EART 160: Planetary Science

06 February 2008

Last Time

• Planetary Surfaces– Summary

• Planetary Interiors– Terrestrial Planets and Icy Satellites– Structure and Composition: What all is inside?– Exploration Geophysics: How can we tell?

Today

• Homework 3 graded• Projects – Have you got a topic yet?• Midterm Friday! – details• Paper Discussion: Stevenson (2001)

– Mars Magnetic Field

• Planetary Interiors– Pressure and Temperature– Heat Sources and Cooling Mechanisms– Rheology

Histogram

0

1

2

3

40 5 10 15 20 25 30 35 40 45 50

Mor

e

Bin

Freq. Mean: 35

St. Dev.: 8

Homework Issues

• Please talk to me if you have difficulties– Before class is not usually a good time– No more Monday due dates

• Units: – Pressure: 1 Pa = 1N m-2 = 1 kg m-1 s-2

– Energy: 1 J = 1 kg m2 s-2

– Power: 1 W = 1 J s-1

• Stress– Tectonic stress is not the Lithostatic Pressure– Normal stress is the Pressure + normal component of Tectonic– Shear stress is the tangential component of Tectonic

Midterm Exam

• Closed-book– I will provide a formula sheet– You may provide an 8.5” × 11” sheet of paper with

whatever you want on it; hand it in with your test.– Formulae won’t help you if you don’t understand

them!

• Several short-answer questions, descriptive• 3 quantitative problems, pick 2 to answer

– Similar to Homework, but less involved– Show your work!

• Review Session? What say ye?

Exam Topics

• Orbital Mechanics– Kepler’s Laws, Newton’s Laws– Conservation of Energy, Momentum, Angular Momentum– Escape Velocity

• Solar System Formation– Composition of the Solar Nebula– Jeans Collapse– Accretion and Runaway Growth– Frost Line

• Meteorites and Asteroids– Chondrites: Remnants from Early Solar System– Role of collisions– Radiometric Dating

Impacts

• Crater size depends on impactor size, impact velocity, surface gravity

• Crater morphology changes with increasing size• Simple vs. complex crater vs. impact basin• Depth:diameter ratio• Crater size-frequency distribution can be used to

date planetary surfaces• Energetics, Global effects due to impacts• Atmospheres and geological processes can

affect size-frequency distributions

Volcanism

• Solidus & liquidus• Magmatism when solidus crosses adiabat

– Higher temperatures, reduced pressure or lowered solidus

• Volcanism when buoyant magma erupts

• Conductive cooling time t = d2/• Magma composition controls style of volcanism• Flow controlled by viscosity

– Viscous materials = ddt

Tectonics

• Planetary cooling leads to compression

• Hooke’s law and Young’s modulus– Elastic materials = E

• Contraction and cooling

• Byerlee’s law

• Styles of tectonicsm: compression, extension, shear

Gradation

• Erosion on planets with atmospheres– Aeolian, Fluvial, Glacial

• Mass Wasting, Sputtering everywhere.

• Valley networks, gullies and outflow channels

Planets are like OgresCompositional Layers1. Core: Metal2. Mantle: Dense silicate rock

(peridotite)3. Crust: thin silicate rock

(basalt)4. Ocean: liquid layer5. Atmosphere: gas layer

Mechanical Layers

1. Inner Core: solid metal

2. Outer Core: liquid metal

3. Lower Mantle: High viscosity silicate

4. Aesthenosphere: ductile upper mantle

5. Lithosphere: Brittle uppermost mantle and crust

ON an icy satellite, the ocean will be beneath the icy mantle.

Other ice phases are denser than water. May have ice – ocean -- ice

MercuryVenus Earth

Moon

Mars

Ganymede

Io

Actual Planetary Interiors

Only Earth has an layered coreThe Moon has a TINY core (why?)

Icy satellites may have liquid oceansbeneath the ice shellHigh-Pressure Ices beneath that.

Interior of Europa -- NASA

Stevenson et al., 2001 Nature

Pressures inside planets• Hydrostatic assumption

(planet has no strength) gdr

dP

R

rgrGrg 03

4)(

2

222

2

2

0 13

21

2

1)(

R

rR

G

R

rRgrP

• For a planet of constant density (is this reasonable?)

• So the central pressure of a planet increases as the square of its radius

• Moon: R=1800km, P=7.2 GPa

• Mars: R=3400km, P=26 GPa

Pressures inside planets• The pressure inside a planet controls how materials behave• E.g. porosity gets removed by material compacting and flowing,

at pressures ~ few MPa• The pressure required to cause a material’s density to change

significantly depends on the bulk modulus of that material

K

dPd

The bulk modulus K controls the

change in density (or volume) due to a change in pressure

• Typical bulk modulus for silicates is ~100 GPa• Pressure near base of mantle on Earth is ~100 GPa• So change in density from surface to base of mantle should be

roughly a factor of 2 (ignoring phase changes)

Real planets

• Notice the increase in mantle density with depth – is it a smooth curve?

• How does gravity vary within the planet?

Phase Transitions

• Under pressure, minerals transform to different crystal structure

• How do we detect this?• Transition zone can

sore a LOT of water!• How do the depths

change on other planets?

Temperature

• Planets generally start out hot (see below)• But their surfaces (in the absence of an

atmosphere) tend to cool very rapidly • So a temperature gradient exists between the

planet’s interior and surface• We can get some information on this gradient by

measuring the elastic thickness, Te

• The temperature gradient means that the planet will tend to cool down with time

Heat Sources

• Accretion and Differentiation– U = Eacc

– Eacc = m Cp T– Cp: specific heat

• Radioactive Decay– E = H m– H ~ 5x10-12 W kg-1

– K, U, Th today – Al, Fe early on

• Tidal Heating in some satellites

Specific Heat Capacity Cp• The specific heat capacity Cp tells us how much energy

needs to be added/subtracted to 1 kg of material to make its temperature increase/decrease by 1K

• Energy = mass x specific heat capacity x temp. change

• Units: J kg-1 K-1

• Typical values: rock 1200 J kg-1 K-1 , ice 4200 J kg-1 K-1

• E.g. if the temperature gradient near the Earth’s surface is 25 K/km, how fast is the Earth cooling down on average? (about 170 K/Gyr)

• Why is this estimate a bit too large?– Atmosphere insulates

TmCE p

Energy of Accretion• Let’s assume that a planet is built up like an onion,

one shell at a time. How much energy is involved in putting the planet together?

early later

In which situation is more energy delivered?

Total accretional energy = R

GM 2

5

3

If all this energy goes into heat*, what is the resulting temperature change?

RC

GMT

p5

3

Earth M=6x1024 kg R=6400km so T=30,000KMars M=6x1023 kg R=3400km so T=6,000KWhat do we conclude from this exercise?

* Is this a reasonable assumption?

If accretion occurs by lots of small impacts, a lot of the energy may be lost to space

If accretion occurs by a few big impacts, all the energy will be deposited in the planet’s interior

So the rate and style of accretion (big vs. small impacts) is important, as well as how big the planet ends up

Cooling a planet• Large silicate planets (Earth,

Venus) probably started out molten – magma ocean

• Magma ocean may have been helped by thick early atmosphere (high surface temperatures)

• Once atmosphere dissipated, surface will have cooled rapidly and formed a solid crust over molten interior

• If solid crust floats (e.g. plagioclase on the Moon) then it will insulate the interior, which will cool slowly (~ Myrs)

• If the crust sinks, then cooling is rapid (~ kyrs)• What happens once the magma ocean has solidified?

Cooling• Radiation

– Photon carries energy out into space– Works if opacity is low– Unimportant in interior, only works at surface

• Conduction– Heat transferred through matter– Heat moves from hot to cold– Slow; dominates in lithosphere and boundary layers

• Convection– Hot, buoyant material carried upward, Cold, dense

material sinks– Fast! Limited by viscosity of material

Running down the stairs with buckets of ice is an effective way of getting heat upstairs. -- Juri Toomre

Conduction - Fourier’s Law

• Heat flow F

T1>T0

T1

T0F d

dz

dTk

d

TTkF

)( 01

• Heat flows from hot to cold (thermodynamics) and is proportional to the temperature gradient

• Here k is the thermal conductivity (W m-1 K-1) and units of F are W m-2 (heat flux is power per unit area)

• Typical values for k are 2-4 Wm-1K-1 (rock, ice) and 30-60 Wm-1K-1 (metal)

• Solar heat flux at 1 A.U. is 1300 W m-2

• Mean subsurface heat flux on Earth is 80 mW m-2

• What controls the surface temperature of most planetary bodies?

Diffusion Equation

• Here is the thermal diffusivity (=k/Cp) and has units of m2 s-1

• Typical values for rock/ice 10-6 m2s-1

F1

F2

z

• We can use Fourier’s law and the definition of Cp to find how temperature changes with time:

2

2

2

2

z

T

z

T

C

k

t

T

p

In steady-state, the heat produced inside the planet exactly balances the heat loss from cooling. In this situation, the temperature is constant with time

0t

T

Diffusion length scale• How long does it take a change in temperature

to propagate a given distance?• This is perhaps the single most important

equation in the entire course:

• Another way of deducing this equation is just by inspection of the diffusion equation

• Examples:– 1. How long does it take to boil an egg?d~0.02m, =10-6 m2s-1 so t~6 minutes

– 2. How long does it take for the molten Moon to cool? d~1800 km, k=10-6 m2s-1 so t~100 Gyr.What might be wrong with this answer?

td ~2

Internal Heating• Assume we have internal heating H (in Wkg-1)

• From the definition of Cp we have Ht=TCp

• So we need an extra term in the heat flow equation:

pC

H

z

T

t

T

2

2

• This is the one-dimensional, Cartesian thermal diffusion equation assuming no motion

• In steady state, the LHS is zero and then we just have heat production being balanced by heat conduction

• The general solution to this steady-state problem is:

22 zbzaT

pCH

Example• Let’s take a spherical, conductive planet in steady state• In spherical coordinates, the diffusion equation is:

01 2

2

pC

H

r

Tr

rrt

T

• The solution to this equation is

)()( 226 rRTrT kH

s

• So the central temperature is Ts+(HR2/6k)

• E.g. Earth R=6400 km, =5500 kg m-3, k=3 Wm-1K-1, H=6x10-12 W kg-1 gives a central temp. of ~75,000K!

• What is wrong with this approach?

Here Ts is the surface temperature, R is the planetary radius, is the density

Convection

• Convective behaviour is governed by the Rayleigh number Ra

• Higher Ra means more vigorous convection, higher heat flux, thinner stagnant lid

• As the mantle cools, increases, Ra decreases, rate of cooling decreases -> self-regulating system

Image courtesy Walter Kiefer, Ra=3.7x106, Mars

Stagnant lid (cold, rigid)

Plume (upwelling, hot)

Sinking blob (cold)

3Tdg

Ra

Viscosity

• Ra controls vigor of convection. Depends inversely on viscosity, .

• Viscosity depends on Temperature T, Pressure P, Stress , Grain Size d.

mndAe RTPVE

A – pre-exponential constant E – Activation EnergyV – Activation Volume R – Gas Constantn – Stress Exponent m – Grain-size exponent

Viscosity relates stress and strain rate

Viscoelasticity

• A Maxwellian material has a viscous term and an elastic term.

• If h is high, we get an elastic behavior. If h is low, we get a viscous behavior.

• Depends also on the rate of stress. Materials are elastic on a short timescale, viscous on a long one.

• There are other types of viscoelasticity, but Maxwell is the simplest

Elastic Flexure• The near-surface, cold parts of a planet (the

lithosphere) behaves elastically• This lithosphere can support loads (e.g. volcanoes)• We can use observations of how the lithosphere

deforms under these loads to assess how thick it is• The thickness of the lithosphere tells us about how

rapidly temperature increases with depth i.e. it helps us to deduce the thermal structure of the planet

• The deformation of the elastic lithosphere under loads is called flexure

• EART163: Planetary Surfaces

Flexural Stresses

• 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)

• The elastic stresses will be both compressional and extensional (see diagram)

• Note that in this example the elastic portion includes both crust and mantle

Elastic plateCrust

Mantle

load

Flexural Parameter• Consider a load acting

on an elastic plate: Te

m

load

3

2

1

4

3 ( )(1 )e

m w

ET

g

w

• The plate has a particular elastic thickness Te

• If the load is narrow, then the width of deformation is controlled by the properties of the plate• The width of deformation is called the flexural parameter and is given by

E is Young’s modulus, g is gravity and n is Poisson’s ratio (~0.3)

• If the applied load is much wider than , then the load cannot be supported elastically and must be supported by buoyancy (isostasy)

• If the applied load is much narrower than , then the width of deformation is given by

• If we can measure a flexural wavelength, that allows us to infer and thus Te directly.

• Inferring Te (elastic thickness) is useful because Te is controlled by a planet’s temperature structure

Example• This is an example of a profile

across a rift on Ganymede• An eyeball estimate of would

be about 10 km• For ice, we take E=10 GPa,

=900 kg m-3 (there is no overlying ocean), g=1.3 ms-2

Distance, km

10 km

• If =10 km then Te=1.5 km

• A numerical solution gives Te=1.4 km – pretty good!

• So we can determine Te remotely

• This is useful because Te is ultimately controlled by the temperature structure of the subsurface

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

110 K270 K

elastic

viscous

190 K•E.g. for ice the base of the elastic layer is at about 190 K• The measured elastic layer thickness is 1.4 km (from previous slide)• So the thermal gradient is 60 K/km• This tells us that the (conductive) ice shell thickness is 2.7 km (!)

Depth

1.4 km

Temperature

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 (approx.) values for solar system objects:Body Te (km) dT/dz

(K/km)Body Te dT/dz

(K/km)

Earth (cont.)

30 15 Venus (450oC)

30 15

Mars (recent)

100 5 Moon (ancient)

15 30

Europa 2 40 Ganymede

2 40

Next Time

• Paper Discussion – Stevenson (2001)

• Planetary Interiors– Cooling Mechanisms– Rheology: How does the material deform?– Magnetism