A rocket engine in the vacuum will experience efficiency loss due to under-expansion in which the pressure of the exhaust is greater than the ambient pressure, which in a vacuum is near zero.

How great is this efficiency loss in real-world terms? The thrust equation has a term in which the ambient pressure is subtracted from the exhaust pressure and multiplied by the cross-sectional nozzle exit plane area. This actually increases the thrust, yet we speak of under-expansion being an efficiency loss. Where is that loss and how great is it in real-world terms.

I tend to think of it, perhaps too simplistically, in terms of geometric vectors with a diagonal force broken down into X and Y vectors. In an ambient engine where exhaust pressure equals ambient, the exhaust gas is longitudinal behind nozzle mostly X vector. As plume expansion occurs with under-expansion, force that could have been imparted in X direction is wasted in Y direction. The question is how much force, as a percentage of thrust?

If we have a small engine less than 25,000lbf of thrust, suppose it fires, in the vacuum of space, say with like Chang'e5 landing, with 25,000 lbs of thrust, doesn't this mean it actually only produces a smaller amount of thrust due to plume expansion, the loss of force in Y vector, or is this already factored into the rating? So if I need to propel say 10,000 lbs, in a vacuum, launch or land in the vacuum scenario, don't I actually need an engine with more thrust than this because of the Y-vector thrust losses? I want to understand theoretically, not for any one specific scenario. How great is the efficiency loss in the real world? How do we explain or quantify the loss when the rocket thrust equation shows a theoretical gain in thrust? Does the efficiency loss due to Y-vector expansion exceed the theoretical thrust gain from rocket thrust equation? Thank you for anyone that can answer.

  • $\begingroup$ Some rocket nozzles are optimized for atmospheric pressure, some for vacuum. Aerospikes are an odd design that gets OK performance in both. See Are Aerospikes Better Than Bell Nozzles? for more $\endgroup$
    – mmesser314
    Commented Jan 20 at 23:03

1 Answer 1


Think of it terms of a gun barrel, with the recoil of the gun representing the trust on a rocket. In a pistol the pressure in the barrel might be say 10 atm as the bullet leaves the nozzle. Once the bullet has left the nozzle that pressure rapidly dissipates in all directions and is mostly wasted. This is a case of under expansion as not all the potential energy of expansion has been utilised. If the barrel is much longer the pressure continues to accelerate the bullet for longer increasing muzzle velocity and the wanted recoil. The pressure drops as the gap between the chamber and the bullet increases. Ideally to impart maximum velocity to the bullet (or maximum recoil if the gun is the rocket) the pressure inside the barrel should be atmospheric just as the bullet reaches the the nozzle. This is not practical usually because of the great length of barrel that would be required, but this is also why high velocity sniper rifles have longer barrels. Now if we had a stupidly long barrel, the pressure inside the barrel would drop below atmospheric effectively creating a vacuum. The pressure on the on butt of the barrel would be greater than the pressure in the chamber reducing the recoil that we want. This is over expansion. When fired in space, it is not possible to have over expansion, because it is not possible to drop the barrel pressure below the external pressure, even with an infinitely long barrel. As long as there is even a few psi left inside the barrel some greater muzzle velocity can be milked by making the barrel longer, but that is assuming no friction. Taking friction into account, there comes a point where the additional friction of a longer barrel negates the theoretical gains from increased expansion.

Edit in response to the OP's comments: As far as I can tell, a rocket rated at 25 kbf in a vacuum has the same rating in an atmosphere. I think what is different is how the efficiency is rated. If we loosely rate efficiency as 1 - f (Exhaust pressure - ambient pressure) where f is some factor, then if the exhaust gas has a pressure of 1 atm and the ambient pressure is zero, the efficiency is less than one. A block of gas that exits the nozzle at one atm in a vacuum will continue to expand after it has left the nozzle and since this external expansion contributes nothing to the thrust it is wasted energy that could have used to accelerate the rocket. The same rocket operating in the atmosphere could claim to have 100% efficiency, because exhaust pressure = ambient pressure and that is the best we can do under atmospheric conditions. How the efficiency is calculated is more of a marketing exercise than a practical consideration.

  • $\begingroup$ Thank you for reply. Follow up question. In a vacuum, in a rocket, there is no bullet. Just the exhaust gas itself. Any molecule leaving the nozzle in a trajectory not directly rearward does not impart as much energy as it might have. Yet you seem to be implying that by the time it gets to the exit nozzle, Newtonian reaction has already occurred, and zipping outward laterally isn't a direct loss of thrust. Adding the bullet to explanation made easier to understand, but made application to rocket difficult. If a rocket is rated say 25K lbf, in vacuum, and is producing that rearward thrust. . . $\endgroup$
    – Nomadicus
    Commented Jan 22 at 17:54
  • $\begingroup$ ... thrust, isn't it actually producing some larger thrust to get the 25K rearward thrust, because of lateral exhaust bleed? Or is that waste already factored into specific impulse of engine and other ratings? $\endgroup$
    – Nomadicus
    Commented Jan 22 at 17:57

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