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EDITED: According to the answer of @asdfex, since the ions leave with a velocity = escape velocity so when they approach an infinite distance the net thrust is 0, this means that:

  1. The total net thrust is 0 only when the ions approach infinite distances?

  2. The ion thruster expels the ions for a few seconds or minutes (depending on the mission) but after this event the satellite returns to a potential 0, this means that the ions are no longer attracted to the satellite, so there is constant net thrust without sending the electrons with the ions?

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From the book "Fundamentals of Electric Propulsion: Ion and Hall Thrusters " the ion ion velocity leaving the thruster is:
$$v_i=\sqrt{\frac{2eV_b}M}.$$

But this is equal to the escape velocity (that is, the minimum velocity that the ion should take to avoid returning to the satellite), demonstration:

First you set the potential energy at $R$ equal to the kinetic energy at infinity:
$$KE(r=∞)=PE(r=R)$$ Or,
$$\frac{mv^2}2=\frac{k_eQq}R$$ (assuming escape from a spherical charge)
$$v=\sqrt{\frac{2k_eqQ}{mR}}$$ assuming escape from a spherical charge, the potential grid of the accelerator is $V=\frac{kQ}R$.

This means that the objective of the electron gun is not to neutralize the ions so that they don't return to the satellite (as I have seen and read in several articles and videos) but rather to expel them from the satellite so as not to charge the satellite negatively (since in the ionization process, 2 electrons are obtained per ion and the accumulated electrons would charge the satellite negatively), am I correct?

If I am correct, that could mean that it doesn't matter to expel the electrons on the opposite side from which the ions come out, the objective is not to charge the satellite.

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  • $\begingroup$ In this question I changed your screen shot of an equation to a real equation in MathJax Screenshots and images of equations are discouraged in Stack Exchange for several reasons, including searchability and readability, especially for those using electronic screen readers which may include those who are visually impaired. Would you like to give it a try yourself this time and write your equations in MathJax? You can use your now-edited previous question as a starting example. Thanks! $\endgroup$ – uhoh Dec 13 '20 at 23:28
  • $\begingroup$ Why do you get 2 electrons per ion? Doubly ionizing is more than twice as hard as singly ionizing. $\endgroup$ – Jon Custer Dec 14 '20 at 17:05
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You're right that one of the main objective is to remove charge from the satellite to keep it neutral (to avoid ions returning to the satellite).

But, and here comes the important point: You calculated that the ions just reach escape velocity. That means, when approaching infinite distance, their speed approaches zero! And, with the ions being at rest with the satellite, the total momentum transfer is zero as well.

To list a few aspects that are good reasons to send the electrons along with the ions:

  • The power requirements are lower: The electrons need a certain amount of acceleration to avoid having them coming back to the satellite. If the electrons are close to the ions, they are attracted and pulled away "for free".

  • If the beam ions stay ionized, all following ions are repelled by their charge, reducing their speed and thrust.

  • The highest thrust can be reached if the ion beam is perfectly focused. If it's charged, the ions will repel each other, broadening the beam.

  • If electrons are ejected in the same direction the satellite accelerates to, it will catch some of them again.

  • And lastly, at least in Hall-effect thrusters, the same cathode is used for ejecting neutralizing electrons and to provide those that ionize the fuel in the first place - hence it has to be placed close to the exit of the engine either way.

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