If I have a propeller with a continuous thrust of $1 \times 10^{-9}$ Newtons, can I take the satellite from LEO to the moon?

How can I calculate the time it would take for a satellite of a certain mass to reach an orbit knowing the thrust?


The answer is essentially here where the give a figure of 8 km/s for the delta-V needed to get from LEO to LLO (Low Lunar Orbit) as 8 km/s. So, if the mass of your satellite is $M kg$ (and it is small enough that light pressure and drag in LEO aren't a concern) your acceleration will be $a = 10^{-9}/M ms^{-2}$ so the time required is $8000/a$ which is $8\times 10^{12} M s$. So a 1 kilo satellite will take about $200\,000$ years.

Editing in response to a comment. Light pressure (from sunlight) near earth is about 10 $\mu Pa$. Atmospheric drag depends on just how low your LEO is and on "space weather" but it is more than this up to altitudes of 900km or so (because light sails are no use below there). So actually, given that you have constant thrust it isn't small size that makes you vulnerable, but large area. If your probe with the $1nN$ engine exposes more than about $1 cm^2$ of cross-section, the forces from drag and/or light pressure are likely to exceed the thrust.

Getting from LLO to the lunar surface requires a much higher thrust, since once you drop out of orbit, you have to get rid of your orbital velocity before you crash.

| improve this answer | |
  • 1
    $\begingroup$ Thank you! just one more question, where does 8000 come from? $\endgroup$ – Ricardo Casimiro Jul 26 at 16:53
  • 2
    $\begingroup$ 8km/s (from the wikipedia page) converted to meters/second. $\endgroup$ – Steve Linton Jul 26 at 17:11
  • $\begingroup$ Roughly how small is "small enough that light pressure and drag in LEO aren't a concern" in this context? $\endgroup$ – uhoh Jul 27 at 9:42
  • 1
    $\begingroup$ @CarlWitthoft since we know the force of the thruster in Newtons, and this answer mentions "size" in reference to a specified mass of 1 kilogram, in this case let's keep it simple and just estimate if that size is a millimeter or a centimeter or a 1 U cubesat. We don't know their densities, but cubesats are very roughly 1 kg/U Cubesat mass density (kg/U) statistics? $\endgroup$ – uhoh Jul 27 at 13:25
  • 1
    $\begingroup$ @CarlWitthoft I think in LEO drag is almost entirely determined by cross-sectional area. The incoming atoms/molecules are more or less ballistic impacts, fluid dynamics doesn't really enter into it. There is a probably a factor of 2, as for light, but my answer is still the right order of magntude -- a 1 nN thruster will not lift the orbit of anything much bigger than a marble from LEO $\endgroup$ – Steve Linton Jul 27 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.