How to calculate the thickness of the ice layer of a frozen ocean planet/moon?

Question

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.

Answer 1

For determining the ice layer thickness you can use a simplified 1D model, as follows:

• You want the ice to be present above a layer of water. This sets the temperature $T_b$ at the ice-water interface. This would be 0 Celsius for pure water a 1 bar.
• You know the thermal flow coming from the core, $Q$
• you want the condition to be stationary, this means that the flow is entirely dissipated to the outer environment.

Therefore you have that $Q=$$\lambda \cdot A \cdot (T_b - T_a)\over d$, where $d$ is the thickness of ice layer, $\lambda$ is the thermal conductivity of ice, $A$ the surface and $T_a$ the surface temperature.

If you have the possibility of setting the other interface temperatures as well, you can use the same equation to determine the thicknesses of all the layers.

Mind that this model can be used for a spherical shell as long as the thickness is small with respect to the radius of the shell. Else a spherical model would be more appropriate.

In that case it would be, assuming only radial flux, $Q=4 \pi \lambda R_a \cdot R_b$$T_a - T_b \over R_b - R_a$

Answer 2

While I'm sure there's hard-set rules, and someone with a degree in astrophysics may answer this question more thoroughly, I think you'll find that theres an incredible amount of factors to something like this.

Take Europa for example.

We estimate anywhere between 19 to 25 kilometers of ice, but mind you. this is an estimate based on a bunch of data we have compiled over time, and even then, the estimate is only down to 6km, which, all-in-all is a big distance (not on an astrophysical scale, but you get what I mean).

We assume there's quite a bit of Liquid water under Europa's surface because of observed ejection of liquid and also how impact craters look/are formed. We also know that if there is liquid water, it is kept liquid simply due to Jupiter's gravitation [citation needed]. We can measure its size and can infer to how much water there is around it, regardless of state. But even in the real world, we come up short answering hard-set questions. We don't know how much not-water the inside of the moon is.

We also have evidence that the ice sheet around Europa might have been different thicknesses throughout its life.

And then we still haven't talked about tectonics or other geological events.

My advice? Let the user of your app, be it you, or anyone else, define themselves, how much ice they want, it's probably better that way, and if someone's worlds have different rules, then your app can accommodate to that by not setting any hard-set rules.

Interested to see what the app looks like/what it will be able to do though!

Answer 3

The layer under ice is the layer of possibly dirty water. Really, if you have tidal and residual warmth and temperature on the surface, "all" you need is "merely" the thermal conductivity of the ice and that of the water layer. The problem is, that the thermal conductivity of water is defined by convection cells structure and is unknown. And it cannot be known without inner observations. You cannot guess it theoretically. The contemporary science cannot predict the behaviour of the turbulent flows. Not only counting by formulae is impossible yet, but even the computer modelling. It is simply one of the unsolved tasks.

Edit. The situation is even worse. Here (https://en.wikipedia.org/wiki/Turbulence_modeling) you can see the maximum where the recent science reached. Only statistical evaluation of the evolution of flows. So, maximally you can guess how the structure of flows could evolve. And that structure is dynamic, and because of that the thickness of the ice layer will dynamically change, and again it will influence that turbulence structure. So, you cannot say what the thickness is, because it simply changes.

Using the equations from the reference, you could guess the dynamical range of the changes. But, I am afraid, that is a great work for a doctor of Sc. degree.

Answer 4

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.

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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).