The Solid Earth Chapter 4 Answers to selected questions. (1) Love waves involve transverse motion, generally arrive before Rayleigh waves.

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1 The Solid Earth Chapter 4 Answers to selected questions (1) Love waves involve transverse motion, generally arrive before Rayleigh waves. () (a) T = 10 s, v ~4 kms -1, so wavelength is ~40 km. (b) T = 100 s, v ~4 kms -1, so wavelength is ~400 km. (a) T = 00 s, v ~4 kms -1, so wavelength is ~800 km. Not necessarily a good choice for resolution of small-scale inhomogeneities. (3) Explosion may have magnitude m b as much as units greater. So body wave amplitude would be times greater. (4) (a) body waves (P, S). (b) body waves (P, SV), Rayleigh waves. (5) For #1, t S-P = 5.8 s, distance = 50 km For #, t S-P =8. s, distance = 71 km For #3, t S-P =.9 s, distance = 5 km. Draw arcs to locate the epicentre. (6) Use Eq. 4.7: log 10 E = M S For an increase of M S from 5 to 6, the energy increases by joules. For an increase of M S from 6 to 7, the energy increases by joules. For an increase of M S from 7 to 8, the energy increases by joules. Use Eq. 4.7: log 10 E = m b For an increase of m b from 5 to 6, the energy increases by joules. For an increase of m b from 6 to 7, the energy increases by joules. For an increase of m b from 7 to 8, the energy increases by joules. (7) Use Eq (8) Energy released by an earthquake with an M s of 8 is joules.

2 (a) This energy would be released by 10 3 earthquakes with M s = 6 (b) This energy would be released by earthquakes with M s = 5 (b) This energy would be released by 10 6 earthquakes with M s = 4 Many small earthquakes would be required each day in order to release the same amount of energy. (14) Earthquake 1 is normal faulting planes: strike ~146, dip ~6 E, strike ~170, dip ~30 W. Earthquake is strike-slip planes: strike ~148, dip ~90, strike ~058, dip ~90. (15) Leaky transform fault to the west and compressive to the east and in the Mediterranean. (a) (b) (Sketches, not accurate). (16) Layer 00 m thick, with α 1 = kms -1 underlain by rock with α =3 kms -1. z Critical distance = 1 α 1 α α 1 = 0. 3 = 0.36 km = 360 m. A shallow seismic experiment over such a structure would require seismometer spacing of ca ten(s) of metre(s) over a few kilometres. Cross - over distance = = 0.89 km.

3 (17) Intercept (s) Velocity (km s -1 ) (a) Incercept = α 1 α 1 α = 6.08 km s -1, assume α 1 = km s -1 : 1.00 = 6.08 = 3.35 km. For other values α 1 = 4 kms 1, =.66 km; α 1 = 5 kms 1, = 4.39 km. α 1.90 = 3.35 z = 7.1 km z = 3.35 z 3 = 30.3 km Crustal thickness is 40.7 km z (b) If there is a layer with intercept 5.50 s and velocity 7.36 km s -1, the layer thicknesses are: 8.70 = 3.35 z 4 =14. km = 3.35 z 3 = 0.8 km Crustal thickness is 45.4 km, greater by 4.7 km z z (18) Three lines can be fitted to the data: Intercept (s) Velocity (km s -1 ) (a) Incercept = α 1 α 1 α α =.7 km s -1, α 1 = 1.5 km s -1, =3 km.

4 3 1.5 = 3.33 seconds. This is very close to 3.36 s so we could ignore any hidden low velocity layer at the seabed. However, if the seabed material had a velocity of.5 km s -1, it would have a thickness of 0.1 km. 4.3 = z z 3 = 0.6 km 4.8 = z 4 =1.6 km z (b) A normal upper mantle velocity of 8.1 km s -1, would have to have an intercept of 5. s, if it is always to be a second arrival (a hidden layer out to 30 km range). This would mean that the 7.5 km s -1 layer would have a thickness of.7 km. (c) If the upper mantle velocity was 7.6 km s -1, the minimum intercept for it to be a second arrival out to 30 km would be 5.0s. In this case the 7.5 km s -1 layer would have a thickness of.1 km. (d) In order to determine the upper mantle velocity the refraction line should be at least 50 km long and ideally considerably longer since the minimum depth for the upper mantle calculated above is km beneath the seabed. (1) Use the method of Sect with the maximum and minimum values as specified in the question = =1.8 α = =.08 = z = =1.175 (a) travel time error of ±0.05s, gives min and max values: 1.71 and 1.89 α 4.00 and 5.59 z 0.80 and 1.68 (b) travel time error of ±0.01s, gives min and max values: 1.78 and 1.8 α 5 and 4.86 z 1.09 and 1.6

5 (c) rms velocity error of ±0.1kms -1, gives min and max values: 1.75 and 1.85 α 4.7 and 5.09 z 1.07 and 1.7 () (a) (b) (c) = = = (3) (a) Sandstone λ = =150 m λ = =100 m λ = = 75 m (c) Gabbro λ = = 350 m λ = = 33 m λ = =175 m (4) Use Sect Plot t against x. Eq shows that for a wide-angle reflection the data will plot as a straight line with intercept 4z /α and slope 1/α The best fitting straight line for this data has intercept 1.00 and slope Hence α=4.00kms -1 and z=km.

6 (7) (a) The four phases have travel times as follows: t = x t = x 3 t =1.8 + x 5.5 t =.56 + x 7 (b) Receivers need to be positioned out to at least 35 km range. The 7 km s -1 refractor should be observable as a first arrival beyond ~0 km. (c) Shoot a reversed line. (d) See discussion on hidden layers and low-velocity zones amplitude of second arrival and delays in travel times. (8) (a) Peak frequencies are 75 and 5 Hz. (b) Coverage for Survey 1 is 7/( 0.5)=7. Coverage for Survey is 100/( 1)=50. (c) The first survey is to determine the shallow structure: shorter streamer, lower moveout, higher frequency. The second survey is to image the crust: greater moveout, longer streamer, lower frequency. (9) Use Eqs and Example in Sect (a) α 1 =.1, α =.55, α 3 = 3.51 (b) =1.06 km, α 1 =.1 kms -1 z =1.46 km, α 1 =.9 kms -1 =.45 km, α 1 = 4.89 kms -1 (31) Sediment / Basement 1 contact Basement 1 / Basement contact = = 0.019

7 (33) Let the distance to train be x. The velocity of sound in air is ~0.33 km s -1 ; estimate the velocity of vibration along the track ~5 km s -1. x 0.33 x 5 = 0 x = 0 = 7.1 km.8 If the train is travelling at 60 km per hour, you have almost 7 minutes. If the train is travelling at 100 km per hour, you have just over 4 minutes. If the train is travelling at 150 km per hour, you have less than 3 minutes.