1. Gravity I:
Gravity anomalies.
Earth gravitational field.
Isostasy.
Moment density dipole.
Practical issues.

2. The basics
Newton’s law of 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 (fortunately  is a small number).
Units of F are N=kg m s-2 .

3. 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:

4. The basics
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.
Question: Is earth gravitational field a constant?
Question: Is earth gravitational field greater at the poles or at the equator?

5. 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.
We now know that it is the Earth that is rotating. Centrifugal acceleration reduces gravitational attraction. Thus, the further you are from the rotation axis, the greater the centrifugal acceleration is.

6. The basics
A scalar field associates a single number (or scalar) to every point in space.
Question: Examples for a scalar field?
Answer: Temperature, topography and pressure.

7. The basics
A vector field associates a vector to every point in space.
Question: Examples for a vector field?
Answer: Speed, magnetic field and forces, such as the gravitational force.

8. The basics
How are scalar and vector fields related to one another?
The gradient of a scalar field is a vector field.

9. The basics
Now g is a vector field:
where r is a unit vector pointing towards the earth’s center.
The gravitational potential, U, is a scalar field:
(verify that U is the potential field of g.)
Note that earth’s gravitational potential is negative.
Potentials are additive, and this property makes them easier (than vectors) to work with.

10. Surface gravity anomalies due to some buried bodies
The general equation is:
where:
 is the gravitational constant
 is the density contrast
r is the distance to the observation point
a is the angle from vertical
V is the volume
Question: Why a cosine term?
Question: Express the gravity anomaly in Cartesian coordinates.
Question: What’s the gravity anomaly due to a sphere?

11. Surface gravity anomalies due to simple-shape bodies
Simple analytical solutions may be derived for simple-shaped buried bodies with uniform density contrast.
A sphere: z a
x/z

12. Surface gravity anomalies: forward versus inverse proplem
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?

13. Surface gravity anomalies: forward versus inverse problem
Inspection of the solution for a buried sphere reveals a non-uniqueness of that problem. The term a3 introduces an ambiguity to the problem (because there are infinite many combinations of  and a3 that give the same a3). Therefore, a maximum gravity anomaly of a certain magnitude can be generated by different combinations of densities and radii.
This point highlights the importance of adding geological and geophysical constraints!
While the maximum anomaly is ambiguous, the shape of the anomaly is not.

14. Surface gravity anomalies: forward versus inverse problem
The solution for a sphere:
can be normalized as follows:
where the hat denotes non-dimensional (or unitless) parameter.
While the 1st term affects only the amplitude of the anomaly, the 2nd affects its shape.
Question: how far should we go to go either side in order to see the shape of the anomaly.

15. Surface gravity anomalies: spatial extent
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:

16. Surface gravity anomalies: spatial extent
What should be the spatial extent of the surveyed region?
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 km), gravity measurements must be made over an extensive area.

17. Gravity anomaly due to a simple-shape buried body
An infinitely long horizontal cylinder
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.
cylinder
sphere

18. Surface gravity anomalies due to simple-shape bodies
An infinite horizontal slab of finite thickness:
Setting (y)= c and integration with respect to r from zero to infinity and with respect to y between 0 and h leads to:
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.

19. Surface gravity anomalies due to complex-shape bodies
The graphical approach: This is done using templates superimposed on a cross section that is divided into elementary areas, each of which contributes equally at the surface station.
surface
station
Station of two
Dimensional structure z 
The gravity effect of a cell at the chart origin is:
with  being the rock density and  the universal gravity constant.