3Gravity and the Figure of the Earth

Two satellites fly one behind the other round the Earth, at an altitude of more than $500\,\mathrm{km}$ and about $220\,\mathrm{km}$ apart, and measure the distance between them all the time, to within a micrometre.

When they pass over a place where mass is concentrated, the leading satellite is pulled ahead first and the gap opens; then the trailing one is pulled in turn and the gap closes. From those tiny changes, month after month since 2002, the Earth’s gravity field has been mapped as it changes: the water stored in river basins rises and falls with the seasons, and the great ice sheets of Greenland and Antarctica lose mass year after year. Gravity weighs what the eye cannot see. This chapter builds the gravity field of a rotating planet from its potential, finds in it the flattening and the moment of inertia, follows its anomalies up to orbit, and then turns to one of the things gravity reveals at the surface: the bending of the lithosphere under the weight of a volcano.

You already know

Newton’s law of gravitation, the gravitational potential energy and the centrifugal force in a rotating frame are the physics books’; the moments of inertia of a solid and the Legendre polynomials $P_2(x) = (3x^2 - 1)/2$ are in the physics and mathematics books’ university volumes, Young’s modulus $E$ and Poisson’s ratio $\nu$ in the physics books’ Year 3 volume. Geodesy and the flattening $f = (a - b)/a$ of an ellipsoid of semi-axes $a > b$ were met in grade 10. Isostasy, Airy compensation, normal gravity, the geoid, the geoid height and the gravity anomalies were defined in the Year 1 volume, in its chapter on isostasy and gravity anomalies.

3.1 The geopotential

On a rotating planet a body at rest feels the attraction of the planet’s mass and, in the frame that turns with it, the centrifugal force. Both derive from a potential.

Definition 3.1 (Geopotential)

The geopotential is the potential energy per unit mass, in the frame rotating with the Earth at the angular velocity $\omega$, of gravitation and the centrifugal force together:

$$W(\mathbf r) = V(\mathbf r) + \Phi(\mathbf r),\qquad V(\mathbf r) = -G\int_{\text{Earth}}\frac{\mathrm dm'}{|\mathbf r - \mathbf r'|},\qquad \Phi(\mathbf r) = -\tfrac12\omega^2 s^2,$$

where $s$ is the distance from the rotation axis. Gravity is $\mathbf g = -\nabla W$, and the surfaces $W = \text{const}$ are its level surfaces: gravity is everywhere normal to them.

A plumb line hangs along $-\nabla W$, and still water settles on a level surface. A rotating fluid planet therefore takes the shape of one of its own level surfaces: the centrifugal potential, deepest at the equator, swells the equator and flattens the poles.

Outside the planet the gravitational potential obeys Laplace’s equation, and its solutions regular at infinity are the solid spherical harmonics.

Proposition 3.2 (Spherical-harmonic expansion of the geopotential)

Outside a body of mass $M$ and equatorial radius $a$, with the origin at its centre of mass and the $z$ axis along its axis of rotation, the gravitational potential at the distance $r$, colatitude $\theta$ and longitude $\lambda$ is

$$\begin{gather*} V = -\frac{GM}{r}\Biggl[1 - \sum_{n=2}^{\infty}\left(\frac ar\right)^nJ_nP_n(\cos\theta)\\ + \sum_{n=2}^{\infty}\sum_{m=1}^{n}\left(\frac ar\right)^nP_{nm}(\cos\theta) (C_{nm}\cos m\lambda + S_{nm}\sin m\lambda)\Biggr], \end{gather*}$$

with the Legendre polynomials $P_n$ and the associated functions $P_{nm}$; the coefficients $J_n$, $C_{nm}$, $S_{nm}$ are numbers, fitted to observations. For a body symmetric about its axis, only the zonal terms $J_n$ remain. No term of degree 1 appears.

Definition 3.3 (Dynamical form factor)

The dynamical form factor $J_2$ of a planet is the coefficient of its degree-2 zonal harmonic: the largest departure of its external potential from that of a sphere, caused by its oblateness. For the Earth, $J_2 = 1.082\,63 \times 10^{-3}$.

Theorem 3.4 (MacCullagh’s formula)

Let $A$, $B$, $C$ be the principal moments of inertia of a body of mass $M$ about axes through its centre of mass, and $I$ its moment of inertia about the line from the centre of mass to an external point at the distance $r$. To order $r^{-3}$ the gravitational potential at that point is given by MacCullagh’s formula

$$V = -\frac{GM}{r} - \frac{G(A + B + C - 3I)}{2r^3} .$$

For a body symmetric about the $z$ axis ($A = B$), $V = -\dfrac{GM}{r} + \dfrac{G(C - A)}{r^3}P_2(\cos\theta)$, so that

$$J_2 = \frac{C - A}{Ma^2} .$$

The degree-2 term of the potential therefore measures the difference of the moments of inertia, not the moments themselves. The precession of the Earth’s axis, driven by the torques of the Sun and Moon on the equatorial bulge, measures the dynamical ellipticity $H = (C - A)/C$; together, $J_2$ and $H$ give $C/(Ma^2)$, the moment-of-inertia factor, $0.3308$ for the Earth,

well below the $0.4$ of a uniform sphere: the mass is concentrated towards the centre.

3.2 The rotating ellipsoid

Definition 3.5 (Reference ellipsoid)

A reference ellipsoid is an ellipsoid of revolution, of equatorial semi-axis $a$ and flattening $f$, chosen so that it is itself a level surface of a model potential: the potential of a rotating body of mass $M$ with the angular velocity $\omega$ and the dynamical form factor $J_2$ of the Earth. Four constants define it; the Geodetic Reference System 1980 takes

$$\begin{gather*} a = 6\,378\,137\,\mathrm{m},\qquad GM = 3.986\,005 \times 10^{14}\,\mathrm{m}^{3}\,\mathrm{s}^{-2},\\ J_2 = 1.082\,63 \times 10^{-3},\qquad \omega = 7.292\,115 \times 10^{-5}\,\mathrm{rad}\,\mathrm{s}^{-1}, \end{gather*}$$

and derives from them the flattening $f = 0.003\,352\,810\,68$ ($1/f = 298.257\,222$) and the semi-minor axis $b = 6\,356\,752.3141\,\mathrm{m}$.

The potential of the model, the normal potential, is known in closed form, and so is its gravity on the ellipsoid, the normal gravity of the Year 1 volume.

Proposition 3.6 (Somigliana’s formula)

On a reference ellipsoid with semi-axes $a$, $b$ and normal gravity $\gamma_e$ at the equator and $\gamma_p$ at the poles, normal gravity at the geodetic latitude $\varphi$ is given by Somigliana’s formula

$$\gamma(\varphi) = \frac{a\gamma_e\cos^2\varphi + b\gamma_p\sin^2\varphi} {\sqrt{a^2\cos^2\varphi + b^2\sin^2\varphi}} .$$

For the Geodetic Reference System 1980, $\gamma_e = 9.780\,326\,771\,5\,\mathrm{m}\,\mathrm{s}^{-2}$ and $\gamma_p = 9.832\,186\,368\,5\,\mathrm{m}\,\mathrm{s}^{-2}$, and the series $\gamma = 9.780327(1 + 0.0053024\sin^2\varphi - 0.0000058\sin^22\varphi)$ $\mathrm{m}\,\mathrm{s}^{-2}$ reproduces the formula to $0.1\,\mathrm{m}\mathrm{Gal}$.

How flat should a rotating planet be? If it is a fluid in hydrostatic equilibrium, its flattening is fixed by its rotation and by how its mass is distributed in depth.

Proposition 3.7 (Clairaut and Darwin–Radau relations)

For a planet in hydrostatic equilibrium, with $m = \omega^2a^2b/(GM)$ the ratio of the centrifugal to the gravitational acceleration at the equator, to first order in $f$ and $m$:

$$J_2 = \frac{2f - m}{3}\qquad\text{(Clairaut)},\qquad \frac{C}{Ma^2} = \frac23\left[1 - \frac25\sqrt{\frac52\frac mf - 1}\right] \qquad\text{(Darwin--Radau)} .$$

A uniform planet ($C/Ma^2 = 0.4$) has $f = 5m/4$; a planet whose mass is concentrated towards the centre is less flattened.

The Clairaut relation, with the Earth’s $m = 0.003\,449\,79$, turns its measured $J_2$ into $f = 0.003\,348\,8$, or $1/298.6$: within $0.12\,\%$ of the flattening of the ellipsoid. The Darwin–Radau relation, with the moment-of-inertia factor, gives the flattening the Earth would have if it were a fluid in equilibrium: $1/299.9$, against the measured $1/298.26$. The Earth is a little flatter than a fluid planet with its rotation and its mass distribution would be: its equatorial bulge stands about $116\,\mathrm{m}$ higher than hydrostatic equilibrium predicts. The Earth is not exactly a fluid in equilibrium, and the small difference is itself a measurement of the stresses its interior sustains.

3.3 The geoid and the disturbing potential

The geoid is the level surface of the real geopotential that coincides with mean sea level; the reference ellipsoid is a level surface of the normal potential $U$ with the same value, $U_0 = W_0$. Their separation, the geoid height $N$, is set by the difference of the two potentials.

Proposition 3.8 (Bruns’ formula)

Let $T = W - U$ be the disturbing potential. The geoid height at a point is given by Bruns’ formula

$$N = -\frac{T}{\gamma},$$

where $T$ is evaluated on the geoid and $\gamma$ is normal gravity there. A mass excess, which deepens the potential ($T < 0$), raises the geoid.

A geoid height of $50\,\mathrm{m}$ is a potential difference of only $490\,\mathrm{J}\,\mathrm{kg}^{-1}$, one part in $10^{5}$ of the geopotential. The geoid heights of the Earth range from about $+85\,\mathrm{m}$ at Iceland to $-106\,\mathrm{m}$ in southern India,

a range of under $200\,\mathrm{m}$ on a planet whose equatorial and polar radii differ by $21\,\mathrm{km}$.

3.4 Satellite gravimetry

Definition 3.9 (Satellite gravimetry)

Satellite gravimetry determines the Earth’s gravity field from the motion of satellites: from the long-term perturbations of their orbits, from the changing distance between two satellites on the same orbit, or from gradiometers that measure the differences of gravity across the spacecraft.

The largest perturbation of a near-Earth orbit comes from $J_2$. The equatorial bulge exerts a torque on the inclined orbit, which turns slowly about the Earth’s axis.

Proposition 3.10 (Nodal precession)

To first order in $J_2$, the line of nodes of a circular orbit of radius $r$ and inclination $i$ turns at the rate

$$\dot\Omega = -\frac32\,nJ_2\left(\frac ar\right)^2\cos i,\qquad n = \sqrt{GM/r^3} ,$$

westward for a prograde orbit ($i < 90{}^{\circ}$), eastward for a retrograde one.

The rate is several degrees per day for a low orbit near the equator; at an inclination a little beyond $90{}^{\circ}$ it matches the Sun’s yearly motion, and the orbit keeps the same orientation to the Sun all year (exercise 12). Measured on many satellites at different inclinations, such precessions give $J_2$ and the higher zonal harmonics.

A pair of satellites does better: the changing distance between them measures the difference of the gravity they feel along the orbit. The GRACE pair flew from 2002 to 2017, and its successor, GRACE Follow-On, since its launch on 22 May 2018, its microwave link measuring the distance to within a micrometre and a laser link testing a precision up to twenty times better.

The field changes from month to month because water moves: snow and soil moisture, groundwater, lakes, the oceans, the ice sheets.

Definition 3.11 (Equivalent water height)

A change of the gravity field that varies slowly over distances much larger than the thickness of the layer where the mass moved can be represented by a change of surface density $\sigma$ ($\mathrm{kg}\,\mathrm{m}^{-2}$) on a thin shell at the Earth’s surface. The equivalent water height is the thickness of fresh water with the same surface density:

$$h_w = \frac{\sigma}{\rho_w} .$$

A metre of water over a basin of $1 \times 10^{6}\,\mathrm{km}^{2}$ is $1 \times 10^{15}\,\mathrm{kg}$, or $1000\,\mathrm{Gt}$. Sea level is the other side of the ledger: $360\,\mathrm{Gt}$ of ice added to the ocean raises it by about a millimetre.

Satellite gravimetry, satellite altimetry and the balance of snowfall against ice discharge each measure the mass balance of the ice sheets; reconciled, they give the series below.

A gravity anomaly measured at the surface is a sum of harmonics; at the altitude of a satellite, each is weakened by a factor that depends on its wavelength.

Proposition 3.12 (Upward continuation)

Above the masses that cause it, a gravity anomaly that varies along the surface as $\Delta g_0\cos kx$, with $k = 2\pi/\lambda$, is at the height $z$ (its upward continuation)

$$\Delta g(x, z) = \Delta g_0\,e^{-kz}\cos kx .$$

At $400\,\mathrm{km}$ altitude an anomaly $1000\,\mathrm{km}$ long keeps $8\,\%$ of its amplitude, one $100\,\mathrm{km}$ long about $1 \times 10^{-11}$ of it: satellites see the long wavelengths, and surveys at the surface the short ones.

3.5 Flexure of the lithosphere

The Year 1 volume compensated a load locally, each column floating on its own. A volcano on an oceanic plate is not compensated beneath itself alone: the plate bends under it over a width of hundreds of kilometres, and the seafloor around the volcano sinks into a moat, beyond which it rises in a low arch.

Definition 3.13 (Lithospheric flexure, flexural rigidity, effective elastic thickness)

Lithospheric flexure is the bending of the lithosphere under a load, modelled as the bending of a thin elastic plate floating on a fluid mantle. The plate’s resistance to bending is its flexural rigidity

$$D = \frac{ET_e^3}{12(1 - \nu^2)},$$

and the thickness $T_e$ of the uniform elastic plate that bends like the real lithosphere is its effective elastic thickness.

Theorem 3.14 (Flexure of a thin plate)

A thin elastic plate of rigidity $D$, bent along one direction by a load $q(x)$ per unit area, deflected downward by $w(x)$ and floating on a mantle of density $\rho_m$, the hollow being filled with material of density $\rho_i$, obeys the flexure equation

$$D\frac{\mathrm d^4w}{\mathrm dx^4} + (\rho_m - \rho_i)g\,w = q(x) .$$

Definition 3.15 (Flexural parameter, forebulge)

The flexural parameter of a plate is the length

$$\alpha = \left(\frac{4D}{(\rho_m - \rho_i)g}\right)^{1/4},$$

which sets the horizontal scale of its flexure. The forebulge is the low arch of upward flexure that a bent plate raises beyond the moat or basin next to its load.

Proposition 3.16 (Line load on a continuous plate)

A line load $V_0$ per unit length along $x = 0$ deflects the plate by

$$w(x) = w_0e^{-|x|/\alpha}\left(\cos\frac{|x|}{\alpha} + \sin\frac{|x|}{\alpha}\right), \qquad w_0 = \frac{V_0\alpha^3}{8D} .$$

The deflection vanishes at $|x| = 3\pi\alpha/4$, the edge of the moat; the forebulge crest is at $|x| = \pi\alpha$, with the height $w_0e^{-\pi} = 0.0432\,w_0$. The buoyancy of the whole deflection balances the load: $\int(\rho_m - \rho_i)gw\,\mathrm dx = V_0$.

Proposition 3.17 (Broken plate)

A plate broken at $x = 0$ and pushed down at its free end, as at a deep-sea trench, deflects as $w = w_0e^{-x/\alpha}\cos(x/\alpha)$ for $x \ge 0$: the deflection vanishes at $x = \pi\alpha/2$, and the crest of the forebulge, the outer rise, is at $x = 3\pi\alpha/4$, with the height $w_0e^{-3\pi/4}/\sqrt2 = 0.067\,w_0$.

Distances are thus the key to the strength of the plate: measure where the moat ends or where the bulge crests, and $\alpha$ follows, then $D$, then $T_e$. Since $T_e$ enters as $\alpha^{4/3}$, a fair estimate of a distance gives a good one of the thickness.

Definition 3.18 (Regional isostatic compensation)

Regional isostatic compensation is the support of a load by the buoyancy of a deflection spread over a region wider than the load, through the bending strength of the lithosphere; in Airy or Pratt compensation, by contrast, each column is compensated beneath itself.

Local and regional compensation balance the same total mass, but distribute it differently: under a narrow load on a stiff plate the root is shallow and wide, and the gravity anomaly over the load is strongly positive, flanked by negative anomalies over the moat. Around the islands of the Hawaiian Ridge, gravity and seismic profiles give effective elastic thicknesses of 17 to $37\,\mathrm{km}$, and 10 to $20\,\mathrm{km}$ along the older seamounts of the Emperor chain, a difference expected from the ages of the volcanoes and of the plate they loaded; the values follow the depth of an isotherm near $350{}^{\circ}\mathrm{C}$ in the cooling oceanic plate.

The same bending, under the weight of a mountain belt, opens the foreland basins of the Year 2 volume, in its chapter on the Alpine orogen.

In the field — A relative gravity survey

An absolute gravimeter drops a corner-cube mirror in a vacuum chamber, again and again, and times its fall with a laser interferometer and an atomic clock; it gives gravity at one station, but it is heavy and slow to set up. The survey itself is done with a relative gravimeter, a small spring balance that measures differences of gravity. The geophysicist levels it on a base plate at each station of a line, reads it, notes the time and the height, and moves on; every few hours the loop returns to a base station, so that the slow drift of the spring and the tides of the solid Earth can be measured and removed. A satellite receiver on a pole nearby gives the position and the height, because gravity changes by about $0.3\,\mathrm{m}\mathrm{Gal}$ for every metre of height.

3.6 Exercises

Exercise 3.1 ★

From the Geodetic Reference System 1980 values $a = 6\,378\,137\,\mathrm{m}$ and $f = 0.003\,352\,810\,68$, compute the semi-minor axis $b$ and $a - b$. With $J_2 = 1.082\,63 \times 10^{-3}$ and $m = 0.003\,449\,79$, compute the flattening that the Clairaut relation predicts, and compare.

Solution

Solution of Exercise 3.1.

$b = a(1 - f) = 6378137\times0.99664719 = 6\,356\,752.3\,\mathrm{m}$, and $a - b = 21\,384.7\,\mathrm{m}$, about $21.4\,\mathrm{km}$. Clairaut: $f = (3J_2 + m)/2 = (0.00324789 + 0.00344979)/2 = 0.003\,348\,8$, or $1/298.61$, against $1/298.26$: $0.12\,\%$ smaller, the size of the second-order terms the relation neglects.

Exercise 3.2 ★

In a dry year, a river basin of $5.0 \times 10^{5}\,\mathrm{km}^{2}$ loses $60\,\mathrm{Gt}$ of water. Compute the change of equivalent water height. By how much would the same mass raise sea level, taking $360\,\mathrm{Gt}$ per millimetre?

Solution

Solution of Exercise 3.2.

$\sigma = 6.0 \times 10^{13}\,\mathrm{kg}/5.0 \times 10^{11}\,\mathrm{m}^{2} = 120\,\mathrm{kg}\,\mathrm{m}^{-2}$, so $h_w = 120/1000 = 0.12\,\mathrm{m}$: $12\,\mathrm{cm}$ of water. Delivered to the ocean, $60/360 = 0.17\,\mathrm{mm}$ of sea level.

Exercise 3.3 ★

With the series $\gamma = 9.780327(1 + 0.0053024\sin^2\varphi - 0.0000058\sin^22\varphi)$ $\mathrm{m}\,\mathrm{s}^{-2}$, compute normal gravity at the equator, at $45{}^{\circ}$ and at the poles. Express the difference between the pole and the equator in milligals.

Solution

Solution of Exercise 3.3.

At the equator $\gamma = 9.780\,327\,\mathrm{m}\,\mathrm{s}^{-2}$. At $45{}^{\circ}$, $\sin^2\varphi = 0.5$ and $\sin^22\varphi = 1$: $\gamma = 9.780327\times1.0026454 = 9.806\,200\,\mathrm{m}\,\mathrm{s}^{-2}$. At the poles, $9.780327\times1.0053024 = 9.832\,186\,\mathrm{m}\,\mathrm{s}^{-2}$. The difference is $0.051\,859\,\mathrm{m}\,\mathrm{s}^{-2}$, or $5186\,\mathrm{m}\mathrm{Gal}$.

Exercise 3.4 ★

Compute the flexural rigidity and the flexural parameter of oceanic plates with $T_e = 10\,\mathrm{km}$ and $T_e = 30\,\mathrm{km}$ ($E = 100\,\mathrm{GPa}$, $\nu = 0.25$, $\rho_m - \rho_w = 2270\,\mathrm{kg}\,\mathrm{m}^{-3}$, $g = 9.81\,\mathrm{m}\,\mathrm{s}^{-2}$). Where are the edge of the moat and the forebulge under a line load?

Solution

Solution of Exercise 3.4.

$D = ET_e^3/(12(1 - \nu^2)) = 10^{11}T_e^3/11.25$: $8.89 \times 10^{21}\,\mathrm{N}\,\mathrm{m}$ for $10\,\mathrm{km}$, $2.40 \times 10^{23}\,\mathrm{N}\,\mathrm{m}$ for $30\,\mathrm{km}$. With $(\rho_m - \rho_w)g = 22\,269\,\mathrm{N}\,\mathrm{m}^{-3}$, $\alpha = (4D/22269)^{1/4}$: $35.5\,\mathrm{km}$ and $81.0\,\mathrm{km}$. The moat ends at $3\pi\alpha/4$, $84\,\mathrm{km}$ and $191\,\mathrm{km}$; the forebulge crests at $\pi\alpha$, $112\,\mathrm{km}$ and $255\,\mathrm{km}$.

Exercise 3.5 ★★

(a) A geoid height of $+50\,\mathrm{m}$: compute the disturbing potential there, taking $\gamma = 9.80\,\mathrm{m}\,\mathrm{s}^{-2}$. (b) Over $100\,\mathrm{km}$ the geoid rises by $10\,\mathrm{m}$. By what angle does the plumb line there differ from the normal to the ellipsoid? Give it in seconds of arc.

Solution

Solution of Exercise 3.5.

(a) $T = -\gamma N = -9.80\times50 = -490\,\mathrm{J}\,\mathrm{kg}^{-1}$. (b) The geoid tilts by $10/100\,000 = 1.0 \times 10^{-4}$ rad relative to the ellipsoid, and the plumb line, normal to the geoid, by the same angle relative to the ellipsoid’s normal: $1.0 \times 10^{-4}\times206\,265 = 20.6''$.

Exercise 3.6 ★★

From $J_2 = 1.082\,63 \times 10^{-3}$ and the moment-of-inertia factor $C/(Ma^2) = 0.3308$, compute the dynamical ellipticity $H = (C - A)/C$ and $1/H$. Which of the two numbers would a uniform Earth with the same $J_2$ change?

Solution

Solution of Exercise 3.6.

$H = J_2/(C/Ma^2) = 1.082\,63 \times 10^{-3}/0.3308 = 0.003\,272\,8$, and $1/H = 305.55$. $J_2$ is the same by hypothesis; a uniform Earth would have $C/(Ma^2) = 0.4$ and so $H = J_2/0.4 = 0.002\,706\,6$: precession would be slower. Measuring both $J_2$ and $H$ is what reveals the concentration of mass towards the centre.

Exercise 3.7 ★★

A satellite flies a circular orbit $500\,\mathrm{km}$ above the equatorial radius ($a = 6378.137\,\mathrm{km}$, $GM = 3.986\,005 \times 10^{14}\,\mathrm{m}^{3}\,\mathrm{s}^{-2}$), inclined at $89{}^{\circ}$. Compute its period and the rate of nodal precession in degrees per day. Why do gravity missions choose near-polar orbits?

Solution

Solution of Exercise 3.7.

$r = 6878.137\,\mathrm{km}$; $n = \sqrt{GM/r^3} = 1.106\,78 \times 10^{-3}\,\mathrm{rad}\,\mathrm{s}^{-1}$, a period of $2\pi/n = 94.6\,\mathrm{min}$. $\dot\Omega = -1.5\times1.106\,78 \times 10^{-3} \times1.082\,63 \times 10^{-3}\times(6378.137/6878.137)^2\times\cos89{}^{\circ} = -2.70 \times 10^{-8}\,\mathrm{rad}\,\mathrm{s}^{-1}$, or $-0.134{}^{\circ}$ per day, westward. A near-polar orbit passes over every latitude, so the whole Earth is mapped; it also precesses slowly, so that the ground tracks fill in regularly.

Exercise 3.8 ★★

Two anomalies of $10\,\mathrm{m}\mathrm{Gal}$ at the surface have wavelengths of $100\,\mathrm{km}$ and $1000\,\mathrm{km}$. Compute their amplitudes at $500\,\mathrm{km}$ altitude.

Solution

Solution of Exercise 3.8.

$e^{-2\pi\times500/100} = e^{-31.4} = 2.3 \times 10^{-14}$: $2 \times 10^{-13}\,\mathrm{m}\mathrm{Gal}$, nothing. $e^{-2\pi\times500/1000} = e^{-3.14} = 0.0432$: $0.43\,\mathrm{m}\mathrm{Gal}$.

Exercise 3.9 ★★

Seaward of a trench, the crest of the outer rise lies $120\,\mathrm{km}$ from the trench axis. Taking the broken-plate model with water above the plate, compute $\alpha$, $D$ and $T_e$, and the distance where the plate crosses its unflexed level.

Solution

Solution of Exercise 3.9.

$3\pi\alpha/4 = 120\,\mathrm{km}$ gives $\alpha = 50.9\,\mathrm{km}$. $D = (\rho_m - \rho_w)g\alpha^4/4 = 22269\times(50\,930)^4/4 = 3.75 \times 10^{22}\,\mathrm{N}\,\mathrm{m}$, and $T_e = (11.25D/E)^{1/3} = 16.2\,\mathrm{km}$. The plate crosses its unflexed level at $\pi\alpha/2 = 80\,\mathrm{km}$ from the trench.

Exercise 3.10 ★★★

(a) Check that $w = w_0e^{-u}(\cos u + \sin u)$, $u = x/\alpha$, satisfies $Dw'''' + (\rho_m - \rho_i)gw = 0$ for $x > 0$, with $\alpha$ as defined. (b) Show that the bending moment $-Dw''$ is greatest under the load and changes sign at $x = \pi\alpha/4$. (c) Where is the plate most likely to crack at its upper surface, and why?

Solution

Solution of Exercise 3.10.

(a) With $u = x/\alpha$, $\mathrm d^4/\mathrm dx^4 = \alpha^{-4}\,\mathrm d^4/\mathrm du^4$. Writing $\cos u + \sin u = \sqrt2\cos(u - \pi/4)$, the function is $\sqrt2w_0\,\mathrm{Re}\,e^{(-1 + i)u - i\pi/4}$, and $(-1 + i)^4 = -4$: so $w'''' = -4w/\alpha^4$. Then $Dw'''' + (\rho_m - \rho_i)gw = (-4D/\alpha^4 + (\rho_m - \rho_i)g)w = 0$, since $\alpha^4 = 4D/((\rho_m - \rho_i)g)$. (b) $w'' = -(2w_0/\alpha^2)e^{-u}(\cos u - \sin u)$, so $M = -Dw'' = (2Dw_0/\alpha^2)e^{-u}(\cos u - \sin u)$: largest in magnitude at $u = 0$, zero where $\tan u = 1$, $x = \pi\alpha/4$. (c) Under the load the plate is concave upward, its upper surface in compression; beyond $\pi\alpha/4$ it is convex upward, its upper surface stretched. There $M' \propto -e^{-u}\cos u$ vanishes at $u = \pi/2$: the stretching is greatest at $x = \pi\alpha/2$, on the outer slope of the moat towards the forebulge, where $|M|$ is $e^{-\pi/2} = 21\,\%$ of its value under the load. That is where normal faults and cracks would open at the top of the plate.

Exercise 3.11 ★★★

With $m = 0.003\,449\,79$ and $C/(Ma^2) = 0.3308$, invert the Darwin–Radau relation for the hydrostatic flattening. Compare with $f = 0.003\,352\,81$: compute the difference and the excess height of the equator above the hydrostatic shape. What does it say about the mantle?

Solution

Solution of Exercise 3.11.

$\sqrt{1 + \eta} = 2.5(1 - 1.5\times0.3308) = 1.2595$, so $1 + \eta = 1.5863$ and $\tfrac52m/f = 2.5863$: $f_h = 2.5\times0.00344979/2.5863 = 0.003\,334\,6$, $1/f_h = 299.9$. The difference $f - f_h = 1.82 \times 10^{-5}$; times $a$, the equator stands $116\,\mathrm{m}$ above the hydrostatic figure. The Earth is not quite a fluid in equilibrium: its interior sustains stresses, from the strength of the lithosphere or the flow and density differences of the mantle, that hold the shape a little away from it.

Exercise 3.12 ★★★

A satellite on a circular orbit $700\,\mathrm{km}$ high is to keep the same orientation relative to the Sun all year: its node must advance by $360{}^{\circ}$ in a tropical year of $365.242$ days. Compute the inclination required.

Solution

Solution of Exercise 3.12.

The node must advance by $2\pi/(365.242\times86\,400) = 1.991 \times 10^{-7}\,\mathrm{rad}\,\mathrm{s}^{-1}$. At $r = 7078.137\,\mathrm{km}$, $n = 1.0602 \times 10^{-3}\,\mathrm{rad}\,\mathrm{s}^{-1}$ and $\tfrac32nJ_2(a/r)^2 = 1.398 \times 10^{-6}$, so $\cos i = -1.991 \times 10^{-7}/1.398 \times 10^{-6} = -0.1424$ and $i = 98.2{}^{\circ}$: a slightly retrograde, sun-synchronous orbit.

3.7 Problem: The Moat around the Oru Ridge

Problem 3.1

Weekend problem — the figure of the Earth, a volcanic load, its flexural moat and what satellites see of it

Oru is an invented volcanic ridge in an ocean: a straight chain of coalesced volcanoes $400\,\mathrm{km}$ long, built on old oceanic lithosphere. Its cross-section is taken as a triangle $80\,\mathrm{km}$ wide at its base and $6\,\mathrm{km}$ high, of density $2800\,\mathrm{kg}\,\mathrm{m}^{-3}$, standing on a sea floor $5\,\mathrm{km}$ deep. Mantle $3300\,\mathrm{kg}\,\mathrm{m}^{-3}$, sea water $1030\,\mathrm{kg}\,\mathrm{m}^{-3}$, $E = 100\,\mathrm{GPa}$, $\nu = 0.25$, $g = 9.81\,\mathrm{m}\,\mathrm{s}^{-2}$, $G = 6.674 \times 10^{-11}\,\mathrm{m}^{3}\,\mathrm{kg}^{-1}\,\mathrm{s}^{-2}$.

Part I — The figure of the Earth. Use the Geodetic Reference System 1980: $a = 6\,378\,137\,\mathrm{m}$, $b = 6\,356\,752.3\,\mathrm{m}$, $GM = 3.986\,005 \times 10^{14}\,\mathrm{m}^{3}\,\mathrm{s}^{-2}$, $J_2 = 1.082\,63 \times 10^{-3}$, $\omega = 7.292\,115 \times 10^{-5}\,\mathrm{rad}\,\mathrm{s}^{-1}$, $f = 0.003\,352\,81$; and $C/(Ma^2) = 0.3308$.

1. Compute $m = \omega^2a^2b/(GM)$.
2. Compute the flattening that the Clairaut relation gives from $J_2$ and $m$, and its inverse.
3. Compute the hydrostatic flattening from the Darwin–Radau relation, and its inverse.
4. Compare both with the observed flattening. By how many metres does the equator stand above the hydrostatic figure?
5. Compute the dynamical ellipticity $H = (C - A)/C$.

Part II — The load.

6. Compute the area of the ridge’s cross-section.
7. Compute the line load $V_0$, taking the whole ridge as submerged, so that it weighs $(\rho_v - \rho_w)g$ per unit volume.
8. If the ridge were compensated locally, in Airy fashion, by a root of its own rock in the mantle, how deep would the root be under the crest?
9. Under local compensation, the root would end at the foot of the ridge, $40\,\mathrm{km}$ from its axis. Bathymetry shows a moat out to $165\,\mathrm{km}$. What does this tell you?

Part III — The flexural moat. The moat is filled with water ($\rho_i = \rho_w$). A seismic profile across the ridge finds the top of the oceanic crust back at its regional depth $165\,\mathrm{km}$ from the axis, and the crest of the forebulge $220\,\mathrm{km}$ from the axis.

10. Compute the flexural parameter from the edge of the moat.
11. Compute it again from the forebulge. Are the two consistent?
12. Taking $\alpha = 70.0\,\mathrm{km}$, compute the flexural rigidity.
13. Compute the effective elastic thickness.
14. Taking $V_0 = 4.17 \times 10^{12}\,\mathrm{N}\,\mathrm{m}^{-1}$ and $D$ rounded to three figures, compute the deflection under the axis.
15. Compute the height of the forebulge.
16. Check that the buoyancy of the whole deflection, $2w_0\alpha(\rho_m - \rho_w)g$, balances $V_0$.
17. Compare $T_e$ with the values found along the Hawaiian Ridge, 17 to $37\,\mathrm{km}$. Compare the deflection under the axis with the Airy root of question 8.

Part IV — Gravity over the moat. Take one harmonic of the flexure, of wavelength $2\pi\alpha$ and amplitude $W = 500\,\mathrm{m}$, deflecting both the sea floor (crust $2900\,\mathrm{kg}\,\mathrm{m}^{-3}$ under water, $5\,\mathrm{km}$ below sea level) and the Moho (mantle under crust, $12\,\mathrm{km}$ below sea level). A thin sheet of surface density $\sigma$ produces the anomaly $2\pi G\sigma$ just above it.

18. Compute the wavelength and the wavenumber $k$ of the harmonic.
19. Compute the amplitude of the anomaly at sea level due to the deflected sea floor, continued up from $5\,\mathrm{km}$ depth.
20. Compute that of the deflected Moho, and the total.
21. Compute the total at the altitude of a satellite, $400\,\mathrm{km}$ above sea level.
22. By what factor is a harmonic of wavelength $80\,\mathrm{km}$, the width of the ridge, weakened at that altitude? What can the satellite resolve?
23. To what spherical-harmonic degree $n \approx 2\pi R/\lambda$ does the harmonic of question 18 correspond ($R = 6371\,\mathrm{km}$)?
24. Give the effective elastic thickness of the plate beneath the Oru Ridge.

Solution

Solution of Problem 3.1.

1. $m = (7.292\,115 \times 10^{-5})^2\times6378137^2\times6356752.3/ 3.986\,005 \times 10^{14} = 0.003\,449\,79$.

2. $f = (3J_2 + m)/2 = 0.003\,348\,8$, $1/f = 298.61$.

3. $\sqrt{1 + \eta} = 2.5(1 - 1.5\times0.3308) = 1.2595$, $\tfrac52m/f_h = 2 + 0.5863$, so $f_h = 0.003\,334\,6$ and $1/f_h = 299.88$.

4. The Clairaut value, $1/298.61$, is within $0.12\,\%$ of the observed $1/298.26$; the hydrostatic one, $1/299.88$, is smaller by $\Delta f = 1.82 \times 10^{-5}$, and $a\,\Delta f = 116\,\mathrm{m}$: the equator stands $116\,\mathrm{m}$ above the hydrostatic figure.

5. $H = 1.082\,63 \times 10^{-3}/0.3308 = 0.003\,272\,8$ ($1/305.6$).

6. $\tfrac12\times80\,\mathrm{km}\times6\,\mathrm{km} = 240\,\mathrm{km}^{2} = 2.40 \times 10^{8}\,\mathrm{m}^{2}$.

7. $V_0 = (2800 - 1030)\times9.81\times2.40 \times 10^{8} = 4.17 \times 10^{12}\,\mathrm{N}\,\mathrm{m}^{-1}$.

8. Airy: $(\rho_m - \rho_v)r = (\rho_v - \rho_w)h$, so $r = 6\times1770/500 = 21.2\,\mathrm{km}$ under the crest.

9. The sea floor is deflected far beyond the foot of the ridge: the load is not compensated column by column but by the bending of the plate over a region four times wider. The plate has strength: regional compensation.

10. $\alpha = 165/(3\pi/4) = 70.0\,\mathrm{km}$.

11. $\alpha = 220/\pi = 70.0\,\mathrm{km}$: the same, as a single elastic plate requires.

12. $D = (\rho_m - \rho_w)g\alpha^4/4 = 2270\times9.81\times(7.00 \times 10^{4})^4/4 = 1.34 \times 10^{23}\,\mathrm{N}\,\mathrm{m}$.

13. $T_e = (12(1 - \nu^2)D/E)^{1/3} = (11.25\times1.34 \times 10^{23}/ 1 \times 10^{11})^{1/3} = 24.7\,\mathrm{km}$.

14. $w_0 = V_0\alpha^3/(8D) = 4.17 \times 10^{12}\times(7.00 \times 10^{4})^3/ (8\times1.34 \times 10^{23}) = 1334\,\mathrm{m}$.

15. $w_0e^{-\pi} = 1334\times0.0432 = 58\,\mathrm{m}$.

16. $2\times1334\times7.00 \times 10^{4}\times2270\times9.81 = 4.16 \times 10^{12}\,\mathrm{N}\,\mathrm{m}^{-1}$, equal to $V_0$ to within the rounding of $D$.

17. $24.7\,\mathrm{km}$ lies inside the range of the Hawaiian Ridge. The plate sinks $1.3\,\mathrm{km}$ under the axis instead of growing a root of $21\,\mathrm{km}$: the same buoyancy is spread over the $330\,\mathrm{km}$ of the moat.

18. $\lambda = 2\pi\times70.0\,\mathrm{km} = 440\,\mathrm{km}$, $k = 1/\alpha = 1.43 \times 10^{-5}\,\mathrm{m}^{-1}$.

19. $\sigma = (2900 - 1030)\times500 = 9.35 \times 10^{5}\,\mathrm{kg}\,\mathrm{m}^{-2}$, so $2\pi G\sigma = 3.92 \times 10^{-4}\,\mathrm{m}\,\mathrm{s}^{-2} = 39.2\,\mathrm{m}\mathrm{Gal}$ at the sea floor; times $e^{-k\times5\,\mathrm{km}} = 0.931$, $36.5\,\mathrm{m}\mathrm{Gal}$ at sea level (negative where the plate is pushed down).

20. $\sigma = (3300 - 2900)\times500 = 2.0 \times 10^{5}\,\mathrm{kg}\,\mathrm{m}^{-2}$: $8.39\,\mathrm{m}\mathrm{Gal}$, times $e^{-k\times12\,\mathrm{km}} = 0.843$, $7.07\,\mathrm{m}\mathrm{Gal}$. The total is $43.6\,\mathrm{m}\mathrm{Gal}$.

21. $e^{-k\times400\,\mathrm{km}} = e^{-5.71} = 0.0033$: $0.14\,\mathrm{m}\mathrm{Gal}$.

22. $e^{-2\pi\times400/80} = e^{-31.4} = 2.3 \times 10^{-14}$: the satellite sees nothing of the ridge itself, only the broad flexure, and even that weakened three-hundred-fold; the ridge’s own anomaly must be measured at sea.

23. $n \approx 2\pi\times6371/440 = 91$.

24. From questions 10 to 13, the effective elastic thickness of the plate beneath the Oru Ridge is $24.7\,\mathrm{km}$.