## Energy requirements to launch to LEO (from vacuum zeppelin)

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idobox
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### Energy requirements to launch to LEO (from vacuum zeppelin)

In the other thread, davidstarlingm wrote:
idobox wrote:Getting on top of the atmosphere doesn't require much energy, it's getting to orbital speed that is really difficult.

If you get on top of the atmosphere (and can stay there for any significant length of time), you can accelerate to orbital speed without drag. Which....I imagine....would save you a lot of energy.

Is it fair to say that if we can get to the top of the atmosphere using less energy than we would waste by accelerating to orbital speed in-atmo, it's more economical to do the former?

Of course, there's the pesky issue of maintaining altitude while you accelerate, but surely that wouldn't take very long. Just before entering LEO, the Space Shuttle engines have to be throttled down to 3 gs for the sake of the astronauts; at this acceleration, the 7.8 km/s of LEO would be reached in 4 minutes 25 seconds. Of course, the mechanics of transitioning from hovering-in-space to full orbit in less than five minutes is going to be incredibly complex. I don't want to even try to model that, but I know the vertical thrust you need decreases over time.

Until we can come up with a true SSTO craft, I'm predicting that the most efficient/economical orbital launch system would be a LTA "spacecraft carrier" which would carry the launch vehicle to a high altitude. The vehicle would execute a vertical takeoff transitioning to horizontal flight, with the altitude and velocity chosen specifically so that the gradually decreasing aerodynamic lift is progressively replaced by centrifugal force as the craft moves into LEO. Again, the important question is whether the cost of ascent is recouped by not having to push through the lower atmosphere.

Giving an object of 1kg a speed of 8km/s requires 32MJ. Bringing that same mass to 100km altitude requires about .9 MJ.

I might be wrong, but I think the main reason for launching rockets from planes or balloons is the engine design. The optimal shape of the nozzle depends on the atmospheric pressure, so if you launch from sea level, your engine becomes inadapted pretty quickly.
Drag might also play, but I suspect the 300m/s or more a plane can fly are more important.
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### Re: Vacuum Zeppelin

Yeah, but you're ignoring the atmosphere there. Plus, it's power rather than total energy that's really the issue.

We use big inefficient rockets because we need a high thrust-to-weight ratio to get off the ground and out of the atmosphere as quickly as possible. Outside most of the atmosphere, and held up by something other than rockets, you could use a high specific impulse rocket running on low power, such as solar, to get up to orbital speed at your liesure.
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Schrollini
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### Re: Vacuum Zeppelin

idobox wrote:Giving an object of 1kg a speed of 8km/s requires 32MJ. Bringing that same mass to 100km altitude requires about .9 MJ.

Most of the energy in a LEO object is kinetic, rather than gravitational potential. I was curious about the energy cost of actually putting it there, though, so I ran some numbers.

Wikipedia suggests that a LEO has a velocity of about 8 km/s, but a typical launch requires a delta-v of up to 10 km/s, with the remainder lost to atmospheric and gravity drag. So in terms of delta-v, the atmosphere is still pretty negligible.

However, the rocket equation tells us that this additional delta-v requirement can be costly in terms of fuel or, equivalently, energy. The mass of propellant to launch a fixed mass to Δv goes like exp(Δv/ve) - 1. Using an exhaust velocity of v[sub]e[/sup] = 4.5 km/s (a rough estimate for the Space Shuttle's main engines), I find that getting to a delta-v of 8 km/s (just LEO velocity) takes only about 60% of the energy of getting to 10 km/s (actually getting to LEO). Of course, a rocket that only holds 60% of the fuel will be less massive, leading to further gains. I'm also ignoring things like staging. Still, a decent rule of thumb seems to be that it takes twice as much energy (in terms of fuel) to get to LEO from the ground than it would to accelerate horizontally after getting to altitude.

I'm not sure how the drag breaks down between atmospheric and gravity terms. But if they're the same order of magnitude, then idobox may indeed be right that the dominant cost of getting to LEO is getting the horizontal velocity. But the relative costs are not nearly as lopsided as the energy budget for a LEO orbiter suggest.

And as gmalivuk points out, if you find a way to negate the drag terms, you can use a high impulse / low thrust engine to beat these chemical rocket costs by a large margin.
Last edited by Schrollini on Tue Jul 23, 2013 9:24 pm UTC, edited 1 time in total.
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davidstarlingm
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### Energy requirements to launch to LEO (from vacuum zeppelin)

Schrollini wrote:I'm not sure how the drag breaks down between atmospheric and gravity terms. But if they're the same order of magnitude....

The following might surprise you.

gmalivuk wrote:Yeah, but you're ignoring the atmosphere there.

Right, that's what I was thinking. That 0.9 MJ needed to get 1 kg to 100 km gets much, much bigger when you factor in drag.

We can roughly estimate the total energy lost to drag by summing the work done by the two main stages, then subtracting the gravitational-potential and kinetic energy of each component at the end of each stage.

Stage 0: The SRBs each provide 14 MNs of thrust for most of their 124-second burn time. Plugging-and-chugging with these equations tells us that the altitude at that point is ~151,000 feet. This is within 3% of the altitude given here, so it seems like a good estimate. Some quick-and-dirty calculus with the help of Wolfram tells me that the definite integral of those two parametric equations from 0 to 124 (seconds) is 8.73e6 feet, or an arc length of 2660.9 km, equating to 7.451e13 joules of work done by the two SRBs.

Stage 1: The three SSMEs provide an average combined thrust of 6.207 MN (averaging SL performance with vacuum performance) for the entire eight minutes of main launch time. During the first 124 seconds of the launch, that's 1.652e13 joules of work (using the same arc length calculated above). They stop firing when the external launch tank is jettisoned at an altitude of 113 km, 61.2 km higher than SRB jettison. It's difficult to determine the horizontal displacement at this point, but given that it's supposed to land in the Indian Ocean when launched from Florida, the peak should occur around 7915 km, 7863 km past the 124 second point (as determined using the parametric equations above). Adding conservatism by using a diagonal instead of the actual curved trajectory, that's a total Stage 1 displacement of 7863.2 km, equating to 4.88e13 joules of additional work done by the SSMEs.

The total work, then, is approximately 1.398e14 joules.

We subtract....

Gravitational potential of two empty 91,000 kg SRBs at 45 km: 7.82e10 joules
Kinetic energy of two empty 91,000 kg SRBs at 1.34 km/s: 1.63e11 joules

Gravitational potential of empty 30,000 kg ET at 113 km: 3.32e10 joules
Kinetic energy of empty 30,000 kg ET just under LEO (7.8 km/s): 9.73e11 joules

Gravitational potential of 110,000 kg orbiter at 113 km: 1.22e11 joules
Kinetic energy of 110,000 kg orbiter at just under LEO (7.8 km/s): 3.57e12 joules

....giving us total combined kinetic and gravitational energy of 4.94e12 joules, meaning that the estimated drag is (1.398e14 - 4.94e12) joules = 1.349e14 joules, over 96% of the total energy cost.

I think those numbers are correct. Are they correct?

Given that the external tank reaches near-orbit along with the orbiter, that's 140 tonnes of payload. So each 1 kg we put in orbit may only cost 32 MJ for its speed and 0.9 MJ for its altitude, but it costs 964 GJ in drag. Safe to say we'd do pretty well if we could cut this out.

Outside most of the atmosphere, and held up by something other than rockets, you could use a high specific impulse rocket running on low power, such as solar, to get up to orbital speed at your liesure.

It's the "held up by something other than rockets....at your leisure" bit that poses the biggest challenge.

Schrollini
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### Re: Vacuum Zeppelin

davidstarlingm wrote:Gravitational potential of two empty 91,000 kg SRBs at 45 km: 7.82e10 joules
Kinetic energy of two empty 91,000 kg SRBs at 1.34 km/s: 1.63e11 joules

Gravitational potential of empty 30,000 kg ET at 113 km: 3.32e10 joules
Kinetic energy of empty 30,000 kg ET just under LEO (7.8 km/s): 9.73e11 joules

Gravitational potential of 110,000 kg orbiter at 113 km: 1.22e11 joules
Kinetic energy of 110,000 kg orbiter at just under LEO (7.8 km/s): 3.57e12 joules

....giving us total combined kinetic and gravitational energy of 4.94e12 joules, meaning that the estimated drag is (1.398e14 - 4.94e12) joules = 1.349e14 joules, over 96% of the total energy cost.

I think those numbers are correct. Are they correct?

I don't see any accounting for the kinetic and potential energy put into the exhaust. Given that there's probably 10 times as much propellant as vehicle, by mass, and that it's moving at more than half orbital velocity, that's a big term.

What you've shown is that very little of the initial energy is put into the payload. This is true even in the case of the no-drag rocket that I worked out. The question I was trying to answer is how much more energy do you need to get into orbit given air and gravity, and the answer seems to be roughly twice.
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Tass
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### Re: Vacuum Zeppelin

Could you take the rocket discussion to another thread? It doesn't really have to do with vacuum zeppelins.

Copper Bezel
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### Re: Vacuum Zeppelin

Sure, but this topic was brought back up in regard to zeppelins as launch platforms, and the rocket discussion is still about whether or not launching from a somehow-elevated platform really helps. I'm having trouble finding any older threads on the subject of not-rockets as first-stage launch - I found this one, but it's fairly short and not very involved. It does raise the issue that zeppelins with enough lift to actually take a launch vehicle anywhere would need to be rather absurdly huge.

Still, regardless of how stuff gets moved around, I want to ask - isn't your percentage lost to "gravity drag" just 1 / gs of thrust? I mean, the fraction of the thrust that's countering the constant acceleration of gravity?
So much depends upon a red wheel barrow (>= XXII) but it is not going to be installed.

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davidstarlingm
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### Re: Vacuum Zeppelin

Schrollini wrote:I don't see any accounting for the kinetic and potential energy put into the exhaust. Given that there's probably 10 times as much propellant as vehicle, by mass, and that it's moving at more than half orbital velocity, that's a big term.

The energy in the exhaust is (thankfully) out of the equation, because it's all contained in the thrust terms. Thanks to the Oberth effect, the increase in propellant velocity is incorporated in the thrust efficiency.

I'm still not sure about my distances, though. Did I calculate the arc length right?

Tass wrote:Could you take the rocket discussion to another thread? It doesn't really have to do with vacuum zeppelins.

We're trying to figure out if vacuum zeppelins would offer a more efficient launch platform for orbital rockets.

Copper Bezel wrote:regardless of how stuff gets moved around, I want to ask - isn't your percentage lost to "gravity drag" just 1 / gs of thrust?

Yeah, unless he's talking about relativistic effects or something, I'm assuming "gravity drag" is just gravitational potential energy. Though I could be wrong. I'm calculating on the basis of displacement, not time, which might take care of that particular concern; if I was doing the calculations based on time, lost thrust due to accelerative drag would be a factor.

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### Re: Vacuum Zeppelin

http://en.wikipedia.org/wiki/Gravity_drag

Keep in mind that the acceleration due to gravity "g" depends on altitude, and is something like ten percent less at the altitude of the ISS. It's an r-squared term, so it sneaks up on one even if its fairly small compared to the radius of the Earth.

Also: http://www.jpaerospace.com/

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Copper Bezel
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### Re: Vacuum Zeppelin

davidstarlingm wrote:Yeah, unless he's talking about relativistic effects or something, I'm assuming "gravity drag" is just gravitational potential energy.

The link is a better explanation, but my point was that it's exactly not that. If your rocket thrusts at 1.1g, only the 0.1g is contributing to the craft's gravitational potential energy, while the 1 g is counteracting the whole "falling down" thing.
So much depends upon a red wheel barrow (>= XXII) but it is not going to be installed.

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Schrollini
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### Re: Vacuum Zeppelin

davidstarlingm wrote:The energy in the exhaust is (thankfully) out of the equation, because it's all contained in the thrust terms. Thanks to the Oberth effect, the increase in propellant velocity is incorporated in the thrust efficiency.

The Oberth effect, as I understand it, says that the energy you invested in accelerating your fuel is not wasted -- you can use some of it to accelerate the payload further. It does not say that the exhaust carries away no energy. In fact, that page explains the Oberth effect by noting that at high vehicle velocity, the exhaust carries away less energy than it would at low vehicle velocity, since the exhaust velocity, relative to our accountant, is less.

The energy balance should look like
Work by motor = energy of craft + energy of exhaust + energy lost to drag
You've calculated the first two, but lumped the third in with drag. Wikipedia says (though I haven't checked) that in the best case scenario, the energy in the exhaust will be 54% of the energy in the craft. (My sample cases aren't that far off from the idea exhaust velocity, though, so my statement about the exhaust energy being much greater than the payload energy may be overly pessimistic.)

Copper Bezel wrote:
davidstarlingm wrote:Yeah, unless he's talking about relativistic effects or something, I'm assuming "gravity drag" is just gravitational potential energy.

The link is a better explanation, but my point was that it's exactly not that. If your rocket thrusts at 1.1g, only the 0.1g is contributing to the craft's gravitational potential energy, while the 1 g is counteracting the whole "falling down" thing.

The link has more details, but the worst case for gravity drag is thrusting straight upwards, in which case the drag is g * thrust time. Using the 8.5 min SSME burn, I get a worst-case gravity drag of 5 km/s. We'd need details of the actual launch trajectory to work it out for a real launch, but an order of 1 km/s is probably not unreasonable. If this is right, (and the 10 km/s to LEO is right), this suggests that air drag is also on the order of 1 km/s.

Tass wrote:Could you take the rocket discussion to another thread? It doesn't really have to do with vacuum zeppelins.
Sorry. I reported myself, but the mods declined to split the thread. So I guess they think the rocket talk is okay? I await the red text of scolding.

As for the real topic: I prefer the Hoover model XR-7 when vacuuming my zeppelin, but my friend swears by his Electrolux.
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davidstarlingm
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### Re: Vacuum Zeppelin

As it states on that wiki page, gravity drag is a function of time. I specifically avoided dealing with anything that was a function of time so I wouldn't have to deal with all the things that were changing as a function of time, like the mass of everything being launched. I never do any more calculus than I have to.

If we had never dealt with an actual Space Shuttle launch, we would need to be calculating everything as a function of time, from the ground up, using an appropriate transonic drag equation, and linear fuel consumption, and so forth. Thankfully, we don't have to. We know the trajectory of an actual launch, and we know that the thrust of the various engines is constant as long as they are in use. So the work done on the spacecraft by the engines (which is all we're interested in) is simply thrust times distance.

Exhaust velocity is the source of the work being done, so it shouldn't be subtracted from the total work done. And, as stated, gravity drag is a function of time. That's why multiplying by distance lets us avoid it. Notice that the effect of gravity drag is in delta-v, not in units of energy. If I have a rocket that provides 6 gs of acceleration for 25 seconds, I need to subtract gravity drag in order to get my final velocity. But if I don't care about my final velocity and I already know my displacement and I simply want to know the actual amount of work done by the engine, I only need to multiply engine thrust by net displacement.

Schrollini
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### Re: Vacuum Zeppelin

davidstarlingm wrote:As it states on that wiki page, gravity drag is a function of time. I specifically avoided dealing with anything that was a function of time so I wouldn't have to deal with all the things that were changing as a function of time, like the mass of everything being launched. I never do any more calculus than I have to.

That's fine. But I wanted, for no reason other than my own curiosity, to try to separate out the atmospheric from the gravity drag. So I made an overestimate of the gravity drag, and then pointed out that the actual value would be difficult to calculate. But the general order of magnitude seems reasonable when compared to the 2 km/s quoted for total drag.

davidstarlingm wrote:So the work done on the spacecraft by the engines (which is all we're interested in) is simply thrust times distance.

Sorry, I misunderstood what you were trying to calculate. I thought you were trying to calculate the total work done and equating that to the energy of the ship + drag. But you were calculating the total work on the ship, which should be okay.

I'm a bit suspicious of your work-done figure, though. Let me try to estimate the energy content of the fuel. The two SRBs each have about 4e5 kg of propellant, with an effective exhaust velocity of 2500 m/s. This gives them a total energy of 2 * 1/2 * 4e5 kg * (2500 m/s)2 ~ 2.5e12 J. The external tank has about 7e5 kg of propellant. The main engines have an exhaust velocity of 4500 m/s, so this system has a total energy of 7e12 J. So the total fuel energy of the SST on the launchpad is around 1e13 J. (Or 10 PJ, which I point out just for the opportunity of using a unit synonymous with bedclothes.) This is a order of magnitude less than your work estimate, which suggests that at least one of us screwed up.

The one niggle I have with your work calculation is that both the SRB and the SSME throttle down towards the ends of their respective burns, to keep the acceleration of the lightening craft below 3g. Since this is when the craft is moving the fastest, this period has an outsized influence on the work done. I have trouble believing that it gives us a factor of 10, though.
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davidstarlingm
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### Re: Vacuum Zeppelin

Schrollini wrote:I wanted, for no reason other than my own curiosity, to try to separate out the atmospheric from the gravity drag. So I made an overestimate of the gravity drag, and then pointed out that the actual value would be difficult to calculate. But the general order of magnitude seems reasonable when compared to the 2 km/s quoted for total drag.

Atmospheric drag is an instantaneous force vector which varies with speed, while gravity drag is (if I understand it correctly) a sum velocity vector which accumulates over time. So I'm not entirely sure how one would go about comparing the two, unless atmospheric drag can be represented as a sum velocity vector. But wouldn't that be the same as using the total energy lost to drag, dividing by the mass, multiplying by 2, and taking the square root (i.e., determining the kinetic energy "stolen" by drag and using that to come up with a velocity)? If we're doing that, it might be easier to just stick with units of work rather than converting back and forth.

I thought you were trying to calculate the total work done and equating that to the energy of the ship + drag. But you were calculating the total work on the ship, which should be okay.

Right, the energy lost to drag is transmitted from the ship to the atmosphere, so it gets contained nicely within the work done on the ship. There's no use in trying to calculate the total energy of the system when we already have the figures for trajectory and whatnot.

I'm a bit suspicious of your work-done figure, though.

Honestly, so was I. When I looked back over my work, I was getting around 61 km of vertical displacement and 52 km of horizontal displacement at SRB jettison using those parametric equations, but Wolfram was giving me an arc length of 2661 km, which makes no sense whatsoever. If those displacement figures are correct, the arc length has to be between 79 and 113 km, which fits with this parametric plot.

Using 100 km as a rough estimate of arc length (midway between the diagonal and a right-angle curve), that gives us 2.8e12 joules of work done by the SRBs, more than a full order of magnitude less.

The one niggle I have with your work calculation is that both the SRB and the SSME throttle down towards the ends of their respective burns, to keep the acceleration of the lightening craft below 3g.

I don't believe the SRBs cannot be throttled down, though their thrust is dramatically reduced in the last 4 seconds before jettison. However, the SSMEs are throttled down to 65-72% around the max-Q point (maximum dynamic pressure due to drag), which lasts for 32 seconds and is roughly linear in terms of trajectory, or 8.2 km. This is a loss of 1.6e10 joules of work, meaning the first stage represents 3.05e11 joules of work done by the SSMEs.

The throttling-down at the very end of burn only happens for ten or twenty seconds; it shouldn't be a major factor.

Calculating the displacement of the second stage is difficult, though. According to the original source of those parametric equations, they are roughly accurate for the first five minutes (300 seconds) of launch, equating to this plot. The solution for dS/dt given shows what seems to be a near-linear velocity plot, climbing to orbital speed within 270 seconds (which doesn't seem right, as these are parametric equations of motion and should represent actual velocity).

I don't want to use the same displacement estimated by the halfway-to-Indian-Ocean approach, but I don't know what other choice I have. The horizontal displacement given by the equation at 300 seconds is well over 8000 km, and that's supposedly accurate, but would give an even higher figure.

This trajectory (Ctrl+F for "STS-30 TRAJECTORY SEQUENCE") gives us relative velocity and altitude at each point along the launch, up to main engine cutoff. The velocity points fit fairly well to a third-order polynomial:

so I'll try integrating from 124 to 511 seconds to find total displacement. Wolfram gives 6.02e6 feet, or 1835 km. This seems more reasonable, and gives a second-stage work product of a much more manageable 1.14e13 joules.

All told, the work is reduced from 1.398e14 joules to 1.45e13 joules. Still higher than your estimated total fuel energy of 1e13 joules, but much closer.

Note that this is an energy-lost-to-drag of 9.5e12 joules, still 65% of total work. For each 1 kg we put in near-orbit this way, that's 6.7e7 joules of energy lost to drag. Visually:

So while we can't expect to save much by merely eliminating the cost of gravitational potential, drag is a much bigger issue.

Schrollini
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### Re: Vacuum Zeppelin

davidstarlingm wrote:Using 100 km as a rough estimate of arc length (midway between the diagonal and a right-angle curve), that gives us 2.8e12 joules of work done by the SRBs, more than a full order of magnitude less.

And this fits fairly nicely with my estimate of 2.5e12 J available energy. I suspect there's a factor of that one of us has missed, since this energy will be split between the exhaust and the vehicle.

davidstarlingm wrote:I don't believe the SRBs cannot be throttled down, though their thrust is dramatically reduced in the last 4 seconds before jettison.

Throttled may be a bad word. But they do shape the fuel to give a non-trivial thrust profile.

davidstarlingm wrote:Note that this is an energy-lost-to-drag of 9.5e12 joules, still 65% of total work.

My initial estimate was that only 60% of the energy used in launch was "useful", and of that, roughly 60% goes into the vehicle. Figuring very roughly, the energy put into the vehicle (36%) would be the same as the energy lost to drag (40%). I think this is close enough to your 1:2 ratio for us to declare agreement.
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Schrollini
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### Re: Vacuum Zeppelin

gmalivuk wrote:I will be splitting the thread when I get to a computer, since recent discussion is entirely about launch mechanics, and is interesting enough not to be lost in a thread nominally about ineffective airship designs.

Thanks. Sorry for the hijack. If I had known this would cause so much discussion, I would have put it in a separate thread to begin with.

If anyone cares to do a better model of the work done by the SSME, here's a plot (from here) giving the thrust profile. It looks like there's a good minute or so of downthrottling at the end.
ssme.png (41.53 KiB) Viewed 35028 times
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davidstarlingm
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### Re: Vacuum Zeppelin

Schrollini wrote:(2.8e12 joules of work done by the SRBs) fits fairly nicely with my estimate of 2.5e12 J available energy. I suspect there's a factor of that one of us has missed, since this energy will be split between the exhaust and the vehicle.

Maybe the missing factor is the non-trivial thrust profile?

Or maybe the physics is getting too messy. I may be wrong, but combining the exhaust velocity and the fuel mass should give you the total available work, not the total chemical energy. The total chemical energy would be an entirely different (and not particularly relevant) consideration.

Prediction: when factoring in the non-trivial thrust profile, my 2.8e12 joules will be reduced to around 2.5e12 joules.

they do shape the fuel to give a non-trivial thrust profile.

I see.

Unfortunately these data points are all given in seconds elapsed, rather than in distance traveled. I'll have to do some Excel magic with those old parametric equations to make the conversions.

There's the same data, mirrored into Excel with the distance provided by the parametric equations. Theoretically, we can multiply the thrust at each step with the displacement at each step to get the work done in that duration. However, the sum of all these figures is 5.7e11 J, which is sizeably less than what we'd previously estimated. Not sure how much we should rely on that.

If anyone cares to do a better model of the work done by the SSME, here's a plot (from here) giving the thrust profile. It looks like there's a good minute or so of downthrottling at the end.

Excellent!

I've already more-or-less accounted for the downthrottling at the beginning, so all we need to do is subtract the work lost in downthrottling at the end.

The point of this downthrottling is to maintain 3 gs of acceleration. Since we're nearly in LEO at this point, the acceleration of gravity shouldn't be a factor. Given that we know the final velocity (7.8 km/s), this allows us to determine the position at each point from ~450 seconds to 511 seconds. 61 seconds at 3 g is a delta-V of 1.79 km/s, meaning that the speed at the start of downthrottling was 6.01 km/s. Using a basic kinematic equation, displacement was 18,600 km. The thrust goes from 104% (2370.16 kN) to 65% (1481.35 kN) during those 61 seconds, meaning dF/dt is -14.57 kN/s.

Of course, that doesn't do us much good, because the decreasing thrust is linear with respect to time, not with respect to distance. Converting from dt to dx isn't very pleasant, but you end up with a loss of 2.032e11 joules, which is noticeable-but-not-terribly-so.

Summing our new work-done-by-engines....

SRBs: 5.7e11 J
SSMEs(pre-jettison): 3.05e11 J
SSMes(post-jettison): 1.12e13 J

Total work: 1.207e13 J

Even closer to your estimated total fuel energy. This reduces energy-loss-to-drag from 65% to 61%.

I think these numbers check out properly.

Schrollini
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### Re: Vacuum Zeppelin

davidstarlingm wrote:There's the same data, mirrored into Excel with the distance provided by the parametric equations. Theoretically, we can multiply the thrust at each step with the displacement at each step to get the work done in that duration. However, the sum of all these figures is 5.7e11 J, which is sizeably less than what we'd previously estimated. Not sure how much we should rely on that.

Is that figure for one or both SRBs? (The thrust graph is for individual ones.) If that's only one, that would be 1.1e12 J total work done on the spacecraft. This would mesh very nicely with my estimate of 2.5e12 J plus a roughly equitable split of energy between the vehicle and the exhaust.

davidstarlingm wrote:I think these numbers check out properly.

They look pretty good.

One other thing we could do to check: The SRBs are jettisoned at 150,000 ft (45 km). There's essentially no air up there, so atmospheric drag losses should be minimal for the rest of the ascent. Because of this, I'm sure the shuttle is thrusting almost horizontally from here on out, so gravity drag losses should also be minimal. So the work done on the vehicle after separation should be almost equal to the energy gained between separation and MECO. This appears to be somewhat off: 1.1e13 J of work after separation and somewhat less than 5e12 J energy gained. Either there's more drag than I expect, or one of these numbers is a bit off.

(I'm a little suspicious that the SSMEs do 30x the work after SRB separation as before. But they are essentially release energy stored in the fuel during the first phase of liftoff, so it's not crazy.)
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### Re: Vacuum Zeppelin

Schrollini wrote:
davidstarlingm wrote:There's the same data, mirrored into Excel with the distance provided by the parametric equations. Theoretically, we can multiply the thrust at each step with the displacement at each step to get the work done in that duration. However, the sum of all these figures is 5.7e11 J, which is sizeably less than what we'd previously estimated. Not sure how much we should rely on that.
Is that figure for one or both SRBs? (The thrust graph is for individual ones.) If that's only one, that would be 1.1e12 J total work done on the spacecraft. This would mesh very nicely with my estimate of 2.5e12 J plus a roughly equitable split of energy between the vehicle and the exhaust.
That's the figure for just one of them, so it looks like the SRB figures do indeed match up pretty well.

(I'm a little suspicious that the SSMEs do 30x the work after SRB separation as before. But they are essentially release energy stored in the fuel during the first phase of liftoff, so it's not crazy.)
Also remember how the v2 term affects kinetic energy. According to the trajectory described earlier, with the SRBs it goes from 0 to about 1.28km/s, for about 825kJ/kg of kinetic energy. At MECO, it's going 7.4km/s, for a total of 27.4MJ/kg, and a ratio of about 32 times more specific energy gained after SRB separation than before, from the actual recorded data.
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:Is that the figure for one or both SRBs?

Oh, don't I feel stupid. Yeah, that piecewise integration was each of the two SRBs, so the total work is 1.14e12 J.

I confess that I'm not sold on how you arrived at the value for total energy, though. For one thing, the exhaust velocity relative to the ground should decrease over time as the rocket gains speed. And even then, I'm not so sure that total energy should be represented as the kinetic energy of the propellant. Wouldn't the total chemical energy be the kinetic energy of the propellant plus the work done to the rocket?

The SRBs are jettisoned at 150,000 ft (45 km). There's essentially no air up there, so atmospheric drag losses should be minimal for the rest of the ascent.

The pressure in the stratosphere is around 1/1000th of sea level pressure, so yeah, there shouldn't be much more drag. But still some. For reference to the larger question of a lighter-than-air high-altitude launch platform, note that the highest unmanned balloon went to 173,900 feet, so it's unlikely that our launch platform would be able to go any higher than the point of typical SRB jettison. The major difference, of course, would be that our launch vehicle would have to start at this altitude from negligible velocity.

The advantage is that we don't have to spend energy on the drag from ground to (this) altitude.

The sum work pre-jettison is 1.45e12 J. At jettison, we've lifted the two empty SRBs, the orbiter, and a tank that is still around 75.6% full (a sum total of 876 tonnes) to 45 km at a velocity of 1.34 km/s. The combined KE and GPE here comes to 1.17e12 J, meaning this first stage costs 2.8e11 J in drag, or 19% of total work done up to that point.

However, note that the KE and GPE of the SRBs (182 tonnes, for a total energy of 2.44e11 J) is also wasted. All in all, this is 5.24e11 J wasted, or 36% of our work-done. Which means our actual launch vehicle can be that much lighter.

It's not currently feasible to launch a SSTO ship using a hybrid turbojet-scramjet engine as the primary stage. The efficiency is higher, but the weight of the fuel to get up to altitude/velocity is just too much. By eliminating the drag and lost energy of getting up to altitude, perhaps a hybrid turbojet-scramjet engine would be more feasible. One could envision a couple of small boost rockets (much cheaper than the SRBs) used to launch from the high-altitude platform, then a ramjet/scramjet to push up to near-orbital speeds before using a short burn from the OMS to jump into orbit. This would almost completely eliminate gravity drag, as the rockets would never be thrusting vertically except during orbital insertion.

I'm sure the shuttle is thrusting almost horizontally from here on out, so gravity drag losses should also be minimal. So the work done on the vehicle after separation should be almost equal to the energy gained between separation and MECO. This appears to be somewhat off: 1.1e13 J of work after separation and somewhat less than 5e12 J energy gained. Either there's more drag than I expect, or one of these numbers is a bit off.

One consideration: at SRB jettison, the thrust-to-weight ratio of the SSME is less than 1, and so the altitude actually declines for a bit before the mass of the system decreases enough to begin climbing. That may be the source of the discrepancy.

But for the purposes of a vacuum zeppelin launch platform, it's that first stage that we're most worried about.

Our real goal, I think, is to make the launch vehicle as small and as reusable as possible.

I'm a little suspicious that the SSMEs do 30x the work after SRB separation as before. But they are essentially release energy stored in the fuel during the first phase of liftoff, so it's not crazy.

Yeah, it seems odd, but if you're looking at the total remaining chemical energy, you have to factor in the kinetic energy of the propellant at SRB jettison. With 75.6% of its propellant remaining, that's 554 tonnes with a kinetic energy of 4.98e11 J, which is hardly shabby.

The one number I'm still suspicious of is the distance covered during the downthrottling phase. 18,600 km is nearly half the diameter of the Earth, and it's an order of magnitude more than the total displacement calculated earlier for the total launch period. But since this number goes into subtracting from the work done by the SSMEs, it's not a major factor; even if it's seriously off, that's only adding conservatism.

So let's say we can build a floating launch platform at 150,000 feet. What then?

We could consider a turbofan-to-ramjet-to-scramjet approach, since our fuel requirements are going to be much lower, but carrying a turbofan into orbit will probably be more trouble than it's worth. I think a mini-SRB is a better bet. Can we do a disposable-solid-rocket-to-scramjet-to-OMS launch? We need to get to at least Mach 5 before a hybrid sc/ramjet will function. Thanks to the stratopause, the speed of sound at 45 km decreases significantly, so that Mach 5 is only 1.64 km/s. Limiting acceleration to 3 gs would require a solid-rocket burn time of at least 55 seconds, at which point the boosters would break away and the scramjet engine would take over. The ship's body need only provide sufficient lift (at 1/1000th of an atmosphere) to counteract gravity during that boost phase.

How small of a launch vehicle can we realistically construct? If we want to use this to ferry either people or cargo into orbit (with the ultimate aim of dramatically reducing the cost/kg of orbital launch), what's a rule of thumb on payload size? Should we try to replicate the Shuttle's payload, or go with something more modest?

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### Re: Energy requirements to launch to LEO (from vacuum zeppel

davidstarlingm wrote:I confess that I'm not sold on how you arrived at the value for total energy, though. For one thing, the exhaust velocity relative to the ground should decrease over time as the rocket gains speed. And even then, I'm not so sure that total energy should be represented as the kinetic energy of the propellant. Wouldn't the total chemical energy be the kinetic energy of the propellant plus the work done to the rocket?

Yep. When the rocket's at rest, all the kinetic energy produced goes into the exhaust. When the rocket's traveling at exhaust velocity, the exhaust is left with zero kinetic energy. When the rocket's faster than the exhaust velocity, the work done on the exhaust is actually negative, since the force and displacement are anti-parallel. But in each case, the extra energy turns out to be ve2/2 per unit mass of propellant spent, so we call that the fuel energy.

To see this, consider a rocket of mass m and a bit of propellant dm traveling at velocity v. The bit of propellant is burnt and expelled backwards, so the rocket has velocity v+dv and the exhaust velocity v-ve. Conservation of momentum gives us dv = dm/m ve. The initial energy is 1/2 (m+dm) v2 + Efuel, while the final energy is 1/2 m (v+dv)2 + 1/2 dm (v-ve)2. These are the same, so a bit of algebra gives you Efuel = 1/2 dm ve2(1 + dm/m). This last term vanishes in the continuum limit, giving us an energy density of Efuel/dm = 1/2 ve2.

In some sense, it had to work out this way: All observers will agree on conservation of energy, so at every point we can choose an observer co-moving with the rocket who notices that all the energy went into moving the exhaust, which has energy density ve2/2. Other observers will disagree where the energy is going, but they all have to agree that that much more kinetic energy appeared after burning the fuel.

The nice thing about this approach, as opposed to doing chemistry, is that we don't have to know anything about the efficiencies of the reaction or the engine. We just see what the exhaust speed is and work out the usable energy in the fuel.

davidstarlingm wrote:One consideration: at SRB jettison, the thrust-to-weight ratio of the SSME is less than 1, and so the altitude actually declines for a bit before the mass of the system decreases enough to begin climbing. That may be the source of the discrepancy.

How about that! It must be odd to experience a period of less than 1g during the ascent -- a bit of a fall in the middle. Depending the orientation of the thrust here (do you lose altitude but try to gain speed, or do you try to keep altitude?), there could be significant gravity drag terms.
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:It must be odd to experience a period of less than 1g during the ascent -- a bit of a fall in the middle. Depending the orientation of the thrust here (do you lose altitude but try to gain speed, or do you try to keep altitude?), there could be significant gravity drag terms.

At this point, the SSMEs are running at 104% rated capacity and the orbiter + tank + propellant weighs in at 694 tonnes, so the acceleration is still 9.58 m/s2. Since I'm assuming most of that is directed horizontally, it's still enough that they aren't going to feel like they're in free fall.

At some point, centrifugal force is going to begin counteracting gravity, so that's why I'm guessing most of the thrust is still directed horizontally. Reduce gravity drag as much as possible, right?

The nice thing about this approach, as opposed to doing chemistry, is that we don't have to know anything about the efficiencies of the reaction or the engine. We just see what the exhaust speed is and work out the usable energy in the fuel.

Yep, that's definitely nice. I'm not sure what it can tell us about the overall costs, though.

What's a reasonable payload requirement for our final stage? Should we be trying to compete with the Shuttle (24 tonnes payload) or the Soyuz spacecraft (7 tonnes total craft mass; very low cargo payload)? Perhaps a 10-tonne payload plus the weight of the craft and engines is enough? Or is enough to do a 5-tonne payload plus the weight of the craft and engines, given that the Soyuz can't carry anywhere near 5 tonnes of actual payload?

EDIT: What about re-entry? I know re-entry shields aren't THAT heavy, but given that we don't have to deal with the mass of the fuel necessary to overcome drag or gravitational potential energy for the first stage of ascent, would it be more weight-efficient to take extra fuel for a de-orbit burn rather than carrying a dangerous and heavy thermal shield for a fiery re-entry?

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### Re: Energy requirements to launch to LEO (from vacuum zeppel

How much do you need to slow down to avoid the need for heat shields? I suspect it's pretty significant.

The Space Shuttle has a mass ratio of about 16 from the ground, which as previously computed gets it about 10km/s. Let's suppose that by starting outside the atmosphere and some other improvements that can get us, we can cut it down to a mass ratio of 8 for the 8km/s needed for orbit.

Unfortunately, a ratio of delta-v translates into the exponent on the mass ratio. So if we want to kill only one-third of orbital velocity, we're back up to a mass ratio of 16. And if 5+ km/s is still too fast, which I rather suspect it is, we have to increase the ratio further.
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

I'm wondering if it's possible to get the mass ratio even lower. We're potentially talking about a partially air-breathing engine here: a sc/ramjet stage that fires from Mach 5 to Mach 17. Plus, it's not truly single-stage; we've still got our booster rockets to kick us up to Mach 5 at our 40-45 km launch in the first place, so the Tsiolkovsky equation doesn't entirely describe the situation.

But perhaps a craft robust enough to handle repeated scramjet flight would only require very slight modifications for heat shielding. I'm just trying to think about reducing the risk of re-entry while making a craft that's as small and reusable as possible.

We need to come up with a hypothetical payload in order to get an idea of what we're actually dealing with. Want to say something the size of the Soyuz? Being able to carry a complete smaller manned orbiter seems like a good baseline. At 7.15 tonnes, the Soyuz is 29% the mass of the Space Shuttle orbiter's payload, meaning we can basically just scale down the orbiter to 66% along each axis and to 29% of mass.

We can scrap each of the 3.5-tonne SSMEs, because the scramjet (which we'll assume to be roughly the size of the OMS engines) is doing the main body of work. This places the orbiter mass at around 29 tonnes excluding fuel and payload.

So that we can design a scramjet for optimal performance, the mass flow (product of air density and velocity) needs to remain roughly constant; this gives us an ideal trajectory. 40 km has an atmospheric pressure of 0.007 atm or a density of 8.7 g/m3 so mass flow is 13.1 kg/s*m2. This flow becomes unsustainable at approximately 5.3 km/s and an elevation of ~50 km, as the Mach number exceeds 17.

Supposing that our ~36-tonne loaded orbiter is released from the boost rockets at 40 km with a velocity of ~1.5 km/s, burns a scramjet to get to 5.3 km/s and an elevation of 50 km, and goes the rest of the way on liquid-fueled engines, what do our fuel requirements look like? The jump from 50 km to 100 km is fairly negligible compared to the requisite kinetic energy; we really just need the delta-V, which is 2.5 km/s. Using the SSME exhaust velocity (4.5 km/s) for a rough estimate of specific impulse, that yields a mass ratio at rocket ignition of only 1.74. Which is pretty darn good. That means we only need 27 tonnes of fuel (at that point).

The mass ratio for the scramjet stage isn't too hard to calculate. Specific impulse for a scramjet at these speeds is around 800 seconds, or 7.85 km/s. delta-V is, as previously calculated, 3.8 km/s, so our mass ratio for this stage is just 1.62. However, about 80% of the fuel is oxidizer, which a scramjet doesn't need; this reduces the effective mass ratio to a miniscule 1.12 and the total mass-at-ignition to 71 tonnes; we need only carry 35 tonnes of fuel, making our net mass ratio at scramjet ignition still under 2.

Using mini-SRBs for our boost should work fairly well. Here our mass ratio is higher due to the lower specific impulse of SRB fuel, but it's still only 1.82, meaning we need about 60 tonnes of fuel, or about 12% as much as a single SRB. Of course, the combined mass of the mini-SRBs will be a factor, but at only 11 tonnes, it's not going to dramatically change the mass ratio. The total launch mass would be ~140 tonnes, a payload fraction of 5.5%. This will remain roughly proportional to the desired payload. In comparison, the Space Shuttle had a payload fraction of 1.2%. The final mass ratio is 3.6.

If we wanted to do a controlled de-orbit burn of 4 km/s, our net mass ratio would jump to 8.7, which is still a good deal better than the space shuttle (the payload fraction drops to 2.3%).

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### Re: Energy requirements to launch to LEO (from vacuum zeppel

davidstarlingm wrote:EDIT: What about re-entry? I know re-entry shields aren't THAT heavy, but given that we don't have to deal with the mass of the fuel necessary to overcome drag or gravitational potential energy for the first stage of ascent, would it be more weight-efficient to take extra fuel for a de-orbit burn rather than carrying a dangerous and heavy thermal shield for a fiery re-entry?

I don't follow your reasoning here. The weight cost of your re-entry system is independent of how you got it into orbit. The fact that everyone is using heat shields means that they're more efficient.

I also question the characterization of heat shields as "dangerous and heavy". Looking over the list of spaceflight mishaps, I only see one obviously attributable to thermal failure: Columbia. And the lesson there isn't "heat shields are dangerous", it's "don't try to be a clever engineer". (That's pretty much what the whole space shuttle program teaches us.) Also note that there have been a number of incidents of incorrect reentry trajectories and orientations, as well as reentries with service modules or retro rockets still attached, where heat shields still managed to do their thing. So not only are they safer than rockets, they provide us with margin of error for other components.

As for heavy, while that certainly was true (I saw that it was 1/3 of the Apollo capsule's weight), I'm not so sure it is still the case. I couldn't find any direct figures for the SpaceX Dragon, which I figure to be state-of-the-art, but I did find that it's heat shield material (PICA-X) has a density of 0.27g/cm3. Another site claims that it's only 3in thick on the Dragon. If so, with an surface area of about 10m2, this would give us a mass of under 300kg for the shield. The Dragon has a dry mass of 4100kg and can return up to 3310kg of cargo, which means we're using 300kg to stop 7410kg. (This, by the way, is a fun number to type on your numpad.) You ain't getting 8km/s (or even 4km/s) with a fuel fraction of 4%. Nor for that matter will you get it with the 33% of Apollo.
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

The maths on this seems pretty complicated, and fairly easy to make a mistake with.

Would this be a situation where double-checking by coding a simple simulation would help? None of the formulas seem to be terribly complicated to find out (eg. air density by height, drag by air density etc.). Then you could try different flight profiles (or match the one the shuttle used) and see if launching from a floating waypoint is significantly cheaper.

(Yes, you'd have to watch for gotchas like: Hey, this flight profile looks great! ...Well, apart from the fact it would raise the temperature of the nosecone past 10,000 degrees that is..!)

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### Re: Energy requirements to launch to LEO

Schrollini wrote:I don't follow your reasoning here. The weight cost of your re-entry system is independent of how you got it into orbit. The fact that everyone is using heat shields means that they're more efficient.

The concept of nearly burning up every time you want to land is just rather unpleasant, particularly with respect to sustainability and reusability. Safety checks are one thing, but if every passenger jet had to have its tires serviced and/or replaced between flights, that would seriously hamper things.

In any case, I'm talking about a mass ratio that's potentially less than 2 at main engine ignition. So perhaps we could afford at least a partially thrust-assisted de-orbit. Is there any way to run a scramjet in reverse?

The SpaceX Dragon is nice, but it uses an ablative shield, which is completely useless for the purposes of reusability. It should be noted that SpaceX is aiming for the development of fully and rapidly reusable rockets, with intact propulsive landing abilities.

A thermal re-entry DOES add an excellent safety factor. But all we really need is to have a ejectable cockpit with a lightweight ablative shield built-in, for use only in extreme emergencies. Typical operation would use the already-robust scramjet underbelly for a partial heat shield while using a controlled burn to decrease the thermal load, adding to reusability.

Keep in mind, a purely delta-V approach isn't the only one, after all. Instead of spending fuel on decreasing velocity, we can spend fuel on gravity drag, maintaining a higher altitude so we spend more time in the upper atmosphere. Lower g-forces mean less of a thermal load. Morever, our spaceplane is designed to produce lift at high velocity; using the right inclination and thrust profile, we could go a LOT farther in exchange for lower thermal loading and a more robust craft. What's cheaper, maintaining a higher altitude at the cost of gravity drag to keep the thermal load at manageable levels, or cutting delta-V outright with a direct burn?

elasto wrote:The maths on this seems pretty complicated, and fairly easy to make a mistake with. Would this be a situation where double-checking by coding a simple simulation would help? None of the formulas seem to be terribly complicated to find out.

It's mostly a lot of algebra. I put everying into Excel yesterday so I could twerk a couple of numbers to automatically adjust everything else. The biggest issue is probably addressing drag.

It's my understanding that scramjet operation depends solely on Mach number, not on the density of the medium. Mach 5 is 2.3% higher at 40 km than at the 35 km where the X-43A has typically flown, making drag forces potentially 4.6% higher, but the density of air is only 54% as high, so the net drag will be 43.5% less. At the end of scramjet burn (~50.5 km), Mach 17 is 7.1% higher, making the drag force potentially 14.8% more, but atmospheric density has dropped to only 14.7%, meaning net drag has dropped by 83.1% in comparison to flight at 35 km.

Here's the breakdown on the speed of sound as a function of altitude:

Thanks to the ozone layer, air temperature increases from the mid-stratosphere up to the stratopause. Though this initially makes Mach 5 harder to obtain, it also allows for a longer scramjet burn, as Mach 17 gets larger and larger the higher up you go. Meanwhile, density drops by half every 5.6 km you ascend. Using our launch profile, instantaneous drag at the end of scramjet burn is only 3.46 times larger than at the beginning of scramjet burn.

Not only does the decreased drag (relative to X-43A altitude) help with fuel costs, but it also reduces the thermal load on the scramjet, making the whole package less heavy.

The current problem with a jet-assisted SSTO is that jets have an incredibly low thrust-to-weight ratio, meaning their only way of producing lift is through aerodynamic forces. This in turn means high drag and long flight times in the lower atmosphere, which eats up fuel even more quickly than the drag on a more vertical rocket; it's a runaway mass ratio. By using a LTA lift to the upper atmosphere and a high-thrust booster, we can neatly evade issues of both lift and thrust.

In order to provide 3 g to the vehicle at launch, (assuming, for the moment, that we're going with a purely thermal re-entry later on), the two boost rockets need to each provide 2.1 MN of thrust (shaped to taper off in order to stay at 3 g), about 15% of the peak thrust of the Space Shuttle SRBs. Does this tell us anything more about the launch mechanics?

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### Re: Energy requirements to launch to LEO (from vacuum zeppel

elasto wrote:The maths on this seems pretty complicated, and fairly easy to make a mistake with.

Would this be a situation where double-checking by coding a simple simulation would help? None of the formulas seem to be terribly complicated to find out (eg. air density by height, drag by air density etc.). Then you could try different flight profiles (or match the one the shuttle used) and see if launching from a floating waypoint is significantly cheaper.

(Yes, you'd have to watch for gotchas like: Hey, this flight profile looks great! ...Well, apart from the fact it would raise the temperature of the nosecone past 10,000 degrees that is..!)
Most of the math being done now uses real flight data for the real atmosphere, and as such is likely to be loads more accurate than anything that tries to model the atmosphere by approximation.
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### Re: Energy requirements to launch to LEO

davidstarlingm wrote:The concept of nearly burning up every time you want to land is just rather unpleasant, particularly with respect to sustainability and reusability. Safety checks are one thing, but if every passenger jet had to have its tires serviced and/or replaced between flights, that would seriously hamper things.

Yes. But if the other option was giving passenger jets tank treads, I'm sure every airport gate would have a set of jacks and pneumatic wrenches.

davidstarlingm wrote:The SpaceX Dragon is nice, but it uses an ablative shield, which is completely useless for the purposes of reusability.

Sure. But from Soyuz and the Space Shuttle, we know that reusability and economy are two wildly different things.

To get your 8 km/s (or even 4km/s), you're going to need a fairly substantial engine. Do you bring it back to Earth with you? If not, replacing the engine has got to be more expensive than replacing a heat shield. If you do, that's more weight to land safely and gently, and you still have to do a lot of refurbishment. More than you think, probably: The Space Shuttle was supposed to be turned around in 14 days; the best actual time was 54 days (88 post-Challenger). Is that really more efficient than just replacing a heat shield?

davidstarlingm wrote:It should be noted that SpaceX is aiming for the development of fully and rapidly reusable rockets, with intact propulsive landing abilities.

I'm not sure what on that page you wanted me to read, but my understanding of their reusable rocket stages was to give the stages heat shields, let them aerobrake, and then use the rockets only at the very end to land the things.

davidstarlingm wrote:A thermal re-entry DOES add an excellent safety factor. But all we really need is to have a ejectable cockpit with a lightweight ablative shield built-in, for use only in extreme emergencies.

I think this thing is called a "space capsule". Apparently it works well enough that people are considering using one in non-emergency situations as well.
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:I think this thing is called a "space capsule". Apparently it works well enough that people are considering using one in non-emergency situations as well.

Heh. Yes, indeed.

I don't think a purely-propulsive re-entry is feasible, but a reduced-thermal-load one might be. So having a purely-thermal re-entry capsule is a good safety factor still.

To get your 8 km/s (or even 4km/s), you're going to need a fairly substantial engine. Do you bring it back to Earth with you? If not, replacing the engine has got to be more expensive than replacing a heat shield. If you do, that's more weight to land safely and gently, and you still have to do a lot of refurbishment. More than you think, probably: The Space Shuttle was supposed to be turned around in 14 days; the best actual time was 54 days (88 post-Challenger). Is that really more efficient than just replacing a heat shield?

Probably not. But we don't have to eliminate a heat shield altogether; we just need to reduce the thermal load until the heat shield can be fully reusable. If we can stay in the upper atmosphere for longer using thrust and aerodynamic lift, we can extend the period of aerobraking to the point that it isn't so dangerous, costly, and so on.

It will be difficult to estimate the energy cost of maintaining a higher-altitude re-entry, but I can give it a shot.

Let's say, just for the sake of setting high goals, that we want to limit the thermal load to the peak drag experienced during the termination of the scramjet stage. In other words, Mach 17 at 50.5 km (5.61 km/s) where the density of the air is 0.002 atm. In order to match that drag at orbital velocity (where the v2 term will increase drag by a factor of 1.93), the air density needs to be 51.7% as high, for an elevation of just 55.8 km. So that's the altitude we'd want to hover/glide at in order to maintain the same thermal load. The velocity necessary for orbit at ~56 km would be 19 km/s, meaning we're not getting ANY effective lift from centrifugal forces.

Of course, we'd have a different angle of attack to increase aerodynamic lift and redistribute the thermal load. Which will alter a lot of things. We know the mass of the craft, but without a good idea of the shape or lift coefficient or other parameters, it's tough to know how everything would work. Even without dealing with the different angle of attack, I really don't know how much force drag would exert at the peak, so it's impossible to know how much time we'd need to spend decelerating, how much drag contributes to lift, the gravity drag we'd need to deal with, and so forth.

The scramjet stage should be designed to obtain the maximum delta-V possible; its whole purpose is to get closer to orbital velocity on a higher-impulse air-breathing engine. So orientation should minimize drag, with exactly enough aerodynamic lift to climb steadily in opposition to gravity drag, maintaining the increasing altitude necessary for constant mass flow (as previously discussed) so the scramjet can have a constant ideal burn rate.

Still doesn't get us much closer to the actual scramjet burn time, though. The limiting factor for a scramjet engine seems to be the thrust-to-weight ratio, which has an upper limit of 3-5. I don't know how much of this is based on the need for bulky active cooling; given our higher-altitude launch, perhaps a thrust-to-weight ratio as high as 6 could be attained.

With this in mind, and presuming that we can afford a combined hybrid-rocket-scramjet-engine weight of 16% (approximately the same as the SSME), that gives us an engine mass of 4.56 tonnes and a peak thrust of 270 kN. At an ignition mass of 71 tonnes and a termination mass of tonnes, that's an initial acceleration of 0.39 g ramping up to a peak acceleration of 0.76 g as propellant is consumed. Of course, this fails to address drag. But we can still get a lower bound on burn time using the equations of motion for uniform jerk:

af = a0 + j*t

vf = v0 + a0*t + 0.5*j*t2

Solving for jerk in the first equation, substituting, and solving for t:

t = 2(vf - v0)/(af + a0)

With:

a0 = 3.83
af = 7.46

v0 = 1500
vf = 5300

this comes out to a time of 673 seconds, or just over eleven minutes. Lower-bound. Rather extended, but not impossible. Jerk is 0.0054 m/s3. Velocity with respect to time (which we'll need for estimating drag) is described by this equation:

v(t) = 1500 m/s + (3.83 m/s2)*t + (0.0027 m/s3)*t2

Lift is proportional to the square of velocity, which goes up, and to air density, which goes down. This plot would suggest a near-linear increase in lifting force with respect to time. This is where things start to break down, though.

If lift starts off providing 9.81 m/s2 of acceleration, I'm pretty sure that the linear increase in force plus the decrease in mass will be plenty to get it up to 50.5 km. So the minimum lifting force we need from the wings is 9.81 x 71 tonnes, or 697 kN. Our ship needs to have a MUCH better lift-to-drag ratio than the Space Shuttle. Using the Concorde's 7.14 supersonic L/D ratio as a rough estimate, that means our ship is built to handle 97 kN worth of thermal loading drag.

If we have the same overall shape as a Concorde, we can probably triple our drag coefficient without stalling on re-entry (the stall is the point at which the thermal load begins to increase due to pressure drag overcoming form drag). ~300 kN of force will decelerate our 29-tonne re-entry vehicle at a cool 1.05 g, taking about 12 minutes to drop down to Mach 1. Pretty sure that allows for enough aerodynamic lift to maintain the necessary altitude drop for a constant deceleration.

Schrollini
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

I don't have anything useful to add to these calculations. But I would like to point out a tidbit I ran across while looking for info on heat shields: to work out the aerodynamics of a reentering spacecraft, you have to take into account chemistry! I'm pretty sure solving reaction equations coupled to compressible fluid flow around an immersed body is one of Dante's lower circles of hell.
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davidstarlingm
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:I don't have anything useful to add to these calculations. But I would like to point out a tidbit I ran across while looking for info on heat shields: to work out the aerodynamics of a reentering spacecraft, you have to take into account chemistry! I'm pretty sure solving reaction equations coupled to compressible fluid flow around an immersed body is one of Dante's lower circles of hell.

All the more reason to avoid thermal re-entry if we can.

"Wait, why aren't we doing it the straightforward way?"

"The math was too hard."

In all seriousness, though, anyone have an idea why they haven't built re-entry vehicles with a hypersonic lift-to-drag ratio that's high enough to maintain altitude for an extended period -- long enough that the drag never gets high enough to set it on fire?

Schrollini
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

davidstarlingm wrote:In all seriousness, though, anyone have an idea why they haven't built re-entry vehicles with a hypersonic lift-to-drag ratio that's high enough to maintain altitude for an extended period -- long enough that the drag never gets high enough to set it on fire?

That's a good question. I'm not coming up with any obvious answers.

Browsing the intarwebs, I found this site on hypersonic vehicles. I can't vouch for its veracity, but it seems to make sense. One thing they say is that there seems to be a maximum possible lift/drag ratio that decreases with Mach number, going to 4, or maybe 6, as Mach number goes to infinity. This may be part of the answer -- you simply can't build something that has L/D > 6 at Mach 25.

But even that doesn't really answer the question. Even if L/D = 6 isn't enough to save you from fire completely (I don't know whether it is or not), it would reduce the thermal load. But the Space Shuttle didn't do this -- it plowed into the atmosphere at an angle of attack of up to 70o, IIRC.

Another possibility is that a hypersonic lifting body may not produce significant lift at subsonic, or even supersonic, speeds. But I'm not convinced that this is such a problem either. If you stall as you come out of the hypersonic regime, why not increase your angle of attach to break further, and then throw out some parachutes once you've dropped to subsonic speeds? You won't get the made-for-TV shuttle landing, but that's okay.
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davidstarlingm
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:
davidstarlingm wrote:In all seriousness, though, anyone have an idea why they haven't built re-entry vehicles with a hypersonic lift-to-drag ratio that's high enough to maintain altitude for an extended period -- long enough that the drag never gets high enough to set it on fire?

That's a good question. I'm not coming up with any obvious answers.

Browsing the intarwebs, I found this site on hypersonic vehicles. I can't vouch for its veracity, but it seems to make sense. One thing they say is that there seems to be a maximum possible lift/drag ratio that decreases with Mach number, going to 4, or maybe 6, as Mach number goes to infinity. This may be part of the answer -- you simply can't build something that has L/D > 6 at Mach 25.

Yeah, looking at those pages on hypersonic vehicles, it seems that conventional aerodynamic lift starts to fail at high speeds (probably when the body is moving far too quickly for Bernoulli effects to propogate); the maximum conventional L/D ratio at a given Mach number is 4(M + 3)/M. Using a waverider design with compressive lift, L/D can be improved to 6(M + 2)/M. At Mach 24, this means a maximum L/D of 4.5 for aerodynamic lift and 6.5 for waverider compressive lift, the latter increasing gradually to 8.4 by the time you hit Mach 5 and 12 at Mach 2.

For the ascent phase, this might work quite well; as the L/D ratio drops off, so would the vehicle mass, meaning a more gradual, linear ascent.

What's a good way to estimate the peak thermal load (thus giving us peak deceleration and a general idea of gravity drag) we can handle without having to resort to a super-robust tiled heat shield? The SR-71 had a pretty high thermal load -- on touchdown, the canopy was well over 500oF -- but it didn't have to have its skin serviced or replaced after every flight. Is thermal load going to be linearly proportional to mass flow? If so, the SR-71's ~1 km/s at ~26 km would provide an upper bound. Atmospheric density drops by 50% every 5.6 km you ascend, meaning that you can double your airspeed with every 5.6 km you ascend while retaining the same thermal load. For a re-entry at 7.8 km/s, it would imply a minimum altitude of ~43 km. Then again, Wikipedia says the Space Shuttle started encountering significant atmospheric density as high as 120 km. According to graphs in this paper, the Space Shuttle encountered peak g-forces and peak thermal loading around 53 km with an airspeed of 3 km/s.

Will a L/D of 5 (increasing as we decelerate) or so give us enough lift to maintain a safe altitude through the re-entry phase? If not, what would the delta-V cost be to overcome gravity drag during that phase to supplement aerodynamic/compressive lift?

EDIT: Looks like we aren't the only ones to think of a lift-assisted re-entry. Here's a pretty in-depth paper on the subject. It implies (on page 6) that the thermal load is proportional to the square root of atmospheric density multiplied by the cube of velocity, providing a "thermal barrier" of velocity and altitude below which thermal loading would exceed design specifications.

Schrollini
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

davidstarlingm wrote:Yeah, looking at those pages on hypersonic vehicles, it seems that conventional aerodynamic lift starts to fail at high speeds (probably when the body is moving far too quickly for Bernoulli effects to propogate); the maximum conventional L/D ratio at a given Mach number is 4(M + 3)/M. Using a waverider design with compressive lift, L/D can be improved to 6(M + 2)/M. At Mach 24, this means a maximum L/D of 4.5 for aerodynamic lift and 6.5 for waverider compressive lift, the latter increasing gradually to 8.4 by the time you hit Mach 5 and 12 at Mach 2.

First, a bit of pedantry: The Bernoulli effect that you're probably thinking of (ρv2/2 + p = C) is only for incompressible flows. There are equivalent formulae for compressible flows, but they aren't as simple. That said, what breaks down isn't anything to do with propagation speed. The Bernoulli equation comes from finding a constant of motion for each fluid element. For a steady-state flow, this turns into a constant along each streamline, and for a uniform source flow, a constant throughout the fluid. But this is not the result of some instantaneous communication amongst the fluid elements.

None of this alters your conclusion that lift works differently in the hypersonic regime than in the subsonic regime.

One thing to note is that those plots of L/D were giving the best ratio for a vehicle optimized for that Mach number. You can't count on a given shape following those trends in general. I'd expect that the ratio would fall off on either side of the design Mach number. Unless you want to go variable geometry, I think you have to pick the Mach number where you need the most lift, optimize for that, and then accept lower L/D ratios in the rest of the flight profile.

davidstarlingm wrote:What's a good way to estimate the peak thermal load (thus giving us peak deceleration and a general idea of gravity drag) we can handle without having to resort to a super-robust tiled heat shield? The SR-71 had a pretty high thermal load -- on touchdown, the canopy was well over 500oF -- but it didn't have to have its skin serviced or replaced after every flight. Is thermal load going to be linearly proportional to mass flow?

A way to get a rough estimate would be to note that drag*velocity gives you a power being dissipated. Some of this will go into heating the air, some into the aircraft. Dividing the power by the surface area gives you some idea of the heat flow rate; combining this with the blackbody equations and emissivity would tell you the temperature assuming we can treat it as a steady-state system. I'm not sure how good that assumption is, but we know that uniform heating isn't a good assumption, so at best this will give you an order-of-magnitude estimate for an "average" part of the vehicle.

A better approach, for the time being, would be to take the floor and ceiling from that paper you've found and see if you can thread an approach through the middle.
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davidstarlingm
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Schrollini wrote:First, a bit of pedantry: The Bernoulli effect that you're probably thinking of (ρv2/2 + p = C) is only for incompressible flows. What breaks down isn't anything to do with propagation speed.

None of this alters your conclusion that lift works differently in the hypersonic regime than in the subsonic regime.

My mistake.

One thing to note is that those plots of L/D were giving the best ratio for a vehicle optimized for that Mach number. You can't count on a given shape following those trends in general. I'd expect that the ratio would fall off on either side of the design Mach number. Unless you want to go variable geometry, I think you have to pick the Mach number where you need the most lift, optimize for that, and then accept lower L/D ratios in the rest of the flight profile.

Well, where do we need the highest L/D ratio? At the beginning of the scramjet stage, when we need lift to balance out gravity? Or during re-entry? And if we have a craft with a geometry optimized for the early scramjet stage (Mach 5-8), how can we estimate L/D at other Mach numbers?

A better approach, for the time being, would be to take the floor and ceiling from that paper you've found and see if you can thread an approach through the middle.

Any idea what that ceiling actually represents? I'm not sure what it's supposed to be.

In any case, the maximum compressive-lift L/D for Mach 5 is 8.4, meaning our 71-tonne launch vehicle would initally have to overcome 83 kN of drag if it had sufficient lift to counterbalance gravity. It seems prohibitively complicated to try and numerically model the changes in velocity, altitude, mass, thrust, drag, and lift which take it to scramjet termination at 50.5 km, 5.3 km/s, and 36 tonnes. Our ascent trajectory should be optimized to maintain the constant mass flow rate of 13.1 kg/s*m2, as previously stated. All with a constant thrust of 270 kN.

Moving over to the re-entry stage....

Page 7 of that paper provides the following equation for hypersonic aerodynamic heating:

q' = Cρ0.5v3

where q' is watts/meter2, C is a constant, and ρ is atmospheric density. ρ, of course, is 1.225*0.5h/5.6. However, it would appear that their thermal limit is about 17 times as high as the thermal loading on the SR-71 (using this equation), which throws a wrench in any attempt to make the two line up.

I think I found a problem, though. Using this equation, the thermal load at scramjet ignition is only 42% more than the load on the SR-71. However, at scramjet termination it would be over 30 times as high, well below the thermal barrier given in that paper. You can see how the velocity/altitude trajectories for constant mass flow (blue) and critical thermal load (red) cross at 4.1 km/s and 48 km (y is altitude in km; x is velocity in hundreds of meters per second). So that may be the maximum speed we can squeeze out of scramjetting, unless the scramjet can survive on a lower mass flow at a higher altitude.

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### Re: Energy requirements to launch to LEO (from vacuum zeppel

I've been answering this type of question other places, and I want to add some detail here. I'm not sure if anyone has pointed this out, but the graph that shows that 68 MJ of the launch energy goes to drag is not correct at all. The energy required is much greater than the orbital energy kinetic energy, but this is due to the rocket equation. Reaction rockets take more energy to get to some particular energy than it would if you had traction with the ground (for instance). That's where all the energy goes. You could say this energy is spent delivering the propellant to arbitrary places and velocities that don't help you.

Gravity drag is significant, but I believe the climb to altitude itself takes more energy. Since orbits tend to be at 200 km or 300 km, that results in a contribution to the Delta v budget of nearly 2 km/s. That explains the large part of how we go from 8 km/s orbital velocity to 10 km/s Delta v budget.

Air resistance isn't as significant as one would predict, simply because the atmosphere is thickest when the rocket is going the slowest. On the other hand, space guns have a big problem with atmospheric drag, since they start out punching through at orbital velocity.

Tass
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Zassounotsukushi wrote:I've been answering this type of question other places, and I want to add some detail here. I'm not sure if anyone has pointed this out, but the graph that shows that 68 MJ of the launch energy goes to drag is not correct at all. The energy required is much greater than the orbital energy kinetic energy, but this is due to the rocket equation. Reaction rockets take more energy to get to some particular energy than it would if you had traction with the ground (for instance). That's where all the energy goes. You could say this energy is spent delivering the propellant to arbitrary places and velocities that don't help you.

Thanks. I remember reading this thread and thinking that there was a lot of misunderstandings being thrown around. But I didn't have a lot of time at that time, so I abstained from joining the discussion.

Zassounotsukushi wrote:Gravity drag is significant, but I believe the climb to altitude itself takes more energy. Since orbits tend to be at 200 km or 300 km, that results in a contribution to the Delta v budget of nearly 2 km/s. That explains the large part of how we go from 8 km/s orbital velocity to 10 km/s Delta v budget.

Air resistance isn't as significant as one would predict, simply because the atmosphere is thickest when the rocket is going the slowest. On the other hand, space guns have a big problem with atmospheric drag, since they start out punching through at orbital velocity.

On the other hand if there was no atmosphere then we could accelerate to orbital velocity at low altitude and then the climb could be done with less delta-V loss because of the Oberth effect.

Schrollini
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Zassounotsukushi wrote:Gravity drag is significant, but I believe the climb to altitude itself takes more energy. Since orbits tend to be at 200 km or 300 km, that results in a contribution to the Delta v budget of nearly 2 km/s. That explains the large part of how we go from 8 km/s orbital velocity to 10 km/s Delta v budget.

The very first post in this thread (after the separation, at least), notes that gravitational potential energy is about 3% of the kinetic energy of a craft at 100 km. At 300 km, it'd be maybe 10%. At the level at which I was working, this is a rounding error.

Now, it's true that an impulsive launch directly up to 200 km (ignoring air resistance) takes a delta-v of 2 km/s. But that's not the "missing" 2km/s -- delta-v isn't additive in the that way, at least if you're smart about your launch profile. An drag-free launch could get you to 200 km and 8 km/s with only 8.2 km/s of delta-v. The remaining delta-v is lost to gravity drag, air drag, and inefficient thrust vectoring. But since that inefficient vectoring is there to prevent even greater losses to the true drag terms, it seems perfectly fair to lump it in under the general heading of "drag".

And if you want to convince me otherwise, it'll take more than your beliefs. Until then, I stand by my original estimate that, very roughly, a third of the launch energy goes into the space craft, a third goes into the exhaust out of necessity, and a third goes to countering drag terms.
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cjameshuff
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### Re: Energy requirements to launch to LEO (from vacuum zeppel

Reaching 300 km altitude requires 2.94 MJ/kg, equivalent to 2425 m/s. Launch at high altitude isn't going to cut more than a tenth or so of this out.
Reaching 8000 m/s requires 32 MJ/kg. Launch at high altitude doesn't change anything here.
For an approximation of gravity drag, thrusting vertically for 200 seconds costs you 1960 m/s of delta-v, about 1.9 MJ. Launch at high altitude takes the biggest chunk out of this, but still only a fraction (even in vacuum, you need to thrust vertically at first to avoid hitting the ground before gaining orbital velocity), and it's the smallest of the major factors. And even 100% of 1.9 MJ is nothing anywhere close to 1/3rd of the total...it's more like 1/20th.

Aerodynamic drag is not an issue. The Saturn V lost only 40 m/s to aerodynamic drag, other rockets tend to be around 100 m/s. The mass of the structure needed to withstand the drag forces during the short period of hypersonic aerodynamic flight is more important.

The biggest mistake, however, is assuming energy requirements are a useful metric. The energy costs work out to around a few dollars per kg, out of a total cost of thousands to tens of thousands of dollars per kg. Rocket propellant is cheap, what makes rocket launches expensive is the disposable rocket and the high operations costs. The most expensive systems ever operated have been air-launch systems (specifically the Pegasus), and the cheapest have been ground-launch systems. SpaceX is launching at around a quarter the price of their competition, achieved by streamlining production and operations, and they are planning on having the first stage fly back for reuse, loading additional propellant and taking a ~30% cut in payload but reducing the cost of the expended hardware by about 75%.