Based on the equations of mass flow rate and exit velocity provided here (https://www.grc.nasa.gov/WWW/K-12/rocket/rktthsum.html), we can see that thrust in the case of chemical engines does not depend on the molecular mass of the exhaust gas because the factor that implicitly contains the molecular mass in the formulas of mass flow rate and exit velocity, $R$ (specific gas constant, which is universal gas constant divided by molecular mass), cancels out when we multiply mass flow rate and exhaust velocity. Despite this, molecular mass does matter because it affects specific impulse.

In the case of an electrostatic engine, a plasma of electrons and ions flow past a strong electric field and are then fired out of the rocket. Take a single electron as an example. Suppose the electrostatic energy fully converts to kinetic energy, we can express exit velocity as $$v_{ex}=\sqrt{2eV/m_e},$$ where $e$ is the charge of an electron, $V$ is the voltage difference between the two plates that form the electric field, and $m_e$ is the mass of an electron. Multiplying the exit velocity by $m_e$, we have the thrust attributed to a single electron. Then we need to multiply the thrust provided by a single electron by the velocity and density of the plasma to get the total thrust, but the end result is still dependent on the square root of $m_e$. Since $m_e$ is very small, the thrust of provided by an electrostatic engine is small.

Is there anything wrong with my understanding?

  • $\begingroup$ can you apply the same calculation to the ions and see if get a larger number? $\endgroup$ – uhoh Nov 5 '20 at 5:08
  • $\begingroup$ as a side note, the available electrical power is often the limiting factor. A lot goes into producing the plasma but a lot can also go into electrostatic acceleration as well. $\endgroup$ – uhoh Nov 5 '20 at 5:11

Your understanding is correct, but perhaps a little narrow. You are focusing on rocket-specific equations, and this leads to some gaps in your reasoning like "that thrust in the case of chemical engines does not depend on the molecular mass of the exhaust gas".

It absolutely does, but it can be concealed by the equations we use for convenience when solving for rocket-specific applications.

Take a step back--the force of accelerating the reaction mass to the exit velocity $v_e$ is $$ F = mv_e^2 $$

You could divide this all to make it mol-specific, giving you $$ \text{$F$ per mol of exhaust} = \text{molar mass}\cdot v_e^2 $$

While it is clear that the molar mass is a factor in any thrust, there are several factors concealing what is here apparent. For example, the exit velocity $v_e$ is actually a function of the molar mass (and chamber pressure & temperature)--this is why most hydrolox runs fuel-rich, as the low-mass, high-velocity unburnt $\text{H}_2$ helps the $I_{sp}$ more than running the engine off-stoichiometric hurts--even though it does harm the thrust. There's a reason solids & kerolox are so much more popular than hydrolox for first stages, where thrust really does matter.

But all other factors the same, the more reaction mass, the more thrust.

  • $\begingroup$ Hi Anton thanks for your answer. I should have realized that I could apply the same approach of breaking total thrust down to thrust per mol of exhaust when evaluating the thrust in the case of chemical engines. Besides, I am still wondering what is the factor in the formulas of mass flow rate and exit velocity except for specific gas constant (R) that is also dependent on molar mass. Otherwise, there is no molar mass left in the equation. $\endgroup$ – Xi Liu Nov 5 '20 at 18:58
  • $\begingroup$ Mass flow rate is non-specific. If you divide everything to make it specific to mols of exhaust, then you'll have a molar mass flow rate ($\text{molar mass}\cdot A \cdot v_e$). But that's not commonly done because the mol number changes in the combustion... $n_\text{reactant} \neq n_\text{products}$. However, in non-relativistic engines, mass is conserved. As a result, it's easiest to not make it specific, because in steady-state, mass flux into the chamber equals mass flux out of the nozzle. Easiest to use mass flow rate without worrying about mols of anything. It's all the same anyways. $\endgroup$ – Anton Hengst Nov 5 '20 at 19:08
  • $\begingroup$ How about the formulas of mass flow rate and exit velocity listed here: grc.nasa.gov/WWW/K-12/rocket/rktthsum.html? By which factor is molar mass concealed except for specific gas constant R? $\endgroup$ – Xi Liu Nov 5 '20 at 19:17
  • $\begingroup$ Mass flow rate is just molar mass times mol flow rate. It's not difficult. Divide anything that has mass in it by mol number and you'll find molar mass. $\endgroup$ – Anton Hengst Nov 5 '20 at 19:41
  • $\begingroup$ Here, it's obfuscated because you are prescribing the throat area and finding the mass flow rate from that--substituting the choked state of the area-mach relation makes for a nasty expression full of gammas. You could easily do the opposite--prescribe the mass flow rate and solve for the throat area. $\endgroup$ – Anton Hengst Nov 5 '20 at 19:43

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