# Thrust of ion thruster when spacecraft is charged

I read in many articles and books that the thrust of an ion thruster is equal to

$$T = \sqrt{\frac{2M}{e}} I_b \sqrt{V_b} \ \ \text{[newtons]}$$

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where $$V_b$$ is the acceleration potential, but this equation is valid when spacecraft potential is 0.

What happens when the surface of the satellite is charged? (The particles that are in space due to the solar wind will charge the surface of the satellite.)
In this case, the complete equation would be something like $$V_b = V_a + V_s$$ where $$V_a$$ is the acceleration potential and the $$V_s$$ is the voltage of the surface charged?

This is a great question!

To first order charging doesn't matter. What matters is the exit velocity of the ions, determined by the acceleration potential difference $$V_b$$.

In a simple situation the thrust is the velocity of the ions with respect to the spacecraft times the mass flow rate $$dm/dt$$ of the ions, and that velocity is determined by (something like) the acceleration potential difference between the plasma or the first grid, and the final acceleration grid.

In the simple situation it's not related to the potential of the spacecraft relative to other things, so the speed of the exiting ions and therefore their momentum won't be affected by the charge of the spacecraft.

However nothing is simple, and yes there can be a small coulomb interaction between the sparse ion plume behind the spacecraft and the spacecraft's residual charge, but that is usually not a problem because the electrons removed from the ions are also shot out the back with an electron gun which is usually aimed into the general direction of the plume. If the velocities are similar, then from a distance it will look roughly neutral and this effect will be minimized.

For more on that see

In your equation the m-dot or $$dm/dt$$ is expressed as the current times mass and some other things. The reason that it looks like the square root of mass is that the velocity term at the end $$\sqrt{V_b}$$ should have an M at the bottom which would then liberate the first M out of its square root and next to $$I$$ where it belongs.

For more on that see:

• This is not an answer, but a question/comment for @uhoh on their answer. If the ions cloud coming out of the back of the spacecraft weren't ionized, say by separating the electron and ion beam emission points, could a meaningfully larger amount of thrust be generated, either by giving the spacecraft a net positive charge, or using the 1/r^3 potential that the drive grid generates? These potentials should push against the ion cloud outside the spacecraft, continuing to accelerate them even after they left. – alessandro Dec 1 '20 at 11:00
• @alessandro that's also an interesting question! I wrote "...yes there can be a small coulomb interaction between the sparse ion plume behind the spacecraft and the spacecraft's residual charge..." I haven't calculated anything but my feeling is that that is such a small effect that it hasn't been considered worth pursuing, but I really don't know. So I think it would be great if you posted that as a new question where the whole community will see it and there's more of a chance of someone coming up with a great answer. Go for it! :-) – uhoh Dec 1 '20 at 22:58