Geophysics, Lecture Notes- Physics - 8, Study notes of Physics

Consequences of Spherical Symmetry, Geoid Effect, Gravity measurement and anomalies Isostasy and earth's interior

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2010/2011

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PX266 Geophysics (2010/11)
Lecture 6 Handout
Gravity Due to Extended Bodies and Gravity Anomalies
Dr. Gavin Bell
Gravity due to a spherical shell
Calculate gravitational potential at P initially Pis outside the shell.
Area of strip =
dbb sin2
Mass of strip =
dbt sin2 2
Contribution to gravitational potential at P:
l
dbtG
dV
sin2 2
Integrate to get V(r) (adding potential scalar):
0
2sin
2)( l
d
btGrV
Cosine rule:
cos2
222 brbrl
Differentiate cosine rule:
br
dll
ddbrdll
sinsin22
Plug this into the definite integral and change the limits carefully:
br br
br
br
l
br
bGt
br
dl
bGtrV
2
22
2)(
bsin
Spherical shell: thickness t, radius b, density
PO
b
Circular strip, width = bd
Q
l
|OP| = r
d
pf3
pf4

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PX266 Geophysics (2010/11)

Lecture 6 Handout –

Gravity Due to Extended Bodies and Gravity Anomalies

Dr. Gavin Bell Gravity due to a spherical shell Calculate gravitational potential at P – initially P is outside the shell. Area of strip =  2  b sin  bd  Mass of strip =   t   2  b^2 sin d  Contribution to gravitational potential at P : l G t b d dV 2   sin  2   Integrate to get V ( r ) (adding potential – scalar):   

0

2 sin

l

d

V r G t b

Cosine rule: 2 cos  2 2 2 lrbbr Differentiate cosine rule:   br l dl 2 l dl  2 br sin d  sin d  Plug this into the definite integral and change the limits carefully:   r b r b r b r b l br Gt b br dl V r Gt b          2 2 2 ( ) 2     b sin  Spherical shell: thickness t , radius b, density 

O P

b

Circular strip, width = b d 

Q

l

|OP| = r d

r t b G b r Gt b V r 2 4 2 2 ( )         But the mass of the shell is just 2 M (^) s  4  tb and so we have r

GM

V r s

i.e. the gravitational potential due to the spherical shell, OUTSIDE the shell, is equivalent to that of a point mass at the centre of the shell with exactly the shell’s mass… the shell acts as a simple point mass at its centre. Easy – but what if P is inside the shell? Consider how we changed the limits on the integral… this time   0  lbr and   lbr

The integral is otherwise the same, and: dl r

b r b r

 

Therefore:  r  t bG CONSTANT

r Gt b V r   

The potential inside the spherical shell is constant, therefore its grad is zero and there is no gravitational field inside the shell. This is true for any inverse square law field (e.g. electromagnetic) and can be shown geometrically (in lecture). Hence, for any point inside a spherically symmetric body (made up of spherical shells, like the Earth) the gravitational potential experienced is due only to the mass ‘ beneath’ the measuring point P as if it were a point mass at the centre.

R M R r rdr 0 2 4   is the total mass beneath P , and ( ) ( ) R 2

GM

g R R

GM

V R R^ R

Q

l

r P

O

b

Adapted from Osman et al. ANNALS OF GEOPHYSICS, VOL. 49, N. 6, December 2006 Measured and simulated gravity anomaly around a salt dome in the Gulf of Mexico. This is an intrusion of older low density sedimentary rock into overlying strata, so the density anomaly is negative. The scale is a few km – such “local” gravity surveys are useful in exploration geophysics since structures such as salt domes can be associated with trapped natural gas, etc. Units: gal or milligal (1 mgal = 10

  • ms - ) (Galileo). A change in gravitational acceleration of 1 mgal is easily measured (1 part in 10 6 ). Note that the authors of the above paper attempted to use a clever technique (FNN = “forced neural network”) to derive a density profile directly from the gravity survey. In general this is very difficult because a given gravitational field could arise from many possible density structures. We normally build a model then calculate  g.