# From the General Thrust Equation towards Tsiolkovsky, how to explain dropping these terms along the way?

The NASA Glen Research Center tutorial page Rocket Thrust Equation links to the General Thrust Equation page which starts with:

Which transcribes into MathJax as:

## $$\ \ \ F = \dot{m}_\mathrm e V_\mathrm e - \dot{m}_0 V_0 + (p_\mathrm e - p_0) A_\mathrm e$$

where $$\mathrm e$$ and $$0$$ indicate nozzle exit and free stream, $$A_\mathrm e$$ is the nozzle exit area, and $$F$$ is thrust, the force on the vehicle.

There are three terms on the right side, and as far as I understand it, the derivation of the Tsiolkovsky rocket equation in a vacuum uses only the first term.

If you had to explain the dropping of the 2nd and 3rd terms in a way that could be understood and believed by beginners to rocket science (like me) but without handwaving, "take my word for it"-ing or "go look it up"-ing, or "go google it"-ing, what would you say while holding the chalk and crossing out each of the last two terms?

Check out the diagram at the top of the page that you got the equation from.

Let's define our terms.

• $$\dot{m}_e V_e$$ is the momentum thrust term
• $$\dot{m}_0 V_0$$ is the incoming momentum term
• $$(p_e - p_0) A_e$$ is the pressure thrust term

The incoming momentum term is important for jet engines because the engine swallows the incoming stream and then accelerates it. It is not important for rocket engines because they don't do that.

If you dropped the incoming momentum term for a jet engine, you could have an empty pipe attached to your airplane and calculate a nice thrust coming out of it! But we know that calculation would be incorrect. To get thrust from your jet engine, it must increase the velocity of the incoming stream. The difference in the inlet and exit velocities gives the thrust.

The pressure thrust term should not be dropped for rocket (or jet) engines. It simply goes to zero when the delta pressure is zero (exit plane pressure matches ambient).

• I agree with your last statement. Rocket engines never really match ambient pressure, especially not in a vacuum. – Paul Oct 21 '18 at 14:14
• I hope that makes it clearer. – Organic Marble Oct 21 '18 at 14:28
• Since Isp = F/m-dot, if the vacuum Isp used is correct, it should take all effects into consideration. So this calculation is only as good as the Isp you use in it. It should be based on the real vacuum thrust, hopefully. I guess what I'm trying to say is Isp is a derived property and for a real world situation should be based on actual measured thrust and flow rate. – Organic Marble Oct 21 '18 at 14:37
• @uhoh Didn’t we like just do this, twice? space.stackexchange.com/q/30500/195 space.stackexchange.com/q/30529/195 – Russell Borogove Oct 21 '18 at 19:27
• I mean this bit of the comment thread, not the overall QA – Russell Borogove Oct 22 '18 at 2:30

Ideally if you could design the nozzle to match the exhaust pressure in a vacuum (i.e. nearly zero), the third term drops automatically. If $$p_0$$ is zero, then $$p_e$$ would have to go to zero as well because an ideally designed nozzle results in no pressure drag (i.e. ambient freestream pressure and exhaust pressure are the same). In reality, such a nozzle could never be built because it would be infinite in length (it takes an infinite length to drop the exhaust pressure to an infinitely small pressure such as a vacuum). But real nozzles are designed to make that exhaust pressure as close to ambient as possible given the constraints of its length while still accelerating the exhaust gas as fast as possible. If I recall correctly, nozzles tend to be optimized for use at ambient pressure close to the launch site on the surface of the earth (because its so hard to lift off), so it makes sense that these terms tend to be included when discussing rocket design.

Also, the freestream velocity of a vaccum would be zero, which would drop the second term. Although, technically speaking, freestream velocity not really well defined. In a vacuum, there is no freestream of anything to begin with, so you can neglect it. The general thrust equation applies more to the case of the presence of a fluid (i.e. air). In a vacuum, those terms just don’t make any sense.

Edit: i dug a little deeper into the meaning of these equations and found that the second term is called the ram drag, which only applies to air-breathing engines like jets. It would have to be dropped for rocket engines because they carry their own fuel/oxidizers. They don’t take air into the engine as part of the combustion process.

So, the second term could be interpreted as a mass flowrate of the intake of air. That flowrate would of course be zero in a vacuum.

• If I understand correctly, $\dot{m}_0$ is the rate at which the engine takes in the outside air. In that case, the second term is 0 for a rocket because $\dot{m}_0=0$. – Litho Oct 21 '18 at 13:11
• @Litho: you’re right! I added that info in the answer. – Paul Oct 21 '18 at 14:02
• "That flowrate would of course be zero in a vacuum." You might want to think about that. Rockets lift off in a non-vacuum. Would that flowrate be non-zero for a rocket at liftoff? – Organic Marble Oct 21 '18 at 17:54
• @OrganicMarble: The OP is asking about Tsiolkovsky’s formulation, which would be in a vacuum. In real life on earth, of course it would be nob-zero. – Paul Oct 21 '18 at 17:58
• I don't think it's non-zero for a rocket. Ever. They don't take in air from the ambient and accelerate it. – Organic Marble Oct 21 '18 at 17:59