Wikipedia gives the formula for the tidal heating $\dot{E}$ as $$\begin{array}{}\text{(1)}& \dot{E}=-\text{Im}(k_2)\frac{21}{2}\frac{R^5{n}^5{e}^2}{G}\end{array}$$ 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 $$\mu =a^3{n}^2⟹n^5={\left(\frac{G{m}_p}{a^3}\right)}^{5/2}$$ where $\mu \equiv G{m}_p$, with $m_p$ the mass of the planet. Therefore, we find that $$\begin{array}{}\text{(2)}& \dot{E}=-\text{Im}(k_2)\frac{21}{2}\frac{G^{3/2}{m}_p^{5/2}{R}^5{e}^2}{a^{15/2}}\end{array}$$ 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, $\text{Eq. }32$: $$k_2=1.5{(1+\frac{19}{2}\frac{{\mu }_c}{\rho g{R}_s})}^{-1}$$ for rigidity ${\mu }_c$, surface gravity $g$ and radius $R_s$. ${\mu }_c$ can be calculated as $$\text{Re}({\mu }_c)=\frac{{\eta }^2{n}^2\mu }{{\mu }^2+{\eta }^2{n}^2},\text{Im}({\mu }_c)=\frac{\eta n{\mu }^2}{{\mu }^2+{\eta }^2{n}^2}$$ 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 $$\text{Re}(z)=a,\text{Im}(z)=b,z=\text{Re}(z)+i\text{Im}(z)=a+bi$$

${\mu }_c$ is an imaginary number, and therefore so is $k_2$. We can simplify this a bit, though. If we set $$a\equiv \frac{19}{2\rho g{R}_s}\text{Re}({\mu }_c),b\equiv \frac{19}{2\rho g{R}_s}\text{Im}({\mu }_c)$$, 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 $\sim {10}^{14}$ Watts. We also know additional parameters:

• $\text{Im}(k_2)\approx 0.015\pm 0.003$
• $R\approx 1800\text{ km}$
• $e\approx 0.0041$
• $a\approx 4.2\times {10}^5\text{ km}$

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 $$\begin{array}{}\text{(3)}& \dot{E}\approx {10}^{14}\left(\frac{\text{Im}(k_2)}{0.015}\right){\left(\frac{m_p}{M_J}\right)}^{5/2}{\left(\frac{R}{1800\text{ km}}\right)}^5{\left(\frac{e}{0.0041}\right)}^2{\left(\frac{a}{4.2\times {10}^{15}\text{ km}}\right)}^{-15/2}\text{ Watts}\end{array}$$ which is hopefully easier to work with than $(2)$.

posted over 8 years ago

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