I wonder whether the Moon and the Jovian moons spin fast enough for a Foucault pendulum to swing on their poles. For instance, I think that it wouldn't swing on Mercury and Venus because these planets rotate too slow.

I think the pendulum does work on Io. Io's rotation period is about 42 hours (or about 1.75 Earth days). And what about Europa and the even-slower-rotating moons?

Perhaps Artemis 3 should take a Foucault pendulum with them and test it on the Moon's south pole.


2 Answers 2



As long as the pendulum can oscillate reliably, the Foucault Pendulum rotation is independent of the strength of gravity.

If your question is about a normal pendulum and you didn't actually mean to talk about the Foucault Pendulum phenomenon, then a normal pendulum will still work even in very weak gravity in a reasonable way, as long as the length of the pendulum is significantly shorter than the diameter of the body and the body is at least roughly spherical and you're at the surface.


Based on comments under @Uwe's answer such as

I'm not asking "how must it be build to work on x planet" but whether it can actually oscillate on extremely-slowly-rotating planets like Venus.


Mercury or the Moon don't have significant atmospheres, therefore there's no friction. Isn't the duration of oscillations also dependent on how strong the gravity is?

I'll elaborate.

The period of one swing of a pendulum is

$$T = 2 \pi \sqrt{\frac{L}{g}}$$

where $L$ is the length of a pendulum and $g$ is the gravitational acceleration. Flip it around and get

$$L = \frac{g T^2}{4 \pi^2}$$

If $T$ is 10 seconds then $L$ is about 25 meters on Earth and somewhat shorter on bodies with smaller surface gravity.

This is almost completely unrelated to the rotation rate of any reasonable body.

The Foucault Pendulum phenomenon is related to the rate at which it very slowly precesses or the plane that the pendulum oscillates seems to slowly rotate relative to the ground.

Imagine you are at the South pole and you set up a really well protected, lossless pendulum. The plane of its oscillations will appear to rotate once a day. It's an apparent rotation, not an "oscillation".

It's not real though, if you viewed from space you'd realized it's really the Earth rotating underneath the pendulum, and the pendulum's plane isn't rotating at all.

If you move to a lower latitude, now the plane will start to rotate a bit even seen from space. If you are on the equator the plane rotates once a day (assuming it's North-South) but from the planet it won't be seen to rotate at all.


The rate of rotation of a Foucault Pendulum as seen on a planet is always some fraction between 0 and 1 times the planet's rotation, and has nothing to do with the strength of the gravity.

As long as the pendulum can oscillate reliably, the rotation is independent of gravity.

  • $\begingroup$ Great explanation. $\endgroup$ Commented Jul 2, 2020 at 15:15
  • $\begingroup$ I'm talking about Focault pendulums of course. You described how it works (I didn't ask for the entire description but thank you regardless) and that it's independent of gravity and rotation but are there really no limits (other than the length of the pendulum relative to the body)? Also, if I understand you well, a pendulum would work well e.g. on the asteroid Lutetia (having 0.003g) if it's smaller than Lutetia's diameter? $\endgroup$ Commented Jul 2, 2020 at 15:20
  • $\begingroup$ @LoveForChrist I explains this much to make sure that a pendulums "oscillations" (what you wrote in your first sentence) were considered separately from the apparent rotation. The shape of asteroid 21 Lutetia might be a problem because it deviates a lot from a sphere. With such an uneven gravity field a pendulums motion might be complicated and start rotating simply due to gravitational perturbations so that would complicate a test of the Foucault pendulum phenomenon, but the pendulum itself would certainly work and oscillate. $\endgroup$
    – uhoh
    Commented Jul 2, 2020 at 23:10
  • $\begingroup$ @uhoh What's wrong with the term "oscillation"? I changed it to swing, that's what I meant. $\endgroup$ Commented Jul 3, 2020 at 4:32
  • $\begingroup$ @LoveForChrist all pendulums swing or oscillate, that's got nothing to to with the effect that a Foucault pendulum demonstrates, which is rotation. Sorry, this has strayed too far from Space Exploration. Questions about the physics of pendulums can be posted in Physics SE. $\endgroup$
    – uhoh
    Commented Jul 3, 2020 at 4:35

A very long Foucault pendulum to show the rotation of Earth should not have any preferred direction of swing. The pendulum weight should be heavy to store enough kinetic energy for a long activity starting with a large amplitude.

Foucault made his most famous pendulum when he suspended a 28-kilogram (62 lb) brass-coated lead bob with a 67-metre long (220 ft) wire from the dome of the Panthéon, Paris.

Besides air resistance, the other main engineering problem in creating a 1-meter Foucault pendulum nowadays is said to be ensuring there is no preferred direction of swing.

From Wikipedia.

So if the Foucault is designed for long oscillation, it will work on Venus, Mercury, Io, Europa and the Moon too. A vacuum will be helpful to remove atmospheric friction. There should be no influence of winds.

But I doubt Artemis 3 may transport a very long pendulum with its mounting to the Moon.

  • $\begingroup$ @uhoh "Besides air resistance, the other main engineering problem in creating a 1-meter Foucault pendulum nowadays is said to be ensuring there is no preferred direction of swing." From wikipedia. So a long pendulum is helpful to avoid a preferred direction of swing. $\endgroup$
    – Uwe
    Commented Jul 2, 2020 at 12:39
  • $\begingroup$ I never thought of that but of course it is right. These would be important points to add to your answer! $\endgroup$
    – uhoh
    Commented Jul 2, 2020 at 12:50
  • $\begingroup$ It would work even on Mercury and Venus? There must be some limit in the oscillation of the pendulum or not? Would a Focault pendulum also swing on a planet whose rotation period is a thousand days? Also, the moons in my question either have no atmosphere or a very thin one (Io) so that doesn't matter much I'd say. :-) $\endgroup$ Commented Jul 2, 2020 at 13:00
  • $\begingroup$ torsional pendulums are a lot easier to build but of course they won't show this effect $\endgroup$
    – uhoh
    Commented Jul 2, 2020 at 13:00
  • 2
    $\begingroup$ @LoveForChrist this is an engineering question about an engineering challenge. I think you are going off-topic now, the answer to all of those is "it depends on how much money you spend and how carefully you design it" and have nothing to do with Space Exploration. $\endgroup$
    – uhoh
    Commented Jul 2, 2020 at 13:01

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