1. Topography of the earth’s surface

2. Depth to the Moho under north america and environs

3. Seismic structure of Greenland margin, and a related interpretive cartoon.

6. Topography in continental mountain chains and plateaus

7. Seismic structure beneath Himalayas

8. First-order topography of the ocean floor

9. Seismic velocity at 100 km depth Fast (blue) = stiff and dense ~ cold Slow (red) = soft and low-density ~ warm

10. Seismic structure near a mid ocean ridge Moho is hiding here at ~6 km Fast (blue) = stiff and dense ~ cold Slow (red) = soft and low-density ~ warm

11. Topography near ocean island chains

12. Seismic structure of the deep mantle near hawaii

13. 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]

14. WM Davis and the Geographic Cycle Incision Erosion Mature Isostatic ‘event’ increases elevation (‘Uplift’)

15. Heat flow at the earth’s surface

16. 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

17. [chalk board notes on heat production and conduction]

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

19. Rayleigh number = Buoyancy Viscous drag X Momentum diffusivity Thermal diffusivity accelerationThermal expansion Kinematic viscosityThermal 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?

20. A numerical model of whole-mantle convection in a 2-D earth

21. 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.

22. 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 balance of surface t0t0 t1t1 t2t2 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)

23. 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 -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(1.5x10 11 ) 2 ; power = 3.8x10 26 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)/(3.8x10 26 ) x E Age ≤ 5000 x E

24. 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 ~ 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 inelastic

25. Case 3: 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 Q.E.D.: Physicists rule; geologists drool