I wrote a quickie answer where I estimate the total rotational angular momentum of Jupiter as about 7E+38 kg m^2/s. However it bothers me because one of the sources gets this number assuming uniform density, and the other two cite 6.9E+38 kg m^2/s but it's not clear to me where this comes from.

Are there better estimates that use reasonable models for Jupiter's density distribution?

update: If not answerable due to uncertainties in density distribution and/or gradients in rotation rate (i.e. non rigid-body rotation) which I sort-of remember reading about recently, then an explanation why a better value isn't available would be an acceptable answer.

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    $\begingroup$ Not sure you can do the $n=3/2$ polytrope analytically. $\endgroup$ – ProfRob Jan 1 '20 at 8:41

A better estimate might be to use the moment of inertia of a $n=3/2$ (fully convective) polytrope, which will be a good approximation in an object like Jupiter, even when it approaches electron degeneracy in its interior. You can look this up and it is given by $I = kMR^2$, with $k=0.205$ (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).

If we use a mass $M$ of $1.898 \times 10^{27}$ kg, the equatorial radius of Jupiter ($R=71500$ km) and the rotation rate determined from its magnetosphere ($P= 9.93$ hours), then the angular momentum ($2\pi I/P$) is $3.496 \times 10^{38}$ kg m$^2$ s$^{-1}$.

If instead, we use the volumetric average radius of $R=69900$ km, this reduces to $3.34\times 10^{38}$ kg m$^2$ s$^{-1}$.

This simple approach neglects the more complicated equation of state of Jupiter, it's non-sphericity and the probable presence of a solid core. Consideration of these, together with constraints provided by the measured even harmonics of the gravitational field (referred to as $J_2$ and $J_4$, led Helled et al. (2011) to suggest that $k=0.264$, with an uncertainty below 1%. Combined with the average radius, this gives $4.30\times 10^{38}$ kg m$^2$ s$^{-1}$.

More recent work by Ni (2018) has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest $k=0.274$ with 0.5% uncertainty.

  • $\begingroup$ I missed the bounty award deadline so have re-bountied. Thanks for the excellent answer! $\endgroup$ – uhoh Jan 1 '20 at 3:53

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