Isostasy: The Concept of Floating Blocks • Isostasy = the concept that large topographic features

effectively float on the asthenosphere – Iso = same – stasis = standstill

– Involves a state of constant pressure at any point • Pressure: P = ρ g h

– If g were measured from a balloon at constant elevation you would expect a mass excess (i.e. larger g) above mountains. • We commonly observe considerably less g over large mountains

– Consistent with low density roots beneath large mountains

Isostasy: The Initial Discovery • The hypothesis that large mountains have low density roots was

first proposed during topographic surveys of India and the Himalayan Mountains

Questions:

• How does this low density root form?

• As a mountain range becomes eroded why are there not large negative anomalies due to the low density roots?

Explanation:

• Large topographic features effectively ‘float’ on the asthenosphere

• Follows Archimedes principle

Archimedes’ Principle • Supposedly, Archimedes was taking a bath one day and noticed

that the tub overflowed when he got in. – He realized that displaced water could be used to

• Detect volumes of odd shaped objects

• Detect density differences (helped determine that a crown wasn’t 100% gold)

Archimedes Principle:

• As an object is partially submerged in a fluid, the weight of fluid that is displaced = the weight reduction of the object

• When the displaced fluid = the weight of the object – It floats

Isostatic Compensation • Take the case of a layered block floating in a fluid

– Remove two layers and… • The bottom is now less deep below the surface of the liquid

• The top is less elevated above the surface of the fluid

– Thus, floating blocks will seek isostatic equilibrium and are said to be isostatically compensated

– In the case of the continental crust, density differences are small, so any significant topography in isostatic equilibrium must be balanced by a thick crustal root • E.g. the Himalayas are < 9 km above sea level, but have roots > 70 km

• So, mountains in isostatic equilibrium are like the tip of an iceberg

Isostasy and Gravity If blocks are in isostatic equilibrium…

• Gravity measurements made at some constant elevation above the blocks would detect (almost) no variation in g – There would be small variations at the edges of blocks due to the nearby

topography

Isostatic Calculations • Take the simple example of two floating blocks A and B

– Each has different layers of varying thickness and density • Includes an asthenospheric layer and an air layer

– If A and B are in isostatic equilibrium • Their total weights are the same

BlockBasthasthairairBlockAasthasthairair hghghghghghghghghg 2211332211

hgWeight

BlockBasthasthBlockAasthasth hhhghhhhg 2211332211

BlockBasthasthBlockAasthasth hhhhhhh 2211332211

These are referred to as the weight equation

BBlock

n

i

ii

ABlock

n

i

ii hh

11

Isostatic Calculations • If A and B are in isostatic equilibrium

– Their total heights are the same

– This is referred to as the height equation

In practice…

• Choose the top to be the highest elevation of rock (or water/ice)

• Choose the bottom to be the lowest elevation of lithosphere

BlockBasthairBlockAasthair hhhhhhhhh 21321

BBlock

n

i

i

ABlock

n

i

i hh

11

Example Isostatic Calculation #1 • Take the case of adding a 2 km thick glacier on top of a continent

– The weight of the ice causes the block to sink deeper until isostatic equilibrium is reached

– Treat the before and after as two separate blocks that are both in equilibrium and use weight equation

(3 x 2.0) + (30 x 2.7) + (70 x 3.1) + (ha x 3.2) = (2 x 0.9) + (3 x 2.0) + (30 x 2.7) + (70 x 3.1)

ha = 0.56 km

So what does this mean?

• Adding 2 km of ice on top of this continent caused 0.56 km of isostatic subsidence

• The elevation of the top of the ice sheet after isostatic equilibrium is reached will be only 1.44 km above the original block (from height equation)

hair + 3 + 30 + 70 + 0.56 = 2 + 3 + 30 +70

hair = 2 – 0.56 = 1.44 km

Example Isostatic Calculation #2 • Suppose that there was a large 2 km deep lake (in equilibrium)

• Gradually, sediments are brought in by rivers that eventually replace the water and fill the lake up to its current water level – How thick would the sediments be? (why not 2 km?) (2 x 1.0) + (3 x 2.0) + (30 x 2.7) + (90 x 3.1) + (ha x 3.2) = (hs x 1.8) + (3 x 2.0) + (30 x 2.7) + (90 x 3.1)

Two unknowns! Need another equation. Use the height equation.

2 + 3 + 30 + 90 + ha = hs + 3 + 30 + 90 hs = ha + 2

Substitute hs = ha + 2 into weight equation 2 + ha x 3.2 = hs x 1.8 2+ ha x 3.2 = (ha + 2) x 1.8 3.2 ha + 2 = 1.8 ha + 3.6 3.2 ha = 1.8 ha + 1.6 1.4 ha = 1.6 ha = 1.142857142 km ha = 1.1 km Use Height equation to figure out hs

hs = ha + 2 = 3.1 km

Example Isostatic Calculation #2 • So, ha = 1.1 km and hs = 3.1 km

– It took 3.1 km of sediment to completely fill up to the previous lake level

– This 3.1 km of sediment added weight and caused an isostatic compensation of -1.1 km (subsidence)

– Likewise, removing material would cause isostatic uplift • But the overall elevation would decrease even after isostatic adjustment

Airy and Pratt Models of Isostasy • Two end member models have been proposed to account for

isostasy.

• Airy Model: All blocks have the same density but different thicknesses

• Pratt Model: All blocks float to the same depth but have different densities

Airy Model of Isostasy • Airy Model: All blocks have the same density but different

thicknesses – Thicker blocks have higher elevation and much thicker roots

– Higher ground is where the lithosphere is thicker

– The weight equation becomes:

BasthasthlithlithAasthasthlithlith hhhh

Pratt Model of Isostasy • Pratt Model: All blocks float at the same depth, but have differing

density – higher elevations indicate less dense rocks

– Higher ground is where the lithosphere is thicker

– The weight equation becomes:

– The height equation is the same

for both Airy and Pratt models

BlithlithAlithlith hh

Airy vs. Pratt: Which is Correct? • The Airy and Pratt are not the only possible models

– They are end member models

– A combination of density changes and lithosphere thickness may occur

But, in general, most data supports…

• The Airy model for continental mountain ranges – Continental mountain ranges have thick crustal roots

• The Pratt model for mid-ocean ridges – Mid ocean ridges have topography that is supported by

density changes • Increased temperature at ridges ---> rocks expand ---> lower density

The Isostatic Anomaly • Recall that earlier we learned that floating blocks (in equilibrium)

should produce no anomaly.

• Gravity measurements at B and D will yield the same values if the free-air correction is applied. – When only the latitude and free-air corrections have been made the result

is called the free-air anomaly

• If the Bouguer correction is made, the extra mass above B is removed and results in a negative Bouguer anomaly

The Isostatic Anomaly • If a region is not isostatically compensated

– The free-air anomaly will be positive

– The Bouguer anomaly will be zero

• Partially compensated regions are common – free-air anomaly > 0

– Bouguer anomaly < 0