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:
Exhaust gas momentum
(M(t) – dm)V(t)
t + dt
(M(t) – dm)(V(t) + dV)
dmV(t) – U
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 ΔPpayload and ΔPfuel 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 Vo and Vfinal, 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 Mpayload and Mo, 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 Mo, 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.
Did you at least understand the bit about staging.
Yes. But I kind of knew about that already.
Rockets are awesomely awful. When are we getting better things Hentzau? When?
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Hang up; shouldn’t the velocity of the fuel at t+dt be “d mV(t) – d mU “, not “d mV(t) – U”? If not, then why does the equation in the picture subtract d mU?
I’m probably just too green at calculus, but I really didn’t get that part.
That’s almost certainly a typo where I forgot to take dm(V(t) – U) out of the brackets correctly, although it’s been two years since I did this and I would need to check more thoroughly to make sure. Impressed that you’re cracking your way through it, though.