Planets, Isostasy and the Seismic Structure of the Earth
2. Isostasy: the Airy and Pratt models
Isostasy is the hydrostatic balance of the lithosphere floating on the weak asthenosphere: columns of different height and density exert the same pressure at a compensation depth, below which the mantle behaves as a fluid with no lateral pressure difference. A famous early clue came when the Great Trigonometrical Survey of India found that the plumb line near the Himalaya was deflected less than the mass of the mountains should cause, so the mountains must have low-density material beneath them. Two models, both put forward in 1855, explain it.
| Feature | Airy | Pratt |
|---|---|---|
| Crustal density | Same everywhere, ρc | Varies: lower under higher topography |
| Column base | Varies: roots under mountains, anti-roots under basins | Level, at one fixed depth D |
| Balance condition | h ρc = r (ρm − ρc) | ρ (D + h) = ρ0 D |
| Best suited to | Mountain belts with thick crust | Oceanic lithosphere: hot, light ridges and cooler, denser old floor |
Airy: a crust of uniform density ρc floats in denser mantle ρm like icebergs of different sizes, and higher peaks have deeper roots. Equating the excess weight above sea level with the buoyancy of the root gives h ρc = r (ρm − ρc), so r = h ρc/(ρm − ρc). For ρc = 2,800 and ρm = 3,300 kg/m³, r = 5.6 h: each kilometre of elevation needs 5.6 km of root. Under an ocean of depth d the anti-root is a = d (ρc − ρw)/(ρm − ρc). Pratt: blocks of different density all reach the same depth D, measured here from sea level. A column rising h above sea level is D + h long and must weigh as much as the reference column of density ρ0 and length D, so ρ = ρ0 D/(D + h): the higher the topography, the lower the density.
Neither model is exact. A real lithosphere has strength, so a load is carried partly by flexure (the regional compensation of Vening Meinesz): broad features are compensated, while a narrow seamount is held up by the rigidity of the plate. Glacial isostatic adjustment shows isostasy in time. A 2 km ice sheet (917 kg/m³) pushes the lithosphere down by ρi hi/ρm, about 0.56 km; when it melts the land rebounds only as fast as viscous mantle flow allows, and Fennoscandia and Hudson Bay are still rising by roughly a centimetre a year. The rebound rate measures mantle viscosity.
3. Internal structure of the Earth
The Earth is layered in two overlapping ways. Compositionally it is crust, mantle and core: the mantle (peridotite, rich in olivine and pyroxene) holds about two-thirds of the mass and the iron–nickel core, with some light elements, about one-third. Mechanically it is lithosphere (rigid, about 100 km, thinner under young ocean floor and much thicker under old cratons), asthenosphere (weak, with a low-velocity zone where a little melt may be present), the stronger lower mantle, the liquid outer core and the solid inner core. Layer boundaries are found as discontinuities, abrupt changes in Vp, Vs or both.
| Boundary | Depth | What changes |
|---|---|---|
| Moho | 5–10 km (ocean), 30–70 km (continent) | Vp jumps to about 8 km/s; compositional, crust to mantle (Mohorovičić, 1909) |
| Lithosphere–asthenosphere | About 100 km, variable | Mechanical and thermal, not compositional; Vs falls in the low-velocity zone |
| 410 km and 660 km | Upper mantle transition zone | Phase changes: olivine to a denser spinel-type form at 410 km; ringwoodite to bridgmanite plus ferropericlase at 660 km |
| Core–mantle boundary (Gutenberg) | About 2,890 km | Vp falls from about 13.7 to 8.0 km/s; Vs drops to 0; density rises from about 5.6 to 9.9 g/cm³ |
| Inner–outer core (Lehmann) | About 5,150 km | Liquid to solid; Vp rises to about 11 km/s; Vs of about 3.5 km/s reappears |
The history of discovery is a small chain of inferences. Mohorovičić found the crust–mantle boundary from a second set of P arrivals in 1909. Oldham (1906) recognised that the Earth has a core, Gutenberg (1914) placed its top near 2,900 km, and Lehmann (1936) explained weak P arrivals inside the P shadow zone by an inner core of higher wave speed, later shown to be solid. Densities rise from about 2.7–3.0 g/cm³ in the crust and 3.3 in the upper mantle to about 5.6 at the base of the mantle, 10–12 in the outer core and about 13 at the centre. The overall velocity structure follows: Vp is about 6 km/s in the upper crust and 8.0 to 8.2 km/s just below the Moho, dips slightly in the low-velocity zone, steps up across 410 and 660 km, rises smoothly to about 13.7 km/s at the base of the mantle, falls to 8.0 km/s in the outer core and rises to about 11 km/s in the inner core; Vs follows the mantle trend to about 7.3 km/s, is zero in the outer core and is about 3.5 km/s in the inner core.
4. Elasticity, body waves and surface waves
For small, fast deformations the Earth behaves as an elastic solid: strain is proportional to stress (Hooke’s law) and is recovered. An isotropic solid needs two independent constants, for example the Lamé parameters λ and μ (rigidity or shear modulus). From them follow the bulk modulus K = λ + 2μ/3, Young’s modulus and Poisson’s ratio ν = λ/[2(λ + μ)]. A fluid has μ = 0.
In an unbounded medium there are two body waves. The P wave is a compression–dilatation: particles move along the direction of travel, volume changes (dilatational) and there is no rotation (irrotational). Vp = √[(K + 4μ/3)/ρ]. The S wave is a shear: particles move across the direction of travel (as SV in the vertical plane or SH in the horizontal), shape changes but volume does not (equivoluminal, rotational), and Vs = √(μ/ρ). S waves cannot pass through a fluid. Since K > 0, Vp/Vs = √(K/μ + 4/3) > 1.15, and in terms of Poisson’s ratio Vp/Vs = √[2(1 − ν)/(1 − 2ν)], which is √3 ≈ 1.73 for a Poisson solid (λ = μ, ν = 0.25) and grows without limit as ν → 0.5. Example: K = 50 GPa, μ = 30 GPa, ρ = 2,500 kg/m³ give λ = 30 GPa, Vp = √(90×10⁹/2,500) = 6,000 m/s, Vs = 3,464 m/s.
Surface waves are guided along the free surface and decay with depth. Rayleigh waves move particles in a vertical ellipse that is retrograde at the surface, exist even in a homogeneous half-space and travel at about 0.92 Vs for a Poisson solid. Love waves are horizontal SH motion at right angles to travel and need a slow layer over faster material; their speed lies between the two Vs values, so they usually arrive before Rayleigh waves. Both are dispersive: where speed rises with depth, long periods sample deeper, faster rock and travel faster. Surface waves spread in two dimensions, so their amplitude falls as 1/√r against 1/r for body waves, and on a record from a shallow earthquake they are the largest arrivals.
| Wave | Particle motion | Passes through | Arrival order |
|---|---|---|---|
| P | Along travel direction | Solids, liquids and gases | First |
| S | Across travel direction | Solids only | Second |
| Love | Horizontal, transverse | Surface layers, needs speed rising with depth | Third |
| Rayleigh | Vertical ellipse, retrograde at surface | Free surface of any solid | Last |
5. Propagation of body waves in the Earth’s interior
Snell’s law at a boundary is sin i₁/v₁ = sin i₂/v₂. In a spherical Earth the quantity that stays constant along a ray is the ray parameter p = r sin i/v (p = sin i/v in flat layers). A ray turns round where i = 90°, at the radius where v = r/p. Because speed generally increases with depth, rays curve back up; a ray entering slower material bends toward the normal. Phase names record the path: p and s for an upgoing leg from the source (pP, sP), c for reflection at the core–mantle boundary (PcP, ScS), K for a P leg in the outer core (PKP, SKS), I for a P leg in the inner core (PKIKP) and i for reflection at its top.
Travel times. The S–P interval fixes distance: D = Δt/(1/Vs − 1/Vp). Take a crust of thickness h and speed v₁ over a mantle of speed v₂ > v₁. At short range the first arrival is the direct wave, t = x/v₁. The head wave (Pn) travels along the Moho after incidence at the critical angle ic = sin⁻¹(v₁/v₂), with t = x/v₂ + 2h cos ic/v₁. Beyond the crossover distance xc = 2h √[(v₂ + v₁)/(v₂ − v₁)] it arrives first; the slope of its t–x line is 1/v₂ and the intercept 2h cos ic/v₁ gives h. For v₁ = 6.0 and v₂ = 8.0 km/s, sin ic = 0.75, ic = 48.6°, and h = 20 km gives xc = 40 √7 ≈ 106 km. If the S–P interval is 25 s with Vp = 6.0 and Vs = 3.5 km/s, D = 25/(1/3.5 − 1/6.0) = 210 km.
Shadow zones. At the core–mantle boundary Vp falls from about 13.7 to 8.0 km/s, so P rays entering the core bend sharply toward the normal and emerge at larger distances. The direct P wave is therefore absent between about 103° and 142° of epicentral distance (the P shadow zone; 1° is about 111 km), while direct S waves are not seen beyond about 103° because the outer core cannot transmit them. Weak P arrivals inside the P shadow zone include diffracted waves and PKIKP waves refracted by the faster inner core, which Lehmann used as evidence for the inner core. A low-velocity layer such as the asthenosphere also bends rays down and leaves a shallow shadow, the same effect on a smaller scale.
Key takeaways
- Terrestrial planets are small and dense with silicate mantles and iron cores: Mercury (huge core), Venus (460 °C CO₂ greenhouse, no moons), Earth (plate tectonics) and Mars (Olympus Mons, thin atmosphere). The Moon has anorthositic highlands and basaltic maria; Io is tidally heated, Ganymede has a magnetic field and Titan a thick nitrogen atmosphere.
- Isostasy is hydrostatic balance at a compensation depth. Airy: constant density, variable roots, r = h ρc/(ρm − ρc) (5.6 h for 2,800 and 3,300). Pratt: constant depth D, variable density, ρ = ρ0 D/(D + h). Flexure and glacial rebound show that real lithosphere has strength and the mantle is viscous.
- The compositional layers are crust, mantle and core; the mechanical ones are lithosphere, asthenosphere, mesosphere, liquid outer core and solid inner core. Key boundaries: Moho (Vp to about 8 km/s), 410 and 660 km phase changes, core–mantle boundary at about 2,890 km (Vp 13.7 to 8.0, Vs to 0) and inner core at about 5,150 km.
- Vp = √[(K + 4μ/3)/ρ] and Vs = √(μ/ρ); P is compressional, irrotational and passes through fluids, S is shear, equivoluminal and cannot. Vp/Vs > 1.15, equals √3 for a Poisson solid (λ = μ, ν = 0.25) and rises with ν.
- Rayleigh waves are retrograde vertical ellipses at about 0.92 Vs; Love waves are SH waves needing a slow layer over a faster base and outrun Rayleigh waves. Both are dispersive and decay as 1/√r, so they dominate distant shallow-event records.
- Snell’s law gives a constant ray parameter. Critical angle sin ic = v₁/v₂; crossover xc = 2h √[(v₂ + v₁)/(v₂ − v₁)]; D = Δt/(1/Vs − 1/Vp). P shadow zone 103° to 142° comes from refraction at the core; direct S is absent beyond 103° because the outer core is liquid.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which terrestrial planet has a very dense carbon-dioxide atmosphere, a surface temperature of about 460 °C and a slow retrograde rotation?
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Answer: B — Venus, whose thick greenhouse atmosphere makes its surface hotter than that of Mercury
Venus holds a CO₂ atmosphere of about 90 times Earth’s surface pressure, and the resulting greenhouse effect gives a surface near 460 °C, hotter than Mercury. Mercury, the tempting answer because it is closer to the Sun, has almost no atmosphere, so its night side cools far below zero. Mars has a thin atmosphere and a cold surface.Which moon is the most volcanically active body known in the solar system, its heat coming mainly from tidal flexing by its parent planet?
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Answer: D — Io, whose orbit around Jupiter is kept slightly eccentric so that its interior is flexed
Io is flexed by Jupiter’s tides because the pull of the other large moons keeps its orbit slightly eccentric; the dissipated energy melts its interior and drives continuous volcanism. Callisto is the opposite, an old unaltered surface. Titan is known for its atmosphere, and Ganymede’s magnetic field does not cause eruptions.Which of the following statements about the Moon are correct?
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Answer: A — The pale lunar highlands are made largely of anorthosite, a plagioclase-rich rock; B — The dark maria are basalt flows that filled large impact basins
Highlands of anorthosite and mare basalts are the two basic lunar terrains. The Moon has no strong global field today, and its radius of about 1,740 km is smaller than Mercury’s 2,440 km.Two mountain ranges, one 2 km high and one 4 km high, are each in isostatic balance. Which description fits the Airy model but not the Pratt model?
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Answer: C — The higher range has a root about twice as deep, and both ranges are made of crust of the same density
In the Airy model the crust has one density, so a range twice as high needs a root twice as deep (r = h ρc/(ρm − ρc)). A fixed compensation depth with density varying with height describes Pratt. Support by plate rigidity is the flexural model, and equal weight with higher density is impossible, since a taller column of denser rock would weigh more.A plateau stands 3.0 km above sea level and is in Airy isostatic balance. Take the crustal density as 2,800 kg/m³ and the mantle density as 3,300 kg/m³, and let the normal crust at sea level, which forms the reference column, be 35.0 km thick. The total crustal thickness beneath the plateau, in km, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 54.8
Root r = h ρc/(ρm − ρc) = 3.0 × 2,800/(3,300 − 2,800) = 16.8 km. The total thickness is the reference crust plus the elevation plus the root: 35.0 + 3.0 + 16.8 = 54.8 km. Quoting only the root (16.8) or the reference crust plus root (51.8) is the common slip.In a Pratt model the compensation depth is D = 100 km below sea level, and the reference column at sea level has density 3,300 kg/m³. Every column extends from depth D up to the surface. The density of a column whose top stands 2 km above sea level, in kg/m³, to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 3235
The balance is ρ (D + h) = ρ0 D, so ρ = 3,300 × 100/(100 + 2) = 330,000/102 = 3,235.3, which is 3,235 kg/m³. Using D − h instead of D + h is the common error, because the column is longer than D, not shorter.A broad high plateau is in isostatic equilibrium, supported by a thick crustal root. Which gravity signature is expected over it?
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Answer: A — A free-air anomaly near zero and a strongly negative Bouguer anomaly
Compensation means the mass excess of the topography is cancelled by the mass deficit of the low-density root, so the free-air anomaly is small. The Bouguer correction removes only the topography, leaving the root’s deficit, so the Bouguer anomaly is strongly negative. A strongly positive free-air anomaly is what an uncompensated load would give.The Mohorovičić discontinuity is the boundary across which:
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Answer: A — P-wave speed rises abruptly to about 8 km/s as crust gives way to mantle
The Moho separates crust from mantle and shows as a rise of Vp to about 8 km/s. The vanishing of S waves and the fall of Vp from 13.7 to 8.0 km/s belong to the core–mantle boundary, and the liquid-to-solid change belongs to the inner-core boundary.Which observation shows most directly that the outer core is liquid?
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Answer: C — S waves are not transmitted through the outer core, because a liquid has zero rigidity
A liquid has μ = 0, so it cannot carry a shear wave: the absence of S in the core is direct evidence. The P shadow zone shows only a drop in Vp at the boundary, which is the tempting but weaker clue, since a velocity drop does not by itself prove a liquid. A mean density of 5.5 shows a dense core, not its state, and the density jump at 2,890 km is mostly a change of composition (silicate to iron alloy).Which of the following statements about surface waves are correct?
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Answer: A — Rayleigh waves move particles in a retrograde ellipse at the surface, in the vertical plane of travel; B — Love waves are horizontally polarised shear waves that need a slow layer lying over faster material
Rayleigh motion is a retrograde vertical ellipse at the surface, and Love waves need a velocity increase with depth, as a slow layer over a faster base provides. A Rayleigh wave is slower than S, about 0.92 Vs for a Poisson solid. Because speed changes with depth, long periods sample faster rock and travel faster, so surface waves are dispersive.A rock has bulk modulus K = 100 GPa, rigidity μ = 45 GPa and density ρ = 2,500 kg/m³. The P-wave speed in this rock, in m/s, is ____.
Numerical answer — type the value.
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Answer: 8000
Vp = √[(K + 4μ/3)/ρ] = √[(100 + 60) × 10⁹/2,500] = √(64 × 10⁶) = 8,000 m/s. Using K alone, or K + μ, gives a wrong speed, because the P wave also shears the rock and needs the 4μ/3 term.A rock has Lamé constants λ and μ that are equal (λ = μ), a Poisson solid. The ratio Vp/Vs of its body-wave speeds is:
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Answer: D — 1.73
Vp/Vs = √[(λ + 2μ)/μ] = √(3μ/μ) = √3 = 1.73, which corresponds to Poisson’s ratio 0.25. The value 3.00 is the square of the ratio (the omitted square root), and 1.15 is the lower limit √(4/3) reached only when K is nearly zero.An earthquake is recorded with an S–P interval of 25 s. Take straight-line paths with Vp = 6.0 km/s and Vs = 3.5 km/s. The travel time of the P wave from the source to the station, in s, is ____.
Numerical answer — type the value.
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Answer: 35
Distance D = Δt/(1/Vs − 1/Vp) = 25/(1/3.5 − 1/6.0) = 25/(2.5/21) = 210 km. The P travel time is 210/6.0 = 35 s (and the S time is 210/3.5 = 60 s, which differs by 25 s). Stopping at the distance, or dividing it by Vs, are the usual slips.A P wave travelling at 5.0 km/s in the crust meets a mantle in which P waves travel at 10.0 km/s. At what angle of incidence, measured from the normal, does the refracted ray travel along the boundary?
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Answer: B — 30°
The critical angle satisfies sin ic = v₁/v₂ = 5.0/10.0 = 0.5, so ic = 30°. The value 60° results from using cos instead of sin, and 26.6° from taking tan⁻¹(0.5). The refracted ray runs along the boundary as a head wave only at this angle.In a refraction survey a crust with P speed 6.0 km/s overlies a mantle with P speed 8.0 km/s. The head wave from the Moho overtakes the direct wave at a crossover distance of 106 km. The crustal thickness, in km, to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 20
From xc = 2h √[(v₂ + v₁)/(v₂ − v₁)], h = (xc/2) √[(v₂ − v₁)/(v₂ + v₁)] = 53 × √(2/14) = 53 × 0.378 = 20.03 km, so 20 km. Writing the root upside down (√7) gives 140 km, a result that is far too thick for a crust and a good warning.Why is no direct P wave recorded between about 103° and 142° of epicentral distance?
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Answer: B — P speed drops sharply at the core–mantle boundary, so rays entering the core bend toward the normal and miss this zone
The sharp drop in Vp at 2,890 km bends rays that enter the core strongly toward the normal. The ray that just grazes the core emerges at about 103°, and the rays that enter it are refracted so strongly that they emerge no nearer than about 142°, leaving a band without direct P. The core transmits P, so it is not absorbed. Conversion is partial and does not remove the P phase, and the asthenosphere produces only a shallow local shadow.Which of the following statements about the layers of the Earth are correct?
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Answer: A — The 660 km discontinuity is a phase change in which ringwoodite breaks down to bridgmanite and ferropericlase; C — The inner core is solid although very hot, because the high pressure raises its melting temperature; D — At the core–mantle boundary P speed drops sharply while density rises
The 660 km boundary is the breakdown of ringwoodite to bridgmanite plus ferropericlase, the inner core freezes because pressure raises the melting point of iron alloy faster than the temperature rises, and across the core–mantle boundary Vp falls from about 13.7 to 8.0 km/s while density rises from about 5.6 to 9.9 g/cm³. The lithosphere–asthenosphere boundary is a change of strength, not of composition, so it is not the crust–mantle change.