Airbreathing hypersonic travel is less energy efficient over long distances than rocket travel

There’s a certain misunderstanding common in aerospace that rockets are horribly inefficient and that long term we need air breathing ramjets or scramjets to efficiently launch things, with the idea that we can thus avoid accelerating oxygen to flight speed, which is considered wasted energy. “Airbreathing hypersonics are five times as efficient as rockets” they say. This, however, is not so.

The misunderstanding comes in part by considering oxygen as just as costly as fuel. Oxygen is not. It can be condensed out of the atmosphere with little energy and is available by the truckload at $100/ton or less. A dedicated production plant can produce it for as low as $10/ton. That compares to $1200 to $3500 per ton for industrial liquid hydrogen which is often the fuel being compared to.

A stoichiometric rocket burns 8 times as much oxygen as it does hydrogen. So if an airbreather consumes a factor of 5 times less propellant than a rocket, that means it consumes nearly twice the hydrogen!

Hydrogen requires the vast bulk of the energy to produce compared to oxygen, a couple orders of magnitude more energy. So for our purposes we can ignore the energy needed to produce liquid oxygen.

Let’s look at LAPCAT II, and airbreathing hypersonic airline concept capable of traveling to the antipodes of the world at Mach 8.

As a percentage of its gross takeoff weight, 45% is hydrogen fuel and 15% is payload:

That means each kg of payload requires 3 kg of liquid hydrogen, which has an energy density of 142MJ/kg, giving an energy cost of 426MJ per kilogram of payload.

Hydrogen with variable mixture from oxygen rich to near stoichiometric would be the best fuel to compare with and the most efficient for rockets, but I will use SpaceX’s ITS from 2016 as a comparison point even though it’s less energy efficient.

ITS has a payload to LEO of 300 tons (more for the tanker variant), and uses a total of 6700 tons of propellant for the first stage and 1950 tons for the second stage ship (both including landing propellant). Given a O:F weight mixture ratio of 3.9, and a specific energy of 55.5MJ/kg for methane, the cost per kg of payload to orbit is just 330MJ, actually less than the hypersonic airliner in spite of using less efficient methane.

You might as well use rockets for long distance transport at high Mach numbers.

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37 Responses to Airbreathing hypersonic travel is less energy efficient over long distances than rocket travel

  1. johnhare john hare says:

    I’d be willing to make a small bet that I could afford to lose that half the response you get is from people that didn’t or didn’t understand your points. That’s been my experience when discussing air breathers.

  2. johnhare john hare says:

    didn’t read or didn’t understand, doh

  3. Jim Davis says:

    I think your conclusion is correct but you have to be very careful drawing conclusions from preliminary designs of different origins. For example if one vehicle is much larger than the other economies of scale might come into play and skew the comparison.

    Another possibility is that different design teams use different assumptions for various reasons. LAPCAT may have always been projected as a commercial airliner and used more conservative assumptions than ITS which was conceived as an interplanetary spacecraft, for example.

  4. gbaikie says:

    The inefficiency of rockets is in the start of orbital trajectory.
    The take off of airliner and rockets is inefficient.
    So that is partially solved by having first stage rocket- so you start with high thrust at take off. And airliner starts with high thrust and using a runway at take off.

    If you going to travel thru vacuum, the fastest and shortest distance to vacuum is the general idea, and the rocket first stages are pretty good at doing this.

    I think with current rockets, the most inefficient part is first 20 seconds of lift off.
    So if you used something like a rocket sled going up steep mountain, that is would what you would be “solving”- basically the first 20 seconds.
    Though if using high elevation mountain and rocket sled you probably also trying to do something about max q.

    I think launching rockets from the ocean is good idea- particularly large rockets.
    Oceans are good places in regard to transporting big and massive stuff. And dry docks are good place to build massive thing.
    And it seems Musk has also reached this conclusion in regards to BFR.
    And it was fundamental aspect of design of the Sea dragon- build in drydock, launch from the ocean.

    But I tend to think one keep the rocket dry [unlike with the Sea dragon], so launch rocket from the ocean from a launch pad. And accelerate the launch pad before launching the rocket, so as to improve the efficiency of rocket in first 20 seconds of lift off. And I call it, a pipe launcher.
    But despite the aspect of launchpad which moves so as to add some velocity to the rocket, the main aspect is that it it launchpad- serves same function as non moving launchpad or not having rocket launched starting from being submerged in water.

    So, it needs a launch tower, and I would have launch tower connected to rocket when delivered to pipe launcher. Pipe launcher is a “pipe” with one capped, capped end at surface, one connects tower and rocket to the cap.
    You add warmed air inside pipe, pipe goes up, and you launch the rocket.
    The pipe is large diameter [a bit bigger than rocket] and long- longer than rocket. And has to strong and lightweight, the tower and pipe launcher would as massive as a fully fuel rocket.
    Now if doubling the mass of rocket how can it use less fuel than a rocket would?
    Well it’s powered by air- one using more LOX [or liquid air which cheaper] as compared to fuel which warms the air. So it could use more liquid air [tonnes] as compared amount LOX [tonnes] used by a rocket. And fuel has warm [liquid] air to about 100 C. Also ocean water can change liquid air into air, so maybe as 1/4 of “fuel” is from warmer ocean water.
    So fuel could be anything, but non cryogenic and not very high pressure natural gas could be used.

  5. Peter says:

    A way to say this that might be easier to grasp:
    A hypersonic airplane is 5 times as efficient as a rocket engine which means that for every ton of propellant that a rocket consumes the airplane would only have to burn 0.2 tons of fuel. However a hydrolox rocket’s propellant is only about 11% hydrogen by mass if that engine is burning stoichiometrically. So this means that in every ton of propellant that this rocket consumes there are only 0.11 tons of (relatively) costly hydrogen.

    In practice hydrolox engines are typically run fuel rich with a ratio closer to 6:1. This would be about 0.14 tons of hydrogen in every ton of propellant but that’s still less than the 0.2 tons for the airplane.

    Both ITS and BFR are methalox. As you and most of your readers know, the stochiometric ratio for methalox is CH4:O2 > 1:2. However methalox engines are typically run lean. So if you have a methalox engine with a O/F ratio of 3.9 then every ton of propellant contains 0.204 tons of methane. That’s an interesting comparison to the 0.2 tons of hydrogen in the original example.

    If a rocket has a payload of 300 tons and a total of 8650 tons of propellant burned at an O/F ratio of 3.9 then every kg of payload requires about 5.88 kg of methane. You also state for LAPCAT II that each kg of payload requires 3 kg of liquid hydrogen. So shouldn’t we be comparing the cost of 3 kg of liquid hydrogen against the cost of 5.88 kg of methane instead of comparing energy costs? When you buy fuel they don’t charge you by the megajoule.

    Overall I think there are other cost considerations when comparing the two designs.

  6. Matterbeam says:

    Scramjets at very high Mach numbers (Mach 8+) suffer from very poor effective Isp (<1000s) unless they switch over the hydrogen fuel (<3000s), in which case they start losing many of their advantages over rockets.

  7. James Walker says:

    Questions from the back of the class (and painfully aware that I’m probably ensuring that John Hare wins his bet):

    Even though the oxygen is of negligible cost, surely the fact that it has to be carried is adding mass to the rocket, which in turn requires more fuel?

    (The question that I suspect will make me look really dumb) Once airbreathers are climbing above 50K, won’t the presence of hydrogen allow them to get fuel from what remains of the atmosphere?

  8. Marshall Eubanks says:

    In both types of fast transportation, you are expending energy to heat up air. In the ballistic (rocket) case, that expense occurs in a short period at the end of the flight (when you need to slow down anyway). In the hyperbolic case, you are heating up 1 km of air for every km you travel. As the hyperbolic velocity increases, at some point hyperbolic flight will become less energy efficient than rocketry

  9. Paul451 says:

    From a military point of view, there is one big benefit for hypersonic over ballistic, the ability to recall the missile/bomber/etc before it’s committed to enemy airspace. In the case of bombers, you can also transit the enemy’s territory, hitting a series of targets. Ballistic is P2P only.

  10. Jim Davis says:

    Ballistic is P2P only.

    Well, point to many points in the case of MIRVed warheads.

  11. Bob Steinke says:


    The Delta-V of antipodal point-to-point is only slightly less than orbital. You could design the system to be “recallable” by accelerating and going once around.

  12. Chris Stelter says:

    The less useful for military applications, the better the case for international commercial deployment. So I count that as a good thing.

  13. gbaikie says:

    Perhaps air breathing works in sense reducing the payload requirement of mothership.
    So have stratolaunch carry air breather rather than carry a rocket {as most mass of rocket is the LOX}.

    Though you have stratolaunch take off with a rocket and add the oxidizer of rocket with a mid-flight refueling.
    And could have stratolaunch take off with 1/2 it’s fuel load, then to fill up with mid-flight fueling, and then mid-flight refuel the rocket being carried with it’s oxidizer.

  14. gbaikie says:

    As understand it, stratolaunch carries about 100 tons of jet fuel, so one could take off with about 50 tons, A tanker plane [ KC-135] could transfer 65 ton [or more- if using it’s own fuel which can transferred from the wings]. Though the rocket might need about 100 tons of LOX- so that need about two tankers provide that much.

  15. Chris Stelter says:

    Yeah, air launch is different due to the limits of lift-off mass. This is similar to military constraints for missiles on jet fighters/bombers and probably why the military is so interested in hypersonics (in spite of not really being more efficient from an on-board energy perspective).

    But again, I am not too interested in military applications, and in fact, I prefer to avoid them if possible.

  16. James Walker says:

    “The less useful for military applications, the better the case for international commercial deployment. So I count that as a good thing.”

    Arguable. The ideal would be that commercial resources can be redirected to the military in crisis, adding to deterrent value, but are so valuable commercially that doing so normally is political suicide.

  17. Archibald says:

    Yes rockets are more efficient and, most importantly, they dot rattle Earth population with sonic booms during the whole trip ! LAPCAT is just NOT realistic.

    This document did the math of suborbital, ballistic travel.

    Admittedly, for trips longer than 10 000 km, you’d better go into orbit.

    Except… if you use ricochet trajectories (Hypersoar !)
    Preston H. Carter got a bunch of papers about this clever trick.

    I think that airbreathing is not needed. What we need is a spaceplane with jet engines and rockets, kerosene and H2O2. Propellant mass fraction 85%, so not SSTO, except with suborbital refueling.

    Such jet-and-rocket plane could do
    – suborbital hops, 10 000 km
    – for longer distance, ricochet trajectories
    – and suborbital refueling to go into orbit

    With a single GE-90 it can take-off from an ordinary airport. Light the rocket at 50 000 ft and goes suborbital, suborbital with ricochet, or in orbit with a refueling.

  18. Paul451 says:


    “Yeah, air launch is different due to the limits of lift-off mass.”

    Only if you carry the fuel on the runway. Aerial refuelling changes the equation. Various proposals have been made (unfortunately fixated on SSTO.)

  19. RobL says:

    No one can make LOX for $0.01/kg ($10/tonne). without factoring in other costs it takes optimistically 0.5kWh of electricity to make 1kg of LOX, which is $0.025/kg at wholesale electricity prices.

  20. Chris Stelter says:

    I’ve seen estimates of down to 0.2kWh per kg of liquid oxygen, and that is four times the thermodynamic limit. Also, it’s possible to get electricity even cheaper than 5 cents per kWh.

  21. John Bucknell says:

    Air-breathing rockets are the lowest cost to orbit – and more so suborbital. The biggest paradigm challenge is propulsive landing – doesnt require wings.

    If you arent aware, I discuss air breathing rockets with greater than 35% in the paper associated with this video:

  22. johnhare john hare says:

    If you have a paper on the subject I am interested. I do not spend hours watching utubes on material I can scan in minutes, especially when the audio is terrible. With a paper, I can go back and check concepts and numbers against other concepts. Can’t do that with a video.

    I’ve done a bit on turborockets. Interested in your concept.

  23. John Bucknell says:

    You are the exception – read the 2015 paper then the 2018 for subsystem and component level analysis. They are in the media section of my profile.

    Paper and presentation are here

  24. John Bucknell says:

    The comment was intended to say “greater than 35% payload fraction to orbit”. Don’t know why that didn’t come through, nor my reply to Mr. Hare.

    All of my work is included in the media section of my linkedin profile:

    Some further discussion on the Space Show:

    Direct links to the paper and presentation are here:

  25. johnhare john hare says:

    I looked up a couple of your papers. I might have been able to buy into some of it until you incorporated Scamjets to mach 15 and apparently ignored atmospheric heating of the whole vehicle at high Mach numbers.

  26. johnhare john hare says:

    Not sure what happened on my end either. Your follow up comments were not showing before my last comment. I don’t do linked in and your other links get 404. At any rate, I found a couple of papers and a thread at nasaspaceflight. I am a skeptic for the moment.

    I am not current on inlets or any of the deeper concepts anymore, too busy trying to keep a construction company together. Bearing in mind that it has been about a decade since I last cracked a book on inlets (intakes in my books), I would be concerned about partial unstart with any AOA misalignment and also lack of pressure recovery in swallowing the boundary layer. I do like the concept of the full diameter turborocket aft using the whole vehicle as center body of the inlet. I would get into how I like using the vehicle as a designed supersonic intake, but topic drift.

    As far as I can tell, insufficient attention has been paid to heat soak of the vehicle during supersonic and hypersonic flight. A long skinny hydrogen tank in Mach several atmospheric flight seems like a serious issue to me, especially when spending several minutes in that regime.

    I don’t believe in scamjets, especially for the acceleration mission.

    Your engine layout is nice. Nuclear heated hydrogen through the fan blades to tip rockets with the exhaust intimately mixed with the compressed air seems like a serious performance enhanced engine. I could see a variation on this going to commercial aircraft and possibly some military. It seems possible that a very low SFC (high Isp) could possibly compensate for the low density fuel, leading to much lower take off weight.

  27. John Bucknell says:

    Thanks for the review so far, John.

    If you wish to see the papers with your spare minutes, drop me an email and I’ll forward.


  28. John Bucknell says:

    I figured out that the links were being truncated – here they are again:



  29. Jim Davis says:


    I have a question about Table 2 in your paper. If you have a rocket Isp of 450 s, and the rocket operates from Mach 15 to Mach 25, the highest possible average Isp, if the airbreathing Isp is infinite, is 1125 s. Yet you show an average of 1574 s.

    How do you get that Isp? What am I missing?

  30. John Bucknell says:

    Hi Jim,

    It is a time-average rather than a velocity average. The air breathing is 1500 to 5500 seconds, so you are averaging across a big number.

    Apologies for the delayed response.


  31. johnhare john hare says:

    I think the problem Jim noticed is that it is not a proper way to use Isp. A proper way would give a clear mass ratio for a given DeltaV. An Isp of 450 would give a mass ratio of about 7.4 for a 9,000 m/s velocity change.

    An analogy might be if you had a trip of 80 miles with 70 highway and 10 in stop and go city traffic with an hour for each section. Average mph is 62.5 by distance, but that tells one nothing about actual duration of the trip. A more accurate measure would put it at 40 mph average. An even more accurate way would be to break down the trip into segment types and calculate from there.

    I think Jim what was pointing out that you are going to use more propellant than is listed in your table, perhaps a lot more. This calls the rest of your paper into question to some extent.

  32. Jim Davis says:

    John Bucknell,

    To expand on John Hare’s comments:

    A time average Isp cannot be used in the rocket equation to get accurate results. An effective Isp must be calculated as follows:

    The rocket equation with variable Isp is:

    Integral_from_0_to_deltaV(dv/g/Isp) = ln(Mi/Mf) Eq.1

    The effective Isp is defined as:

    deltaV = g Isp,eff ln(Mi/Mf) Eq. 2

    Combining Eq. 1 and Eq. 2 gives:

    Isp,eff = deltaV/g/Integral_from_0_to_deltaV(dv/g/Isp)

  33. John Bucknell says:


    You are both right in a way. The reason the rocket equation is so simple is that you are able to largely ignore losses for a launch as the Isp term dominates. So you are able to take orbital velocity and add a small fudge factor for aero and gravity losses and get the right answer.

    For complex systems, you have to do it long hand. A launch model integrates over each time step, calculating the velocity increment based upon losses and propulsion performance up to orbital velocity. As Isp is just thrust divided by propellant mass flow, you can get Mf every time step as well as time average the Isp to get effective Isp. I conveniently run my ascent model at 1 Hz, so the mass flow and propellant mass increment are the same magnitude.

    This is why I indicated effective Isp is a time average. My correlation studies are right there in the 2015 paper – I got Mf for the Falcon 9 Block 2 within a tenth of a percent, as well as other nuclear, lox augmented and airbreathing rockets. I stand by my results.

  34. Bob Steinke says:

    They’re not arguing that your trajectory simulation is wrong. They are saying that a time average of Isp is kind of a useless metric and shouldn’t be called “effective Isp”.

    For example, say your rocket starts with a mass of 7389 kg, and you fire one engine with an Isp of 1500 until the mass is 2718 kg (MR=e, delta-v=1500*g), and then fire a different engine with an Isp of 500 until the mass is 1000 kg (MR=e, delta-v=500*g).

    If the first engine is low thrust and burns for 9 minutes and the second engine is high thrust and burns for 1 minute, then the time averaged Isp is 1400, but if you put 1400 and an overall MR of e^2 into the rocket equation it doesn’t give you the total delta-v that your rocket actually produced.

    The reason to have a metric called “effective Isp” is to say, yes we know Isp is constantly changing, but if you had a different system where Isp was constant, here’s the value it would have to have to produce the same result. For that to work, you have to do a delta-v weighted average, not a time average.

  35. John Bucknell says:

    Thanks Bob,

    I understand now. I would however counter that deltaV in atmospheric flight is also a low value metric, and effective Isp isnt telling you as much. I feel time-average Isp gives a useful frame of reference for magnitude of performance of a launch rocket. Let me go back and recalculate effective Isp separately.

  36. John Bucknell says:

    I get 1,381 seconds for a mass ratio of 1.75 and delta V of 25kft/s.

    Doing this on a bus on my phone, so please check my work.

  37. John Bucknell says:

    The above is for the SCTR in the 5m core diameter, 450s vacuum Isp.

    BTW, I’m on a bus on a rain forest tour in Northern Queensland.

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