# Where can I learn to calculate the physics of an ion thruster?

I am trying to build an ion thruster, but I don’t know where to start learning the actual numbers behind one. I know how they work on a basic level, but not much more.

Basically I want to be able to estimate how much thrust it will output for a given voltage and current all else being equal. Of course I’m sure there is math behind the difference in anode and cathode sizes and shapes, and I would like to learn that too.

What would be the best resource for starting my research?

• What is your timeframe and constraints? Would a masters degree in aerospace engineering violate any of them? Jun 13 '20 at 19:15
• If you just want to calculate the thrust, $F=\frac{dp}{dt}=v\dot{m}$ where $\dot{m}$ is the mass flow rate (kg/sec) and the accelerated velocity is given by $v = \sqrt{2qeV/m}$ where $q$ is usually equal to 1 (singly ionized atoms) $e$ is the electron charge, $V$ is the acceleration voltage and $m$ is the mass of one atom. Have a look at Ion Thruster Thrust Calculation Problem and What could go wrong if someone overpowered an ion engine?
– uhoh
Jun 13 '20 at 22:16
• – uhoh
Jun 13 '20 at 22:17
• @lijat I would hope to get a decent understanding of it within a few weeks or maybe a monthish but I will probably keep learning until it clicks. Unfortunately while a masters in aerospace engineering would be awesome, it's a little pricey. Jun 15 '20 at 3:21
• @uhoh I found the Ion Thruster Thrust Calculation Problem and your response to it when trying to figure it out myself, but I think I might be missing some basic concepts which tie everything together. As for the other questions you linked, I will definitely be checking those out. Thank you! Jun 15 '20 at 3:31

You can get a very approximate answer just from fundamental physics. Your ions probably have $$e$$ unit of charge (ie they are missing one electron), so in dropping through a potential $$V$$ Volts they will acquire $$Ve$$ Joules of energy. So if they have mass $$m$$ and exhaust velocity $$v$$ you will get $$\frac{1}2 mv^2 = Ve,$$ so $$v = \sqrt{\frac{2Ve}m}$$

Now suppose the current is $$I$$ that determines the number $$n$$ of ions per second. $$I = ne$$ where $$e$$ is the charge on an electron, $$1.6 × 10^{-19}$$ coulombs.

So now the thrust is the same as the momentum per second of the exhaust, which is $$nmv$$ so we can assemble all our equations to get $$T = I\times \sqrt{\frac{2Vm}e}$$ For example a xenon ion has mass 131 daltons, which is $$2.2\times 10^{-25} kg$$ so for $$V = 10000 V$$ and $$I = 1 A$$ we get

$$T = \sqrt{\frac{20000 \times 2.2\times 10^{-25}}{{1.6\times 10^{-19}}}} = 150mN$$

This would be a 10kW power consumption if perfectly efficient. Realistically it would use more power and create less thrust because of inefficiencies.

• OK so let me see if I understand this properly. I'm going to use nitrogen instead of xenon because its more accurate to what I would be doing and to make sure I'm not blindly copying your equations. Nitrogen Jun 15 '20 at 3:39
• @saddlepiggie there will be several interesting disadvantages to using N2 in an ion thruster, so I think "What are the technical challenges using nitrogen instead of a noble gas for an ion thruster?" would be an excellent question of its own! Consider posting that as a new question so that it can be answered in a new post rather than in comments.
– uhoh
Jun 15 '20 at 3:45
• @uhoh Whoops I hit enter before I was done. I have looked at some resources about why xenon is used in spacecraft before asking this, and if I understand it correctly it is because: 1. Xenon is very heavy, meaning you get more thrust per atom that lighter elements. 2. Xenon is relatively easy to ionize. 3. Its not radioactive like heavier atoms such as Radon. 4. Noble gases are stable. OK, I need to go to bed now but I'm going to come back tomorrow and finish what I started in my first comment. Also thanks a lot for the information you have put out. Jun 15 '20 at 3:54
• @saddlepiggie new questions should be posted as new questions, not in comments. Comments are considered temporary and can be deleted at any time. If you post it as a new question it provides space for many people to post answers, and these could easily contain information specifically about nitrogen that you didn't read about there.
– uhoh
Jun 15 '20 at 3:59
• Continuing from my first comment (sorry if the formatting is bad): Just to vary the equation, I will use .5 amps and 15000 volts. Nitrogen has a mass of 2.3×10^−26 kg. Meaning the equation would look like: T=.5√30000×2.3×10−25 / 1.6×10^−19 = 33mN rounded. Also, plugging your last equation into a calculator gives me 165mN. Did you round this or did I mess something up? Jun 15 '20 at 23:34