How to calculate the thickness of the ice layer of a frozen ocean planet/moon?
How do I calculate the thickness of the upper ice layer on cold ocean worlds like Europa, Enceladus, and Ganymede? I'm asking this for a programm I'm currently writing.
The given/known quantities are:
- Mass
- Radius
- Average density of metal (assume Fe) core, rock (assume MgSiO3) layer, and ice (assume H2O) layer
- Internal heat from tidal heating, radioactive decay and residual formation heat
- External heating from solar radiation (assume equal constant heating distributed over the entire surface or ignore it if it is irrelevant/minuscule)
- Pressure of the atmosphere above (assume pure N2 with variable pressure or no atmosphere)
I want to compute the thickness of the frozen (crust) and molten layer.
Bonus stuff: If easy adaptation of the formula for the calculation of the rocky layers crust thickness is possible I would appreciate it. Should it be possible, an adaptation to find the rocky crust thickness and the ice crust thickness on wolds where both are needed would be helpful.
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1 answer
There are a whole bunch of different ways to determine the thickness of a planet's ice sheets; over the decade, dozens have been tried on Europa and other icy bodies. Broadly speaking, as we're working from a purely theoretical perspective, our method will have to be thermodynamic in nature - we can't look at cratering patterns or rifts in the ice. I'm going to briefly mention a couple of these treatments, all of which (at least in the case of Europa) appear to return similar results within an order of magnitude of each other.
Quick & Marsh 2015
It turns out that L.Dutch's simple approach returns reasonable results. Let $Q_D$ be the heat transferred to Europa by tidal dissipation. We know that $Q_D$ is proportional to the temperature gradient within the ice, and that the gradient can be approximated as $$\frac{dT}{dz}\approx\frac{T_S-T_M}{d}$$ where $T_S$ is the surface temperature, $T_M$ is the melting point of water, and $d$ is the thickness of the ice sheet. We then find that $$d=\frac{k(T_S-T_M)4\pi R^2}{Q_D}$$ where $R$ is the radius of the planet and $k$ is the thermal conductivity. Now, $Q_D$ itself can be calculated from the physical and orbital properties of the planet: $$Q_D=\frac{21}{2}\frac{k_2(\omega R)^5e^3}{Q_oG}$$ where $\omega$ and $e$ are the tidal forcing frequency and orbital eccentricity, and the other parameters are constant. We then see that $$d\propto \frac{T_S-T_M}{\omega^5 R^3e^3}$$
Ojakangas & Stevenson 1989
You can do a more detailed treatment of thermal equilibrium from a materials science perspective, focusing on how ice behaves at different temperatures. From a rheological approach, Ojakangas and Stevenson derived a solution of the form $$d=\frac{\ln(T_M/T_S)}{\left[\left(\frac{2}{a_1}\right)\int_0^{T_M}\frac{q(T)dT}{T}+\left(\frac{H}{a_1}\right)^2\right]^{1/2}}$$ where $q(T)$ is the volumetric dissipation rate and $a_1$ and $H$ are constants. Calculating $q(T)$ may be non-trivial depending on the rheological model you assume.
In general, actually, the linear temperature gradient approach works fairly well. I would strongly recommend using it for your purposes. Here are some general notes, for whatever method you choose:
- You'll need to know the body's physical and orbital properties to compute the heat generated by tidal dissipation.
- The surface temperature $T_S$ can be determined by standard calculations of effective temperature.
- You can use similar treatments to determine the structure of subsurface layers (see Hussmann et al. 2002).
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