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?


1 Answer 1


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:

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 $$\dot{E}\approx10^{14}\left(\frac{\text{Im}(k_2)}{0.015}\right)\left(\frac{m_p}{M_J}\right)^{5/2}\left(\frac{R}{1800\text{ km}}\right)^5\left(\frac{e}{0.0041}\right)^2\left(\frac{a}{4.2\times10^{5}\text{ km}}\right)^{-15/2}\text{ Watts}\tag{3}$$ which is hopefully easier to work with than $(2)$.

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    $\begingroup$ ...that doesn't seem very simple :o $\endgroup$
    – James
    Commented Dec 20, 2016 at 16:59
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    $\begingroup$ @James I don't really want to just say, "It's complicated", but . . . it's complicated. $\endgroup$
    – HDE 226868
    Commented Dec 20, 2016 at 21:58
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    $\begingroup$ 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. $\endgroup$ Commented Dec 21, 2016 at 16:49
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    $\begingroup$ @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
    Commented Jun 26, 2018 at 15:15
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    $\begingroup$ @KEY_ABRADE You're absolutely right - thanks for the fix! $\endgroup$
    – HDE 226868
    Commented Jul 18, 2022 at 17:47

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