Download Topography of the earth`s surface

Topography of the earth’s surface
Depth to the Moho under north america and environs
Seismic structure of Greenland
margin, and a related interpretive
cartoon.
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 ~ cold
Slow (red) = soft and low-density ~ warm
Seismic structure near a mid ocean ridge
Fast (blue) = stiff and dense ~ cold
Slow (red) = soft and low-density ~ warm
Moho is hiding here
at ~6 km
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
Isostatic ‘event’
increases elevation
(‘Uplift’)
Mature
Heat flow at the earth’s surface
Temperature gradients near 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
[chalk board notes on heat production and conduction]
Note that conduction also leads to a change in rheology between interior and outer shell
What are the dynamics of the hot, viscous (fluid like) interior?
Rayleigh number =
Buoyancy
Viscous drag
acceleration
Momentum diffusivity
X Thermal diffusivity
Thermal expansion
Temperature contrast
Length scale
Kinematic viscosity
Thermal diffusivity
If > ~1000, convection ensues. The mantle is ~106
A numerical model of whole-mantle convection in a
2-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
Melting point of rock
1500
t2
t1
t0
T (˚C)
‘pinned’ by radiative balance
of surface
0
Radial distance
Jheat = k(dT/dx)
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
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:
Total mass M at center-of-mass
location, i
-GMimj
Potential energy =
Rji
(plus any contained in rotation
or other motion of cloud)
Rji
Component particle mass m
at location j
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:
V = (2GMs/R)0.5 = 618 km/s
= 0.5Ms(6.18x105)2
0.5MsxV2
Age ≤ 3.8x1026 J/s
i0.5miv
2
Age ≤ 1015 s ~ 30 Million years
Q.E.D.: Physicists rule; geologists drool