If a jet engine flies with 4 m/s in the air and we want to calculate thrust for two different reference frames, can we prove that the calculated thrust for both reference frames are equal to each other, so either of them is correct?

According to an observer on the ground;

Air velocity at inlet: 0 m/s

Air velocity at nozzle: 5 m/s

Aircraft velocity: 4 m/s (just for an example)

Air density: 1 kg/m^3

Inlet and nozzle area: 1 m^2

Inlet mass flow rate: 0 kg/s

Nozzle mass flow rate: 5 kg/s

If we convert these values to engine-based reference frame;

Air velocity at inlet: 4 m/s

Air velocity at nozzle: 9 m/s

Aircraft velocity: 0 m/s (just for an example)

Air density: 1 kg/m^3

Inlet and nozzle area: 1 m^2

Inlet mass flow rate: 4 kg/s

Nozzle mass flow rate: 9 kg/s

Can you make calculations for engine-based reference frame and ground-based reference frame and prove they are equal to each other?

According to my calculations, the ground-based reference thrust is 25 N, and the engine-based reference thrust is 81-16: 65 N.

Why do I find different thrust values for different frames? Which one is correct? In general, we make these calculations in aerospace engineering for the engine-based reference frame in which the aircraft is considered to be stationary, but why?


  • $\begingroup$ I can't give a full answer on why this is incorrect, but I believe it has something to do with not accounting for the direction of the velocity. If the aircraft is moving at 4m/s (say from left to right) then from the engine's point of view the air moves from right to left, which is in the other direction: -4m/s. Likewise, it makes no sense for the exhaust velocity to be greater than the aircraft's velocity. Here I've used plain positive/negative values to show directions in a 1d space (since that's all we care about), but using vectors to indicate direction should work too. $\endgroup$ Jun 15, 2023 at 10:31
  • $\begingroup$ I believe I didn't make any direction-related mistake. $\endgroup$
    – Jawel7
    Jun 15, 2023 at 10:45
  • 2
    $\begingroup$ If the inlet mass flow is 0 the engine is not running. You seem to be confusing mass flow rate with a velocity. $\endgroup$ Jun 15, 2023 at 12:07

2 Answers 2


The thrust is same in both cases; the anomaly in your calculations is due to the fact that you are not accounting for the engine's own velocity.

In the first case, you say that the inlet mass flow rate is 0 kg/s because the air velocity at the inlet is 0 m/s. However, since the engine itself is moving forward at 4 m/s, the actual mass flow rate through the inlet is 4 kg/s, same as in case two.

Also, the exhaust mass flow shouldn't be greater than the inlet mass flow - that would be against the law of conservation of mass (unless the engine is running a ludicrously rich mixture). Since the exhaust velocity is greater than inlet velocity, the exhaust area must be smaller than inlet area such that the mass flow throughout the engine remains the same.

So the thrust in both cases is:

$$ F \ = \ ma \ = \ \frac{m}{t} × ∆v $$

$$ F = 4×5 $$ $$ F = 20 N $$

  • $\begingroup$ In both cases, thrust can't be the same because you need to look at momentum change of the air. In the engine-based reference frame, inlet and outlet velocities are much higher rather than in the ground-based reference frame. I think you didn't understand complexity of the question. $\endgroup$
    – Jawel7
    Jun 15, 2023 at 11:07
  • 2
    $\begingroup$ @Jawel7 I think you didn't understand the simplicity of the question :) In both cases, ∆v is 5 m/s. Also in both cases, the mass flow through the inlet (and therefore the engine) is 4 kg/s (explained in the 2nd para of my answer). So why should the change in momentum be different in both cases? $\endgroup$ Jun 15, 2023 at 11:15
  • $\begingroup$ Who said inlet and outlet mass flow rate are the same? In no jet engine, they are the same because there's fuel addition. $\endgroup$
    – Jawel7
    Jun 15, 2023 at 11:23
  • 1
    $\begingroup$ @Jawel7 It looks like you didn't bother to read my answer. $\endgroup$ Jun 15, 2023 at 11:24
  • $\begingroup$ But why are we using mass flow rate at inlet and outlet separately for thrust calculations in jet engines then? We are not assuming mass flow rate at the inlet and exhaust as the same? Can you make the same calculation with fuel injection? Let's say mass flow rate at the inlet is 1 kg/s and fuel is 1 kg/s and exhaust flow rate is 2 kg/s. Can you make the same calculation and find the same thrust for both cases? I couldn't find actually. $\endgroup$
    – Jawel7
    Jun 16, 2023 at 11:16

I think there's a mistake in the first case of the "observer on the ground" when it is stated that the "Inlet mass flow rate is 0 kg/s": that would imply that the engine is not ingesting any air i.e. that the inlet is simply closed and the engine not working.


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