1. EART162: PLANETARY INTERIORS

2. This Week – Isostasy and Flexure
See Turcotte and Schubert chapter 3
How are loads supported?
Isostasy – zero elastic strength
Flexure – elastic strength (rigidity) is important
What controls rigidity?
We can measure rigidity remotely, and it tells us about a planet’s thermal structure

3. Airy Isostasy (1854)
Let’s start by assuming that the crust and mantle are unable to support loads elastically
Load
Load
Crust rc
Mantle rm
The crust will deflect downwards until the surface load (mass excess) is balanced by a subsurface “root” (mass deficit – dense mantle replaced by light crust)
90% of an iceberg is beneath the surface of the ocean for exactly the same reason
This situation is called (Airy) isostasy
This requires the mantle to flow.

4. Consequences of Isostasy
In the case of no elastic strength, the load is balanced by the mantle root: hrc=r(rm-rc) h r
Crust rc tc r
This also means that there are no lateral variations in pressure beneath the crustal root
Constant pressure
Mantle rm
So crustal thickness contrasts (Dtc=h+r) lead to elevation contrasts (h), with:
or,
Note that the elevation is independent of the background crustal thickness tc

5. Pratt Isostasy (1855)
Similar principle – pressures below some depth do not vary laterally.
Here we explain h as due to variations in density, rather than crustal thickness h tc
Constant pressure
Mantle rm
Is that mountain high because it’s the same density and big, or because it’s really low density? Mid ocean ridges have higher temperature rocks with lower density, which supports topography.
What’s an example of where this mechanism occurs on Earth?

6. Pratt Isostasy (1855)
Similar principle – pressures below some depth do not vary laterally.
Here we explain h as due to variations in density, rather than crustal thickness
r2 > r1 h tc r1 r2
Constant pressure
Mantle rm
Give example of cork and water. Mid ocean ridges have higher temperature rocks with lower density, which supports topography.
What’s an example of where this mechanism occurs on Earth?

7. Gravity Effects
Because there are no lateral variations in pressure beneath a certain depth, that means that the total mass above this depth does not vary laterally either
So what sort of gravity anomalies should we see?
Very small ones!
big gravity
very small gravity – why? (“mass dipole”)
Uncompensated load – Dg=2prcGh
Crust rc
Compensated load – Dg~0
So we can use the size of the gravity anomalies to tell whether or not surface loads are compensated

8. Example - Mars
The southern half of Mars is about 3 km higher than the northern half (why?)
But there is almost no gravity anomaly associated with this “hemispheric dichotomy”
We conclude that the crust of Mars here must be compensated (i.e. weak)
Pratt isostasy? Say r1=2700 kgm-3 (granite) and r2=2900 kgm-3 (basalt). This gives us a crustal thickness of 45 km

9. Mars (cont’d)
On the other hand, some of the big volcanoes (24 km high) have gravity anomalies of mGal
If the volcanoes were sitting on a completely rigid plate, we would expect a gravity anomaly of say 2.9 x 24 x mGal
We conclude that the Martian volcanoes are almost uncompensated, so the crust here is very rigid
Olympus
Ascraeus
Pavonis
Arsia
Uncompensated load – Dg=2prcG
Crust rc
Remember that what’s important is the strength of the crust at the time the load was emplaced – this may explain why different areas have different strengths

10. The Moon
Free air gravity
Topography

11. The Moon Free air gravity Some basins are positive, some are negative.
The farside has moderately low positive gravity and high topography. Mostly compensated?
Topography

12. Flexure
So far we have dealt with two end-member cases: when the lithosphere is completely rigid, and when it has no strength at all (isostasy)
It would obviously be nice to be able to deal with intermediate cases, when a load is only partly supported by the rigidity of the lithosphere
I’m not going to derive the key equation – see the Supplementary Section (and T&S Section 3-9) for the gory details
We will see that surface observations of deformation can be used to infer the rigidity of the lithosphere
Measuring the rigidity is useful because it is controlled by the thermal structure of the subsurface

13. Flexural Stresses
load
Crust
Elastic plate
Mantle
In general, a load will be supported by a combination of elastic stresses and buoyancy forces (due to the different density of crust and mantle) – examples of each
The elastic stresses will be both compressional and extensional (see diagram)
Note that in this example the elastic portion includes both crust and mantle

14. Flexural Equation (1) x
ρw = water density for oceanic case, crustal density for continental case z
q(x)
w(x) rw P P Te rm
P is force per unit length
Elastic plate
Fluid mantle rm
D is the (flexural) rigidity, Te is the elastic thickness

15. Flexural Equation (2) Here the load q(x)=rlgh We’ll also set P=0rm Te
Here the load q(x)=rlgh
We’ll also set P=0
The flexural equation is: w rm
If the plate has no rigidity, D=0 and we get
This is just the expression for Airy isostasy
So if the flexural rigidity is zero, we get isostasy

16. Flexural Equation (3)
What’s this?
Let’s assume a harmonic variation in loading h=h0eikx
Then the response must also be harmonic w=w0eikx
We can relate h0 to w0 as follows
Here Dr=rm-rl and k=2p/l, where l is the wavelength a
D = 0 is simple isostasy. Long wavelengths, low k, more like isostasy. Vice versa.
What happens if D=0?
What happens at very short or very long wavelengths?

17. Degree of Compensation
The deflection caused by a given load:
We also know the deflection in the case of a completely compensated load (D=0): C
wavelength 1
Short l:
Uncompensated
Long l:
Compensated
0.5
Dk4/Drg=1
The degree of compensation C is the ratio of the deflection to the deflection in the compensated case:
Long l, C~1 (compensated); short l, C~0 (uncomp.)
C~1 gives small gravity anomalies, C~0 large ones
Critical wavenumber: C=0.5 means k=(Drg/D)1/4

18. Example
Let’s say the elastic thickness on Venus is 30 km (we’ll use E=100 GPa, v=0.25, g=8.9 ms-2, Dr=500 kg m-3)
The rigidity D=ETe3/12(1-v2) ~ 2x1023 Nm
The critical wavenumber k=(Drg/D)1/4 ~1.3x10-5 m-1
So the critical wavelength l=2p/k=500 km
1000 km
2000 km
3000 km
0.8 km
1.6 km
Would we expect it to be compensated or not?
What kind of gravity anomaly would we expect?

19. Free Air Gravity for the Moon (again)
What is the longest wavelength topography signal?
Where do you expect the gravity to be highest?
Topography
Gravity
From Zuber et al. 2013

20. Crustal thickness – how?

21. Degree of Compensation (2)
Weaker (small Te)
~0 mGal/km 1
Short l:
Uncompensated
Stronger (large Te) C
0.5
Long l:
Compensated
~120 mGal/km
wavelength
What gravity signals are associated with C=1 and C=0?
How would the curves move as Te changes?
So by measuring the ratio of gravity to topography (admittance) as a function of wavelength, we can infer the elastic thickness of the lithosphere remotely

22. Admittance Example Comparison of Hawaii and Ulfrun Regio (Venus)
Nimmo and McKenzie 1998
Comparison of Hawaii and Ulfrun Regio (Venus)

23. Flexural Parameter (1) Consider a line load acting on a plate:rw
Consider a line load acting on a plate:
w(x) w0 Te
x=0 rm x
Except at x=0, load=0 so we can write:
What does this
look like?
Boundary conditions for an unbroken plate are that dw/dx=0 at x=0 and w as
i.e. tangent is zero a
The solution is
Here a is the flexural parameter
(Note the similarity of a to the critical wavenumber)

24. Flexural Parameter (2)
For a line load, the zero crossing xo = 3πα/4, and the forebulge maximum is at xb = πα.
Therefore, α is ~less than the width of deformation.

25. Example: Mariana Trench
In this model , there is a bending moment in addition to the line load, as the plate is subducted
The Pacific Plate is subducted beneath the smaller Mariana Plate that lies to the west.
From T&S

26. Flexural Parameter (2) Flexural parameter a=(4D/g(rm-rw))1/4
It is telling us the natural wavelength of the elastic plate
E.g. if we apply a concentrated load, the wavelength of the deformation will be given by ~a
Large D gives long-wavelength deformation, and v.v.
If the load wavelength is >> ~a, then the deformation will approach the isostatic limit (i.e. C~1)
If the load wavelength is << ~a, then the deformation will be small (C~0) and have a wavelength given by ~a
If we can measure a flexural wavelength, that allows us to infer ~a and thus D or Te directly. This is useful!

27. Example This is an example of a profile across a rift on Ganymede
Note vertical length scale is small
10 km
This is an example of a profile across a rift on Ganymede
An eyeball estimate of ~3a would be about 10 km
For ice, we take E=10 GPa, Dr=900 kg m-3, g=1.3 ms-2
Load
Distance, km
If ~3a = 10 km then Te=6.5 km
So we can determine Te remotely
This is useful because Te is ultimately controlled by the temperature structure of the subsurface

28. Te and temperature structure
Cold materials behave elastically
Warm materials flow in a viscous fashion
This means there is a characteristic temperature (roughly 70% of the melting temperature) which defines the base of the elastic layer
E.g. for ice the base of the elastic layer is at about 190 K
The measured elastic layer thickness is 6.5 km (from previous slide)
So the thermal gradient is 12 K/km
This tells us that the ice shell thickness is ~13 km
What’s wrong with these assumptions?
Surf. Temp.
273 K
190 K
110 K
6.5 km
Liquid water below. (convection changes geotherm).
Depth
elastic
viscous
Temperature
Liquid!

29. Te and age
Small Te
Large Te
Decreasing age
The elastic thickness recorded is the lowest since the episode of deformation
In general, elastic thicknesses get larger with time (why?)
McGovern et al., JGR 2002
So by looking at features of different ages, we can potentially measure how Te, and thus the temperature structure, have varied over time
This is important for understanding planetary evolution

30. Te in the solar system Remote sensing observations give us Te
Te depends on the composition of the material (e.g. ice, rock) and the temperature structure
If we can measure Te, we can determine the temperature structure (or heat flux)
Typical (very approx.) values for solar system objects:
Body
Te (km)
dT/dz (K/km) Te
Earth (cont.) 30 15
Venus (450oC)
Mars (recent)
100 5
Moon (ancient)
Europa 2 40
Ganymede

31. Summary Flexural equation determines how loads are supported:
The flexural parameter a gives us the characteristic wavelength of deformation
Loads with wavelengths >> a are isostatically supported
Loads with wavelengths << a are elastically supported
We can infer a from looking at flexural topography (or by using gravity & topography together – admittance)
Because the rigidity depends on the temperature structure, determining a allows us to determine dT/dz

32. Supplementary Material follows

33. Plate Bending l
Bending an elastic plate produces both compressional and extensional strain
The amount of strain depends on the radius of curvature R
extension Dl y
compression f R R
Strain f
Note that there is no strain along the centre line (y=0)
The resulting stress is given by (see T&S):

34. Radius of Curvature
What is the local radius of curvature of a deforming plate? Useful to know, because that allows us to calculate the local stresses
It can be shown that Dx a
q2-q1 R R
So the bending stresses are given by q2 w A q1
(y is the distance from the mid-plane) x
Plate shape described by w(x)

35. Bending Moment B C Balance torques: V dx = dM
q(x)
(force per unit area)
q(x)
V+dV V
(shear force per unit length)
M+dM M
(moment) dx y
Balance torques: V dx = dM
Balance forces: q dx + dV = 0
Put the two together:
sxx
Te/2 B
Moment: C
Does this make sense?

36. Putting it all together . . .
Putting together A, B and C we end up with a
Here D is the rigidity
Does this equation make sense?

37. Last Week
Elasticity: Young’s modulus = stress / strain (also Poisson’s ratio – what does it do?)
Another important (why?) variable is the bulk modulus, which tells us how much pressure is required to cause a given change in density. Definition?
Flow law describes the relationship between stress and strain rate for geological materials
(Effective) viscosity is stress / strain rate
Viscosity is very temperature-dependent: exp(-Q/RT)