Sanity check this math for me please (starship fuel use)

[Junghalli]

OK, I was trying to find some figures for fuel consumption for a RAIR-type ship (a Bussard ramjet derivative that carries its own fuel, but not its own propellant) for my uni. I wasn't sure exactly how I should go about this, so I figured I might try calculating the energy that goes into the exhaust. The engine can be modelled as a straightforward nuclear thermal rocket, in this case with an exhaust velocity of 350 km/s (fairly arbitrary number). I figured since the ship doesn't use up onboard propellant the acceleration could be modelled as an efficient momentum transfer between the propellant (interstellar hydrogen) and the ship.

The ship's final speed is .9 c. Since this is around 771 times the exhaust velocity, I figured the ship should have to suck through 771 times its own mass in interstellar hydrogen. Say the ship is 1000 tons, this comes out to 771,000 tons. This is not accounting for relativistic effects, which I compensate for later.

To be perfectly honest I was a bit lazy to plug it into the relativistic KE equation, but I wrote down at an earlier time that the energy needed to accelerate to .9 c would be 3.207 times the number you get by using the simpler newtonian KE equation. So:

(1/2(771,000,000)(350,000^2)) X 3.207 = 1.5 X 10^20 joules, about.

The problem is when I do the same thing with the starship itself I get

(1/2(1,000,000)(270,000,000^2)) X 3.207 = 1.17 X 10^23 joules. Three orders of magnitude more energy.

This doesn't make any sense. Surely kinetic energy transfer would be symmetrical. I can accept an inefficiency, but this is apparently energy from nothing: I must have screwed up. It's also a rather significant engineering consideration, since it makes the difference between the ship having a fuel/ship ratio of 1/4 and 200/1.

What am I messing up?

[Surlethe]

The ship's final speed is .9 c. Since this is around 771 times the exhaust velocity, I figured the ship should have to suck through 771 times its own mass in interstellar hydrogen.

Why do you figure this?

To figure the classical acceleration (1/m)(dp/dt) caused by the constant-velocity continuous exhaust, you need to know the density of the exhaust stream. Put another way: if you're shooting bowling balls out the back of your ship at the same rate I'm shooting peas, you're going to accelerate significantly faster than I do. Relativistically, describe your ship's motion with momentum p since you can easily get its velocity out of that. Then the instantaneous change in momentum dp/dt will be equal to the instantaneous change in momentum of the (interstellar medium) ISM caused by your scoopjetthingy. This in turn depends on how quickly you're already moving with respect to the ISM. This will lead you to a differential equation that I'm too lazy to figure out at the moment; see below for the reason.

Actually, it occurs to me that a constant-velocity exhaust is not going to permit you to move faster than the speed of the exhaust. Why? Well, if you do, your scoopjetthingy will actually be slowing down, on average, the hydrogen it sucks up to spit it out the tail end at 350 km/s. That (positive, from your starship's point of view) change in the momentum of the exhaust has to come from somewhere: it will cause a (negative, again from the starship's frame) change in the starship's momentum. If a hydrogen atom comes flying toward yoiu at 450 km/s, and you catch it and fling it off behind you at 350 km/s, you will have to, in the final analysis, be flying backward with velocity v sugh that (your mass)(your new speed) + (350 km/s)(mass of proton) = (450 km/s)(mass of proton).

[Feil]

Your ship isn't changing mass. Just calc the kinetic energy gain by the ship, find the kinetic energy gain per unit mass of the exhaust (relative to the spaceship, which we take as stationary with an accelerating universe zipping past it, since our power supply is on the spaceship and therefore the accelerated object (the ship) is stationary relative to the reference frame of the power supply), divide the ship gain by the per-unit-mass gain, and the quotient is the mass of hydrogen you use.

T=energy of ship
U=energy per unit hydrogen
T/U=mass of hydrogen

This will be somewhat incorrect, because your ship's mass must change as it expends energy accelerating the hydrogen. If you in stead take T as the energy ship's average mass, that will be fine. Add the mass-equivalent of the energy gained by the ship to the ship's mass and divide by two to get a decent approximation of the average mass.

Note that the mass of the hydrogen is as perceived by the ship frame, where the mass of the hydrogen increases as its velocity relative the ship approaches c. This is why the ship observes itself accelerating at a linear rate (albeit with more and more extra hydrogen slipping through the gaps) while an observer in the ship's original frame observes the ship accelerating at a decreasing rate with c as an asymptote.

[Feil]

Surlethe, I think he means that the exhaust stream is at 350km/s relative to the incident velocity of the hydrogen. Otherwise it doesn't make sense.

[Junghalli]

Why do you figure this?

Basic logic: I figured if the exhaust velocity is 1/700th of the ship's final velocity then logically you'd need to accelerate 700 tons of exhaust in one direction at 350 km/s to get 1 ton of ship to accelerate to 270,000 km/s in the other. I could be very wrong though, I clearly seem to have screwed up somewhere.

Actually, it occurs to me that a constant-velocity exhaust is not going to permit you to move faster than the speed of the exhaust. Why? Well, if you do, your scoopjetthingy will actually be slowing down, on average, the hydrogen it sucks up to spit it out the tail end at 350 km/s. That (positive, from your starship's point of view) change in the momentum of the exhaust has to come from somewhere: it will cause a (negative, again from the starship's frame) change in the starship's momentum.

Ah, I'm aware of that problem with a Bussard ramjet, which is why I don't simply have the ship scoop deuterium out of the interstellar medium and feed it into a fusion drive. The ship is designed to funnel the ISM through itself and heat it up without slowing it down.

Surlethe, Feil, thanks for the help but I'm afraid I have to admit it's a little beyond my level. Pretty much everything I know about rocket mechanics I got from reading Atomic Rocket; I know basic delta V and kinetic energy calculations and that's pretty much the extent of my knowledge. How do I get the energy per unit of hydrogen? I'm afraid I'm not exactly clear about what units you're talking about; hydrogen consumed per second?

[Feil]

The energy per unit (mass) of hydrogen is the relativistic kinetic energy of 1kg traveling at 350,000m/s. However, I wouldn't trust what I said in the light of what Surlethe has pointed out (and even before hand, I'm still not particularly qualified). This board has several math experts; I'm not one of them, and anything I say is the half-blind leading the blind.

[AMX]

Surely kinetic energy transfer would be symmetrical.

Where did you get that idea?
KE can't be equal, unless mass is identical - otherwise you're violating conservation of momentum.
(Not saying that's the only error, but it's the one I noticed at a casual glance.)

[Surlethe]

Surlethe, I think he means that the exhaust stream is at 350km/s relative to the incident velocity of the hydrogen. Otherwise it doesn't make sense.

It still doesn't make sense; what if the ship is traveling at c-350 km/s? It makes more sense, and jives better with what he's describing, for the ship to not accelerate the hydrogen by a certain amount, but instead to add a certain amount of kinetic energy. Classically, adding a constant amount of kinetic energy boils down to adding a fixed amount of speed, so that's what his intuition says to do, but the intuition breaks down at relativistic velocities.

[Surlethe]

Why do you figure this?

Basic logic: I figured if the exhaust velocity is 1/700th of the ship's final velocity then logically you'd need to accelerate 700 tons of exhaust in one direction at 350 km/s to get 1 ton of ship to accelerate to 270,000 km/s in the other. I could be very wrong though, I clearly seem to have screwed up somewhere.

I'm pretty sure it just doesn't work like that. You're dealing with continuous momentum-changes that are relativistic; your approach, though, is classical, and seems to assume discrete objects. You're right that if 1 ton of ship throws 700 tons of hydrogen at 350 km/s in one direction, the ton of ship will go flying off at 270,000 km/s in the other direction, by basic conservation of momentum. But your setup is a little more complex than that, since you assume zero momentum to begin with -- this permits you to equate (1 ton of ship)(270k km/s) = (700 tons of H)(350 km/s) -- which is not true. Instead, you have to deal with the fact that you're traveling relative to the interstellar hydrogen you're trying to use as propellant, which means that, setting the ship's initial speed to zero for convenience, you have to have the final momentum of the ship/hydrogen system, after a short time dt, equal to the initial momentum of the incoming hydrogen. That's why, classically, you can't travel faster than your exhaust velocity if it's constant.

Actually, it occurs to me that a constant-velocity exhaust is not going to permit you to move faster than the speed of the exhaust. Why? Well, if you do, your scoopjetthingy will actually be slowing down, on average, the hydrogen it sucks up to spit it out the tail end at 350 km/s. That (positive, from your starship's point of view) change in the momentum of the exhaust has to come from somewhere: it will cause a (negative, again from the starship's frame) change in the starship's momentum.

Ah, I'm aware of that problem with a Bussard ramjet, which is why I don't simply have the ship scoop deuterium out of the interstellar medium and feed it into a fusion drive. The ship is designed to funnel the ISM through itself and heat it up without slowing it down.

There you go. That's not the same as ejecting it at a constant velocity; instead, you're adding so much kinetic energy to the system. You can set up some equations that will relate the gain in energy to the gain in momentum and use those to calculate the motion of your interstellar rocket.

Surlethe, Feil, thanks for the help but I'm afraid I have to admit it's a little beyond my level.

That's fine; there's always lots of stuff beyond everyone's level (except maybe Kuroneko's).

[Junghalli]

For the sake of simplicity, can I just assume that the kinetic energy I got for the ship itself travelling at .9 c is the bare minimum amount of energy that needs to be expended to get it up to that speed? That seems logical: you'd have to expend as much or more kinetic energy to get the ship up to .9 c as it "holds" when it reaches .9 c.

In which case I may have to go with a different drive system, since a fuel/ship mass ratio of 200/1 for starships won't work for the universe I'm envisioning. Bleh, I'd hoped to avoid a full-on Bussard ramjet: it requires a ridiculous amount of supertech (you've got to use slow difficult to ignite protium-protium fusion, and somehow do that efficiently while the fuel shoots through the reactor at high fractions of c). Maybe I'll use a Bussard ramjet variant where you just fuse the deuterium in the ISM, but building an engine that can work with fuel that's 1% usable and 99% garbage would be pretty impressive too (and you probably can't seperate them out, because you can't slow the stuff down or you impose a terminal velocity of >.12 c on your ramjet). I suppose I'll just have to accept a little more handwavium in my setting than I was hoping I'd have to.

[Surlethe]

For the sake of simplicity, can I just assume that the kinetic energy I got for the ship itself travelling at .9 c is the bare minimum amount of energy that needs to be expended to get it up to that speed? That seems logical: you'd have to expend as much or more kinetic energy to get the ship up to .9 c as it "holds" when it reaches .9 c.

Of course. In fact, it couldn't be any other way: you're simply invoking Conservation of Energy.

In which case I may have to go with a different drive system, since a fuel/ship mass ratio of 200/1 for starships won't work for the universe I'm envisioning. Bleh, I'd hoped to avoid a full-on Bussard ramjet: it requires a ridiculous amount of supertech (you've got to use slow difficult to ignite protium-protium fusion, and somehow do that efficiently while the fuel shoots through the reactor at high fractions of c). Maybe I'll use a Bussard ramjet variant where you just fuse the deuterium in the ISM, but building an engine that can work with fuel that's 1% usable and 99% garbage would be pretty impressive too (and you probably can't seperate them out, because you can't slow the stuff down or you impose a terminal velocity of >.12 c on your ramjet). I suppose I'll just have to accept a little more handwavium in my setting than I was hoping I'd have to.

Yeah. The best thing to do in a sci-fi setting, IMHO, is to not work out a drive in detail, but just present what it does. In this case, that would be a drive that simply injects x MeV of kinetic energy per particle of hydrogen it scoops. It's easier said than done, and it has a hard feel to it, rather than a soft "I can get across the universe in fifty seconds fap fap fap" feel.

[Kuroneko]

The kinetic energy at the final velocity would indeed be a lower bound, but it tends to underestimate the requisite energy.

If the ship has no exhaust of its own and has nonzero velocity relative to the dust, then the problem of a lower bound is actually very simple. It gets some incoming four-momentum from the interstellar dust, accelerates it, and uses it as exhaust, with some outgoing four-momentum. The net change in four-momentum must therefore be null, since the ship did not provide any mass to it, just energy (although that has a mass-equivalent). Thus, the whole scenario is equivalent to the a photon rocket in a perfect vacuum. Another way to think about it is as follows: the exact mechanism is irrelevant; therefore, in the limit of unit efficiency, we can imagine a direct mass-to-light conversion, with the outgoing photons perfectly absorbed by the dust particles passing by.

For a starting mass M at initial velocity V, and payload mass m at final velocity v, conservation of momentum requires that γmv-ΓMV = P/c, where P is the total emitted photon energy in the dust frame, and in turn conservation of energy forces ΓMc² = γmc² + P, or equivalently ΓΜc(c+V) = γmc(c+v). The required energy in the ship frame is just E = (M-m)c², or
[1] E = Mc²[ 1 - (Γ/γ)(c+V)/(c+v) ] = mc²[ (γ/Γ)(c+v)/(c+V) - 1]
Using some algebraic gymnastics, one can let x = Γ(M/m)(1+V/c), so that the final velocity in terms of M/m and initial velocity is
[2] v = c tanh(log x) = c(x²-1)/(x²+1).

If the photon rocket accelerates from rest (V = 0, Γ = 1) to v = 0.900c (γ = 2.29), then the required energy 3.02e17J/kg of payload mass, compared to the specific kinetic energy of 1.16e17J/kg. However, it should be noted that unlike a true photon rocket, the drive you describe is incapable of accelerating from rest, and inefficiencies would drive the energy requirements higher still.