacecraft from Earth orbit at a very particular point in time, which in turn gives rise to the charming concept of the spacecraft launch window. In the case of planets this is the same as the synodic period, so in the case of Mars there will be one launch window every two years,  Venus has one every 1.6 years, and so on.

I may as well end on an anecdote. Most spacecraft heading towards the outer solar system use something similar to a Centaur rocket booster to attempt a direct trajectory approach and get there relatively quickly. The Galileo probe was scheduled to be launched on a Centaur deployed from the Space Shuttle in the late 80s. Unfortunately for Galileo the Challenger disaster happened and not only did the mission get pushed back, but Centaur boosters were banned from being carried on the Shuttle. Without the Centaur the scientists behind Galileo had to display a significant amount of ingenuity to get the probe up to the required velocity to send it out to its destination, Jupiter.

This is Galileo’s final trajectory. It features a Venus gravity assist, an Earth gravity assist, a visit to an asteroid called Gaspra, a return to Earth for a second gravity assist, and then finally – after three years of travel that brought it back to where it started – it was ready to start the trip out to Jupiter. It still took four years. New Horizons used a Centaur booster and a Star-48B engine in an unusual three-stage configuration of the Atlas rocket to get to Jupiter in just nine months. So while the methods of interplanetary travel I have described here are very energy and fuel-efficient, they are very definitely not fast.

The post The Journey Is More Interesting Than The Destination. appeared first on The Scientific Gamer.

I’m back! (Back! Back.) And for the first of many new scienceposts, I bring you a quick précis of what I just spent the last week doing. I feel somewhat tawdry doing this since it’s duplicating my space school research more than a little bit, but what the hell – it’s my research*, and since I had to aim the task at eleven year olds I didn’t get to use all of it.

We’ll start with this really great chart featured on wikipedia’s page on space colonisation.

Delta-v is the fancy space scientist way of saying “change in velocity”. Objects that are stationary with respect to the thing you are trying to reach or get away from have zero delta-v. A spacecraft sitting on the launchpad on Earth has zero delta-v with respect to Earth orbit. To get it up to LEO, we are going to have to change its delta-v by around 10 km s^-1 to accelerate it to the orbital velocity of 8 km s^-1 required to stay in low Earth orbit. Where do the additional 2 km s^-1 of delta-v come from, if we only need to get it up to 8 km s^-1? That’s extra delta-v needed to counteract the effects of atmospheric drag and gravitational forces during the spacecraft’s ascent. Delta-v is basically a measure of the amount of “effort” needed to move from one orbit to another – or one place to another – on the part of the spacecraft engine, which typically burns propellant to create reaction mass to produce delta-v.

The thing about delta-v is that it’s not an absolute measure. It takes no notice of spacecraft attributes like payload weight or fuel weight which make the amount of thrust required from the spacecraft engine different from spacecraft to spacecraft. It tells you nothing about the specific amount of effort you’re going to have to make to get a specific spacecraft into space; in order to get that we’d have to take these specific spacecraft attributes and multiply them by the delta-v required in order to get a specific required thrust value unique to that spacecraft. Once we’ve done that we can’t then take that thrust value and apply it to another spacecraft because its weight, engine, structure etc. will be all different. We’d have to recalculate the thrust value using that spacecraft’s own attributes.

What delta-v does do, though, is provide a quick and easy scalar quantity that allows spacecraft designers to estimate how difficult a particular orbital maneuver will be relative to what it’s already done, or has yet to do. The above chart is a summary of the delta-v budget required to get to various places in the solar system from other places in the solar system. By looking at it we can see that getting from Earth to LEO is rather hard, requiring a delta-v of 9.3 km s^-1 (yeah I rounded up before, so sue me), but that getting from the LEO altitude of just 300 km up all the way out to the Moon requires a total delta-v of just 4.8 km s^-1. In other words, no matter what spacecraft you’re dealing with and how heavy it is, it’s always going to take twice as much effort to cover the first 300 kilometres from the surface of the Earth to LEO as it is to travel the next 380,000 km out to lunar orbit.

The reason I like this chart so much is because it illustrates a key restriction on travelling around the solar system that I’ve brought up more than a few times before on here: by far the hardest part of space travel is getting out of the Earth’s gravity well. Once you’ve done that you can get about on relatively little fuel because the subsequent delta-v requirements are comparatively low, but while the Earth provides a rather pleasant living environment for human beings – it doesn’t burn us, freeze us, crush us, poison us or suffocate us, unlike literally everywhere else in the universe – being stuck at the bottom of its gravity well is actually a crippling handicap as far as space travel is concerned. When the amount of energy you have to expend travelling those first 300 km is only slightly less than the amount of energy you’d expend getting from there to freakin’ Mars, you know you have a problem.

 As long as spacecraft manufacturing and launch facilities remain Earthbound – as they are likely to do for the foreseeable future – human exploration of the solar system and beyond is always going to be held back by this massive delta-v requirement to get out of the gravity well. Earth’s gravity is not something that can be counteracted, cancelled out or removed; as long as we launch stuff from its surface we’re always going to have to use these massive, inefficient rockets to provide that large initial delta-v investment. The only thing we can do about it is move somewhere else, building and launching our spaceships from locations which are a little kinder in terms of delta-v requirements such as geostationary orbit or even the surface of the Moon. As you can imagine, creating the sort of offworld industrial base required to do this is going to be a bit of a stretch considering we can hardly claim to have conquered even LEO with the ISS.

Some additional notes about the chart: “C3” is another way of saying escape velocity, so the Earth C3 and Mars C3 points on the chart are the Earth and Martian escape velocities respectively. The reason the Mars C3 point comes right after the Earth C3 one is because you can follow the chart back in the other direction, using it to go from Mars to low Mars orbit to Mars escape velocity, and that allows you to get a rough value for the amount of delta-v required to get back to Earth.

 Another crucial point about delta-v is that you need it to slow down as well as speed up a spacecraft. Accelerating up to the Martian escape velocity is no good whatsoever if you don’t have the fuel left to slow the spacecraft down again once it arrives back at Earth. Fortunately, if you happen to be going to – or even passing by – a large planetary body with a sufficiently thick atmosphere, you can slow the spacecraft down effectively “for free” expending just a bare minimum of fuel by using a maneuver called aerobraking. Here we make atmospheric drag forces work for us for once; by adjusting the spacecraft trajectory so that it skips through part of the planet’s atmosphere, the drag forces that arise from this will slow the spacecraft down for us with no further effort required on our part to change its velocity. The amount of time it spends within the atmosphere depends on how much we want to slow it down; if we want to stop it entirely then we just do a stock re-entry procedure. This free delta-v is one way only – you can use it to get from LEO back to Earth, but you can’t use it to get from Earth to LEO – and that’s denoted on the chart by the giant red arrows.

Finally, now that we understand how delta-v works it should be obvious why we use the quantities of impulse and specific impulse to describe rocket engines and their thrust efficiency rather than force or energy or any of that jazz. Thrust exerts a force on the spacecraft according to F = mass times acceleration. To achieve a certain change in velocity we’re going to have to accelerate it over a sustained period of time — so to get to 1 km s-1 we accelerate it at 1 m s-2 for 1000 seconds or something — and since impulse equals force multiplied by the time you apply that force for, once you factor out the spacecraft mass (so divide the thrust force F by m) you end up with acceleration multiplied by time, and that is just another way of saying delta-v.

*By which I mean “It’s my shameless cribbing from Wikipedia.”

The post To Vee Or Not To Vee. 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.