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
  1 g h1   2 g h2   3 g h3   air g hair
Weight   g h
  asth g hasth  BlockA    1 g h1   2 g h 2   air g h air   asth g h asth  BlockB
g   1 h1   2 h2   3 h3   asth hasth  BlockA  g   1 h1   2 h 2   asth hasth BlockB
  1 h1   2 h2   3 h3   asth hasth BlockA
These are referred
to as the weight
equation
 n

   i hi 
 i 1
 Block
A
 n

    i hi 
 i 1
 Block
B
   1 h1   2 h 2   asth h asth  BlockB
Isostatic Calculations
• If A and B are in isostatic equilibrium
– Their total heights are the same
h1  h2  h3  hair
 h asth BlockA  h1  h 2  hair  h asth BlockB
– This is referred to as the height equation
 n

  hi 
 i 1  Block
 n

   hi 
 i 1  Block
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
A
B
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:
hlith  lith  hasth  asth  A
 hlith  lith  h asth  asth  B
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:
hlith  lith  A
 hlith  lith  B
– The height equation is the same
for both Airy and Pratt models
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