In the rocket thrust equation, as well as the thrust due to the mass flow rate of the exiting propellant, there is also an additional pressure term included. However I have noticed that this term is not actually included when calculating the thrust produced by an engine in space. I understand that there is no atmosphere, so pressure is zero. So shouldn't P0 term here be zero and therefore the "additional thrust" due to the pressure difference is just Pe * A?

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Source: https://www.grc.nasa.gov/www/k-12/airplane/rockth.html


1 Answer 1


Yes that's correct. The ambient pressure pushes against the exhaust pressure, which is the meaning of the $p_e - p_0$ term. In space the ambient pressure is zero ($p_0 = 0$).

However there is a tradeoff between the $A_e(p_e - p_0)$ term and the exhaust velocity $v_e$ (strictly $\dot m\ v_e$, but $\dot m\ $is fuel flow & is considered constant). Through very complex maths you can show the ideal nozzle design is achieved when $p_e - p_0 = 0$, i.e. the exit pressure equals ambient pressure. This is because the exhaust velocity increases as the exit pressure declines to ambient pressure, but declines again once the exhaust has to push against ambient pressure. This involves considering the $A_e$ term, and the ratio of throat area to nozzle area. (A full analysis is very complex, but heh, that's rocket science.)

A consequence is rocket engines used near sea level and those designed to operate in space have very different nozzle sizes.

These are identical SpaceX Raptor engines. Only the nozzle differs. One is designed to operate at sea level and the other is optimised for space. Can you guess which is which?

Raptor sea level vs space

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    $\begingroup$ Thanks for your reply. The bigger one is for space I’m guessing. So essentially the nozzles are designed to keep the exit pressure equal to ambient pressure and as a result the pressure term disappears from the rocket thrust equation? $\endgroup$
    – G11
    Commented Apr 28 at 0:15
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    $\begingroup$ @G11 Yes, that's correct. Of course you can never get the exit pressure down to a vacuum, but you can try! $\endgroup$
    – Galerita
    Commented Apr 28 at 2:36
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    $\begingroup$ @G11 for boosters what you suggest is not possible, since the atmospheric pressure drops as the vehicle ascends. This means that a fixed nozzle can never "keep the exit pressure equal to ambient pressure" and the pressure term remains important in the equation. Aerospike engines are a theoretical way to address this problem, extendable exit cones have been used in practice. See many questions here about this including space.stackexchange.com/questions/46280/… and space.stackexchange.com/a/46564/6944 $\endgroup$ Commented Apr 28 at 14:44

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