The Solid EarthChapter 2

Answers to selected questions

(1) (a)

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B vD=B vC +CvDB vD

2 = 62 +102 − 2 ×10cos 25( )B vD = 5.22

sin φ( ) =10 × sin 25( )

5.22φ =126o

(b)

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A vJ =A vB+BvJAt 3 cm per year in E - W direction, time taken to move 3000 km is

3000 ×103

3×10−2 =108 a =100 Ma

(Diagrams are sketches and are not exactly to scale.)

(2)

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A vB=A vF+FvB

(Diagrams are sketches and not exactly to scale)

The B-F ridge approaches the trench eastwards at 0.8 cm per year. 2,200 km to thetrench will take 275 Ma.Triple junction J approaches the northern trench at 4.2 cm per year. 1500 km takes36 Ma when the last remnant of the Cooler plate is subducted in this region.

(3)

Lat Long Location Plate pair Azimuth Rate(mm/yr)

54°N 169°E W. Aleutian Trench N Am - Pac 133 7452°N 169°W E. Aleutian Trench N Am - Pac 145 6838°N 122°W San Francisco - San

Andreas FaultN Am - Pac 146 46

26°N 110°W Gulf of California N Am - Pac 129 4613°S 112°W East Pacific Rise Naz - Pac 103 14336°S 110°W East Pacific Rise Pac - Ant 281 9459°S 150°W Antarctic - Pacific Ridge Pac - Ant 303 7145°S 169°E S. New Zealand Pac - Aus 248 3555°S 159°E Macquarrie Island Pac - Aus 212 2752°S 140°E Southeast Indian Ridge Aus - Ant 359 6928°S 74°E Southeast Indian Ridge Aus - Ant 045 587°N 60°E Carlsberg Ridge Ind - Afr 032 2622°N 38°E Red Sea Ar - Afr 012 1055°S 5°E Southwest Indian Ridge Afr - Ant 043 1452°S 5°E Mid-Atlantic Ridge Afr – S Am 068 30

9oN 40oW Mid-Atlantic Ridge Afr – S Am 090 2835oN 35oW Mid-Atlantic Ridge Afr – N Am 104 2166oN 18oW Iceland Eur – N Am 105 1836oN 8oW Gorringe Bank Afr - Eur 310 435oN 25oE E. Mediterranean Afr - Eur 353 912oS 120oE Java Trench Au – Eur 017 7635oN 72oE Himalayas Ind - Eur 003 4235oS 74oW S. Chile Trench Naz – S Am 079 804oS 82oW N. Peru Trench Naz – S Am 082 7020oN 106oW Middle America Trench Co – N Am 034 41

(5) λp = 45°, φp = 0°, ω = 10-10 radians per year, R = 3400 km.

(a) subduction, strike-slip and ridge.(b) Magnetic lineations between a and b, b and c.(c) Use method of Sect. 2.4.2.

a: λx = 0°, φx = 0°b: λx = 0°, φx = 90°c: λx = 0°, φx = 180°d: λx = 0°, φx = -90°

At a:

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a = cos−1 cos 45( )cos 0( )[ ]= cos−1 0.707[ ]= 45o

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C = sin−1 cos 45( )sin 0( )sin 45( )

= 0o

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β = 90o

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v =ωRsin a( )=10−10 × 3400 ×103 ×103 sin 45( )= 0.24 mm a-1, azimuth 090, strike slip

At b:

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a = cos−1 cos 45( )cos 0 + 90( )[ ]= 90o

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C = sin−1 cos 45( )sin −90( )sin 90( )

= −45o

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β = 45o

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v =ωRsin a( )=10−10 × 3400 ×103 ×103 sin 90( )= 0.34 mm a-1, azimuth 45, strike slip/extension

At c:

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a = cos−1 cos 45( )cos 0 −180( )[ ]=135o

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C = sin−1 cos 45( )sin −180( )sin 135( )

= 0o

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β = 90o

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v =ωRsin a( )= 0.24 mm a-1, azimuth 090, strike slip

At d:

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a = cos−1 cos 45( )cos 0 + 90( )[ ]= 90o

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C = sin−1 cos 45( )sin 90( )sin 90( )

= 45o

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β =135o

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v =ωRsin a( )= 0.34 mm a-1, azimuth 135, oblique subduction

(d)

(sketches only)

Stable with hemispheric plates for all pole positions. Plates stay thesame size but are subducted along half of their margin and createdalong the other half.

(6) AωB 3x10-9 radians a-1, 30°N 0°E BωC -6x10-9 radians a-1, 90°N 0°E R=6000 km.

(a) Use method of Sect. 2.4.3.

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AωC=AωB+BωC

xAC = −6 ×10−9 cos 90( )cos 0( ) + 3×10−9 cos 30( )cos 0( )= 2.6 ×10−9

yAC = −6 ×10−9 cos 90( )sin 0( ) + 3×10−9 cos 30( )sin 0( )= 0

zAC = −6 ×10−9 sin 90( ) + 3×10−9 sin 30( )= −4.5 ×10−9

AωC = 2.62 + 4.52 ×10−9 = 5.2 ×10−9

pole position :

λAC = sin−1 4.5 ×10−9

5.2 ×10−9

= −60 = 60oS

φAC = tan−1 02.6

= 0

(b) At b:

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λX = 0, φX = 90. B relative to A: β = 030, v = 18 mm a-1

C relative to B: β = 270, v = 36 mm a-1

C relative to A: β = 300, v = 31 mm a-1

(not to scale)bc would go through C if plate B is subducting beneath plate C.

At d:

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λX = 0, φX = −90. B relative to A: β = 150, v = 18 mm a-1

C relative to B: β = 270, v = 36 mm a-1

C relative to A: β = 240, v = 31 mm a-1

(not toscale)Stable if plates B and C are subducting beneath A, or if plate A issubducting between plates B and C.

Extension is taking place betweenplates A and B and between plates Aand C at b. The magnetic stripes oneach plate are 78 km wide and strikeE-W.

Extension is also taking placebetween plates B and C at d. Themagnetic stripes on each plate are180 km wide and strike N-S.

Ridges: d to North Pole; between a and b, b and c.Strike-slip: at a and c.Subduction: North Pole to b; between c and d, d and a.

(8)

(Sketch) Ridge half-spreading rates are 2.0 cm per year and 3.45 cm per year. Triple junction moves at 3.9 cm per year at 275 relative to plate C.

(10)

(Sketch only)

(a) 11.6 cm per year(b) 6 cm per year parallel to BvN(e) (Sketches only, not to scale)

At A At B

(f) BJN: 6 cm per year for 2 Ma, 120 km spacing, 1.6 cm per year for 2 Ma, 32 km spacing. (Sketch, not to scale)

(11) Similar to Q6.

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λAB = 0, φAB = 90, ωAB = 3.82 ×10−7 deg per yearλBC = −90, φBC = 0, ωBC = 2.86 ×10−7 deg per year

(a)

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xAC = 3.82cos 0( )cos 90( ) + 2.86cos −90( )cos 0( )( ) ×10−7

= 0yAC = 3.82cos 0( )sin 90( ) + 2.86cos −90( )sin 0( )( ) ×10−7

= 3.82 ×10−7

zAC = 3.82sin 0( ) + 2.86sin −90( )( ) ×10−7

= −2.86 ×10−7

AωC = 3.822 + 2.862( ) ×10−7

= 4.77 ×10−7

λAC = sin−1 −2.864.77

= −36.8o

φAC = sin−1 3.820

= 90o

(b) Magnetic lineations formed at ridge along equator between plates A and Cfrom 0° to 90°W. Maximum spreading rate is 5 cm per year at 0°N, 0°W (thetriple junction) - spreading is oblique. On far side of planet the boundarybetween plates A and C is a subduction zone with oblique subduction.

(c) The triple junction on the visible side of the planet is FFR, on the far side thetriple junction is FFT. See Fig. 2.16 for stability of triple junctions. The FFRjunction is not stable. The FFT junction will be stable if plate A is subductingbeneath plate C (since then ac and bc will be coincident).