The continental lithosphere (plate) \"floats\" higher on the Earth\'s asthenosph

Question

The continental lithosphere (plate) "floats" higher on the Earth's asthenosphere than the oceanic lithosphere (plate). Assume that at a depth of 50 km below sea level, the oceanic lithosphere and continental lithosphere are both equally buoyant or "compensated". This case is shown in Figure 3 below, where the oceanic lithosphere is composed of [(sediment+basalt crust+ some mantle) +seawater] and the continental crust is made of (granite crust + mantle). Mathematically, when two columns of rock are equally buoyant, the lithostatic pressure (pressure due to the rocks) at the compensation depth will be equal c. Seawater Sediment Pseawater 1.03 gcm2 psediment-2.4 gcm-3 Granite continental crust Basalt Pbasalt 2.9 gcm3 Peridotite mantle peridotite 3.3 gcm ranite 2.8 gcm3 nsation -50 km 8 3 km Column of seawater, sediment, basalt oceaniC crust and peridotite mantle Column of granite continenta crust and peridotite mantle 39 km 15 km Compensation depth50 km 1338 Wadsworth Publishing Company/ITP Figure 3: Two columns of lithosphere that are compensated i.e.Peoi, lthosphere PeextipentaL lthosphere at 50 km depth below sea level For one rock type, lithostatic pressure can be calculated using where P is pressure, p is density, g is gravity, h is thickness and the 1 refers to the rock or material type For three different rock/material types (1-3) the calculation is Calculate the thickness of the continental granite crust in Figure 3 showing your workin Note that density is given in gcm3 and height is given in km.

Explanation / Answer

Answer : The required thickness of the continental granite crust is = 36.46 km

Explanation : For the oceanic lithosphere,

p1 = 1.03 g/cm3 = 1030 kg/m3; h1 = 3km = 3000m

p2 = 2.4 g/cm3 = 2400 kg/m3; h2 = 1km = 1000m

p3 = 2.9 g/cm3 = 2900 kg/m3; h3 = 6km = 6000 m

p4 =3.3 g/cm3 = 3300 kg/m3; h4 = 39km = 39000 m

Thus, POtotal = g(p1h1+p2h2+p3h3+p4h4)

=g*106(1.03*3 + 2.4*1 + 2.9*6 + 3.3*39)

=9.8*106(3.09+2.4+17.4+128.7)

=9.8*106*151.59

For continental lithosphere,

p1 = 2.8 g/cm3 = 2.8*103 kg/m3; h1 = ?

p2 = 3.3 g/cm3 = 3.3*103kg/m3; h2 = 15 km = 15*103 m

PCtotal = g*106(p1h1 + p2h2) = 9.8*106(2.8*h1 + 3.3*15)

Now according to question, at a compensation depth of 50km below sea level,

POtotal = PCtotal

or, 9.8*106*151.59 = 9.8*106*(2.8h1 + 49.5)

or, 151.59 = 2.8h1 + 49.5

or, 2.8h1 = 102.09

Or, h1 = 36.46 km = 36km approximately.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.