3
$\begingroup$

If the Earth were to be flat, but still rotating with a constant angular frequency, would the Foucault pendulum still show it's precessing motion?

The angular velocity of the precession can be derived to be: $$ \omega = -(\Omega \cos\beta)\mathbf{k} $$ where $\Omega$ is the angular frequency of the earth, $\mathbf{k}$ is the apparent vertical and $\beta$ is the angle between the apparent vertical and the rotation axis.

On a rotating disk, with the same rotational frequency as the Earth, one would expect that the formula for $\omega$ would still be valid, albeit with a different apparent vertical $\mathbf{k}$.

EDIT: Gravity on the disk is assumed to be uniformly perpendicular to the disk, such as most flat-Earth theories suggest.

N.B. I do not believe the Earth to be flat, this question is purely out of interest.

$\endgroup$
  • $\begingroup$ How does gravity work on your flat Earth, is it a homogeneous gravitational acceleration perpendicular to the disk? (Because that is not the field the disk would source itself.) $\endgroup$ – Void Jan 25 '18 at 9:58
  • $\begingroup$ I assumed gravity to be perpendicular to the disk. I made an edit to clarify. $\endgroup$ – Simon Jan 25 '18 at 10:01
  • 1
    $\begingroup$ Think of a horizontal disc rotating on a table underneath a simple pendulum whose point of suspension is fixed relative to the table. $\endgroup$ – Farcher Jan 25 '18 at 10:34
  • $\begingroup$ Do flat-earthers think that the Earth spins like this, or do they assume it tumbles in some way (which would, I suppose, explain the Sun's motion better)? A Foucault pendulum on a tumbling flat Earth would be interesting. $\endgroup$ – tfb Jan 25 '18 at 18:38
  • $\begingroup$ I'm not sure, since I'm not that well informed on flat-earth theory, but i found a couple of posts online where flat earthers 'debunked' the foucault pendulum experiment by saying that the flat earth rotated too. But I agree that a tumbling earth would be interesting.. $\endgroup$ – Simon Jan 26 '18 at 15:02
5
$\begingroup$

Yes, it would work, but $\omega$ would be the same at all points on the disc, and equal to the rate of rotation of the disc itself. So although the pendulum would precess, you could very easily know if you were on a flat disc by moving it around and comparing precession rates,

$\endgroup$
  • 3
    $\begingroup$ For centrifugal forces absolutely negligible with respect to the gravitational force, the above is correct. However, it is not correct when the centrifugal acceleration becomes large enough to change the direction of the apparent vertical. Then there are corrections to the frequency of order $(a_c/a_g)^2 = (\Omega^2 r/g)^2$, so it is a question of how large the disk and the rotation are. $\endgroup$ – Void Jan 25 '18 at 18:23
  • $\begingroup$ @Void: thanks. I think that the normal foucault pendulum computation for a spherical Earth ignores centrifugal forces: is that right? If it is (or in either case) are you happy for me to add your comment to my answer, as it's not really right without it. $\endgroup$ – tfb Jan 25 '18 at 18:35
  • 1
    $\begingroup$ @Void +1. But, on earth, what we commonly call "gravity" is a mix of gravitational and centrifugal forces. That is the only definition under which gravity points "straight down". The same definition could be used on this hypothetical flat disc: the actual gravitational field could point more and more inward the further from the axis of rotation you go, such that "gravity" (which includes centrifugal force) always points down, just as it does on earth. $\endgroup$ – Ben51 Jan 25 '18 at 18:38
-2
$\begingroup$

A rotating flat Earth can't actually exist. The bigger a planet is, the smaller the gravitational constant would have to be for it to have Earth's gravitational field strength and the slower it would have to be rotating for its centrifugal force to be balanced by gravity and be rotating barely slowly enough to form into a stable oblate shape. A flat Earth would be infinitely big and so would necessarily have a zero rotation rate.

Let's suppose it has a nonzero rotation rate and there's also some magic force it exactly balances the centrifugal force in its frame of reference so that the only fictitious force in its frame of reference that's not balanced by and equal and opposite force is the Coriolis force. According to this answer, precession can be caused by something other than the Coriolis force. Under certain conditions, if the Earth were rotating slowly enough, we still wouldn't be able to tell from watching the pendulum that Earth is rotating.

$\endgroup$

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.