11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I honestly think this might be more suited to Astronomy's site. You might find someone here who can answer it, but it seems to be a real-world question, not about worldbuilding. If you can show us how it relates to your world that might at least stop the question getting closed. $\endgroup$ – Tantalus' touch. Mar 25 '19 at 18:33
  • $\begingroup$ @Agrajag It appears to me that the question is specifically about building planets (worlds). I don't think it could be any more on topic. $\endgroup$ – Arkenstein XII Mar 25 '19 at 19:49
  • $\begingroup$ Are you looking for the dynamic or the steady state equation? $\endgroup$ – L.Dutch - Reinstate Monica Mar 25 '19 at 19:52
  • $\begingroup$ @ArkensteinXII Well not about a world at least, or is the "How do I balance 15 variables I'm not going to define, all of which have an undefined error bar?" sort of question on topic, d'y' rekon? $\endgroup$ – Tantalus' touch. Mar 25 '19 at 19:53
  • 1
    $\begingroup$ I usually do the first one when I move questions, although this time there's no need to delete the original. $\endgroup$ – SealBoi Mar 25 '19 at 21:51
2
$\begingroup$

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.

schematic of the 1D model

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$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Not knowing the thermal flow from the core, you cannot evaluate the thickness of ice, because for its counting you need thermal flow through the lower and upper surfaces, (you know only one of them), and their temperatures (about 0 and the measured one). $\endgroup$ – Gangnus Mar 28 '19 at 8:06
3
$\begingroup$

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!

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thes goal is to be realistic. I'm not willing to go down the quick-and-drity approximation road just yet. $\endgroup$ – TheDyingOfLight Mar 25 '19 at 20:45
  • 1
    $\begingroup$ That's fair, and I admire you wanting to do this as realistic as possible. I'd echo the other commenters and say to post it in Astronomy instead. You can either delete the question here or ask to close it. $\endgroup$ – Cristian C. Mar 25 '19 at 20:57
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The question is tagged hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. $\endgroup$ – L.Dutch - Reinstate Monica Mar 26 '19 at 6:29
  • 1
    $\begingroup$ @L.Dutch I think the answer "there are no known equations to describe this" is a valid one in this case. After all, it's hard to provide hard science for a topic of which there is no science (yet). $\endgroup$ – Rekesoft Mar 26 '19 at 9:30
  • $\begingroup$ @Rekesoft, that might be true for the liquid layer. However, OP is asking for the thickness of the solid layer, and also the statement "this is not possible" should be backed up by appropriate citation/reference. Mind, I am not discussing the content, but the form, due to the tag involved. $\endgroup$ – L.Dutch - Reinstate Monica Mar 28 '19 at 7:32
  • $\begingroup$ @L.Dutch I had changed the question, so, formally your requirement is fulfilled. But really, the older variant of the post, equal to the first paragraph, did not need reference, for the information mentioned there belongs to the middle school program. We don't support 2*2=4 by references. It is enough widely known information to be checked by public knowledge. $\endgroup$ – Gangnus Mar 28 '19 at 8:00
2
$\begingroup$

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).
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.