How does one calculate the tidal heating of a satellite?
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I've been searching for hours, and most formulas I can find use complex/imaginary numbers or variables that I don't know or can't find out (such as the imaginary part of the planet's love number, which leads into the complex number thing). What's the "simplest" equation you guys know of?
planets physics hard-science moons
Serban Tanasa
Alexader FergusonAlexader Ferguson
Wikipedia gives the formula for the tidal heating $\dot{E}$ as $$\dot{E}=-\text{Im}(k_2)\frac{21}{2}\frac{R^5n^5e^2}{G}\tag{1}$$ where $R$ is the radius of the satellite, $n$ is something weird called its mean orbital motion, and $e$ is the eccentricity of its orbit. I actually don't like this representation. Another way to rewrite it uses the relation $$\mu=a^3n^2\implies n^5=\left(\frac{Gm_p}{a^3}\right)^{5/2}$$ where $\mu\equiv Gm_p$, with $m_p$ the mass of the planet. Therefore, we find that $$\dot{E}=-\text{Im}(k_2)\frac{21}{2}\frac{G^{3/2}m_p^{5/2}R^5e^2}{a^{15/2}}\tag{2}$$ That's kind of ugly, but it gets rid of $n$, and so all of the other variables are either properties of the moon's orbit, or physical properties of the moon or planet.
Calculating the second Love number
I ignored that $k_2$ - called the second Love number - because it's kind of tricky to calculate. I usually ignore it completely, and substitute in something like $0.02$ or $0.03$ for $\text{Im}(k_2)$ for satellites like our Moon (see 1 and 2). But if you really want to calculate it, go ahead.
My reference is Hussman et al. (2010), specifically, $\text{Eq. }32$: $$k_2=1.5\left(1+\frac{19}{2}\frac{\mu_c}{\rho gR_s}\right)^{-1}$$ for rigidity $\mu_c$, surface gravity $g$ and radius $R_s$. $\mu_c$ can be calculated as $$\text{Re}(\mu_c)=\frac{\eta^2n^2\mu}{\mu^2+\eta^2n^2},\quad\text{Im}(\mu_c)=\frac{\eta n\mu^2}{\mu^2+\eta^2n^2}$$ and $$\mu_c=\text{Re}(\mu_c)+\text{Im}(\mu_c)$$ for elastic rigidity $\mu$, viscosity $\eta$, and mean motion $n$, defined as $2\pi$ divided by the period of the satellite's orbit. $\text{Re}(z)$ and $\text{Im}(z)$ denotes the real and imaginary parts of a complex number. In other words, if $$z=a+bi$$ for real numbers $a$ and $b$, then $$\text{Re}(z)=a,\quad\text{Im}(z)=b,\quad z=\text{Re}(z)+i\text{Im}(z)=a+bi$$
$\mu_c$ is an imaginary number, and therefore so is $k_2$. We can simplify this a bit, though. If we set $$a\equiv\frac{19}{2\rho gR_s}\text{Re}(\mu_c),\quad b\equiv\frac{19}{2\rho gR_s}\text{Im}(\mu_c)$$, then $$k_2=(a+1)\frac{1.5}{(a+1)^2+b^2}-\frac{1.5bi}{(a+1)^2+b^2}$$ and so we have a much better expression for $\text{Im}(k_2)$: $$\text{Im}(k_2)=-\frac{1.5b}{(a+1)^2+b^2}$$ There. I hope that was fun. Again, though - you're much better off just substituting in typical values. $k_2$ has been studied and measured in a lot of detail.
Scaling based on Io
Measurements have done on the relatively significant tidal heating of Io, one of Jupiter's moons. A reasonable value for $\dot{E}$ is $\sim10^{14}$ Watts. We also know additional parameters:
$\text{Im}(k_2)\approx0.015\pm0.003$
$R\approx1800\text{ km}$
$e\approx0.0041$
$a\approx4.2\times10^5\text{ km}$
Therefore, letting $M_J$ be the mass of Jupiter, and plugging in $G^{3/2}$, we find that, assuming a similar internal model as Io, the magnitude of $\dot{E}$ is $$\dot{E}\approx10^{14}\left(\frac{\text{Im}(k_2)}{0.015}\right)\left(\frac{m_p}{M_J}\right)^{5/2}\left(\frac{R}{1800\text{ km}}\right)^5\left(\frac{e}{0.0041}\right)^2\left(\frac{a}{4.2\times10^{15}\text{ km}}\right)^{-15/2}\text{ Watts}\tag{3}$$ which is hopefully easier to work with than $(2)$.
$\begingroup$ ...that doesn't seem very simple :o $\endgroup$ – James♦ Dec 20 '16 at 16:59
$\begingroup$ @James I don't really want to just say, "It's complicated", but . . . it's complicated. $\endgroup$ – HDE 226868♦ Dec 20 '16 at 21:58
$\begingroup$ I have no idea if your equations are correct but they look complex enough to be right. I'll give it a few more days then award you the bounty. $\endgroup$ – Bellerophon Dec 21 '16 at 16:49
$\begingroup$ @Bellerophon Looking back, this was kind of a crappy answer (inaccessible, at least). I've edited it to make it a bit more useful. $\endgroup$ – HDE 226868♦ Jun 26 '18 at 15:15
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