According to NASA

Each Space Shuttle Main Engine operates at a liquid oxygen/liquid hydrogen mixture ratio of 6 to 1 to produce a sea level thrust of 179,097 kilograms (375,000 pounds) and a vacuum thrust of 213,188 (470,000 pounds).

Why does a rocket engine provide more thrust in vacuum than in atmosphere?

Does this hold true for all rocket engines?

  • 1
    $\begingroup$ I would venture the answer lies in the question; effort applied to overcome atmospheric resistance. $\endgroup$
    – Everyone
    Oct 18 '13 at 12:40
  • 1
    $\begingroup$ It is just static thrust (thrust found in lab experiment) I think @Everyone $\endgroup$
    – Hash
    Oct 18 '13 at 12:57
  • 1
    $\begingroup$ While in vacuum, there's an extra 1 bar pressure difference between the combustion chamber and the outside. I don't know how much of a contribution that makes. $\endgroup$
    – SF.
    Dec 1 '14 at 11:50
  • 3
    $\begingroup$ Conduct this thought experiment: Take an empty bottle, i.e. filled with air at 1 atmosphere. Put a cork on. Remove the cork a) in your living room. b) in a vacuum. What will happen in each case? $\endgroup$
    – Jens
    Jun 1 '15 at 16:52

Rocket thrust is given by the equation

$$ F = \dot{m}v_{exit} + A_e(P_1 - P_2) $$

where $\dot{m}$ is the mass flow rate, $v_{exit}$ is the average exit flow velocity across the exit plane, $A_e$ is the cross-sectional area of the exhaust jet at the exit plane, $P_1$ is the static pressure inside the engine just before the exit plane, and $P_2$ is the ambient static pressure (i.e. atmospheric pressure).

Provided that the nozzle is not overexpanded and flow separation does not occur, $A_e$ remains constant, and the thrust difference is realized primarily from the change in $P_2$. If nozzle is overexpanded to the point that flow separation occurs, however, the exhaust jet area drops as well, causing further losses.

  • 11
    $\begingroup$ So to put it in layman's terms - the thrust does not need to push against air-pressure and forms a more optimal shape? $\endgroup$
    – john3103
    Oct 18 '13 at 15:55
  • $\begingroup$ More or less, that is correct. $\endgroup$
    – Tristan
    Oct 18 '13 at 16:17
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    $\begingroup$ @john3103 I'd leave off the "and forms a more optimal shape part." The main consideration is that in vacuum there is no atmospheric pressure fighting the engine thrust. It's true this back pressure changes the optimal expansion ratio, but that's not the fundamental reason that vacuum thrust is always larger than sea-level thrust. $\endgroup$
    – Adam Wuerl
    Oct 19 '13 at 21:32

In addition to the answer of Tristan I would like to add few more points

The thrust in the rocket is equal to $T=\dot m V$ (Assuming the rocket nozzle is operating at its optimum condition)

The Thrust is a strong function of the exhaust velocity

$$V=\sqrt{\frac{2 \gamma R_{{}^{\circ}} T_{{}^{\circ}}}{(\gamma -1) \mu }\left(1-\left(\frac{P_e}{P_c}\right){}^{\frac{\gamma -1}{\gamma }}\right)}$$

This equation gives the exhaust velocity of the rocket

The exhaust velocity is a function of $\left(\frac{P_e}{P_c}\right){}^{\frac{\gamma -1}{\gamma }}$ and for vacuum the $P_e$ is almost equal to zero so the above term reduces to zero hence the exhaust velocity is maximum

For sealevel the above term does not reduces to zero so the exhaust velocity is less compared to that in the vacuum

Hence the thrust in the vacuum is more than that of in sea level (within atmosphere)

  • 1
    $\begingroup$ If the flow is supersonic (as any marginally acceptable rocket engine would have it), then unless the nozzle is overexpanded to the point of flow separation, the flow inside the nozzle would have no way of "knowing" what the ambient pressure is. $P_e$ is only going to be a function of the nozzle shape and the upstream flow conditions. This is why first stage engines do not perform as well as upper stage engines at high altitudes -- the flow is underexpanded, and any expansion after the flow leaves the nozzle contributes nothing to the thrust. $\endgroup$
    – Tristan
    Oct 18 '13 at 16:21
  • $\begingroup$ @Tristan when the flow is overexpanded or underexpanded there is some wastage of thrust because of the formation of shock waves (either oblique or expansion wave) which creates back pressure and reduce the flow speed(increase the stagnant pressure) thereby increasing the thermal energy but this does not happen in vacuum even when the supersonic flow $\endgroup$
    – Hash
    Oct 18 '13 at 16:40

There are many reasons for that...

  1. One is the atmospheric resistance which is very obvious
  2. Expansion of gases passed by the nozzle exit is very important in deciding the thrust produced. In normal atmosphere, pressure of the gas at the exit is negative gauge and hence nozzle is under-expanded which produces minimum thrust. In vacuum, it is over-expanded which produces higher thrust.

Efficiency of the engine bell at the nozzle end.

The bell-shape allows the gas to expand but this shape is usually tuned for the region in which the engine operates, namely low altitude + thick air or high altitude + thinner air.

The bell-shape is usually a compromise, since as the rocket climbs the air thins. So what's best? Efficiency at low altitudes that gets worse as the rocket climbs ever higher... or low efficiency that improves? The mission designer and the engine designer figured that out in the 1950s!

The engines work best when there's no air for the expanding gas to push against, wasting thrust, namely 'in space'.


I'm not sure the premise of the question -- that all rocket engines provide more thrust in vacuum -- is correct.

A design feature of aerospike engines is that they overcome this very problem, to provide nearly uniform levels of thrust both in and out of the atmosphere. AFAIK they have never flown for orbital missions, but they have been built, and fired extensively on test stands, and smaller models have flown on tests.


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