Orbital rockets do not *seem* efficient. They take off in what amounts to a mushroom cloud, using extravagant amounts of propellant to rise even the first 100 meters into the air to clear the tower. Yet, their efficiency rivals that of your modern pickup truck. Let me be clear by what I mean by efficiency, here: Energy efficiency, not necessarily mass efficiency. That is usually a strange assumption, but I hope it will make more sense as we go on.

An efficiency expressed as a percent has a numerator and a denominator. The numerator for orbit is the minimum possible energy that could possibly be required, such that if you did somehow beat it, you could conceivably *harvest* energy. In our cases, it’s the total potential and kinetic energy that the payload needs to be. For orbit, that’s the energy required to lift up a payload to ~100-200km and accelerate it perfectly to orbital velocity. That total is about 32MegaJoules (MJ) per kilogram. (A Joule is 1 Watt for 1 second. 1 kilowatt-hour is 1000*3600 Joules or 3.6MJ, 1055 Joules is equal to a BTU, and 1 Gigajoule is roughly 1 MMBTU.) And the denominator is the chemical (or any other kind) of input energy (and we will use the High Heating Value of any fuel, i.e. all the heat generated, including that it the water vapor).

Take the 2021 F-150, for instance. It has a payload capacity of 1,765 lbs and a curb weight of 4,705 lbs. (Edmunds.com). It has a combined EPA fuel efficiency of about 20mpg (presumably without a full payload, but for our purposes, we’ll assume a full payload). The “Ecoboost” engine by FORD has a peak efficiency of about 33% (Energy.gov), and there’s about a 20% loss of power through the transmission for an automatic transmission to the rear wheels (F150 Forum). The maximum freeway grade is 6% (which is to say, a rise of 6 meters for every 100 meters traveled). So if you were to carry a full payload of dirt up a height of 6 miles, it’d take about 100 miles horizontal distance. The total energy in this case is not just 100 miles divided by 20mpg, because we’re also traveling up in altitude, so we have to take into account the mass of the vehicle plus payload and the efficiency of the engine and drivetrain. As before, the engine is 33% efficient and the 20% losses (80% efficiency) mean a total of .8*.33 =.264 or 26.4% efficient. However, your payload is only a fraction of the total weight! Curb weight plus max payload gives 1765lbs+4705lbs = 6470lbs. Payload is just 27.3% of the total mass. So the overall efficiency is .273*.264= 7.2%. 1765lbs is 803kg, times gravity (9.8m/s^2) is 7867.6 Newtons. 6 miles is 9,656 meters, so multiplying these two together gives us the potential energy of our payload (and usually we want it stationary up there, so no kinetic energy component). 7867.6N*9,656m =75970125 Newton-meters (which is the same as Joules). 76MJ potential energy. As we established earlier, the efficiency of the engine, drivetrain, and the payload mass fraction (we can ignore mass of fuel for this short distance) is 76MJ/0.072072 = 1054MJ. It took 1054MJ in gasoline to go up 6 miles.

But wait! We also didn’t include the energy losses of traveling that 100 miles horizontal distance. To know that, we need to know the energy content of the fuel (as cruise efficiency is given in miles per gallon). That is typically 46.4MJ per kg, with fuel having a density of 0.737kg per liter, or 129.4 MJ / US gallon. 100 miles horizontal distance at 20mpg needs 5 gallons, which at 129.4MJ/gallon, means 647.25MJ in chemical energy in horizontal travel. (This is a restriction of road vehicles… they don’t travel straight up! That reduces achievable efficiency, although perhaps we could improve things with a steeper grade, at the expense of more complication in payload considerations.) The total energy needed is therefore 1055MJ+647.25MJ = 1702MJ in gasoline form. The total efficiency is therefore just: 76MJ/(1702MJ) = 4.5%.

This is for a modern vehicle using its max built-in payload capacity at the typical max freeway grade (other roads can do 7-10% grade or even higher, but there are other issues with such aggressive grades that could violate other assumptions).

How does this compare to a rocket?

To make things simple, we’ll use Falcon 9. I’m using Ed Kyle’s wonderful website (may it rest in peace!) which has some pretty decent estimates for propellant masses for both stages of Falcon 9:

SpaceLaunchReport (Archive.org)

For “Falcon 9 Block 5 Merlin 1D Fuller Thrust,” He estimates 418.7tonnes (t) of propellant for the first stage and 111.5t of prop for the second stage (total of 111.5t+418.7t=530.2t). The quoted expendable payload is 22.8t, so each kg to orbit requires 23.25kg of propellant. Merlin 1D has an oxygen:fuel mass ratio of 2.36 (Merlin Evolution), reducing that 23.25kg down by a factor of (1+2.36)=3.36 to 6.92kgRP-1/kgLEO. RP-1 fuel (kerosene) has an energy density of 43MJ/kg, so Falcon 9 needs just 297.6MJ per kg launched to LEO, expendably. Note that 16.33kg of liquid oxygen is also needed, and it takes about 200kWh per tonne of cryogenic liquid oxygen at modern cryogenic separation plants, or 11.8MJ additional energy per kg to orbit. This is only ~3% of the fuel energy, but we include it here anyway. (Not including cryogenic chilling of the kerosene at this time, but that should be much less than the liquid oxygen, so a rounding error here… although the LOx also takes some energy to subchill as well.) So a total of 309MJ/kgLEO, so given 32MJ theoretical total orbital energy from the ground, the Falcon 9’s energy efficiency in expendable mode is 32MJ/(309MJ) = 10.3%. Over double our maxed out pickup truck!

This isn’t a hard limit, either. Merlin is a gas generator rocket, much less efficient than a staged combustion engine. It also operates fuel-rich, which wastes chemical energy. An ox-rich staged combustion kerolox engine should do even better. In principle, you can improve the efficiency of a rocket by blending in water in the beginning part of the first stage burn and using a hydrolox upper stage (especially if near-stoichiometric or ox-rich). Higher pressure (i.e. from staged combustion) and altitude compensation could further improve efficiency, as could structural mass improvements. Operating with 3 stages instead of 2 could also in principle help, staging off dry mass even earlier.

But how could a rocket be so efficient?

It’s worth looking at the ratio of the upper stage dry mass to the payload mass. In the case of our truck, the payload was just 27.3% of the “burnout” mass, but for Falcon 9, the upper stage burnout mass would be just 4.5t or so, therefore the 22.8t payload would be about 83.5% of the burnout mass, 3 times better than the pickup truck. This is enabled by staging. A single stage rocket would not have this fabulous burnout mass efficiency.

Second, rockets operate at extreme pressures, comparable to a race car’s supercharged engine in the case of Merlin to double that pressure for Raptor. AND rockets expand from that extremely high pressure to vacuum. Those two things combined allow higher thermodynamic efficiency than a car engine’s 33%. Greater than 50% thermodynamic efficiency is possible. Plus, rockets have sufficient thrust to weight ratio to accelerate (initially) straight up and quickly out of the drag-causing atmosphere whereas trucks have to deal with rolling resistance and air resistance throughout the whole trip. All these things together mean that in spite of the fact that rockets have to carry their oxidizer with them and in spite of the rocket equation and the fact that rockets can only push on their own exhaust (and not the whole earth like a wheeled vehicle), a two-stage modern chemical rocket can beat a modern pickup truck in load moving efficiency in terms payload final total (potential+kinetic) energy.

Granted, this is for expendable mode. However, a droneship recovery is now about 80% (18.4t) of that previous 22.8t payload. And even doubling the mass of the upper stage to allow recovery would still leave you at 60% original efficiency, or about 6.3% (~500MJ/kgLEO) for full reuse, still greater than the pickup truck. (Note that my calculations for estimated Starship efficiency give a similar 500MJ/kg, in spite of the higher performance Raptor, and I think that’s due to how non-optimally heavy the Starship upper stage is compared to the first stage… in part because it brings the whole fairing to orbit.)

This all might seem unfair. After all, I’m just counting the altitude, not the distance traveled when making this efficiency calculation. But this might betray a lack of appreciation for orbit: in space, there’s no air resistance to speak of. Kinetic and potential energy are, for most intents and purposes (outside of very low earth orbit), conserved. On Earth’s surface, there is friction everywhere and travel anywhere requires energy expenditure even without a net change in altitude. So rocket travel in space is, if anything, even more efficient than I express here. However, there’s a big entrance fee required: orbital velocity. Rockets don’t require massive energy because they’re efficient. That is largely just the cost admission. Orbit is just fundamentally high-energy, and engineers have had to develop some of the most efficient and remarkable machines ever devised to access this realm.

This energy efficiency comparison applies to aircraft as well, and even to other orbital launch concepts. Single Stage vehicles, regardless of propulsion method, will tend to do worse due to the greater dry mass. And high Isp propulsion used for the first stage is wasted energy. Perhaps I’ll do my next post comparing to Skylon. This topic is a continuation of some of my previous posts on fundamental rocket efficiency considerations.

But let’s leave you with one last comparison: Fuel efficiency of an A380 jumbo jet is about 72 miles per gallon on long haul flights (Wiki: Fuel economy in aircraft). Given the density and specific energy of jet fuel, the energy needed to transport a 160 pound adult halfway across the planet on an A380 (in a seat, stopping to refuel) is the same as launching them to orbit on an expendable Falcon 9 (without a seat, no refueling). So rocket travel is not *inherently* far more energy-intensive than long-haul air travel. It’s the same order of magnitude.

“It’s the same order of magnitude.”

The work necessary to fly halfway around the world is

W = D * s

where

W = work in Joules (J)

D = drag in Newtons (N)

s = distance in meters (m)

If lift is equal to weight then this can be rewritten as

W = m * g * s / (L/D)

where

m = mass in kilograms (kg)

g = acceleration of gravity (m/s^2)

L/D = lift to drag ratio

For a (conservative) L/D of 15 the work required to fly a unit mass halfway around the world (20,000,000 m) is

W = 1 * 9.8 * 20000000 / 15 = 13.1 MJ

compared to 32 MJ for a kg in low earth orbit.

So, yes, within an order of magnitude but still over twice as high.

“So rocket travel is not *inherently* far more energy-intensive than long-haul air travel. It’s the same order of magnitude.”

And if have fast turn around, a rocket could do more trips.

And significant part of a rocket efficiency is lost from gravity loss and significant aspect of rocket assisted launch is related to reducing that gravity loss.

Another aspect is difference of path of route. Airline goes thru polar region to be more efficient. What about in world where the routes is more equatorial {most of world’s population is nearer the equator]?

@Jim Davis:

Indeed, however rockets have the benefit of expanding to vacuum, allowing a thermodynamic efficiency potentially almost twice as good as a jet engine, canceling out much of the difference. Plus airplanes typically get a landing weight to payload weight of at least 3 to 1 (757 has an OEW of 64tonnes, freighter version typically carries 32t of cargo), whereas the Falcon 9 rocket gets a (payload+ upper stage burnout) to payload ratio of 1.2:1, over twice as good.

And besides, once you’re in orbit, you get a “free” return trip… (well, you need a heatshield… but included if you’re assuming an RLV like Starship) so maybe a round-trip to the other side of the world would be a better comparison.

“Indeed, however rockets have the benefit of expanding to vacuum, allowing a thermodynamic efficiency potentially almost twice as good as a jet engine, canceling out much of the difference.”

The *thermodynamic* efficiency is almost twice as good but the *propulsive* efficiency is correspondingly worse with a rocket because the velocity varies so greatly so the overall efficiency is about a wash.

“Plus airplanes typically get a landing weight to payload weight of at least 3 to 1 (757 has an OEW of 64tonnes, freighter version typically carries 32t of cargo), whereas the Falcon 9 rocket gets a (payload+ upper stage burnout) to payload ratio of 1.2:1, over twice as good.”

But the upper stage doesn’t land, at least with Falcon. We’ll see about Starship.

“And besides, once you’re in orbit, you get a “free” return trip…”

If your destination is orbit the rocket is the way to go. Obviously. But if your destination is the other side of the world then the return trip is certainly not free whether the trip is by rocket or airliner.

But I agree with your larger point that rockets are quite reasonably efficient vehicles for low earth orbit operation.

“But I agree with your larger point that rockets are quite reasonably efficient vehicles for low earth orbit operation.”

Rocket could refueled in orbit, with spacefaring civilization {rocket fuel cheap in orbit}

and have crazy idea of refueling suborbital- it would a bit longer for the longer suborbital trip. More likely to do suborbital refueling as way to go someplace other than Earth.

Elon needs a Hemi