The Earth's axis is 23.5 degrees away from othogonality to the ecliptic, and it takes about 26 000 years for it to precess fully. I have neither an intuitive sense nor the formula for precession frequency in front of me for how the latter depends on the former (and other things). So...

  1. What is the general formula for gyroscopic precession? I remember working it out as a function of moment of inertia and torque, but it was fiddly and grad mechanics was seven years ago.

  2. What is the formula substituting gravity for the force and a reasonable moment of inertia for a planet?

  3. Is a uniform density sphere a good enough approximation for B?

  4. Do you know of any empirical measurements of any of the above, for any exoplanets? Particularly, any really fast precessors?

  5. Any bigger conclusions we can draw from the above? See Is it easier to learn more about the seasonal changes in an exoplanet's atmosphere when the exoplanet orbits a binary star system?

  • 1
    $\begingroup$ The precession arises from the asymmetric pull of tidal forces on the equatorial bulge. Tidal torque is linear in primary mass (actual for the Earth the moon contributes at the same order of magnitude as the sun), inverse cubic in distance, goes by sine of the tilt, and I never worked out exactly how it scales with oblateness. The torque works to change the angular momentum of the planet which is linear in mass and angular velocity and square in planetary radius. So the period of precession should go $P \propto R^3/(m M \omega r^2\sin\theta)$. But the details are messy. $\endgroup$ Commented Sep 16, 2011 at 17:52

1 Answer 1


The Detectability of Transit Depth Variations Due to Exoplanetary Oblateness and Spin Precession provides a partial answer to your question. They provide the spin procession period for a circular orbit in useful units for talking about gas giant exoplanets:

$\theta$ = axial tilt (obliquity)

$C = \frac{I}{M_p R_{eq}^2}$, where $I$ is the moment of intertia

$P_{orb}$ = orbital period

$P_{rot}$ = rotational period

$P_{prec} = \left(\frac{13.3\;yr}{cos(\theta)}\right) \left(\frac{C / J_2}{13.5}\right) \left(\frac{P_{orb}}{15\; days}\right)^2\left(\frac{10\; hr}{P_{rot}}\right)$

The factor of 13.5 in the denominator of the second expression on the right hand side of the precession period is that of Saturn. So, if you put in all of the fiducial values, you get a precession period of 13.3 years over the cosine of the axial title. For shorter orbital periods, you get shorter precession periods, but you also get faster tidal spin-orbit synchronization. So they say the optimal exoplanets to look for this are in hot Jupiters with periods are of order 15-30 days. They will not necessarily have undergone tidal spin-orbit syncronization yet, but could have precession period on the order of decades.

This has not yet been observed, so far as I can tell (and I am reasonably confident about that). But, the cool thing is that the authors say in the abstract:

The detectability of the TδV signal would be enhanced by moons (which would decrease the precession period) or planetary rings (which would increase the amplitude)

They say later on that that may introduce some ambiguity in analysis, but more detailed analysis may allow for the presence rings and moons to be inferred around short period gas giants, which would be pretty cool. Of course, they also assumed circular orbits. Eccentric orbits will experience other sources of precession, and slightly modify that spin precession. So, all in all, it is pretty complicated, but there is the prospect for observing some of these effects on human timescales with all of the exoplanets currently being found.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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