What is the approximate influence of atmospheric drag on the cost of rocket launches? Is it beneficial to have launch sites located at higher altitudes?

Cape Canaveral is at sea level, but I've noticed that Chinese launch sites are located on quite high altitudes (although they could still be higher if located in Tibet).

  • 6
    $\begingroup$ There are many considerations more significant than air drag. Logistics of delivering parts by railway, working conditions for the employees, safe buffer zone around the launchpad, proximity to Equator for "orbital speed" kickstart (that extra 1700km/h added to orbital speed at the equator versus zero for near-polar launches!) - it may be more profitable to launch 300km further to the south than 3km higher above sea level... $\endgroup$ – SF. Jul 25 '13 at 12:08
  • $\begingroup$ I agree with SF. The most significant factor is latitude, but it is indeed an interesting question. Some launch systems, such as those that launch from aircraft or balloons, are designed precisely to avoid having to push through the relatively dense atmosphere near the Earth's surface. $\endgroup$ – robguinness Jul 25 '13 at 12:17
  • 1
    $\begingroup$ There is the start of an answer from NASA, see nasa.gov/mission_pages/station/expeditions/expedition30/… $\endgroup$ – PearsonArtPhoto Jul 25 '13 at 13:14
  • 1
    $\begingroup$ I would imagine the Chinese wouldn't want to launch from Tibet even if it was a better launch site. It is a slightly difficult area $\endgroup$ – Rory Alsop Jul 25 '13 at 13:33
  • 1
    $\begingroup$ To develop some intuition on this subject, it may help to keep in mind that the time rockets spend in the low atmosphere (below the height of the tallest mountains) is normally only a few seconds, while the rocket is traveling relatively slowly. These are basically the same reasons why using atmospheric oxygen as a fuel source (as in a ramjet) doesn't help much. $\endgroup$ – RickNZ Oct 7 '18 at 21:21

I'll give you the numbers. I'm breaking this up into 3 different terms. There's atmospheric drag, what I'll call the "hover" term, and the gravitational potential climb. I will more or less assume a flight directly up. You're welcome to use whatever term for velocity you want, as none of them will be representative. I'll take the Shuttle's speed at halfway to max Q. This is 1000 ft/s, or about 300 m/s.

You would think atmospheric drag would be very difficult. It's actually not. In any case, you would probably use the v^2 relationship for drag. But if you think about where that comes from, it basically assumes that all the air in front of you is accelerated to the speed of your craft (minus any departure from unity in the drag coefficient). So for a good approximation, just take the mass-thickness (I call mu) for the entire atmosphere, and multiply by the metric for velocity.

Also, I'll use the numbers for Falcon 9, which is a diameter of 3.66 meters and launch mass of 333,400 kg. Yes, a lot of these numbers change over the course of the flight, but in ways that are fairly obvious if you changed this to do numerical integration.

$$\Delta V (drag) = 1/2 \mu C_d A v / M $$ $$= (0.5)(10 \text{ tonnes} / m^2)(0.5)\pi(3.66/2 m)^2(300 m/s)/(333.4 \text{ tonnes}) $$ $$= 23.7 m/s$$

Wow. That is not much. Maybe velocity should be higher. But still, out of 10 km/s total, this is a tiny amount. Atmospheric drag complicates launches, but not much because of its Delta v value.

Next, the "hover" term. This represents the gravity drag. Again, I'm forced to assume a pretty much upward launch. I'll also compare sea-level to Mt. Everest, at a height of 8,848 m. Not that you'd set up a launchpad there, but we need this to answer the question.

$$\Delta V = g h / v = (9.8 m/s^2)(8,848 m)/(300 m/s) = 298 m/s$$

Now this is much more significant. This isn't all of the gravity drag either. It's still sucking your delta v budget after you're out of the atmosphere, until you get to full orbital velocity.

Let's move on to the gravitational potential itself.

$$\Delta V = \sqrt( g h ) = \sqrt( (9.8 m/s^2) (8,848 m) ) = 294.5 m/s$$

The sum of all these is a ballpark estimate of the benefit you would get from changing your launch location from sea-level to Mt. Everest. Honestly though, you save a comparable amount just by moving it down to the equator, where the Earth's rotation gives you a bigger boost.

Anyway, this is 616.7 m/s out of a total budget of 10 km/s. So it would be less than 10%. By the rocket equation, this can still make a difference. But then again, actual costs are complicated.

  • 2
    $\begingroup$ Something about the drag equation seems wrong... That doesn't take in to account the amount of time that the rocket is in the atmosphere, which should be a critical component to figuring out the delta V from atmospheric drag... $\endgroup$ – PearsonArtPhoto Jul 26 '13 at 12:02
  • 2
    $\begingroup$ @PearsonArtPhoto The time factor is accounted for by the velocity, which is in the equation. Slower velocity, longer time. What I did was remove the length scale. A qualitative model is compressing the atmosphere to a single sheet that the rocket breaks through. Then the sheet's mass within the area the rocket hits is propelled away at half the rocket's speed. This abandonment of the length scale is justifiable from the mathematics. Half the distance at twice the density gives the same impulse (or delta v) by the drag equation. It's still a heavy sheet, like 10 vertical meters of water. $\endgroup$ – AlanSE Jul 26 '13 at 13:04
  • $\begingroup$ Ahhh, that makes sense. Thanks! $\endgroup$ – PearsonArtPhoto Jul 26 '13 at 13:43
  • $\begingroup$ The drag loss calculation seems low and doesn't account for variable speed. Most drag losses are going to occur in the transonic region, which appears to be ignored by this analysis. $\endgroup$ – Adam Wuerl Jul 28 '13 at 6:23
  • $\begingroup$ @AdamWuerl It kind of depends on which question your asking. If you want to know how much Delta V you'll save by moving from Cape Canaveral to Mt Everest, it shouldn't be that far off. That was kind of my intention. I used half of velocity at max Q, so if you're interested in the entire trip, it would be better to use twice that, or more, I'm not sure exactly. $\endgroup$ – AlanSE Jul 28 '13 at 11:30

Let's compare two rockets with somewhat similar specifications, but one very large difference.

  • Falcon 1- Carries about 670 KG to LEO (See the User's Guide) Mass 38555 KG (Wikipedia). Launched from sea level.

  • Pegasus- Carries about 450 KG to LEO (See Wikipedia). Mass 18,500 KG. Launched from 40,000 feet.

This makes a lot of assumptions, but let's just assume you could scale linearly the mass of the Pegasus. That would give the rocket a mass of 27000 KG to lift the payload of the Falcon 1. That is a difference of 40% roughly. Why the difference?

  1. Falcon 1 is a LOX/RP liquid propellant rocket whereas Pegasus is solid rocket based. The technologies have vastly different thrust/weight ratios, specific impulse, and mass fractions. Solid rockets tend to have higher mass fractions and T/W because the engines are less complex (i.e. no pressurization systems, or plumbing, or turbo-pump machinery. But solids are typically more expensive, and typically have fixed impulse burns (i.e. they cannot be turned off). This is less of an issue for early stages, but the reason the Pegasus added the optional HAPS (hydrazine) final stage for precision orbit insertion.

  2. Because it is air-launched Pegasus can fly a different trajectory. Instead of a low angle-of-attack ascent followed by a gravity turn, Pegasus has wings. It flies at positive angle of attack and uses lift to aid in ascent.

  3. Engines at altitude can use a more efficient engine design (i.e. rocket nozzle expansion ratios tuned to atmospheric pressure at drop vice sea level.

  4. There is a small velocity gain from the air launch. This isn't significant (~2% of orbital velocity), but is there.

  5. The Pegasus doesn't have to worry about changing inclinations, as the Falcon 1 does. But the specified numbers for Falcon 1 don't take in to account a varied inclination.

  6. There is considerably less air density at 40,000 feet, which results in lower integrated drag losses.

  7. You are 10 km higher at altitude. This probably isn't significant.

Bottom line is, raising your altitude of your launch site will give you an increase in performance, both by increasing your engine efficiency, and reducing your drag. These numbers wouldn't be as drastic for a launch site of 10,000 feet, but they still would represent a measurable change in performance.

  • 3
    $\begingroup$ +1 especially for point 5. Unless a rocket engine uses a variable 'aperture' nozzle, it is effectively specialized for a particular external pressure. Engine designers would use the variable engine nozzle if it did not add extra weight, complexity and cost.. $\endgroup$ – Andrew Thompson Jul 27 '13 at 2:16

The flight path of a rocket is barely focused on 'up'. It is easy to get high enough to where orbit is possible. Consider the case of sounding rockets, which even smaller organizations or amateur rocket groups can achieve.

The hard part is going fast enough to orbit. I.e. Getting up to speed.

Thus a rocket typically heads straight up quickly to get out the thick part of the atmosphere and then turns and primarily accelerates to orbital speed.

Starting from a higher altitude would be minorly helpful (I cannot quantify the number, alas) for the very first few moments of flight, but after that would be of little benefit.

As noted in the comments the extra cost of logistics for getting fuel, oxidizer, parts, and payloads to the higher altitudes would mostly likely not be worth it.


From reading a little about escape velocity it seems that the higher the atmospheric pressure, the more velocity is gained from the thrust, and the biggest boost to escape velocity is achieved at ground level. It makes perfect sense to launch from sea level where the atmosphere is thickest; it gets the biggest push from the earth & the largest amount of ambient feedback from the atmospheric pressure which lessens the higher it rises. So from the ground up the initial kick has to sustain the velocity on a perpendicular path from the plane of the earth otherwise the rocket will lose velocity and start to arc, and just one degree of arc off perpendicular indicates a proportional reduction in velocity. It makes no logical sense that a rocket would waste the most expensive part of its launch system to then fly around like a jet to whip up speed in the most inefficient part of the sky for a rocket engine. Take note that the rocket starts with the pressure inside it by being able to push, as it rises to the upper atmosphere the pressure starts to flatten until it reaches vacuum where it will invert. The question of friction is difficult to quantify but just as a body entering the earth heats up and goes through a tempering scale, then so should a body leaving the earth, and other factors enter the friction equation in the upper part of the sky like solar particles, cosmic rays etc, it all adds to the loss of efficiency from that initial kick.

  • 3
    $\begingroup$ Erm, no. Escape velocity has nothing to do with atmospheric pressure, thrust of conventional rocket engines is increased with reduction of ambient pressure, and there isn't any such ambient feedback that would be useful on ascent , with small exemptions being Earth's rotation on its axis providing a small but not insignificant boost to achieving orbital velocity the closer to equator you launch, and ground effect on liftoff (good for small vehicles, bad for big ones tho). I think you're misapplying Oberth effect, but am not sure. Also, the question isn't about air launch to orbit. ;) $\endgroup$ – TildalWave Sep 19 '15 at 16:20

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.