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This answer to Does Juno's UVS have any chance to spot Europa plumes? links to Juno OPAG REPORT by Scott Bolton, OPAG, September 2020. The slides detail past and future Juno flybys of Jupiter and some of its moons, and last slide (shown below) says:

  • Juno will constrain $\boldsymbol k_{\boldsymbol 2}$ to help clarify the physical origin of Io’s volcanism
  • Juno will monitor lo volcanic activity, including the polar region. Global mapping addresses where internal dissipation of tidal heating occurs.

Question: What is $k_2$, how does it relate to Io's volcanism and how can Juno constrain its value?


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It is called Tidal love number. The definition is as follows:

In Newtonian gravitational theory, a tidal Love number relates the mass multipole moment created by tidal forces on a spherical body to the applied tidal field. The Love number is dimensionless, and it encodes information about the body's internal structure. (Poisson et.al., 2009)

For Io, the real part of Love number is calculated to be $\mathrm{Re(k_2)=0.09\pm0.02}$, much lower than the predicted value $\mathrm{Re(k_2)\approx0.50}$ (Bierson et. al. 2016).

For more information, refer to these papers:

  • On the degeneracy of the tidal Love number k2 in multi-layer planetary models: application to Saturn and GJ 436b, U. Kramm, N. Nettelmann, R. Redmer and D. J. Stevenson, A&A, Volume 528, April 2011, DOI: 10.1051/0004-6361/201015803
  • Hinderer, T., “Tidal Love Numbers of Neutron Stars”, The Astrophysical Journal, vol. 677, no. 2, pp. 1216–1220, 2008. doi:10.1086/533487.
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  • $\begingroup$ any thoughts on "... and how can Juno constrain its value?" $\endgroup$
    – uhoh
    Sep 27 at 11:05
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    $\begingroup$ $k_2$ is one of many (an infinite number of) tidal Love numbers. There's no need to invoke relativity theory here. Newtonian mechanics work quite nicely with regard to planets and moons. $\endgroup$ Sep 27 at 11:33
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$k_2$ is one of three tidal Love-Shida numbers related to how gravitation of another body (Jupiter in this case) changes a planet-like body's second degree spherical harmonics (Io in this case).

Three Love and Shida numbers exist for each degree of spherical harmonic coefficients. The three Love-Shida numbers for a given degree $n$ are

  • $k_n$, which describes how the other body changes the $n^{th}$ degree spherical harmonic coefficients of the body of interest. This makes the spherical harmonic coefficients time-dependent.
  • $h_n$, which describes how those changes in the $n^{th}$ degree spherical harmonic coefficients affect the height of the body of interest relative to the tide-free shape.
  • $l_n$, which describes how those changes in the $n^{th}$ degree spherical harmonic coefficients result in horizontal shifts of the body of interest relative to the tide-free shape.

A.E.H. Love described the Love numbers $k_n$ and $h_n$ in a 1909 paper to aid in describing the Earth's solid body tides. T. Shida added $l_n$ to the mix a bit later (1912). The zeroth order Love numbers are in a sense baked-in to the spherical harmonics model. The second order Love and Shida numbers are the key numbers of interest as the time-varying changes to the second degree spherical harmonic coefficients strongly dominate over higher degree terms.

I'm not going to attempt to reproduce the math-out equations here. You can find the math-out equations in various papers at iers.org. (The concept was developed to help in explaining the Earth tides, so papers at iers.org make imminent sense.) All of the Love and Shida numbers are unitless. A body made of the perfectly rigid form of unobtanium will have a $k_2$ value of 0, while a body made of the incompressible dust form of unobtanium will have a $k_2$ value of 1.5. Real objects will have a $k_2$ value between 0 and 1.5.

How does it relate to Io's volcanism?

There are multiple models regarding the details of Io's volcanism. Each of these different models results in different plausible range of $k_2$. Estimated values for $k_2$ based on Juno gravity experiments can thus rule out some of these models. By way of analogy, scientists long thought that the Moon and Mars had completely solid cores. Gravity experiments ruled out those concepts. The Moon and Mars have molten or partially molten cores (solid inner cores and a molten outer cores). The observed $k_2$ values from gravity experiments are inconsistent with the solid core hypothesis for those two bodies but are consistent with the molten / partially molten core hypothesis.

How can Juno constrain its value?

Juno by itself cannot do so. Juno's telecommunication system is outfitted with equipment that enables the Deep Space Network to track Juno's range and range rate with extreme precision. (The range rate measurements are the ridiculously precise.) With multiple close fly-bys of Io, those precise measurements give the Juno gravity science team the data needed to estimate Io's second degree spherical harmonics coefficients, and hence Io's $k_2$ value.

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  • $\begingroup$ Unobtanium is a truly remarkable substance that comes in several forms. The rigid forms come in two flavors, massive and massless, both with an infinite speed of sound. The dusty form is highly non-rigid; unobtanium in this form is held together only by gravity. Other gravitating bodies can distort the shape of a dusty unobtanium object, and do so instantaneously, despite the lack of other interactions. The speed of sound in a dusty unobtanium object is once again infinite. $\endgroup$ Sep 27 at 13:11
  • $\begingroup$ Super answer, thank you! One think I haven't been able to figure out; are the spherical harmonics from the expansion of the shape of the body's surface produced by the other body, or the expansion of the resulting gravitation potential of the new mass distribution? $\endgroup$
    – uhoh
    Sep 27 at 13:35
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    $\begingroup$ @uhoh - That's a trick question! The answer depends on how the spherical harmonics model was developed. There are three contenders for how to do it. A mean tide model has both the direct and indirect effects of the zero frequency components due to other bodies baked in. A zero tide model removes the direct effects but leaves the indirect effects. A tide-free model removes both the direct and indirect effects. IERS Technical Note 36 discusses these to some extent. $\endgroup$ Sep 27 at 20:25
  • $\begingroup$ Okay.... I'll make some coffee, strap on my thinking cap and take a look. Thanks! $\endgroup$
    – uhoh
    Sep 27 at 23:38

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