Topography of the earth’s surface. Depth to the Moho under north america and environs.

Topography of the earth’s surface

Depth to the Moho under north america and environs

Seismic structure of Greenlandmargin, and a related interpretivecartoon.

Topography in continental mountain chains and plateaus

Seismic structure beneath Himalayas

First-order topography of the ocean floor

Seismic velocity at 100 km depth

Fast (blue) = stiff and dense ~ coldSlow (red) = soft and low-density ~ warm

Seismic structure near a mid ocean ridge

Moho is hiding hereat ~6 km

Fast (blue) = stiff and dense ~ coldSlow (red) = soft and low-density ~ warm

Topography near ocean island chains

Seismic structure of the deep mantle near hawaii

• High topography = thick crust or warm mantle, and visa versa

• Often crust is thick and mantle cold, and topography is still fairly high; Thus crustal thickness effect ‘trumps’ mantle temperature effect

These observations reflect the role of isostacy in controlling topography

[chalk board notes on isostacy and orographic cycle]

WM Davis and the Geographic Cycle

Incision

Erosion

Mature

Isostatic ‘event’increases elevation

(‘Uplift’)

Heat flow at the earth’s surface

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 surface

Temperature gradients near the earth’s surface

[chalk board notes on heat production and conduction]

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

Rayleigh number = Buoyancy

Viscous drag XMomentum diffusivity

Thermal diffusivity

acceleration Thermal expansion

Kinematic viscosity Thermal diffusivity

Length scale

Temperature contrast

If > ~1000, convection ensues. The mantle is ~106

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

A numerical model of whole-mantle convection in a2-D earth

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.

Lord Kelvin’s measurement of the age of the earth

Take 2: directly determine age of the Earth by inverting the conductive temperature profile observed in its outer few km of crust

T (˚C)

Radial distance

1500

‘pinned’ by radiative balanceof surface

t0t1t2

0

dT/dt = k d2T/dx2

k = thermal diffusivity ~ 5x10-3 cm2/s (= ‘conductivity’/(densityxCv))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

Jheat = k(dT/dx)

Lord Kelvin’s measurement of the age of the earth

Take 3: 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-1m-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 ≤ (2x1030 Kg)/(3.8x1026) x E Age ≤ 5000 x E

Lord Kelvin’s measurement of the age of the earth

Take 3, 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 ~ 5x107 J/Kg

If the sun is a ball of gasoline, it is ≤ 2.5x1011 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 inelastic

Case 3: Sun’s accretion, continued:

Age ≤ 0.5MsxV2

3.8x1026 J/s

Age ≤ 1015 s ~ 30 Million years

Potential energy = -GMimj

Rji

Total mass M at center-of-masslocation, i

Component particle mass mat location j

Rji

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 equalto the escape velocity from the Sun today:

(plus any contained in rotationor other motion of cloud)

V = (2GMs/R)0.5 = 618 km/s

i0.5miv2 = 0.5Ms(6.18x105)2

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