estimating Europa's ice-crust thickness

[Widdekind]

Determining the thickness of Europa's ice-crust

 

Even as the below-water height of an earth ice berg, can be determined, from its above-water height (by the ratio of 6/7ths : 1/7th); so, too, the thickness of Europa's ice-crust could be determined, by measurements, of how high that ice rides, up above the water presumably present, in the fissures, in between the ice-crust 'tectonic plates'. For example, if the water, in those fissures, was -1 km below the ice-crust surface, then the ice-crust would be ~7 km thick [modified by the actual ice density ratio, appropriate to Europa's colder climate].

 

 

Melt-water must be left inside the melt-shaft, to equalize pressure, at depth

 

Pressure always increases, with depth. So, if a hypothetical human probe melted through Europa's ice-crust, completely vaporizing the water in its wake, so that its melt-shaft was left empty; then, upon penetrating to the high pressure deep water below, pressures would push the probe, upon a 'geyser' of water, back towards the surface -- specifically, ~6/7ths of the way (since ice is 6/7ths of the density of water [at least at terrestrial temperatures]).

 

Thus, such a probe should leave the melt-water in the melt-shaft, to weigh down upon the probe, and impute compensating pressures.

 

[Widdekind]

The following fails to consider that Europa is tide-locked, into synchronous revolution-rotation, in its orbit around Jupiter. Thus, the gravity field experienced by Europa is spatially non-uniform, which would distort the equilibrium shape of the moon (making it oblate, squashed down at the poles, and pulled apart in the middle). However, a spatially varying field does not, thereafter, induce tides -- only a time-varying field does, by preventing any single moon shape from 'fitting' all the induced tidal forces. So, what actually matters more, is the time-varying 'differential tidal forces', caused by the orbits eccentricity (which, from the moon's perspective, moves Jupiter inward & outward); or, caused by the moon's http://axial tilt of 1 degree (which, from the moon's perspective, moves Jupiter up above, and down below, the moon's mid-plane). The former effect makes all tidal forces felt by the moon, vary by the cube of the ratio of the orbital periapsis to apoapsis (= (1+e)^3). And thus, the differential tidal force (= (1+e)^3 - 1 ~= 3e), for Europa (e = 0.009), is only a few percent, of the value calculated below -- which would imply experienced surface tides, of only a few meters, a little like the ocean tides experienced on earth.

 

Jupiter induces ~100 meter tides on Europa ?

 

Planetary bodies approximately obey a simple scaling relation, between bulk average density, and surface gravity. (Essentially, gravity squeezes, and compresses, the material.) There-from, the change in surface gravity, caused by tidal forces from Jupiter, creates compressions & decompressions, in Europa, which contractions & expansions are experienced as surface tides. From said simple scaling relation, said tides are predicted to be order-of-magnitude 100 meters:

 

[math]\bar{\rho} \equiv \frac{<\rho>}{<\rho>_{\oplus}}[/math]

 

[math]\bar{g} \equiv \frac{g}{g_{\oplus}}[/math]

 

[math]\bar{\rho} \approx 0.29 + 0.44 \times \bar{g}[/math]

 

[math]\therefore \; <\rho> \; \approx \; 0.29 <\rho>_{\oplus} \; + \; \; 0.44 \frac{<\rho>_{\oplus}}{g_{\oplus}} g[/math]

 

[math]\delta <\rho> \; \approx \; 0.44 \frac{3}{4 \pi G R_{\oplus}} \delta g [/math]

 

[math]\delta <\rho> \; \equiv \; \delta \left( \frac{M}{V} \right) \; = \; - \frac{M}{V^2} \delta V \; = \; - <\rho> \frac{\delta V}{V} \; = \; -<\rho> \frac{3 \delta R}{R}[/math]

 

[math]\therefore \; - < \rho > \frac{3 \delta R}{R} \; \approx \; 0.44 \frac{3}{4 \pi G R_{\oplus}} \delta g [/math]

 

[math]\delta R \; \approx \; - \frac{0.44}{3} \left( \frac{R}{R_{\oplus}} \right) \left( \frac{\delta g}{g} \right) \times R [/math]

 

[math]\delta g \equiv \frac{2 G M_J}{D^3} \times (2 R)[/math]

 

[math]\therefore \; \frac{\delta g}{g} \; = \; 4 \left( \frac{M_J}{m} \right) \left( \frac{R}{D} \right)^3[/math]

 

[math]\therefore \; \delta R \; \approx \; - 0.59 \left( \frac{M_J}{m} \right) \left( \frac{R}{D} \right)^3 \left( \frac{R}{R_{\oplus}} \right) \times R[/math]

 

[math]\therefore \delta R \; \approx \; \pm 110 \, m[/math]

 

Europa's pack-ice produces Pressure Ridges 300 meters high; such 'sails' suggest underlying 'keels' ~10x as thick (~3 km)

 

On Europa, pressure-induced ridges, up to 300 meters high, form at the boundaries, between ice-crust plates. And, on earth, at earth arctic temperatures, these pressure-ridge 'sails' overly downward projecting 'keels', 7-8x as deep, below the water, as the sail is high, above the water. Assuming Europan ice, being colder, is denser, such 'sails' of 300 meters, probably imply corresponding keels 10x as deep, ~3 km (not including the intervening ice-crust thickness itself).

 

Moreover, the largest Europan ice-crust plates are ~30 km across. And, aspect ratios (width : thickness) of ~5 could be common. Such would suggest, that Europa's ice crust could be 30 km / 5 = 6 km thick. If so, Europa's ice-crust is 'glacially thick' (~3-6 km, or several miles).

 

 

 

estimate ice-crust thickness from heat flows ??

 

Europa's expected equilibrium surface temperature is ~90 K:

 

[math](1 - a) \; \pi R^2 \; \frac{L_{\odot}}{4 \pi D_J^2} \; \equiv \; \left( 4 \pi R^2 \right) \; \sigma T^4[/math]

 

[math]\therefore \; \frac{(1 - a)}{\sigma} \; \frac{L_{\odot}}{16 \pi D_J^2} \; \equiv \; T^4[/math]

 

[math]\therefore \; T \approx 93 K[/math]

Europa's actual surface temperatures vary, pole-to-equator, from 50-110 K.

Edited April 18, 2011 by Widdekind

[Widdekind]

If Europa is colder at the poles (55K) than the equator (110K), the ice would be thicker at the poles -- so best to do the drilling/melting, at the equator (at moon noon-day) ?

[Widdekind]

Europa 'ice rafts', near moon equator, jut 300 meters above frozen 'water line', implying 4 km submerged below

 

According to NASA, the resolution, of the Europa Ice Rafts image, is 54 meters per pixel. And, the 'ice cliffs', of the presumed 'ice rafts', which just up above the presumed '(frozen) water line', are roughly 3 pixels wide. Accounting for the viewing angle, therefore, those 'ice cliff edges' are probably about 300 meters high.

 

Now, water ice, at roughly 90 K, has a density of 0.934 g/cm3. And, the ratio of (1-0.934) : 0.934 is roughly 1:14. Doesn't that mean, that there would be about 14x as much ice 'below the water line', than above it, on Europa?

 

And therefore, those ice rafts, when they were rafts, were probably about 300 meters x 14 = 4 km thick. Since then, the surface has obviously frozen over, enclosing the rafts in more ice. Ipso facto, that surface is colder, now, than whenever those rafts were rafts. Er go, that 4 km thickness is a 'minimum thickness when warm' figure.

 

This writer concludes, therefore, that that ice formation, near Europa's equator -- "9.4 degrees north latitude, 274 degrees west longitude" -- is, currently, >4 km thick.