<p>A better estimate might be to use the moment of inertia of a <span class="math-container">$n=3/2$</span> (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 <a href="https://ocw.mit.edu/courses/physics/8-282j-introduction-to-astronomy-spring-2006/assignments/compset2_soln.pdf" rel="nofollow noreferrer">look this up</a> and it is given by <span class="math-container">$I = kMR^2$</span>, with <span class="math-container">$k=0.205$</span> (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).</p>

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

<p>If instead, we use the volumetric average radius of <span class="math-container">$R=69900$</span> km, this reduces to <span class="math-container">$3.34\times 10^{38}$</span> kg m<span class="math-container">$^2$</span> s<span class="math-container">$^{-1}$</span>.</p>

<p>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 <span class="math-container">$J_2$</span> and <span class="math-container">$J_4$</span>, led <a href="https://arxiv.org/abs/1109.1627" rel="nofollow noreferrer">Helled et al. (2011)</a> to suggest that <span class="math-container">$k=0.264$</span>, with an uncertainty below 1%. Combined with the average radius, this gives <span class="math-container">$4.30\times 10^{38}$</span> kg m<span class="math-container">$^2$</span> s<span class="math-container">$^{-1}$</span>. </p>

<p>More recent work by <a href="https://www.aanda.org/articles/aa/abs/2018/05/aa32183-17/aa32183-17.html" rel="nofollow noreferrer">Ni (2018)</a> has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest <span class="math-container">$k=0.274$</span> with 0.5% uncertainty.</p>

CC BY-SA 4.0
2020-01-01T03:53:23.600Z
I missed the bounty award deadline so have re-bountied. Thanks for the excellent answer!
uhoh
CC BY-SA 4.0
2019-12-24T08:23:53.887Z
2019-12-24T08:48:42.653Z
2019-12-24T08:48:42.653Z
ProfRob
<p>I wrote a <a href="https://space.stackexchange.com/a/29697/12102">quickie answer</a> 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.</p>

<p>Are there better estimates that use reasonable models for Jupiter's density distribution?</p>

<p><strong>update:</strong> 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.</p>

CC BY-SA 4.0
2020-01-01T08:41:17.163Z
Not sure you can do the $n=3/2$ polytrope analytically.
ProfRob
<jupiter><planetary-science>
Total rotational angular momentum estimates for Jupiter?
uhoh
# Total rotational angular momentum estimates for Jupiter?

(Asked by: username_0 on 2018-07-26)

I wrote a [quickie answer](https://space.stackexchange.com/a/29697/12102) 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.

**Tags:** jupiter, planetary-science
> **Comments:**
> **username_1**: Not sure you can do the $n=3/2$ polytrope analytically.  

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## 1 answers

### Answer by username_1 on 2019-12-24. Score: 2 | Accepted answer

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](https://ocw.mit.edu/courses/physics/8-282j-introduction-to-astronomy-spring-2006/assignments/compset2_soln.pdf) 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)](https://arxiv.org/abs/1109.1627) 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)](https://www.aanda.org/articles/aa/abs/2018/05/aa32183-17/aa32183-17.html) 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.

> **Comments:**
> **username_0**: I missed the bounty award deadline so have re-bountied. Thanks for the excellent answer!  

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