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GE 11a 2014, Lecture 6 GE 11a 2014, Lecture 6

GE 11a 2014, Lecture 6 - PowerPoint Presentation

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GE 11a 2014, Lecture 6 - PPT Presentation

Conduction the lithosphere and isostacy To zeroth order the earths surface is bimodal in height with respect to sea level Similar things are also true for the moon and Mars though we will end up deciding it ID: 416691

earth age energy sun age earth sun energy lord kelvin

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Slide1

GE 11a 2014, Lecture 6

Conduction, the lithosphere and

isostacySlide2

To

zero’th order, the earth’s surface is bimodal in height with respect to sea level Slide3

Similar things are also true for the moon and Mars, though we will end up deciding it

re

flects something unique (and uniquely important) on Earth

Moon

MarsSlide4

The Catastrophists view of the North Atlantic Slide5

Cartoon of crust and lithosphere on the board…Slide6

A shaggy dog story about the first organized thought on this subject:

Lord

Kelvin

s response to

uniformitarianism+catastrophism

• First quantitative estimates of the ages of celestial objects based on

modern

physical theory (I.e., Newtonian physics, thermodynamics, Fick’s laws and the kinetic theory of gases).• Engaged a mature scientific community and discredited

lax

logic of Uniformitarian dating• Arguments of this kind are still made to date astrophysical events, processes on other planets, and poorly sampled geologic events

Lord Kelvin looking into a boxSlide7

Lord Kelvin

s measurement of the age of the earth

Take 1: a proof was presented in his Ph.D. thesis, but he burned his writings on this work

after his thesis defense. It has never been recovered or reproduced.Slide8

Lord Kelvin

s measurement of the age of the earth

Take 2: determine the age of the Sun using principles of gravitation and thermodynamics;

infer this to be the maximum age of the Earth.

I: Measure flux of energy at earth

s surface

(best above atmosphere directly facing sun)

=1340 Js

-1

m

-2

II: Integrate over area of a sphere with radius

equal to distance from earth to sun (assumes

sun emits energy isotropically)

area = 4π(1.5x1011)2; power = 3.8x1026 Js

-1

If dJ/dt is a constant:

(dJ/dt)xAge ≤ mass of sun x initial energy content (

E

, in J/Kg))

Age ≤ (2x10

30 Kg)/(3.8x1026) x E Age ≤ 5000 x ESlide9

Lord Kelvin

s measurement of the age of the earth

Take 2, continued:

Age of sun ≤ 5000 x initial energy content of sun in J/Kg

Case 1: If sun

s radiance is driven by a chemical reaction, like combustion, then it

s highest

plausible

initial energy content is ~ 5x10

7

J/Kg

If the sun is a ball of gasoline, it is ≤ 2.5x10

11

s, or

8000 years

, old

Case 2: Sun

s radiance is dissipating heat derived from its initial accretion:

Potential energy of pre-accretion cloud…

converts to kinetic energy when cloud collapses…

turns into heat if collisions between accreting material are inelasticSlide10

Case 2: Sun

s accretion, continued:

Age ≤

0.5M

s

xV

2

3.8x10

26

J/s

Age ≤ 10

15

s ~

30 Million years

Potential energy =

-

GM

i

m

j

R

ji

Total mass M at center-of-mass

location, i

Component particle mass m

at location j

R

ji

Solution depends on the distribution of mass and velocity in the cloud before its collapse to form the sun

One simple solution supposes all constituent masses arrived at the sun with a velocity equal

to the escape velocity from the Sun today:

(plus any contained in rotation

or other motion of cloud)

V = (2GM

s

/R)

0.5

= 618 km/s

i

0.5m

i

v

2

= 0.5M

s

(6.18x10

5

)

2

Slide11

Lord Kelvin

s measurement of the age of the earth

Take 3: directly determine age of the Earth by inverting the conductive temperature profile

observed in its outer few km of crust

Measurements from a geothermal area in Iceland

The archetype for the outer 300 km of the Earth

dT/dz ~ 1˚/40 meters, on average, near Earth

s surfaceSlide12

Lord Kelvin

s measurement of the age of the earth

Take 3: directly determine age of the Earth by inverting the conductive temperature profile

observed in its outer few km of crust

Q.E.D.: Physicists rule; geologists drool

T (˚C)

Radial distance

1500

pinned

by radiative balance

of surface

t

0

t

1

t

2

0

dT/dt = k d

2

T/dx

2

k = thermal diffusivity ~ 5x10

-3

cm

2

/s (=

conductivity

/(densityxC

v

))

Solution not simple, but is approximated by x = (kt)

0.5

, where

x = distance from surface to mid-point in T profile.

x ~ 30 km; t ~ 20 million years

Melting point of rock

J

heat

= k(

dT

/dx)Slide13

Note that conduction also leads to a change in rheology between interior and outer shellSlide14

Rayleigh number =

Buoyancy

Viscous drag

X

Momentum diffusivity

Thermal diffusivity

acceleration

Thermal expansion

Kinematic viscosity

Thermal diffusivity

Length scale

Temperature contrast

If > ~1000, convection ensues. The mantle is ~10

6

What are the dynamics of the hot, viscous (fluid like) interior? Slide15

A numerical model of whole-mantle convection in a

2-D earth