Is This Figure Accurate (Acceleration Energy)?

[Junghalli]

I've been trying to calculate how much energy a spacecraft would consume in accelerating for my SF setting. I looked around the internet and I got this formula:

1/2m(v^2)

Presumably m=mass and v=(change in) velocity

Is this equation correct?

[General Soontir Fel]

1/2 mv^2 is the kinetic energy (if the speeds are non-relativistic). If the mass of the spacecraft is constant, the energy difference is, in fact, 1/2 m(v2^2 - v1^2). Things get rather more complicated if you're dealing with changes in spacecraft mass or if you have relativistic velocities.

[Kwizard]

The equation K=1/2mv^2 gives the kinetic energy of an object of mass m with instantaneous velocity v. You'll want to find the difference between the kinetic energy of the vessel before accelerating and the kinetic energy after accelerating.

Of course, that's the amount of energy required to change velocity if your engines are 100% efficient at converting the energy content of fuel into KE of the ship. Any spacecraft engine is bound to produce waste heat and have a less-than-ideal fuel efficiency, so in practice you'll need more energy than the amount given by that crude estimate. There are also relativistic effects to be reckoned with if we're trying to speed up to a significant fraction of c.

[Kuroneko]

For a particle, the spent energy would be the difference in the kinetic energy [1/2]mv² plus the difference in gravitational potential. If gravity is not an issue, this means the spend energy is (m/2)Δ(v²) rather than (m/2)(Δv)². However, rockets do not have a constant mass, and therefore should not be treated as particles.

Going by the rocket equation v = -u ln R for an idealized Newtonian rocket, where R is the mass fraction remaining, u is the exhaust velocity and v is the change in rocket velocity, we have exp[-v/u] = R = m/M, where M is the initial mass and m is the final mass. Meanwhile, the rocket burns fuel at a constant rate dm/dt = F/u, so that M = m + [F/u]t. Substituting and solving gives F = [mu/t][exp(v/u)-1], so the energy is E = Pt = [(1/2)Fu]t = (1/2)mu²[exp(v/u)-1]. Alternatively, E = [1/2]Mu²[1-exp(-v/u)] in terms of the initial mass M instead.

[Junghalli]

So, if I treat the initial velocity as zero, then plug the ship's mass with fuel and the final velocity into 1/2m(v^2), then repeat that with the ship's final mass and use the number halfway between the two results, would that give me a decent ballpark estimate for how much energy the engine must generate minus inefficiencies?

[Kuroneko]

In general, no; because the rocket is changing its reference frame as it accelerates, it's not that simple. It would seriously underestimate the required energy for moderately sized mass fractions. The exact formulae (for ideal efficiency) are as follows:
E = (m/2)u²[1/R-1] = (m/2)[v/ln R]²[1/R-1] = [(1/2)mv²][(1-R)/R]/[ln R]² = [(1/2)Mv²][1-R]/[ln R]²,
where R<1 is the fraction mass remaining. The substitution of u = -v/ln R from the rocket equation removes the apparent dependence on engine parameters. So you can see your proposed upper bound of [(1/2)Mv²] is actually less than the true energy requirement when [1-R]/[ln R]² > 1, which occurs for R > 0.49 or so.

[Junghalli]

Did I get this right?

The ship being modeled is a freighter with a delta V of 1,730 km/s, and I am assuming it carries 708,300 tons of iron. I'm assuming an AM-plasma drive, which has an exhaust velocity of 7,840 km/s and therefore, according to the equation gives by Atomic Rockets (Delta V = , it should have a mass fraction of 1.247, which should give it a total mass of around 1.6 million tons with fuel, engines and everything.

E = (m/2)u^2 [1/R-1]

M=1,600,000,000 kg
U=7,840,000 m/s
R= 1,600,000,000 kg/1.247 = .442

E = ((1,600,000,000/2)(7,840,000^2))(1/.442-1)
E = 6.21 X 10^22 joules

Divided over 48 hours (time taken to accelerate to 1,730 km/s at 10 m/s^2) that comes out to 3.59 X 10^17 watts.

That's... one massive buttload of energy. I think I'm going to have to either massively reduce its performance or cargo capacity. Letting alone the fact this thing would apparently burn up more than 800 kg of antimatter with every trip I can't see how its engine fails to instantly melt itself; that's like an 83 megaton bomb going off in its engine room every second.

Weirdly enough, when I try to work the calculation the other way and calculate the energy it consumes per second of acceleration (10 m/s) I get a much lower number:

(1/2(1,600,000,000))(10^2) = 80 GJ

Then when I make it 20 m/s instead I get 320 GJ. My knowledge of physics is pretty much high school level (which is to say close to nonexistant), but I'd think it taking 3 times more energy to accelerate the second 10 m/s than the first really doesn't make sense (yes, I know about it getting harder to accelerate as you go faster, but I thought that was only a significant factor at relativistic speeds), so I'm pretty sure I'm doing something wrong.

I'm actually kind of hoping I cataclysmically fucked up, because otherwise I'm going to have to either massively re-evaluate the energy production of my civilization, the level of trade of my civilization, or both, or introduce more handwavium into the setting's technology. Especially since the numbers don't get much better when I reduce the ship's mass to a few tens of thousands of tons (they still have 16 digits attached).

[Feil]

For a ballpark number, treat the ship's mass as a constant,call the momentum of the exhaust Pdt, and then solve this formula for Pdt

Vship=tPdt/Mship

Where Vship is the velocity of the ship relative to its strating velocity
Mship is its mass, a constant
t is time since the start
Pdt is the rate at which momentum is imparted to the exhaust.



Edt = sqrt[Pdt^2*c^2+(Mship*c^2)^2)]

Integrate Edt with respect to time for the energy expended.




I think that should work. Somebody please correct me if I'm wrong.

[Junghalli]

Hmm, I found a website that says that at take-off the space shuttle's rockets generate 12 GW. Maybe if I could find out the mass of the orbiter/EFT/booster complex and its acceleration at take-off I could just scale that up.

[Junghalli]

Addenda to previous post. I've been able to find the following figures for the space shuttle at take-off:

Mass (w/ boosters+EFT): 2 million kg
Acceleration: 29 m/s^2
Engine power: 12 GW

That would indicate that it takes 2.1 GW to accelerate 1000 kg of matter at 10 m/s^2. Would that work?

[Junghalli]

Ghetto edit: that last figure should be 2.1 MW, not 2.1 GW.

[Kuroneko]

The ship being modeled is a freighter with a delta V of 1,730 km/s, and I am assuming it carries 708,300 tons of iron. I'm assuming an AM-plasma drive, which has an exhaust velocity of 7,840 km/s and therefore, according to the equation gives by Atomic Rockets (Delta V = , it should have a mass fraction of 1.247, which should give it a total mass of around 1.6 million tons with fuel, engines and everything.

Note that Atomic Rockets' R is M/m, whereas the R in the formulae above is the fraction remaining, m/M; they're reciprocals of one another. So for v = 1.73e6 m/s, m = 7.083e8 kg, u = 7.84e6 m/s and 1/R = 1.247, we have E = (1/2)[mu²][1/R-1] = (1/2)[mu²][1.247-1] = 5.39e21 J. Alternatively, you could use the one of the other formulae, e.g., E = (1/2)[mu²][exp(v/u)-1] = 5.38e21 J, which as expected gives the same answer.

If the acceleration is over 48 hours, the power then P = E/(48 hrs) = 3.11e16 W. You can also get thrust from P = (1/2)Fu, giving F = 2P/u = 7.93e9 N. This means that at the end of the acceleration phase the acceleration is F/m = 11 m/s², and at the beginning F/(1.247m) = 9.0m/s². This is actually a fairly comfortable near-Earth gravity environment to be in, if the engine can actually provide it without complications. That's a stretch, but at least it's an order of magnitude less than your previously calculated value.

Weirdly enough, when I try to work the calculation the other way and calculate the energy it consumes per second of acceleration (10 m/s) I get a much lower number:
(1/2(1,600,000,000))(10^2) = 80 GJ

Kinetic energy is not constant across different frames of reference, and the energy spent by the engine is not the kinetic energy gained by the payload. What you'll see in the original frame is that the rocket gains not only obtains kinetic energy from the engine but so does the exhaust, which has a variable speed in that frame. The exhaust velocity is constant as measured from the rocket, but unfortunately the rocket doesn't stay in a single inertial frame, giving rise to the complicated expressions for E above.

[Sikon]

Did I get this right?

Well, no.

The following is more verbose than just giving a large equation, but thinking it out like this can make one confident of understanding the calculation and not making a mistake:

Let's break down the problem into a series of small calculations. That's particularly since one is unlikely to really remember a long equation unless one uses it all the time. A complicated equation with a narrow field of application is the kind of material many people memorize and then forget years after a physics course. Yet, in contrast, one can better remember the simpler fundamental relationships that are the more broadly applicable foundations of physics, like F = MA, KE = 0.5 MV^2, P = MV, etc.

To provide 10 m/s^2 acceleration to 1.6 billion kilograms (1.6 million metric tons) of mass means 1.6E10 N of force, as force equals mass times acceleration (F = MA).

Each kilogram per second of exhaust expelled at 7.84E6 m/s (7840 km/s) provides 7.84E6 N of force.

As a result, the rocket must initially expel ~ 2.04E3 kilograms per second of propellant for that thrust force (a figure which decreases as rocket mass decreases from propellant consumption, if acceleration is to remain 10 m/s^2 constantly).

Then 2.04E3 kilograms of exhaust accelerated to 7.84E6 m/s (relative to the rocket engine) has kinetic energy of ~ 6.27E16 joules, as kinetic energy is half of mass times velocity squared (KE = 0.5 MV^2). So the engine must deliver that much kinetic energy to its exhaust per second.

So the engine has a power of ~ 6E16 watts, except inefficiencies would make the total figure higher.

That's to accelerate the rocket at 10 m/s^2 with such high velocity exhaust when the rocket is initially fully fueled with 1.6 million metric tons mass.

Letting alone the fact this thing would apparently burn up more than 800 kg of antimatter with every trip I can't see how its engine fails to instantly melt itself; that's like an 83 megaton bomb going off in its engine room every second.

Yes.

Although that's not exactly the number of megatons per second (see above), looking at the power to mass ratio like this to check plausibility is good, in contrast to many sci-fi authors. With real world materials, this would be probably breaking plausibility with such a power to mass ratio. Although not analyzing in depth, I definitely suspect even a several-hundred-thousand-ton external pulse detonation engine like some nuke-pulse concepts would be unable to survive so many thousands of terawatts.

*********

Aside from reducing an unnecessary velocity goal, a high velocity ship would be more plausible if it didn't have such relatively high acceleration, reducing the power to mass ratio required for the fictional engine.

With 10 m/s^2 acceleration, it looks like you are aiming for 1g acceleration. That's unneeded, really.

Even if there are unaugmented biological crew onboard, artificial gravity is so easy that there is no comparison, versus the unlikelihood of this one ship's engine handling the raw power equivalent of many thousands of times modern earth's total 2 TWe power generation capability ... especially considering the limits of real world materials on power to mass ratio in a real engineering design.

Not a large area is strictly needed. Although for permanent residence like walking around and working in 1g gravity 24 hours a day, it would be preferable to have 1 rpm to 3 rpm rotation rate, corresponding to hundreds of meters diameter, that's not needed.

If one's really concerned about size, a small rotation device can provide sufficient exposure to 1g gravity for a brief portion of a day.

To illustrate the principle, here's an example of a NASA short radius centrifuge fitting within a moderate sized room:



The rotation rate experienced by the person in it is, of course, many rpm but not too uncomfortable when it only needs to be for a limited portion of time.

That's not very "high tech" but does work.

Also, aside from temporary exposure to 1g gravity daily in an exercise chamber or the like for health reasons with a human crew, lesser gravity is sufficient elsewhere. Even if most of the rest of the living quarters are to have artificial gravity, that doesn't have to be as high as 1g.

A fraction of 1g is probably more fun and at least as practical anyway. For example, although 100 meter radius is involved in providing 1g gravity at 3 rpm, the same rotation rate can provide fractional-g gravity with a far smaller chamber inside the ship.

Indeed, for example, 15 meters in radius is suitable for providing 0.15g gravity. That's like lunar gravity, sufficient for all purposes aside from the temporary 1g gravity exposure provided elsewhere for health reasons.

Alternately, even within real-world laws of physics, temporary exposure to the equivalent of 1g artificial gravity may be possible in a small chamber without rotation. That's through the same principle as how this frog is levitated with a sufficiently intense magnetic field, a diamagnetic effect which is the closest thing to practical antigravity or artificial gravity possible within known physics:



There's not been detailed study of health effects, so there's some uncertainty with that, but at least the frogs appear unharmed enough. Such is mainly not practical in the real world today because it would take a large mass (e.g. tons) of expensive superconductor to create a 1g-equivalent field large enough for a small room, but it is physically possible.

Of course, if the crew was genetically engineered or non-biological or if the cargo ship had no crew, there could be no concern about 1g gravity anyway.

In any case, one way or another, there's not real need for 1g acceleration of the ship.

The ship being modeled is a freighter with a delta V of 1,730 km/s, and I am assuming it carries 708,300 tons of iron. I'm assuming an AM-plasma drive, which has an exhaust velocity of 7,840 km/s and therefore, according to the equation gives by Atomic Rockets (Delta V = , it should have a mass fraction of 1.247, which should give it a total mass of around 1.6 million tons with fuel, engines and everything.

Also, one must question why 708000 tons of iron is being shipped so fast at such relative astronomical energy expenditure, literally 2+ orders of magnitude more velocity and 4+ orders of magnitude more kinetic energy than involved in transporting it without such at hurry at a few km/s instead.

It only takes a few kilojoules per gram to refine iron, although in space there may be use of nickel-iron that isn't even previously oxidized (e.g. nickel-iron asteriods, almost like giant blocks of pre-made steel). Also, it only takes a few kilojoules per gram to give a shipment a few km/s velocity such as to travel across a solar system. In contrast, though, the OP's 1730 km/s is more than a million kilojoules per gram, many orders of magnitude different and not needed.

For interplanetary travel, a delta v like 1730 km/s isn't needed for bulk cargo when at most a handful of km/s are sufficient for shipment within a solar system. It is not like the time in transit should matter much when it is just raw material, cheap due to the almost unlimited amount available. The situation may be more like shipping iron ore by slow oceanic freighters on earth than like rushed, expensive air package delivery.

Also, probably antimatter engines would not be economically the best choice for bulk cargo even if the technology was available, nor engines with such extreme energy-intensive exhaust velocity, with such inefficient, relatively expensive transport reserved for more special applications if developed.

Probably this does not reference interstellar travel, but, definitely, large quantities of bulk material like iron shouldn't be shipped over interstellar distances in preference to using the iron in asteroids already in the target star system.

I'm actually kind of hoping I cataclysmically fucked up, because otherwise I'm going to have to either massively re-evaluate the energy production of my civilization, the level of trade of my civilization, or both, or introduce more handwavium into the setting's technology.

Yes.

A high level of trade within a solar system is plausible, just not quite in the manner of the earlier freighter's performance figures. A transport carrying 710000 tons of iron would more likely slowly journey across part of the solar system over months, analogous to bulk freighters carrying iron ore on earth's oceans today that can take weeks to make their journey. After all, the cheap material isn't worth shipping in a hurry.

Taking advantage of what is possible in space, the bulk transport ship itself would probably mass a small fraction of the huge amount of raw material and propellant it slowly accelerated. If there was any crew at all, it would probably be relatively small, a little analogous to mostly automated 200,000-ton supertankers of today with a handful of crewmembers.

In contrast, a fast passenger ship might use more energy for well-paying passengers, making the journey in days to weeks, a little analogous to terrestrial airline aircraft carrying passengers at greater speed than cargo ships for greater expense.

[Kuroneko]

Sikon, I rather think the rocket equation v = -u ln R is simple enough to be worth remembering. From there, the rocket engine accelerates the exhaust mass M-m to the exhaust velocity u, thus spending E = (1/2)[M-m]u² of energy. The only thing the rocket equation does is to allow us to express this in terms of the change in velocity v: it is equivalent to m = Mexp(-v/u), so E = (1/2)[mu²][exp(v/u)-1]. It really doesn't get simpler than that (and this is slightly cleaner than the off-the-cuff derivation in my first post here, since it bypasses the thrust F).

[Edit: minus sign]

[Junghalli]

You're right Sikon, a slow ship with a low acceleration and delta V definitely makes much more sense for an ore hauler. An Orion could probably get something close to the performance I originally had in mind (supposedly it could do 1 G with a delta V of 1000 km/s), but unfortunately its future looks bleak at the moment.

What sort of drive system do you think would be good for this? I'm thinking either nuclear-thermal or antimatter-catalyzed fusion.

Having Earth-normal gravity in the ship actually wasn't a big concern to me, it just "felt right" for a civilization capable of building large numbers of interstellar ships. Though that's really apples and oranges; just because you can build interstellar ships it doesn't necessarily mean it's practical or intelligent to make every ship throw around energy extravagantly. A cautionary lesson in going with gut feeling and not checking the numbers.

[Junghalli]

OK, here's what I got so far. Did I get the numbers right?

I scaled the cargo down to 2000 tons and gave it a mass ratio of 4. Then I threw in the odd 200 tons of engine, spaceframe, food supplies for the crew etc. I'm using a D-T fusion drive, which has an exhaust velocity of 22,000 m/s.

The Tsiolovski equation for Delta V is Delta V = Exhaust Velocity ln (Initial Mass in kg/Final Mass in kg). So:

Dv = 22,000 ln (8,800,000 / 2,200,000)
Dv = 30.5 km/s

http://en.wikipedia.org/wiki/Tsiolkovsk ... t_equation

The energy production necessary to get a certain delta V is determined by the equation 1/2m(v^2) where m is the mass of propellant and v is the exhaust velocity.

http://en.wikipedia.org/wiki/Spacecraft ... efficiency

So in this case that's 1/2(6,600,000)(22,000^2) = 1.6 X 10^15 joules.

If it accelerates at 1 m/s^2 that's 52.5 GW assuming perfect energy efficiency (unrealistic, obviously). That... sounds feasible.

This ship would take around 6 months to get from the belt to Earth.

Hmm, I don't know this will leave enough shipping volume to make mining belt iron very profitable (iron production today amounts to 1 billion tons, which would equal 500,000 times the payload of one of these little ships), but rarer metals like aluminum should still be profitable.

[Kuroneko]

You did the math correctly, although you if you have a fixed final velocity v and final mass m, in mind you can further bring down the energy requirement. The minimum occurs when u/v = α = 2+W(-2exp(-2)) = 1.5936, where W() is the Lambert W. Therefore:
E_min = (1/2)[mu²][exp(1/α)-1] = 0.4365mu² = 0.7721mv².
[Edit: the actual calc was in error. It doesn't benefit if exhaust speed is constrained.]

[Junghalli]

Looking over Atomic Rockets, low-gear VASIMR actually looks like a better alternative than D-T fusion. It gets you a Delta V a little less than 10 km/s higher for the same amount of fuel, with an energy expenditure of 132.4 GW including inefficiences at 1 m/s^2. It shaves the trip to an average of 4-10 months.

I think 132.4 GW should be feasible, right? Isn't the Saturn V's engine power something like 80 GW?

It's a little weird to think that my FTL drive will probably be able to get to a lot of stars in less time. But that's sort of apples and oranges, I guess.

[Sikon]

While most raw materials including iron and aluminum are unlikely to ever be shipped to earth's surface, being available in practically unlimited quantities in earth's crust as a large percentage of even the average rock and most cheaply obtained through local production, there is great value in returning extraterrestrial material to earth orbit, from metals to ices.

Even near earth objects would amount to probably trillions of tons of total resources, before needing to go as far as the asteroid belt with its quadrillions of tons. For example, if a gravity assist is used, delta v for capture of material from the comet 1979 VA is 4.6 km/s, and it alone has up to on the order of 100 billion tons of water ice. Mining objects is eased by the low gravity; for example, for 1979 VA with 1/10000th of earth's gravity, it only takes 1 ton of rocket thrust to lift 10000 tons of the surface, though transport to earth orbit is a separate situation.

As one illustration, to return material to earth orbit (a high, elliptic orbit) from that object with a nuclear-reactor steam rocket can be done with 800 MWt reactor power per 500 metric tons returned, and a tanker can return around 25 times its own mass. For example, a 200 ton tanker could return 5000 tons, or a 20000-ton tanker could return a half million tons per trip. See here and here.

The former 200-tons-empty tanker would require no more than ~ 8 GW or less reactor power.

The latter giant tanker returning a half million tons per trip would require no more than ~ 800 GWt reactor power, which may seem like a large amount by terrestrial standards, but it is better than it seems, as that is thermal power, a lot easier to get per gigawatt than electricity, in terms of far lesser mass required with a nuclear system.



For perspective, the NERVA nuclear rocket program of the 1960s obtained 4 MW thermal power per liter of reactor, proportionally like 2 cubic meters or 200 cubic meters for 0.8 GWt or 800 GWt respectively, although actually a little less power density and lower operating temperature can be sufficient for nuclear steam rockets. Of course the total volume and mass of the whole system would be much greater than that alone, but the general situation is that at least hundreds of megawatts thermal power per ton of nuclear reactor is obtainable.

Such is not difficult technology by sci-fi standards. Nuclear steam rockets have been estimated to be 150 sec to 235 sec Isp, operating at 1200 degrees Celsius in the latter case (much less than NERVA actually, as not the same performance is needed for the relatively easier task of moving material around in near-zero-g instead of launching material from earth).

Over a period of decades, each transport could make many trips, and millions of tons could be returned by a relatively small mass of transports.

The preceding is not necessarily optimal, just an illustration. For example, 1988TA has just a 3.6 km/s capture delta v for returning material from it to earth orbit instead, corresponding to under 60% as much energy involved per unit mass of material returned, and it is a hydrated NEO. There may be suitable NEOs still closer in delta v recently discovered or yet to be discovered since more are found each decade.

Of course, this is not a high tech method by sci-fi standards, and, alternately, there are other plausible options, including the various forms of nuclear external pulse propulsion that can be much higher performance, which could be used with ice or other extraterrestrial material as a large quantity of inert propellant blown out by the nuclear explosions. Or, for example, a sufficiently developed future space civilization might mainly skip propellant consumption for a lot of commercial transport within a star system because of having a vast network of mass driver accelerators and capturing deaccelerators almost everywhere. There are a number of possibilities.

What sort of drive system do you think would be good for this? I'm thinking either nuclear-thermal or antimatter-catalyzed fusion.

Those are definitely possibilities.

There's more to cover, but that will be left for a later post.

[Sikon]

for 0.8 GWt or 800 GWt respectively

... should be

for 8 GWt or 800 GWt respectively