19Geophysical Fluid Dynamics

Stir a cup of tea and the swirl dies in a minute; the leaves gather in the middle of the bottom and nothing else happens. The ocean and the atmosphere are stirred too, by the Sun and the wind, but on a rotating planet and in layers of different density. Rotation and stratification make them obey rules that a teacup never shows: winds blow along the isobars instead of across them, columns of fluid refuse to tilt, and the wind of the upper troposphere is written in the temperatures below it. This chapter sets up the equations of these large, slow, thin flows and the few numbers that decide which terms matter.

You already know

The Coriolis and centrifugal forces in a rotating frame, the Navier–Stokes and continuity equations, viscosity, the Reynolds number and the hydrostatic relation are in the physics books. The Coriolis parameter $f = 2\Omega \sin\varphi$, the pressure-gradient force, geostrophic balance, the geostrophic and gradient winds and the jet streams are in the Year 1 volume, in its chapter on the general circulation of the atmosphere; the Ekman layer in its chapter on the circulation of the ocean; potential temperature, static stability and the oscillation of a lifted air parcel in its chapter on the structure of the atmosphere and the ocean.

19.1 Equations in a rotating frame

In a frame that turns with the Earth at the angular velocity $\vec\Omega$, a fluid parcel of velocity $\vec u$ obeys

where $\mathrm D/\mathrm Dt = \partial/\partial t + \vec u\cdot\nabla$ is the derivative following the parcel and $\nu$ the kinematic viscosity. The centrifugal force has been added to gravitation: their sum $\vec g$ is the effective gravity, normal to the level surfaces of the geopotential, and the vertical is defined by it. Only the Coriolis term, $-2\vec\Omega\times \vec u$, remains of the rotation, and it does no work: it turns a parcel without changing its speed. The Earth turns at $\Omega = 7.292\,115 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$, once a sidereal day of $23.9345\,\mathrm{h}$.

For motion over distances much smaller than the Earth’s radius $a$, a local Cartesian frame is enough: $x$ eastward, $y$ northward, $z$ upward, at the latitude $\varphi$, the tangent plane of the figure below. There $\vec\Omega = \Omega\,(0, \cos\varphi, \sin\varphi)$, and the three components of the momentum equation are

with $\vec F$ the friction. For the thin, nearly horizontal flows of the ocean and the atmosphere the terms in $\tilde f$ are dropped: $\tilde f w$ is small beside $fv$ because $w \ll v$, and $\tilde f u$ is a ten-thousandth of $g$ (Exercise 19.5). The Coriolis force then acts only in the horizontal, through the local vertical component of the rotation, $\Omega\sin\varphi$.

Definition 19.1 ($f$-plane and $\beta$-plane)

Near a reference latitude $\varphi_0$, the $f$-plane approximation takes the Coriolis parameter as constant, $f = f_0 = 2\Omega\sin\varphi_0$. The $\beta$-plane approximation keeps its first-order change with the northward distance $y$: $f = f_0 + \beta y$, with $\beta = \mathrm df/\mathrm dy = 2\Omega \cos\varphi_0/a$.

At $45{}^{\circ}$, with the mean radius $a = 6371\,\mathrm{km}$, $f_0 = 1.031 \times 10^{-4}\,\mathrm{s}^{-1}$ and $\beta = 1.62 \times 10^{-11}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}$: over $1000\,\mathrm{km}$ of latitude $f$ changes by $16\,\%$. The $f$-plane suits motions much smaller than that; the $\beta$-plane is needed for basin-wide currents and planetary waves.

Two more approximations simplify the equations without losing what matters. The first concerns density.

Definition 19.2 (Boussinesq approximation)

The Boussinesq approximation writes the density as $\rho = \rho_0 + \rho'$, with $\rho_0$ a constant and $|\rho'| \ll \rho_0$, and keeps $\rho'$ only where it is multiplied by $g$. The continuity equation becomes $\nabla\cdot\vec u = 0$, and the momentum equation

$$\frac{\mathrm D\vec u}{\mathrm Dt} + f\,\vec k\times\vec u = -\frac{1}{\rho_0}\nabla p' + b\,\vec k + \nu\nabla^2\vec u , \qquad b = -g\,\frac{\rho'}{\rho_0} ,$$

where $p'$ is the departure of the pressure from the hydrostatic pressure of the density $\rho_0$, $\vec k$ the upward unit vector and $b$ the buoyancy.

In the ocean $\rho'$ is a few $\mathrm{kg}/\mathrm{m}^{3}$ on $1025$, a fraction of a per cent, and the approximation is excellent. In the atmosphere density falls by a factor $e$ in about $8.5\,\mathrm{km}$; the approximation holds for layers much thinner than that, and deeper flows replace $\rho_0$ by a mean profile $\rho_0(z)$, with the same physics.

The second concerns the vertical. A fluid at rest is in hydrostatic equilibrium; a fluid in motion keeps it to a very good approximation when the motion is much wider than it is deep.

Definition 19.3 (Hydrostatic approximation)

The hydrostatic approximation replaces the vertical momentum equation of a moving fluid by the hydrostatic relation, $\partial p/\partial z = -\rho g$.

Proposition 19.4 (When the hydrostatic approximation holds)

For a motion of horizontal scale $L$, depth $H$ and speed $U$, with a Rossby number $U/(fL)$ at most of order one, the vertical acceleration is smaller than the vertical gradient of the pressure perturbation by a factor of order $(U/fL)\,(H/L)^2$.

A weather system $1000\,\mathrm{km}$ wide in a troposphere $10\,\mathrm{km}$ deep has $(H/L)^2 = 10^{-4}$ and a Rossby number of about 0.1: the error of the hydrostatic approximation is of order $10^{-5}$. It fails only in narrow, deep motions: a thunderstorm updraught, a convective plume in the ocean.

19.2 Scales of rotating flow

Which terms of the momentum equation matter is decided by two ratios.

Definition 19.5 (Rossby and Ekman numbers)

For a flow of speed $U$, horizontal scale $L$ and depth $H$ on a planet of Coriolis parameter $f$, the Rossby number is $\mathrm{Ro} = U/(fL)$, the ratio of the acceleration, of order $U^2/L$, to the Coriolis force, of order $fU$. The Ekman number is $\mathrm{Ek} = \nu/(fH^2)$, the ratio of the viscous force, of order $\nu U/H^2$, to the Coriolis force; in a turbulent flow $\nu$ is replaced by an eddy viscosity.

A bathtub draining $0.1\,\mathrm{m}/\mathrm{s}$ over $0.3\,\mathrm{m}$ has a Rossby number of some thousands: the Earth’s rotation does not decide the sense of its swirl. A depression $1000\,\mathrm{km}$ across with winds of $15\,\mathrm{m}/\mathrm{s}$ has $\mathrm{Ro} \approx 0.14$, and an ocean gyre $\mathrm{Ro} \sim 10^{-4}$. Viscosity is weaker still: the molecular viscosity of water is $10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$, and the eddy viscosities that stand for the turbulence of the ocean, $10^{-5}\text{ to }10^{-3}\,\mathrm{m}^{2}/\mathrm{s}$ in the vertical, give $\mathrm{Ek} \sim 10^{-5}$ over $1000\,\mathrm{m}$ of depth. The two numbers are related to the Reynolds number by $\mathrm{Re} = UL/\nu = (\mathrm{Ro}/\mathrm{Ek})(L/H)^2$.

When both numbers are small, the steady horizontal momentum equation keeps only the Coriolis force and the pressure gradient: the flow is in geostrophic balance, $f\,\vec k\times\vec u_g = -(1/\rho)\nabla_h p$, which the Year 1 volume read from weather maps. Geostrophy is the limit $\mathrm{Ro}, \mathrm{Ek}\to 0$; its corrections are of order $\mathrm{Ro}$, and friction acts only in thin layers of relative thickness $\sqrt{\mathrm{Ek}}$, the Ekman layers at the bottom of the atmosphere and the top and bottom of the ocean. On an $f$-plane the geostrophic velocity has no horizontal divergence: $\partial u_g/\partial x + \partial v_g/\partial y = -(1/\rho f)(\partial^2 p/\partial x\partial y - \partial^2 p/\partial y\partial x) = 0$ for uniform $\rho$.

The opposite limit, rotation with no pressure gradient, is the simplest motion a rotating fluid can make.

Definition 19.6 (Inertial oscillation)

An inertial oscillation is the horizontal motion of a fluid on which no horizontal force acts but the Coriolis force.

Theorem 19.7 (Inertial circles)

On an $f$-plane, a parcel set moving at speed $U$ with no horizontal force but the Coriolis force keeps its speed and describes a circle of radius $U/|f|$, clockwise in the Northern Hemisphere and anticlockwise in the Southern, in the inertial period $2\pi/|f|$, which is half a sidereal day divided by $\sin\varphi$.

At $45{}^{\circ}$ the inertial period is $16.9\,\mathrm{h}$, and a current of $0.2\,\mathrm{m}/\mathrm{s}$ turns on a circle of $1.94\,\mathrm{km}$; at $30{}^{\circ}$ the period is a full sidereal day, and at the equator there are no inertial circles. After a storm has set the surface layer of the ocean moving and then died, the water does exactly this. Drogued drifting buoys at $15\,\mathrm{m}$ in the North Pacific traced loops for days after a storm in October 1987, carried along by the mean current: the third path of the figure below, after the circle and the spiral.

Between geostrophy and the inertial circle lies the flow round a centre of high or low pressure, where the isobars are curved and the acceleration towards the centre is no longer negligible.

Definition 19.8 (Gradient wind)

The gradient wind is the steady, frictionless flow along curved isobars in which the pressure-gradient force, the Coriolis force and the centripetal acceleration balance.

Proposition 19.9 (The curvature correction)

Let the isobars be circles of radius $r$ in the Northern Hemisphere, and $V_g$ the geostrophic speed that their pressure gradient would give, with $\mathrm{Ro}_g = V_g/(fr)$. Round a low, the gradient wind is $V = 2V_g/(1 + \sqrt{1 + 4\mathrm{Ro}_g})$, slower than $V_g$; round a high, $V = 2V_g/(1 + \sqrt{1 - 4\mathrm{Ro}_g})$, faster, and it exists only if $\mathrm{Ro}_g \le 1/4$. To first order, $V = V_g(1 \mp \mathrm{Ro}_g)$.

The Rossby number of the isobars measures the correction: a tenth for a depression of moderate size. The condition $\mathrm{Ro}_g \le 1/4$ is why the centre of an anticyclone has weak gradients and light winds, while that of a depression can hold steep gradients and gales.

19.3 Rotation without stratification

A rotating fluid of uniform density, in slow motion, behaves as if it were made of rigid columns parallel to the rotation axis.

Theorem 19.10 (Taylor–Proudman)

In a steady flow of an inviscid fluid of uniform density, rotating at $\vec\Omega$, with small Rossby number, the velocity does not vary along the direction of the rotation axis: $(\vec\Omega\cdot\nabla)\vec u = 0$.

With $\vec\Omega$ along $z$, $\partial\vec u/\partial z = 0$: all three velocity components are the same at every height. If a rigid bottom forbids $w$ there, $w = 0$ everywhere, and the horizontal flow is the same at every level. An obstacle on the floor of a rotating tank is therefore felt all the way to the surface: the fluid above it cannot climb over it, since that would stretch or squeeze a column, and goes round it at every height. The column of fluid that stays above the obstacle is a Taylor column.

The atmosphere and the ocean are stratified, and stratification relaxes the constraint: a stratified fluid can vary with height, in the way the thermal wind theorem prescribes.

19.4 Stratification and the thermal wind

The Year 1 volume lifted an air parcel in a stable atmosphere and found that it oscillates about its level. The same holds in any stratified fluid, and the frequency of the oscillation is the measure of stratification.

Definition 19.11 (Buoyancy frequency)

The buoyancy frequency $N$ of a fluid at rest is defined by

$$N^2 = \frac{g}{\theta}\frac{\mathrm d\theta}{\mathrm dz}\ \text{(air)}, \qquad N^2 = -\frac{g}{\rho_0}\frac{\mathrm d\rho_\theta}{\mathrm dz}\ \text{(seawater)},$$

where $\theta$ is the potential temperature of the air and $\rho_\theta$ the potential density of the water, the density it would have if brought adiabatically to a reference pressure.

Theorem 19.12 (Oscillation of a displaced parcel)

A parcel displaced vertically by a small distance $\zeta$ from its level in a fluid at rest, adiabatically and without mixing, obeys $\ddot\zeta = -N^2\zeta$. If $N^2 > 0$ it oscillates at the angular frequency $N$; if $N^2 < 0$ the displacement grows and the fluid is unstable.

In the troposphere of the standard atmosphere, which cools by $6.49\,\mathrm{K}/\mathrm{km}$ against the $9.77\,\mathrm{K}/\mathrm{km}$ of the dry adiabat ($g/c_p$, with $c_p = \gamma R/(\gamma - 1)$, $\gamma = 1.4$ and $R = 286.9\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$), $N$ averages $0.0113\,\mathrm{s}^{-1}$, a period of $9.3\,\mathrm{min}$; in the isothermal lower stratosphere it is $0.0210\,\mathrm{s}^{-1}$. In the ocean, the water column of the North Atlantic at $40.5{}^{\circ}$ N, $30.5{}^{\circ}$ W in the annual mean has its strongest stratification in the seasonal pycnocline, $N = 0.0106\,\mathrm{s}^{-1}$ at about $30\,\mathrm{m}$, six cycles per hour, and much weaker values below: a few cycles per hour at most, as observed across the upper ocean.

Stratification lets the flow vary with height, and how it does so depends on whether the surfaces of constant density follow those of constant pressure.

Definition 19.13 (Barotropic and baroclinic fluids)

A fluid is barotropic if its density is a function of pressure alone, so that surfaces of constant density coincide with surfaces of constant pressure; it is baroclinic if they cross.

In the atmosphere, a horizontal gradient of temperature along a pressure surface makes the air baroclinic. Hydrostatic balance then makes the pressure surfaces slope more steeply with height, and geostrophic balance turns the growing slope into a growing wind.

Theorem 19.14 (Thermal wind)

In an atmosphere in hydrostatic and geostrophic balance, the geostrophic wind varies with pressure as

$$\frac{\partial u_g}{\partial\ln p} = \frac{R}{f}\left(\frac{\partial T}{\partial y}\right)_p , \qquad \frac{\partial v_g}{\partial\ln p} = -\frac{R}{f}\left(\frac{\partial T}{\partial x}\right)_p .$$

Between two pressure levels $p_b > p_t$, the thermal wind, the difference of the geostrophic winds, is

$$\vec u_g(p_t) - \vec u_g(p_b) = \frac{R}{f}\,\ln\frac{p_b}{p_t}\;\vec k\times\nabla_p\bar T ,$$

where $\bar T$ is the mean temperature of the layer: it blows along the isotherms of the layer with the cold air on its left in the Northern Hemisphere.

Corollary 19.15 (Thermal wind in height)

In height coordinates, for the atmosphere, $\partial u_g/\partial z = -(g/fT)\,\partial T/\partial y + (u_g/T)\,\partial T/\partial z$; for the ocean in the Boussinesq approximation, $\partial u_g/\partial z = (g/f\rho_0)\,\partial\rho/\partial y$ and $\partial v_g/\partial z = -(g/f\rho_0)\,\partial\rho/\partial x$.

The second term of the atmospheric form, from the decrease of temperature with height, is smaller than the first and vanishes in an isothermal layer; integrated across a layer with its mean temperature, the first term alone gives nearly the pressure form, since $\Delta z \approx (R\bar T/g) \ln(p_b/p_t)$ (Problem, Part IV). In the ocean the same theorem turns a section of density across a current into its vertical shear: the basis of the dynamic method of the oceanographers.

The jet streams of the Year 1 volume are thermal winds. In the January zonal mean of 1991–2020, the westerly wind at $30{}^{\circ}$ N grows from $4.3\,\mathrm{m}/\mathrm{s}$ at $850\,\mathrm{hPa}$ to $40.4\,\mathrm{m}/\mathrm{s}$ at $250\,\mathrm{hPa}$; with $f$ at $30{}^{\circ}$ and $R$, this shear is the signature of a layer whose mean temperature falls northward by $7.5\,\mathrm{K}$ per $1000\,\mathrm{km}$. At $45{}^{\circ}$ N the shear from $6.0\text{ to }23.2\,\mathrm{m}/\mathrm{s}$ gives $5.0\,\mathrm{K}$ per $1000\,\mathrm{km}$. The figure below shows the profiles and, for each latitude, the poleward drop of temperature read from the shear: largest in the subtropics of the winter hemisphere, which is where the subtropical jet stands, $43.9\,\mathrm{m}/\mathrm{s}$ at $200\,\mathrm{hPa}$ and $30{}^{\circ}$ N in January and $41.0\,\mathrm{m}/\mathrm{s}$ at $30{}^{\circ}$ S in July. Above $200\,\mathrm{hPa}$ at $30{}^{\circ}$ N the wind weakens again: the thermal wind theorem read backwards says that there, in the lower stratosphere, the temperature rises towards the pole.

19.5 The Rossby radius of deformation

Rotation and stratification each impose a scale; together they impose a length.

Definition 19.16 (Rossby radius of deformation)

The Rossby radius of deformation of a layer of depth $H$ and buoyancy frequency $N$ is $L_R = NH/|f|$.

Proposition 19.17 (Potential and kinetic energy of balanced flow)

In a hydrostatic, geostrophic flow of horizontal scale $L$ in a layer of depth $H$ and buoyancy frequency $N$, the ratio of the potential energy stored in the displaced density surfaces to the kinetic energy is of order $(L/L_R)^2$.

Motions much wider than $L_R$ hold most of their energy in tilted density surfaces, motions much narrower in their velocity; at $L_R$ the two are equal, and the instability of baroclinic flows, which feeds on the potential energy of the tilted surfaces, grows fastest near that scale. For the troposphere, with the mean $N$ of the standard atmosphere, $H = 11\,\mathrm{km}$ and $f$ at $45{}^{\circ}$, $L_R = 1210\,\mathrm{km}$: the scale of the depressions and anticyclones on a weather map. For the upper kilometre of the North Atlantic column at $40.5{}^{\circ}$ N, $L_R = 38\,\mathrm{km}$, thirty times smaller: the ocean has its own weather, eddies tens of kilometres across, and a model that resolves them needs a much finer grid. On scales near $L_R$, $f$ changes by $\beta L_R/f$, $19\,\%$ for the atmosphere and under one per cent for the ocean: ocean eddies live on an $f$-plane, weather systems feel the $\beta$-plane.

In the lab — A planet on a turntable

In a laboratory of geophysical fluid dynamics, a tank of water stands on a turntable; a camera turns with it and sees the water as an observer on a planet would. Before each run the water is left to spin up until it turns as a solid body, which takes a few minutes, many turns of the table. Then the experimenters inject dye, or move an obstacle slowly along the floor: the dye rises in a vertical column above the obstacle and the flow parts round it at every level. With a layer of fresh water over salt water, or a heated outer wall and a cooled centre, the tank becomes baroclinic, and jets, eddies and meandering fronts appear, whose size the Rossby radius of the tank predicts.

19.6 Exercises

Exercise 19.1 ★

Take $\Omega = 7.292\,115 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$. (a) Compute the Coriolis parameter and the inertial period at $30{}^{\circ}$, $45{}^{\circ}$ and $60{}^{\circ}$ N. (b) Compute the radius of the inertial circle of a current of $0.20\,\mathrm{m}/\mathrm{s}$ at the same latitudes. (c) Compare the inertial period at $30{}^{\circ}$ with the sidereal day, $23.9345\,\mathrm{h}$.

Solution

Solution of Exercise 19.1.

(a) $f = 2\Omega\sin\varphi$: $7.292 \times 10^{-5}$, $1.031 \times 10^{-4}$ and $1.263 \times 10^{-4}\,\mathrm{s}^{-1}$; $2\pi/f$: 23.93, 16.92 and $13.82\,\mathrm{h}$. (b) $U/f$: 2.74, 1.94 and $1.58\,\mathrm{km}$. (c) At $30{}^{\circ}$, $\sin\varphi = 1/2$ and $2\pi/f = 2\pi/\Omega$: the inertial period is the sidereal day, $23.93\,\mathrm{h}$.

Exercise 19.2 ★

At $45{}^{\circ}$ N, compute the Rossby number of a dust devil ($10\,\mathrm{m}/\mathrm{s}$, $10\,\mathrm{m}$), a sea breeze ($5\,\mathrm{m}/\mathrm{s}$, $20\,\mathrm{km}$), a depression ($15\,\mathrm{m}/\mathrm{s}$, $1000\,\mathrm{km}$) and an ocean gyre ($0.05\,\mathrm{m}/\mathrm{s}$, $2000\,\mathrm{km}$). Which of them are close to geostrophic balance?

Solution

Solution of Exercise 19.2.

With $f = 1.031 \times 10^{-4}\,\mathrm{s}^{-1}$: dust devil $9.7 \times 10^{3}$, sea breeze 2.4, depression 0.15, gyre $2.4 \times 10^{-4}$. The depression and the gyre are close to geostrophic balance; the dust devil turns either way, unaware of the Earth’s rotation, and the sea breeze feels it only over a day.

Exercise 19.3 ★

At $45{}^{\circ}$ N the mean temperature of the troposphere, $250\,\mathrm{K}$, falls northward by $5\,\mathrm{K}$ per $1000\,\mathrm{km}$. Using the first term of the thermal wind in height, compute the vertical shear of the westerly wind and the increase of the wind over $10\,\mathrm{km}$.

Solution

Solution of Exercise 19.3.

$\partial u_g/\partial z = (g/fT)\,|\partial T/\partial y| = 9.81\times 5 \times 10^{-6}/(1.031 \times 10^{-4}\times250) = 1.90 \times 10^{-3}\,\mathrm{s}^{-1}$; over $10\,\mathrm{km}$ the westerly wind increases by $19\,\mathrm{m}/\mathrm{s}$.

Exercise 19.4 ★

The stability $E = -(1/\rho)\,\mathrm d\rho_\theta/\mathrm dz$ of the upper kilometre of the ocean lies between $50 \times 10^{-8}$ and $1000 \times 10^{-8}\,\mathrm{m}^{-1}$; in deep trenches it is $10^{-8}\,\mathrm{m}^{-1}$. Compute $N = \sqrt{gE}$, the period of the buoyancy oscillation and $N$ in cycles per hour for these three values.

Solution

Solution of Exercise 19.4.

$E = 50 \times 10^{-8}$: $N = 2.21 \times 10^{-3}\,\mathrm{s}^{-1}$, a period of $47\,\mathrm{min}$, $1.27$ cycles per hour. $E = 1000 \times 10^{-8}$: $N = 9.90 \times 10^{-3}\,\mathrm{s}^{-1}$, $10.6\,\mathrm{min}$, 5.7 cycles per hour. $E = 10^{-8}$: $N = 3.13 \times 10^{-4}\,\mathrm{s}^{-1}$, $5.6\,\mathrm{h}$, 0.18 cycles per hour.

Exercise 19.5 ★★

A weather system has $U = 10\,\mathrm{m}/\mathrm{s}$, $L = 1000\,\mathrm{km}$, $H = 10\,\mathrm{km}$, at $45{}^{\circ}$ N. (a) Estimate its vertical velocity and its vertical acceleration, and compare the latter with $g$ and with the vertical gradient of the pressure perturbation, $fUL/H$. (b) Compute the dropped Coriolis term $\tilde fu$ and compare it with $g$.

Solution

Solution of Exercise 19.5.

(a) $W \sim UH/L = 0.1\,\mathrm{m}/\mathrm{s}$ and $\mathrm Dw/\mathrm Dt \sim UW/L = 10^{-6}\,\mathrm{m}/\mathrm{s}^{2}$, $1.0 \times 10^{-7}$ of $g$. The pressure term, $fUL/H = 0.103\,\mathrm{m}/\mathrm{s}^{2}$, is larger by $1 \times 10^{5}$: the ratio, $9.7 \times 10^{-6}$, is $\mathrm{Ro}(H/L)^2$ with $\mathrm{Ro} = 0.097$. (b) $\tilde fu = 2\Omega\cos45{}^{\circ}\times10 = 1.03 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$, $1.05 \times 10^{-4}$ of $g$.

Exercise 19.6 ★★

Take $f = 10^{-4}\,\mathrm{s}^{-1}$ and circular isobars of radius $300\,\mathrm{km}$. (a) The pressure gradient round a low gives $V_g = 20\,\mathrm{m}/\mathrm{s}$: compute the gradient wind. (b) Round a high, $V_g = 5\,\mathrm{m}/\mathrm{s}$: compute the gradient wind. (c) What is the largest $V_g$ possible round the high, and the wind then?

Solution

Solution of Exercise 19.6.

$fr = 30\,\mathrm{m}/\mathrm{s}$. (a) $\mathrm{Ro}_g = 0.667$, $V = 40/(1 + \sqrt{3.667}) = 13.7\,\mathrm{m}/\mathrm{s}$, a third slower than geostrophy. (b) $\mathrm{Ro}_g = 0.167$, $V = 10/(1 + \sqrt{0.333}) = 6.34\,\mathrm{m}/\mathrm{s}$. (c) $V_g \le fr/4 = 7.5\,\mathrm{m}/\mathrm{s}$, and then $V = fr/2 = 15\,\mathrm{m}/\mathrm{s}$, twice the geostrophic speed.

Exercise 19.7 ★★

A tank $20\,\mathrm{cm}$ deep turns once every $6\,\mathrm{s}$; its water has $\nu = 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$. (a) Compute $f = 2\Omega$, then the Rossby number of a flow of $1\,\mathrm{cm}/\mathrm{s}$ past a cylinder $5\,\mathrm{cm}$ across, and the Ekman number of the tank. Will a Taylor column form? (b) Theory gives a spin-up time $H/\sqrt{\nu\Omega}$: compute it in minutes and in turns of the table.

Solution

Solution of Exercise 19.7.

(a) $\Omega = 2\pi/6 = 1.047\,\mathrm{rad}/\mathrm{s}$, $f = 2.094\,\mathrm{s}^{-1}$. $\mathrm{Ro} = 0.01/(2.094\times0.05) = 0.095$ and $\mathrm{Ek} = 10^{-6}/(2.094\times0.04) = 1.19 \times 10^{-5}$: both small, so the flow obeys the Taylor–Proudman theorem and a Taylor column forms. (b) $0.2/\sqrt{10^{-6} \times1.047} = 195\,\mathrm{s}$, $3.3\,\mathrm{min}$, 33 turns of the table.

Exercise 19.8 ★★

In the January zonal mean of 1991–2020 the westerly wind is $4.27\,\mathrm{m}/\mathrm{s}$ at $850\,\mathrm{hPa}$ and $40.39\,\mathrm{m}/\mathrm{s}$ at $250\,\mathrm{hPa}$ at $30{}^{\circ}$ N, $5.99\,\mathrm{m}/\mathrm{s}$ and $23.17\,\mathrm{m}/\mathrm{s}$ at $45{}^{\circ}$ N. With $R = 286.9\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$, compute the meridional gradient of the layer-mean temperature at both latitudes, and say why the jet is stronger at $30{}^{\circ}$ N.

Solution

Solution of Exercise 19.8.

$\partial\bar T/\partial y = -f\,\Delta u/(R\ln(850/250))$, with $\ln(850/250) = 1.224$. At $30{}^{\circ}$ N, $-7.29 \times 10^{-5}\times36.12/(286.9 \times1.224) = -7.50\,\mathrm{K}$ per $1000\,\mathrm{km}$; at $45{}^{\circ}$ N, $-1.031 \times 10^{-4}\times17.18/351.1 = -5.05\,\mathrm{K}$ per $1000\,\mathrm{km}$. The shear is proportional to the gradient divided by $f$: the gradient is 1.5 times larger at $30{}^{\circ}$ and $f$ 1.4 times smaller, so the shear is 2.1 times larger.

Exercise 19.9 ★★

(a) Compute the Rossby radius of the troposphere ($N = 0.0113\,\mathrm{s}^{-1}$, $H = 11\,\mathrm{km}$, $45{}^{\circ}$ N) and of the upper ocean ($N = 0.003\,57\,\mathrm{s}^{-1}$, $H = 1000\,\mathrm{m}$, $40.5{}^{\circ}$ N). (b) With $a = 6371\,\mathrm{km}$, compute $\beta$ at both latitudes and the relative change $\beta L_R/f$ of the Coriolis parameter over each radius.

Solution

Solution of Exercise 19.9.

(a) $L_R = 0.0113\times11\,000/1.031 \times 10^{-4} = 1206\,\mathrm{km}$; $0.00357\times1000/9.47 \times 10^{-5} = 37.7\,\mathrm{km}$, 32 times smaller. (b) $\beta = 2\Omega\cos\varphi/a$: $1.62 \times 10^{-11}$ and $1.74 \times 10^{-11}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}$. $\beta L_R/f = 0.19$ for the atmosphere, 0.007 for the ocean.

Exercise 19.10 ★★★

A parcel of the surface ocean set moving at $U$ is slowed by a linear friction: $\mathrm du/\mathrm dt = fv - ru$, $\mathrm dv/\mathrm dt = -fu - rv$. (a) Solve with $w = u + \mathrm iv$ and describe the path. (b) The parcel is also carried by a steady current $u_0$ eastward: give its position. (c) At $45{}^{\circ}$ N with $U = 0.2\,\mathrm{m}/\mathrm{s}$, $r^{-1} = 4\text{ days}$ and $u_0 = 0.05\,\mathrm{m}/\mathrm{s}$, give its speed relative to the current after four days and the distance it has drifted.

Solution

Solution of Exercise 19.10.

(a) $\mathrm dw/\mathrm dt = -(r + \mathrm if)w$, so $w = U\mathrm e^{-rt}\mathrm e^{-\mathrm ift}$: the velocity turns at the rate $f$ while its magnitude decays as $\mathrm e^{-rt}$. The position $x + \mathrm iy = U(1 - \mathrm e^{-(r + \mathrm if)t})/(r + \mathrm if)$ spirals into the point $U/(r + \mathrm if)$. (b) $x + \mathrm iy = U(1 - \mathrm e^{-(r + \mathrm if)t})/(r + \mathrm if) + u_0t$: the spiral is drawn out into loops. (c) $0.2\,\mathrm e^{-1} = 0.074\,\mathrm{m}/\mathrm{s}$; the centre has drifted $u_0t = 0.05\times4\times86\,400 = 17.3\,\mathrm{km}$ eastward.

Exercise 19.11 ★★★

Across an ocean front at $40{}^{\circ}$ N the density of the water increases northward by $1\,\mathrm{kg}/\mathrm{m}^{3}$ over $100\,\mathrm{km}$, through the upper $1000\,\mathrm{m}$; below, the water is uniform and still. With $\rho_0 = 1025\,\mathrm{kg}/\mathrm{m}^{3}$, compute the vertical shear of the current and the current at the surface. In which direction does it flow?

Solution

Solution of Exercise 19.11.

$f = 9.37 \times 10^{-5}\,\mathrm{s}^{-1}$; $\partial u/\partial z = (g/f\rho_0)\,\partial \rho/\partial y = 9.81\times10^{-5}/(9.37 \times 10^{-5}\times1025) = 1.02 \times 10^{-3}\,\mathrm{s}^{-1}$. With $u = 0$ at $1000\,\mathrm{m}$ the surface current is $1.02\,\mathrm{m}/\mathrm{s}$, eastward: in the Northern Hemisphere the current flows with the light water on its right.

Exercise 19.12 ★★★

The annual-mean column at $40.5{}^{\circ}$ N, $30.5{}^{\circ}$ W has potential densities $\sigma_\theta = \rho_\theta - 1000\,\mathrm{kg}/\mathrm{m}^{3}$ of 26.029 at the surface, 26.697 at $100\,\mathrm{m}$, 27.079 at $500\,\mathrm{m}$ and 27.628 at $1000\,\mathrm{m}$. (a) Compute $N$ in each of the three layers, in $\mathrm{s}^{-1}$ and in cycles per hour. (b) Which layer is the most stable, and why? (c) Compare the values with the range of the upper ocean in Exercise 19.4.

Solution

Solution of Exercise 19.12.

(a) $N^2 = (g/\rho)\,\Delta\sigma_\theta/\Delta z$. 0–$100\,\mathrm{m}$: $9.81\times0.668/(1026.4\times100)$, $N = 7.99 \times 10^{-3}\,\mathrm{s}^{-1}$, 4.6 cycles per hour; 100–$500\,\mathrm{m}$: $3.02 \times 10^{-3}\,\mathrm{s}^{-1}$, 1.7 cycles per hour; 500–$1000\,\mathrm{m}$: $3.24 \times 10^{-3}\,\mathrm{s}^{-1}$, 1.9 cycles per hour. (b) The top $100\,\mathrm{m}$: it contains the warm surface water above the seasonal pycnocline. Between 100 and $500\,\mathrm{m}$ the water is more uniform than below. (c) $E = N^2/g$ is $650 \times 10^{-8}$, $93 \times 10^{-8}$ and $107 \times 10^{-8}\,\mathrm{m}^{-1}$, all within the range of the upper ocean.

19.7 Problem: Winds over the Steppe of Orrin

Problem 19.1

Weekend problem — the rotation, the stratification and the winds of an invented plain at $48{}^{\circ}$ N

The Steppe of Orrin is a flat plain at $48{}^{\circ}$ N with a lake, Lake Orrin, $12\,\mathrm{km}$ wide, and a small laboratory with a rotating tank. Take $\Omega = 7.292\,115 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$, $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$, $R = 286.9\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$ and $c_p = 1004\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$. All values of the steppe, its lake and its winds are invented.

Part I — Does rotation matter?

1. Compute the Coriolis parameter and the inertial period at Orrin.
2. Compute the Rossby number of five flows: water draining from a bathtub ($0.1\,\mathrm{m}/\mathrm{s}$ over $0.3\,\mathrm{m}$); the tank of the laboratory, which turns once every $10\,\mathrm{s}$ (with $f = 2\Omega_{\text{tank}}$; $5\,\mathrm{mm}/\mathrm{s}$ over $0.2\,\mathrm{m}$); the evening breeze off the lake ($5\,\mathrm{m}/\mathrm{s}$, $20\,\mathrm{km}$); a depression ($15\,\mathrm{m}/\mathrm{s}$, $1000\,\mathrm{km}$); the gyre of an ocean basin at the same latitude ($0.05\,\mathrm{m}/\mathrm{s}$, $2000\,\mathrm{km}$).
3. Which of these flows feel the rotation that drives them, and which are close to geostrophic?
4. Compute the Ekman number of the tank ($0.2\,\mathrm{m}$ deep, $\nu = 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$), of the depression (eddy viscosity $10\,\mathrm{m}^{2}/\mathrm{s}$, depth $10\,\mathrm{km}$) and of the gyre (eddy viscosity $10^{-3}\,\mathrm{m}^{2}/\mathrm{s}$, depth $1000\,\mathrm{m}$).
5. The Ekman layer has the thickness $\delta = \sqrt{2\nu/f}$. Show that $\delta/H = \sqrt{2\,\mathrm{Ek}}$ and compute $\delta$ for the tank and for the air of the steppe.
6. What do these numbers say about where friction matters?

Part II — The wind at the ground.

7. At the ground the pressure is $1012.9\,\mathrm{hPa}$ and the temperature $288.1\,\mathrm{K}$. Compute the density of the air.
8. The sea-level pressure falls northward by $4\,\mathrm{hPa}$ per $300\,\mathrm{km}$. Compute the geostrophic wind and give its direction.
9. The same gradient surrounds a depression whose isobars have a radius of $600\,\mathrm{km}$. Compute the Rossby number of the isobars and the gradient wind.
10. The same gradient surrounds an anticyclone of the same radius. Compute the gradient wind, and the largest pressure gradient, in $\mathrm{hPa}$ per $300\,\mathrm{km}$, that the anticyclone can hold.

Part III — Stratification.

11. The temperature of the troposphere falls by $6.49\,\mathrm{K}/\mathrm{km}$. Compute the dry adiabatic lapse rate $g/c_p$, and $N$ and the period of the buoyancy oscillation at $5\,\mathrm{km}$, where the temperature is $255.7\,\mathrm{K}$.
12. Above $11\,\mathrm{km}$ the air is isothermal at $216.6\,\mathrm{K}$. Show that $N^2 = g^2/(c_pT)$ there, and compute $N$ and its ratio to the value of question 11.
13. Compute the Rossby radius of the troposphere of Orrin, with the $N$ of question 11 and $H = 10\,\mathrm{km}$, and compare it with the size of the depression of Part I.
14. In summer, the upper $20\,\mathrm{m}$ of Lake Orrin are lighter than the deep water by $1.5\,\mathrm{kg}/\mathrm{m}^{3}$ ($\rho_0 = 1000\,\mathrm{kg}/\mathrm{m}^{3}$). Compute the reduced gravity $g' = g\,\Delta\rho/\rho_0$, the speed $\sqrt{g'h}$ of the slow internal waves of the upper layer, and the Rossby radius $\sqrt{g'h}/f$.
15. Will the rotation of the Earth shape the internal motions of the lake?

Part IV — The thermal wind.

16. Two radiosonde stations stand $600\,\mathrm{km}$ apart on the same meridian. The mean temperature of the layer from the ground, $1013\,\mathrm{hPa}$, to $265\,\mathrm{hPa}$, about $10\,\mathrm{km}$, is $253.0\,\mathrm{K}$ at the southern station and $249.4\,\mathrm{K}$ at the northern. Compute the meridional temperature gradient in $\mathrm{K}$ per $1000\,\mathrm{km}$.
17. The thickness of a layer between two pressures is $\Delta z = (R\bar T/g)\ln(p_b/p_t)$. Compute it at both stations, and the slope of the $265\,\mathrm{hPa}$ surface relative to the ground-level one.
18. Deduce the increase of the geostrophic wind from the ground to $265\,\mathrm{hPa}$, from that slope.
19. Compute the same increase with the thermal wind theorem in pressure.
20. Is the air of the steppe barotropic or baroclinic? What would the wind at $265\,\mathrm{hPa}$ be if it were barotropic?
21. Compute the increase from the first term of the thermal wind in height, with the mean temperature of the standard troposphere between 0 and $10\,\mathrm{km}$, $255.7\,\mathrm{K}$, and compare.
22. Compute the wind at $541\,\mathrm{hPa}$, about $5\,\mathrm{km}$, assuming the temperature gradient is the same at every level.
23. The Richardson number $N^2/(\partial u/\partial z)^2$ must fall below 0.25 for the shear to drive turbulence. Compute it for the troposphere of Orrin.
24. In January the zonal-mean westerly at $250\,\mathrm{hPa}$ is $23.2\,\mathrm{m}/\mathrm{s}$ at $45{}^{\circ}$ N and $40.4\,\mathrm{m}/\mathrm{s}$ at $30{}^{\circ}$ N. Compare with the wind of Orrin, and explain why a local wind can exceed the zonal mean.
25. Give the geostrophic wind at $10\,\mathrm{km}$ above the Steppe of Orrin.

Solution

Solution of Problem 19.1.

1. $f = 2\times7.292\,115 \times 10^{-5}\sin48{}^{\circ} = 1.084 \times 10^{-4}\,\mathrm{s}^{-1}$; $2\pi/f = 16.1\,\mathrm{h}$.

2. Bathtub $3.1 \times 10^{3}$; tank ($f = 4\pi/10 = 1.257\,\mathrm{s}^{-1}$) 0.020; breeze 2.3; depression 0.14; gyre $2.3 \times 10^{-4}$.

3. The bathtub feels nothing of the Earth’s rotation; the breeze feels it only after some hours. The tank, the depression and the gyre have small Rossby numbers and are close to geostrophic.

4. Tank $10^{-6}/(1.257\times0.04) = 2.0 \times 10^{-5}$; depression $10/(1.084 \times 10^{-4}\times10^8) = 9.2 \times 10^{-4}$; gyre $10^{-3}/(1.084 \times 10^{-4}\times10^6) = 9.2 \times 10^{-6}$.

5. $\delta/H = \sqrt{2\nu/(fH^2)} = \sqrt{2\,\mathrm{Ek}}$. Tank $\sqrt{2\times10^{-6}/1.257} = 1.3\,\mathrm{mm}$; air $\sqrt{20/1.084 \times 10^{-4}} = 430\,\mathrm{m}$.

6. Friction acts in layers a few thousandths of the depth in the tank and the ocean, a few hundredths in the atmosphere; above them the flow is nearly frictionless.

7. $\rho = p/(RT) = 101\,290/(286.9\times288.1) = 1.225\,\mathrm{kg}/\mathrm{m}^{3}$.

8. $u_g = (1/\rho f)\,|\partial p/\partial y| = (400/300\,000)/(1.225 \times1.084 \times 10^{-4}) = 10.0\,\mathrm{m}/\mathrm{s}$, from the west, with the low pressure to the north on the left.

9. $\mathrm{Ro}_g = 10.04/(1.084 \times 10^{-4}\times600\,000) = 0.154$; $V = 2\times10.04/(1 + \sqrt{1.618}) = 8.84\,\mathrm{m}/\mathrm{s}$, $12\,\%$ slower.

10. $V = 2\times10.04/(1 + \sqrt{0.382}) = 12.4\,\mathrm{m}/\mathrm{s}$. The largest $V_g$ is $fr/4 = 16.3\,\mathrm{m}/\mathrm{s}$, a gradient $\rho fV_g = 2.16 \times 10^{-3}\,\mathrm{Pa}/\mathrm{m}$, $6.5\,\mathrm{hPa}$ per $300\,\mathrm{km}$.

11. $g/c_p = 9.77\,\mathrm{K}/\mathrm{km}$. $N^2 = (9.81/255.7)\times(9.77 - 6.49)\times10^{-3} = 1.26 \times 10^{-4}\,\mathrm{s}^{-2}$, $N = 0.0112\,\mathrm{s}^{-1}$, a period of $9.3\,\mathrm{min}$.

12. With $\Gamma = 0$, $N^2 = (g/T)(g/c_p) = g^2/(c_pT) = 4.43 \times 10^{-4}\,\mathrm{s}^{-2}$, $N = 0.0210\,\mathrm{s}^{-1}$, 1.9 times the tropospheric value.

13. $L_R = 0.0112\times10^4/1.084 \times 10^{-4} = 1035\,\mathrm{km}$, the size of the depression.

14. $g' = 0.0147\,\mathrm{m}/\mathrm{s}^{2}$; $\sqrt{g'h} = 0.54\,\mathrm{m}/\mathrm{s}$; $L_R = 0.542/1.084 \times 10^{-4} = 5.0\,\mathrm{km}$.

15. Yes: the lake, $12\,\mathrm{km}$ wide, is more than twice its Rossby radius, so its internal waves and currents are turned by the Coriolis force.

16. $(249.4 - 253.0)/600 = -6.0\,\mathrm{K}$ per $1000\,\mathrm{km}$.

17. $\ln(1013/265) = 1.3409$. South: $286.9\times253.0\times1.3409/9.81 = 9922\,\mathrm{m}$; north $9781\,\mathrm{m}$. The $265\,\mathrm{hPa}$ surface lies $141\,\mathrm{m}$ lower in the north, a slope of $2.35 \times 10^{-4}$ relative to the lower surface.

18. $\Delta u = (g/f)\times\text{slope} = 9.81\times2.35 \times 10^{-4}/ 1.084 \times 10^{-4} = 21.3\,\mathrm{m}/\mathrm{s}$.

19. $(R/f)\ln(p_b/p_t)\,|\partial\bar T/\partial y| = (286.9/1.084 \times 10^{-4})\times1.3409\times6.0 \times 10^{-6} = 21.3\,\mathrm{m}/\mathrm{s}$, the same: the two calculations are one.

20. Baroclinic: the temperature, hence the density, changes along the pressure surfaces. Barotropic, the wind would not change with height: $10.0\,\mathrm{m}/\mathrm{s}$.

21. $(g/f\bar T)\,|\partial T/\partial y|\,\Delta z = 9.81\times 6.0 \times 10^{-6}\times10^4/(1.084 \times 10^{-4}\times255.7) = 21.2\,\mathrm{m}/\mathrm{s}$, within $0.3\,\%$.

22. $10.04 + (286.9/1.084 \times 10^{-4})\times6.0 \times 10^{-6}\times \ln(1013/541) = 10.04 + 9.96 = 20.0\,\mathrm{m}/\mathrm{s}$.

23. $\partial u/\partial z = 21.3/10^4 = 2.13 \times 10^{-3}\,\mathrm{s}^{-1}$; $\mathrm{Ri} = 1.26 \times 10^{-4}/4.54 \times 10^{-6} = 28$, far above 0.25: the shear does not make the free troposphere turbulent.

24. The wind of Orrin, $31\,\mathrm{m}/\mathrm{s}$, is stronger than the zonal mean at $45{}^{\circ}$ N and weaker than at $30{}^{\circ}$ N. The polar-front jet wanders north and south; at any moment its gradient, and its wind, is concentrated over a band, and averaging round a latitude circle spreads it out.

25. $10.0 + 21.3$: the geostrophic wind at $10\,\mathrm{km}$ above the steppe is $31\,\mathrm{m}/\mathrm{s}$, from the west ($31.3\,\mathrm{m}/\mathrm{s}$ by the calculation).