I've been calculating the needed mass flow rate for a hypothetical ion rocket, lifting off from Earth's surface and propelled with a Dual-Stage 4-Grid ion thruster. Data given: mass of the spaceship /inkl. propellant/ is 1000 t, Isp=19300s. So, the thrust I need is F=mg=9,81m/s2 x 1000000kg=9810000 N. For the exhaust velocity I come to Ve=Isp*g=19300s x 9,81m/s2=189333 m/s. Finally, for the mass flow rate I come to F/Ve=(9810000kgm/s2)/(189333m/s)=51.813 kg/s. This mass flow rate of approx. 52 kg/s seems pretty low to me, even considering the high Isp. Is there any error in the logic and calculations? Thanks guys!

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    $\begingroup$ "Is there any error in the logic and calculations?": Many. 1) Ion drive thrust to weight ratio in space is very low. In the 1/10000 range. 4 to 5 orders of magnitude too low to liftoff in earth gravity. 2) Ion drives only work in vacuum. At atmospheric pressure, your ISP 19300s ion drive will be more like ISP=0.2. Yes, it will emit a gentle breeze, exactly in proportion to using its propellant as a cold gas thruster that is not at all designed to be a thruster. $\endgroup$ Dec 18, 2021 at 17:42
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    $\begingroup$ Assuming you DO get an ion drive with ISP 19300, and with a thrust-to-weight ratio of more than one. and you pump earth's atmosphere away, so the ion drive can even operate. Yes, that mass rate sounds about right. But you might also want to bend a though to the power requirement of your drive. For that 9810000N thrust at ISP of 19300, your ion drive is requires a power input of 2.75e13 watt. That's a bit much, just under 10 times the global electricity production. Your utilities bill will be scary. $\endgroup$ Dec 18, 2021 at 17:51
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    $\begingroup$ I knew something has to be wrong :D Cheers CuteKItty_pleaseStopBArking, thanks for the prompt reply! $\endgroup$
    – sea_for
    Dec 18, 2021 at 18:28
  • $\begingroup$ As a note: It is possible to make a form of electrostatic air-breathing thruster, but I am not under the impression that these are able to lift their power source. $\endgroup$
    – ikrase
    Apr 2, 2023 at 7:37
  • $\begingroup$ You could use a thruster that runs on the atmospheric mass, with an initial boost, of course, then switch to onboard mass when in vac. The other consideration for this would be to use a boost vehicle comprised of a solid or liquid fuel booster, with an atmospheric thruster around it in rings, then detatch a vacuum optimised one. However you might be better to consider using a high TWR Horizontal takeoff booster and a normal vacuum second stage. $\endgroup$ Apr 3, 2023 at 1:25

1 Answer 1


Thrust needed for Ion Thruster Lift off from Earth surface... Is there any error in the logic and calculations?


$$m = 1 \times 10^6 \ \text{kg}$$ $$F = 9.8 \times 10^6 \ \text{kg m/s}^2 \approx v_E \frac{dm}{dt}$$

If "Isp=19300s" then:

$$v_E = 9.8 \times 1.93 \times 10^4 \approx 1.89 \times 10^5 \text{m/s}$$

That definitely gives $dm/dt$ of 51.8 kg/s, so your math checks out.


You're going to need a whole lot of electrical power to make this work. Usually the biggest power hog is the production of the plasma from neutral gas, not the acceleration. It's a highly inefficient process to heat electrons enough and contain them efficiently enough to ionize a flowing gas with a high ionization efficiency.

The traditional xenon is way way to expensive and rare to use in a launch vehicle at this rate (52 kg/s times say 154 seconds is about 8 tons of propellant) The price of xenon varies dramatically with purity, so you'll spend some time checking how well your particular engine design can deal with impurities that try to reabsorb your free electrons.

Instead you'll walk up the table of elements like SpaceX to krypton and then to argon which is about 1% of our atmosphere so much cheaper to produce and purify. It's probably a sweet spot though you may consider other alternatives.

The atomic mass of argon is about 40, so you'll be releasing about 1.3 moles per second or $1.3 \times 6 \times 10^{23} = 7.8 \times 10^{23} \ \text{atoms/sec}$. Each atom will be charged +1e so divide that by 1 Coulomb; that's $7.8 \times 10^{23} / 6.24 \times 10^{18} = 125,000 \ \text{amps}$ of charged argon!

I'm not going to calculate the coulomb repulsion in that ion beam, but I have a hunch it's not going to be held together by a normal nozzle. I think we're talking super-duper superconducting solenoids.

Let's calculate the voltage needed to accelerate our ions to $1.9 \times 10^5 \text{m/s}$. For one atom, the mass is $40 \times 938 \times 10^6 \text{eV/c}^2$. With $v/c$ of $6.3\times 10^{-4}$ and an atomic mass of $40 \times 938 \times 10^6 \text{eV/c}^2$ I get a kinetic energy of about 7500 eV per atom, or an accelerating voltage of 7.5 kV which is the right ballpark for an ion engine.

Note that 125,000 amps times 7500 volts is about a gigawatt of electrical power just to do the acceleration. I have a hunch you'll need a few orders of magnitude more than that to produce and control an absurdly large plasma and flow rate.

  • $\begingroup$ Great stuff! Will such a drive work at all in the atmosphere? $\endgroup$ Apr 2, 2023 at 15:22
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    $\begingroup$ @OrganicMarble between the huge amount of electrical power required and the coulomb self-repulsion of such a high density of charged particles, it's hard to imagine that this could exist at all. I don't know what happens when this hits atmosphere; it's a totally different regime than anything that's been done in the past. If I had to venture a wild guess, this much energy would "make a hole" in the atmosphere. $\endgroup$
    – uhoh
    Apr 2, 2023 at 20:44

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