Ridge Push and the Drag Force

Geodynamics

Plate-driving forces Lecture 10.6 - Ridge push and the drag force

Geodynamics www.helsinki.ﬁ/yliopisto 1 Goals of this lecture

• Calculate the ridge push and drag forces acting on a lithospheric plate

2 The driving forces of plate motion 165 3 RP F

Downloaded from http://gji.oxfordjournals.org/ atDalhousieUniversityonSeptember1,2013 Forsyth and Uyeda, 1975 Uyeda, and Forsyth 3 F 2 F ) of the oceanic lithosphericplate: oceanic the )of 3 we must consider the forces acting on on acting forces the consider must we � 1 �� F � = RP F ) and side ( side )and 2 � results from the elevation of oceanic ridges relative to the to relative ridges oceanic of elevation the from results ), bottom ( bottom ), 1 Ridge push Ridge �

This motion may also be viewed as gravitational sliding gravitational as viewed be also may motion This

the top ( top the To calculate the ridge push force force push ridge the calculate To The difference in elevation results in a pressure head that drives the plate away away plate the drives that head pressure ina results inelevation difference The ridge the from Ridgepush seaﬂoor

•

• • The driving forces

Ridge push of

FRP plate motion

6.22 The Forces that DriveForsyth Plate and Uyeda, Tectonics 1975 519

Fig. 6.44, Turcotte and Schubert, 2014 165

• With this force balance in mind, we

http://gji.oxfordjournals.org/ from Downloaded

can see 2013 1, September on University Dalhousie at 1. The horizontal force on the base of the plate must be equal to the integrated lithostatic pressure in the mantle along RD 2. The horizontal force on the top of the plate must be equal to the integrated hydrostatic pressure along AB Figure 6.44 Horizontal forces acting on a section of the ocean, lithosphere, 3. The horizontal force actingand mantleon the at lithospheric an ocean ridge. section BC is equal to the integrated pressure in the lithosphere The integrated pressure force on the upper surface of the lithosphere F2 is • Note that this pressureequal to shouldF4,thenetpressureforceon include that resulting ABfrom,becausethesectionofwater the overlying RAB oceanic water must be in equilibrium. Thus we can integrate the hydrostaticpressurein the water to obtain 4 w F2 = F4 = ρwgy dy, (6.396) !0 where ρw is the water density. The horizontal force F3 acting on the section of lithosphere BC is the integral of the pressure in the lithosphere PL

yL F3 = PL dy,¯ (6.397) !0 where y¯ PL = ρwgw + ρLgdy¯′ (6.398) !0 and ρL is the density in the lithosphere. Substituting Equation (6–398) into Equation (6–397) gives

yL y¯ F3 = ρwgw + ρLgdy¯′ dy.¯ (6.399) !0 " !0 #

The net horizontal force on the lithosphere adjacent to an ocean ridge FR is obtained by combining Equations (6–394), (6–396), and (6–399) w F = F F F = g (ρ ρ )ydy R 1 − 2 − 3 m − w !0 yL y¯ + g (ρ ρ )w + ρ y¯ ρ dy¯′ dy.¯ m − w m − L !0 " !0 # (6.400) The driving forces

Ridge push of

FRP plate motion

6.22 The Forces that DriveForsyth Plate and Uyeda, Tectonics 1975 519

Fig. 6.44, Turcotte and Schubert, 2014 165

at Dalhousie University on September 1, 2013 1, September on University Dalhousie at http://gji.oxfordjournals.org/ from Downloaded

• At a constant depth �

P = ⇢gy assuming constant density of the overlying material Figure 6.44 Horizontal forces acting on a section of the ocean, lithosphere, • Integrated over a depthand range mantle �1 to at � an2 ocean ridge.

y2 Pint = ⇢gy dy The integrated pressure force on the upper surface of the lithosphere F2 is y1 Z equal to F4,thenetpressureforceonAB,becausethesectionofwaterRAB must be in equilibrium. Thus we can integrate the hydrostaticpressurein the water to obtain 5 w F2 = F4 = ρwgy dy, (6.396) !0 where ρw is the water density. The horizontal force F3 acting on the section of lithosphere BC is the integral of the pressure in the lithosphere PL

yL F3 = PL dy,¯ (6.397) !0 where y¯ PL = ρwgw + ρLgdy¯′ (6.398) !0 and ρL is the density in the lithosphere. Substituting Equation (6–398) into Equation (6–397) gives

yL y¯ F3 = ρwgw + ρLgdy¯′ dy.¯ (6.399) !0 " !0 #

Ridge push of

FRP plate motion

6.22 The Forces that DriveForsyth Plate and Uyeda, Tectonics 1975 519 Going back to the original force

• Fig. 6.44, Turcotte and Schubert, 2014 165

balance equation for ridge push,

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F = F F F RP 1 2 3 • After some mathematical substitutions and integrations we ﬁnd

2 ⇢m↵v(T1 T0) FRP = g⇢m↵v(T1 T0) 1+ t Figure 6.44 Horizontal⇡ (⇢ forces⇢ acting) on a section of the ocean, lithosphere, and mantle at an oceanm ridge. w

�� Mantle density �0 Temperature at plate surface The integrated pressure force on the upper surface of the lithosphere F2 is equal to F ,thenetpressureforceonAB,becausethesectionofwaterRAB �� Water density 4 � Age of oceanic plate must be in equilibrium. Thus we can integrate the hydrostaticpressurein 6 �1 Mantle temperaturethe water to obtain w F2 = F4 = ρwgy dy, (6.396) !0 where ρw is the water density. The horizontal force F3 acting on the section of lithosphere BC is the integral of the pressure in the lithosphere PL

yL F3 = PL dy,¯ (6.397) !0 where y¯ PL = ρwgw + ρLgdy¯′ (6.398) !0 and ρL is the density in the lithosphere. Substituting Equation (6–398) into Equation (6–397) gives

yL y¯ F3 = ρwgw + ρLgdy¯′ dy.¯ (6.399) !0 " !0 #

Ridge push of

FRP plate motion

6.22 The Forces that DriveForsyth Plate and Uyeda, Tectonics 1975 519

Fig. 6.44, Turcotte and Schubert, 2014 165

at Dalhousie University on September 1, 2013 1, September on University Dalhousie at http://gji.oxfordjournals.org/ from Downloaded

• Using typical values, we ﬁnd the ridge push force is

12 -1 �� = ~4×10 N m • Note that this is about an order of magnitude smaller thanFigure the 6.44 slab Horizontal forces acting on a section of the ocean, lithosphere, and mantle at an ocean ridge. pull force

The integrated pressure force on the upper surface of the lithosphere F2 is equal to F4,thenetpressureforceonAB,becausethesectionofwaterRAB must be in equilibrium. Thus we can integrate the hydrostaticpressurein the water to obtain 7 w F2 = F4 = ρwgy dy, (6.396) !0 where ρw is the water density. The horizontal force F3 acting on the section of lithosphere BC is the integral of the pressure in the lithosphere PL

yL F3 = PL dy,¯ (6.397) !0 where y¯ PL = ρwgw + ρLgdy¯′ (6.398) !0 and ρL is the density in the lithosphere. Substituting Equation (6–398) into Equation (6–397) gives

yL y¯ F3 = ρwgw + ρLgdy¯′ dy.¯ (6.399) !0 " !0 #

Drag force of plate FDF motion

416 FluidForsyth Mechanics and Uyeda, 1975

• The drag force on the base of the oceanic 165

lithosphere can both drive and resist plate

http://gji.oxfordjournals.org/ from Downloaded tectonics, depending on the relative motion 2013 1, September on University Dalhousie at between the plate and the underlying mantle

• If we assume that the underlying mantle resists or drives plate motion by viscous Fig. 6.2, Turcotte and Schubert, 2014 ﬂow across a ﬁxed-thickness layer, the drag force on the plate is simply

u FDF = ⌘as L h

�� Viscosity of asthenosphere ℎ Thickness of viscous layer �� Velocity difference � Length of the plate Figure 6.2 One-dimensional channel ﬂows of a constant8 viscosity ﬂuid.

To evaluate the constants, we must satisfy the boundary conditions that u =0 at y = h and u = u0 at y =0.Theseboundaryconditionsareknownas no-slip boundary conditions.Aviscousﬂuidincontactwithasolidbound- ary must have the same velocity as the boundary. When these boundary conditions are satisﬁed, Equation (6–11) becomes 1 dp u y u = (y2 hy) 0 + u . (6.12) 2µ dx − − h 0

If the applied pressure gradient is zero, p1 = p0 or dp/dx =0,thesolution reduces to the linear velocity profile y u = u0 1 . (6.13) ! − h" This simple flow, sketched in Figure 6–2a,isknownasCouette flow.Ifthe velocity of the upper plate is zero, u0 =0,thevelocityprofileis 1 dp u = (y2 hy). (6.14) 2µ dx − When we rewrite this in terms of distance measured from the centerline of the channel y′,where h y′ = y , (6.15) − 2 The driving forces

Drag force of plate FDF motion

416 FluidForsyth Mechanics and Uyeda, 1975

165

at Dalhousie University on September 1, 2013 1, September on University Dalhousie at http://gji.oxfordjournals.org/ from Downloaded

• Again using typical values, we ﬁnd the drag force is

13 -1 Fig. 6.2, Turcotte and Schubert, 2014 �� = ~1×10 N m • Note that this value is similar in magnitude to the slab pull force

Figure 6.2 One-dimensional channel ﬂows of a constant9 viscosity ﬂuid.

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• Reference(s): Forsyth, D., & Uyeda, S. (1975). On the Relative Importance of the Driving Forces of Plate Motion*. Geophysical Journal International, 43(1), 163–200. doi:10.1111/j.1365-246X.1975.tb00631.x