The basics. Newton ’ s law of universal gravitation : where: F is the force of gravitation. m 1 and m 2 are the masses. r is the distance between the masses. is the gravitational constant that is equal to 6.67x10 -11 Nm 2 kg -2 (fortunately is a small number). - PowerPoint PPT Presentation
Newton’s law of universal gravitation: where: • F is the force of gravitation. • m 1 and m 2 are the masses. • r is the distance between the masses. • is the gravitational constant that is equal to 6.67x10 -11 Nm 2 kg -2 . Units of F are N=kg m s -2 . The basics Newton’s second law of motion: where: • m is the mass. • a is acceleration. By combining the universal law of gravitation with Newton’s second law of motion, one finds that the acceleration of m 2 due to its attraction by m 1 is:
Transcript
Newton’s law of universal gravitation:where:• F is the force of gravitation.• m1 and m2 are the masses.• r is the distance between the masses.• is the gravitational constant that is equal to 6.67x10-11 Nm2kg-2.
Units of F are N=kg m s-2 .
The basics
Newton’s second law of motion:where:• m is the mass.• a is acceleration.
By combining the universal law of gravitation with Newton’s second law of motion, one finds that the acceleration of m2 due to its attraction by m1 is:
Gravitational acceleration is thus:where:• ME is the mass of the Earth.• RE is the Earth’s radius.
Units of acceleration are m s-2, or gal=0.01 m s-2.
The basics
• The Earth is an oblate spheroid that is fatter at the equator and is thinner at the poles.
• There is an excess mass under the equator.
• Centrifugal acceleration reduces gravitational attraction. Thus, the further you are from the rotation axis, the greater the centrifugal acceleration is.
where r is a unit vector pointing towards the earth’s center. The gravitational potential, U, is a scalar field:
• Note that Earth’s gravitational potential is negative.• Potentials are additive, and this property makes them easier (than vectors) to work with.• To verify that U is the potential field of g take its derivative with respect to R.
The gradient of a scalar field is a vector field.
Surface gravity anomalies due to some buried bodies
The general equation is:
where: is the gravitational constant is the density contrastr is the distance to the observation pointa is the angle from verticalV is the volumeQuestion: Why a cosine term?
Solution for a sphere:
z
a
x/z
Surface gravity anomalies due to some buried bodies
Infinitely long horizontal cylinder:
Surface gravity anomalies due to some buried bodies
Buried infinite slab:
Surface gravity anomalies due to some buried bodies
Buried infinite slab:
An infinitely long horizontal cylinder
cylinder sphere
The expression for a horizontal cylinder of a radius a and density :
It is interesting to compare the solution for cylinder with that of a sphere.
This highlights the importance of a 2-D gravity survey.
Surface gravity anomalies due to some buried bodies
What should be the spatial extent of the surveyed region?
To answer this question it is useful to compute the anomaly half-distance, X1/2, i.e. the distance from the anomaly maximum to it’s medium. For a sphere, we get:
Surface gravity anomalies due to some buried bodies
• The signal due to a sphere buried at a depth Z can only be well resolved at distances out to 2-3 Z.
• Thus, to resolve details of density structures of the lower crust (say 20-40 km), gravity measurements must be made over an extensive area.
The ambiguity of surface gravity anomalies
In the preceding slide we have looked at the result of a forward modeling also referred to as the direct problem:
In practice, however, the inverse modeling is of greater importance:
Question: Can the data be inverted to obtain the density, size and shape of a buried body?
• Inspection of the solution for a buried sphere reveals a non-uniqueness of that problem. The term a3 introduces an ambiguity to the problem, and different combinations of densities and radii can produce identical anomalies.
1. Near surface very elongated body2. Shallow elongated body3. Deep sphere
• The same gravity anomaly may be explained by different anomalous bodies, having different shapes and located at different depths: measured gravity anomaly
In summary, we want to know:
But actually, gravity anomaly alone cannot provide this information.
The ambiguity of surface gravity anomalies
The observed anomaly may be explained equally well with deep models with small density contrast or shallow models with greater density contrast.
Question: is the MAR in isostatic equilibrium?
Here’s an example from a Mid-Atlantic Ridge (MAR).
The ambiguity of surface gravity anomalies
Geoid
• the observed equipotential surface that defines the sea level.• the shape a fluid Earth would have if it had exactly the gravity field
of the Earth• roughly the sea-level surface - dynamic effects such as waves, and
tides, must be excluded• geoid on continents lies below continents - corresponds to level of
nearly massless fluid if narrow channels were cut through continents
• geoid highs are gravity highs
The geoid
The vector gravity (g) is perpendicular to the geoid.
Reference geoid is a mathematical formula describing a theoretical equipotential surface of a rotating (i.e., centrifugal effect is accounted for) symmetric spheroidal earth model having realistic radial density distribution.
The geoid
The international gravity formula gives the theoretical gravitational acceleration on a reference geoid:
observed geoidreference geoid
The geoid height anomaly is the difference in elevation between the measured geoid and the reference geoid.
Note that the geoid height anomaly is measured in meters.
The geoid
Geoid anomaly
Map of geoid height anomaly:
Note that:• The differences between observed geoid and reference geoid are as large as 100 meters• In continental regions, they do not correlate with topography because of isostatic
compensation
Figure from: www.colorado.edu/geography
Question: what gives rise to geoid anomaly?
Geoid anomaly
Differences between geoid and reference geoid are due to:
• Topography
• Density anomalies at depth
Figure from Fowler
Geoid anomaly
Figure from McKenzie et al., 1980
Two competing effects:
1. Upwelling brings hotter and less dense material, the effect of which is to reduce gravity.
2. Upwelling causes topographic bulge, the effect of which is to increase gravity.
What is the effect of mantle convection on the geoid anomaly?
Flow
Temp.
upwellingdownwelling
Geoid anomaly
SEASAT provides water topography
Note that the most prominent features on most geoid maps (depending on filtering used) are subduction zones.
Geoid anomaly
Free-air gravity anomaly from satellite altimetry for the Tonga-Kermadec region
Comparison of topography along east west profiles across the subduction zone at 20, 25 and 30°S (thick/blue) to observed topography (thin/black)
Cross-sections across subduction-zone geoid anomalies show an asymmetric anomaly low (trench) and an anomaly high (presence of cold, dense slab in lighter asthenospere):
(From: Billen and Gurnis, EPSL, 2001)
Geoid anomaly
GEOID99: a refined model of the geoid in the United States
Heights range from a low of -50.97 meters (magenta) in the Atlantic Ocean to a high of 3.23 meters (red) in the Labrador Strait
Geoid anomaly
A model for mass change due to post-glacial rebound and the reloading of the ocean basins with seawater.
Blue and purple areas indicate rising due to the removal of the ice sheets. Yellow and red areas indicate falling as mantle material moved away from these areas in order to supply the rising areas, and because of the collapse of the forebulges around the ice sheets.
Geoid anomaly and corrections
Geoid anomaly contains information regarding the 3-D mass distribution. But first, a few corrections should be applied:• Free-air (required)• Bouguer (required)• Terrain (optional)
Geoid anomaly and corrections
Free-air correction, gFA:
This correction accounts for the fact that the point of measurement is at elevation H, rather than at the sea level on the reference spheroid.
Geoid anomaly and corrections
Since:
where:• is the latitude • h is the topographic height• g() is gravity at sea level• R() is the radius of the reference spheroid at
The free-air correction is thus:
This correction amounts to 3.1x10-6 ms-2 per meter elevation.
Question: should this correction be added or subtracted?The free-air anomaly is the geoid anomaly, with the free-air correction applied:
Geoid anomaly and corrections
Geoid anomaly and corrections
Bouguer correction, gB:
This correction accounts for the gravitational attraction of the rocks between the point of measurement and the sea level.
Geoid anomaly and corrections
An infinite horizontal slab of finite thickness:
Note that the gravity anomaly caused by an infinite horizontal slab of thickness h and density c is independent of its distance b from the observer.
Geoid anomaly and corrections
Setting (y)= c and integration with respect to r from zero to infinity and with respect to y between 0 and h leads to:
Geoid anomaly and corrections
Geoid anomaly and corrections
The Bouguer correction is:
where: is the universal gravitational constant is the rock densityh is the topographic height
For rock density of 2.7x103 kgm-3, this correction amounts to 1.1x10-6 ms-2 per meter elevation.
Question: should this correction be added or subtracted?
The Bouguer anomaly is the geoid anomaly, with the free-air and Bouguer corrections applied:
Geoid anomaly and corrections
Terrain correction, gT:
This correction accounts for the deviation of the surface from an infinite horizontal plane. The terrain correction is small, and except for area of mountainous terrain, can often be ignored.
Geoid anomaly and corrections
The Bouguer anomaly including terrain correction is:
Bouguer anomaly for offshore gravity survey:• Replace water with rock• Apply terrain correction for seabed topography
After correcting for these effects, the ''corrected'' signal contains information regarding the 3-D distribution of mass in the earth interior.
Isostasy
The deflection of plumb-bob near mountain chains is less than expected. Calculations show that the actual deflection may be explained if the excess mass is canceled by an equal mass deficiency at greater depth.
A plumb-bobPicture from wikipedia
Isostasy: the Airy hypothesis (application of Archimedes’ principal)
• Two densities, that of the rigid upper layer, u, and that of the substratum, s.• Mountains therefore have deep roots. A mountain height h1 is underlain by a root of thickness:
• Ocean basin depth, h2, is underlain by an anti-root of thickness:
r1
u
s
h1
r3
d
• Ocean basin whose depth is h2 is underlain by a high density material, 2, that is given by:
Isostasy: the Pratt’s hypothesis
• The depth to the base of the upper layer is constant.• The density of rocks beneath mountains is less than that beneath valleys.• A mountain whose height is h1 is underlain by a root whose density 1 is:
Isostasy:
Isostasy
Questions:
• Which is the correct hypothesis?
• Does isostatic equilibrium apply everywhere?
Is the person resting on top of a spring-mattress in a state of isostatic equilibrium?
Isostasy: elastic flexure
Like the springs inside the mattress, the elastic lithosphere can also support excess mass.
Thick plates can support more excess mass than thin plates.
Isostasy: elastic flexure
The response of the lithosphere to a vertical load depends on the lithosphere elastic properties as follows:
where D is the flexural rigidity, that is given by:
with:E being Young Modulush being the plate thickness being Poisson’s ratio
Isostasy: elastic flexure
The figure below shows the solution for the case of a line load:
Note the flexural bulge on either side of the depression.
Of course in reality the boundary conditions are more complex…
Figure from Fowler
Isostasy: example from the Hawaii chain
bathymetry
free-air
Two effects:• Elastic flexure due to island load.• A swell due to mantle upwelling.
Figure from Fowler
Isostasy: example from the Mariana subduction zone
defle
ction
[km
]
distance [km]
• The accretionary wedge loads the plate edge causing it to bend.• A flexural bulge is often observed adjacent to the trench.• Topography of Mariana bulge implies a 28 km thick plate.
Fluxural bulge
Figure from Fowler
Isostasy example from the Tonga subduction zone
defle
ction
[km
]
distance [km]
• The Tonga slab bends more steeply than can be explained by an elastic model.• It turned out that an elastic-plastic model for the lithosphere can explain the bathymetry data.
Figure from Fowler
Isostasy: local versus regional isostatic equilibrium
According to Pratt and Airy hypotheses, excess mass is perfectly compensated everywhere. This situation is referred to as local isostasy.
The situation where some of the load is supported by the strength of the lithosphere is referred to as regional isostasy. In this case, isostatic equilibrium occurs on a larger scale, but not at any point.
Isostasy
Questions:
1. Isostatic equilibrium means no excess mass. Does this mean no gravitational anomaly.
2. Can we distinguish compensated from uncompensated topographies?
Isostasy: gravity
100% compensated
A rule of thumb: A region is in isostatic equilibrium if the Bouguer anomaly is a mirror image of the topography.
Figure from Fowler
Isostasy: gravity
Uncompensated
A rule of thumb: A region is NOT in isostatic equilibrium if the Bouguer anomaly remains flat under topographic highs and lows.
Figure from Fowler
Isostasy: isostatic rebound
Figure from Fowler
The rate of isostatic rebound depends on the elastic properties of the lithosphere (including its thickness) as well as the mantle viscosity.
Isostatic rebound can be observed if a large enough load has been added or removed fast enough.
Small loads, a few km in diameter, can tell us about the elastic properties of the crust.
Isostasy: isostatic rebound
The Dead Sea, Israel:• During the past few decades, the Dead Sea water-level is dropping at a rate of 1 m/yr.
Isostasy: isostatic rebound
Data
Model
Residual
Ground displacement due to water-level changes can be reproduced using a homogeneous elastic half-space model.
Medium size loads, say ~100 km diameter, can tell us about the viscosity of the asthenosphere.
Isostasy: isostatic rebound
Lake Bonneville, Utha:• A lake 300 m deep dried up 10,000 years ago.• Lake center has risen by 65 m.
Fennoscandia:• Removal of 2.5 km thick ice at the end of the last ice age 10,000 years ago.• Current peak uplift rate is 9 mm/yr.
Large loads, say ~1000 km diameter tell us about the upper and lower mantle viscosity.
Isostasy: isostatic rebound
Great Britain:• Glaciation
affected Scotland, but not Southern England.
• Uplift rate of up to 10cm per century.
Dipole moment of density anomaly: dipole moment of density distribution
We have seen that the gravity anomaly due to a horizontal layer of thickness y is:
thus the gravity potential of this layer is:
The dipole moment of density distribution is just:
We shall see that it is the dipole moment of density distribution, which contains information regarding the mass distribution, and may help to discriminate between the two isostatic models.
Dipole moment of density anomaly: gravity potential
Combining this with the expression for the gravity potential of a Bouguer slab (see previous slide) leads to:
U=Uobs
U=Uref
Dipole moment of density anomaly: Airy versus Pratt
Airy (positive topography):
Pratt (positive topography):
In areas of isostatic equilibrium, we would wish to know whether mass is distributed according to Airy or Pratt models.
Dipole moment of density anomaly: Airy versus Pratt
Pratt (positive topography):
Replacing with leads to:
Note the linear relation between N and h.
+y
Dipole moment of density anomaly: Airy versus Pratt
Airy (positive topography):
+y
Replacing with leads to:
Note the NON-LINEAR relation between N and h.
Dipole moment of density anomaly: Airy versus Pratt
Pratt
Airy
• N versus h for Airy and Pratt models.
• Airy model implies a nearly factor of 3 difference between N/h on-land and off-shore.
Figure from Turcotte and Schubert
At what direction does the Pacific plate moves?
Dipole moment of density anomaly: Airy versus Pratt
Dipole moment of density anomaly: Airy versus Pratt
Figure from Turcotte and Schubert• Dependence of the observed geoid anomaly on bathymetry across the Hawaiian swell and across the Bermuda swell compared with the predicted anomaly according to Airy and Pratt models.
• Fair (or good?) agreement is obtained for Pratt model with a compensation depth of 100 km.
• If we accept the Pratt model to be applicable, the conclusion is that the mantle rocks beneath these swells have anomalously low density down to a depth of 100 km.
Dipole moment of density anomaly: Airy versus Pratt
• A comparison between Airy-predicted and measured geoid anomaly across the Atlantic continental margin of N. America.
• It follows from this comparison that the continental margin is in a state of isostatic equilibrium according to Airy model.
Figure from Turcotte and Schubert
Dipole moment of density anomaly: Airy versus Pratt
Further reading:
* Turcotte, D. L. and G. Schubert, Geodynamics, Cambridge University Press.
* Fowler, C. M. R., The solid Earth, Cambridge University Press.
Practical issues
(This lecture is based largely on: http://www.earthsci.unimelb.edu.au/ES304/)
The shape of the gravity anomaly depends not on the absolute density, but on the density contrast, i.e. the difference between the anomalous density and the “background density”.
Practical issues
Here’s a list of densities associated with various earth’s materials:
Note that:• Density differences are quite small.• There's considerable overlap in the measured densities.
Practical issues
Consider the variation in gravitational acceleration due to a spherical ore body with a radius of 10 meters, buried at a depth of 25 meters below the surface, and with a density contrast of 500 kg per meter cubed.
The maximum anomaly for this example is 0.025 mGal.
(keep in mind that 9.8 m/s2 is equal to 980,000 mGal !!!)
Practical issues
• Owing to the small variation in rock density, the spatial variations in the observed gravitational acceleration caused by geologic structures are quite small
• A gravitational anomaly of 0.025 mGal is very small compared to the 980,000 mGals gravitational acceleration produced by the earth as a whole. Actually, it represents a change in the gravitational field of only 1 part in 40 million.
• Clearly, a variation in gravity this small is going to be difficult to measure.
Practical issues
How is gravity measures:
• Falling objects
• Pendulum
• Mass on a spring
Practical issues
Falling objects:
The distance a body falls is proportional to the time it has fallensquared. The proportionality constant is the gravitationalacceleration, g:
g = distance / time2 .
To measure changes in the gravitational acceleration down to 1 part in 40 million using an instrument of reasonable size, we need to be able to measure changes in distance down to 1 part in 10 million and changes in time down to 1 part in 10 thousands!! As you can imagine, it is difficult to make measurements with this level of accuracy.
Practical issues
Pendulum measurements:
The period of oscillation of the pendulum, T, is proportional to one over the square root of the gravitational acceleration, g. The constant of proportionality, l, is the pendulum length:
Here too, in order to measure the acceleration to 1 part in 50 million requires a very accurate estimate of the instrument constant l, but l cannot be determined accurately enough to do this.
Practical issues
But all is not lost:
• We could measure the period of oscillation of a given pendulum by dividing the time of many oscillations by the total number of oscillations.
• By repeating this measurement at two different locations, we can estimate the variation in gravitational acceleration without having to measure l.
Practical issues
Mass on a spring measurements:
The most common type of gravimeter used in exploration surveys is based on a simple mass-spring system.
According to Hook’s law:
X = mg / k ,
with k being the spring stiffness.
Practical issues
• Like pendulum the measurements, we can not determine k accurately enough to estimate the absolute value of the gravitational acceleration to 1 part in 40 million.
• We can, however, estimate variations in the gravitational acceleration from place to place to within this precision.
Under optimal conditions, modern gravimeters are capable of measuring changes in the Earth's gravitational acceleration down to 1 part in 1000 million.
Practical issues
Various undesired factors affect the measurements:
• Temporal (time-dependent) variations:
1. Instrumental drift2. Tidal effects
• Spatial variations:
1. Latitude variations2. Altitude variations3. Slab effects4. Topography effect
Practical issues
Instrumental drift:
The properties of the materials used to construct the spring change with time. Consequently, gravimeters can drift as much as 0.1 mgal per day.
What causes the oscillatory changes superimposed on the instrumental drift?
Practical issues
Tidal effect:
In this example, the amplitude of the tidal variation is about 0.15 mGals, and the amplitude of the drift appears to be about 0.12 mGals over two days. These effects are much larger than the example gravity anomaly described previously.
Practical issues
• Since changes caused by instrumental drift and tidal effects do not reflect the mass distribution at depth, they are treated as noise.
• Strategies to correct for instrumental drift and tidal effects are discussed in: www.earthsci.unimelb.edu.au/ES304/MODULES/GRAV/NOTES/tcorrect.html
Practical issues
Regional and local (or residual) gravity anomalies:
Consider a spherical ore body embedded in a sedimentary unit on top of a (denser) Granitic basement that is dipping to the right.
Practical issues
The strongest contribution to the gravity is caused by large-scale geologic structure that is not of interest. The gravitational acceleration produced by these large-scale features is referred to as the regional gravity anomaly.
Practical issues
The second contribution is caused by smaller-scale structure for which the survey was designed to detect. That portion of the observed gravitational acceleration associated with these structures is referred to as the local or the residual gravity anomaly.
Practical issues
There are several methods of removing unwanted regional gravity anomalies. Here's an example for a graphical approach:
Smoothing in 1 dimension Smoothing in 2 dimensions
Practical issues
Variations in gravity around the globe are inferred from satellite orbit.
The balance between the gravitational attraction and the centrifugal force is written as:
This leads to:
where T is the satellite’s period, 2r / V .
Practical issues
Yet, the highest resolution whole earth gravity maps are derived from radar measurement of the height of the sea surface.
