I am interested in knowing how many newton's of force is needed to keep 1 kg of mass hovering above the moon's surface.

And whether the current most powerful ion thrusters would be up to the task? I believe ESA tested an ion thruster recently which achieved 100 kilo newton's. (I massively mis-remembered this part. I was thinking of the X3 thruster, and the figure is 5.4 Newtons.)

Thank you in advance!

  • $\begingroup$ About 16,6% of the earth. $\endgroup$ Jun 4 '19 at 15:51
  • Look up the surface gravity of the moon. It'll be in units of meters per second squared.
  • Look at the definition of the Newton.
  • Use those two things to determine the thrust needed to keep 1kg of mass aloft
  • Look up the specifications for current ion thrusters. Pay attention to mass vs. thrust ratio, or maximum acceleration.
  • Do some math.

I think you'll find that even current ion thrusters have significantly less acceleration available than the surface acceleration of the moon. Ion thrusters are mass efficient, and they're a good candidate for interplanetary travel because small continuous acceleration will get the job done as well as short bursts of big acceleration. But takeoff & landing will still require chemical rockets.


You don't need any energy or thrust, if your 1kg mass is already in orbit.

Basically an object orbiting the moon relatively near the surface, say one hundred miles, stays on orbit with no energy needed and rotates with a speed of v = SQRT(G*M/R).

In this case approximately 1.022 km/s.

Lifting up some mass and placing it on mons orbit is a different story! you need to calculate the difference between the potential energy of the mass on the moon and in orbit altitude

$E1-E2 = -GMm*((1/R)-(1/(R+d))$

R= moon radius and d is the orbital altitude.

  • 1
    $\begingroup$ But orbiting isn't the same as hovering. $\endgroup$ Jun 5 '19 at 1:15
  • $\begingroup$ I think “hovering” implies maintaining its position relative to the moon’s surface. $\endgroup$
    – Eric S
    Jun 6 '19 at 1:19
  • $\begingroup$ I may have understood hovering in it's wider sense. Otherwise as we know the surface gravity of the moon is roughly 16% of the earth and any thing to stay afloat need that much thrust. $\endgroup$
    – kamran
    Jun 6 '19 at 1:24

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