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Introducción a la Geofísica
2010-01
TAREA 6
1) FoG. Calculate the bulk modulus (K), the shear modulus (µ) and Poisson’s
ratio (ν) for the lower crust, upper mantle and lower mantle, respectively,
using Eqs. (3.153) and (3.156) and the values for the P-wave (α) and S-wave
(β) velocities, and density (ρ) in the following table.
The seismic velocities α and β are related to the elastic parameters as follows:
µ
#
"=
!
"2 =
µ = #" 2
4
µ
&
K
4
4 )
3
= + $ 2 , from which K = #(" 2 % $ 2 +
'
#
#
3
3 *
K+
In terms of the Lamé constants:
!
!
2
K ="+ µ
3
2
" = K # µ = $(% 2 # 2& 2 )
3
and the Poisson ratio is
!
"=
!
!
"=
#
2( # + µ)
#($ 2 % 2& 2 )
(
2 # ($ 2 % 2& 2 ) + & 2
($ % 2& )
=
) 2($ % 2& )
2
2
2
2
Inserting the values given in the table, we obtain the following values for the elastic
constants in different depth regions:
2) FoG The table below gives the densities and seismic P- and S-wave velocities
at various depths in the Earth.
(a) From these quantities calculate the rigidity modulus (µ), bulk modulus (K),
and Poisson’s ratio (ν) at each depth.
The equations for computing the elastic parameters are the same as in the
previous exercise. The values given in the table for various depths in the Earth
are converted to depth-profiles of K, µ and ν, as in the following table:
(b) Discuss in your own words the information that these data give about the
deep interior of the Earth.
The Earth’s shell-like internal structure of mantle, fluid outer core, and solid
inner core are evident from the velocities and elastic parameters ratios. The
inner core is solid, but the high values of α/β, K/µ and Poisson’s ratio show
that it is less rigid than the mantle.
3) FoG A strong earthquake off the coast of Japan sets off a tsunami that
propagates across the Pacific Ocean (average depth d = 5 km).
(a) Calculate the velocity of the wave in km hr-1 and the corresponding
wavelength, when the wave has a dominant period of 30 min.
The phase velocity and group velocity of a tsunami are equal when the water
depth is
much less than the wavelength. In this case, the velocity V of the tsunami is
related to the ocean depth d by the equation.
V = gd
!
The wavelength λ and period T of the wave are related to the velocity by
" = VT
For an ocean depth of 5 km, the velocity of the tsunami over the open ocean is
!
V=
(9.81)(5000) = 221 ms-1 = 797 km hr -1
The wavelength corresponding to a period of 30 min (1800 s) is:
λ = (221)(1800) = 397800 m = 398 km
!
(b) How long does the wave take to reach Hawaii, which is at an angular
distance of 54° from the epicenter?
The time taken for the tsunami to reach Hawaii is the great circle distance
divided by the velocity of the wave.
The great circle distance is s = R θ where R = 6371 km is the Earth’s radius
and θ is the angular distance in radians, in this case (54/180)π = 0.3 π .
s = 6371(0.3) π = 6005km
The time for the tsunami to reach Hawaii is therefore
t = 6005 / 797 = 7.5 hr
4) FOG In a two-layer Earth the mantle and core are each homogeneous and the
radius of the core is one-half the radius of the Earth. Derive a formula for the
travel-time curve for the arrival time t of the phase PcP at epicentral distance
Δ. Verify the formula for the maximum possible value of Δ in this model.
The geometry of the problem is shown in the diagram above. The distance
traveled by the PcP phase is (2s), where
" R %2
" R% ") %
s2 = R 2 + $ ' ( 2R$ ' cos$ '
#2&
#2& #2&
s=
") %
R
5 ( 4 cos$ '
#2&
2
The travel-time for the arrival at epicentral distance Δ is
!
t=
!
$#'
2s R
=
5 " 4 cos& )
%2(
V V
The maximum travel-time tm occurs when the PcP path is a straight line
tangential to the core at the point P. Triangle OPS0 has sides R and R/2
enclosing angle (Δ/2), which is then equal to 60°, because
# " & OP ( R 2) 1
cos% ( =
=
=
$ 2 ' OS0
R
2
The maximum travel-time is then equal to
!
tm =
!
S0 P + PG0
=2
V
(
)=R
3R 2
V
V
3
This is the same value as obtained from the travel-time equation with Δ equal
to 120°, the maximum possible epicentral angle for a PcP phase in this
problem.
The travel-time equation can be written in terms of tm as
t = tm
!
5 4 $# '
" cos& )
3 3 %2(
The travel-time curve looks like the following diagram:
There are no PcP arrivals at epicentral distances greater than 120°.
5) FoG The following table gives arrival times of P-waves (tp) and S-waves (ts)
from a nearby earthquake:
(a) Plot the arrival-time differences (ts – tp) against the arrival times of the Pwave to produce a Wadati-diagram.
%# (
The equation of the Wadati plot is t s " t p = t p ' "1*
&$ )
!
(b) Determine the ratio α/β of the seismic velocities.
The best-fit straight line to the points on the Wadati diagram in this exercise is
y = 0.737x - 37.0 , where x = tp is the arrival time of the P-wave, and y = (ts – tp)
is the difference in travel times of the P- and S-waves. The coefficient of x
gives
%" (
' $1* = 0.73 from which the velocity ratio is
&# )
"
= 1.737
#
(c) Determine the time of occurrence (t0) of the earthquake
!
The intercept of the best-fit straight line on the y-axis is –37.0; using the slope
of the line gives the intercept on the x-axis.
0.737x = y + 37.0
y
37.0
x=
+
= 1.36y + 50.5
0.737 0.737
The earthquake occurred at 50.5 seconds after 22:36.
!