Chapter 7
7.7 At a certain location along the ITCZ, the surface wind at 10◦N is
blowing from the east—northeast (ENE) from a compass angle of 60◦at a
speed of 8 m s−1and the wind at 7◦N is blowing from the south—southeast
(SSE) (150◦) at a speed of 5 m s−1. (a) Assuming that ∂/∂y >>
∂/∂x, estimate the divergence and the vorticity averaged over the belt
extending from 7◦N to 10◦N.
(b) The meridional component of the wind drops off linearly with
pressure from sea level (1010 hPa) to zero at the 900-hPa level. The
mixing ratio of water vapor within this layer is 20 g kg−1. Estimate the
rainfall rate under the assumption that all the water vapor that
converges into the ITCZ in the low level flow condenses and falls as
rain.
or, in Cartesian coordinates,
uΨ = −∂Ψ/∂y; vΨ = ∂Ψ/∂x
(a) Pure zonal flow: eastward if m < 0 and westward if m >
0.
(b) A wavy zonal flow in which the wavelength of the waves is L. (c)
Solid body rotation: counterclockwise if m > 0.
7.11 Apply Eq. (7.5), which describes the advection of a passive
tracer ψ by a horizontal flow pattern to a field in which the initial
conditions areψ = −my.
(a) Prove that at the initial time t = 0,
Proof: At the initial time,
| and |
∂t |
|
|
| d |
µ∂ψ |
¶ |
= |
|
|
| = |
(2)
|
| d |
µ−∂ψ |
¶ |
= |
|
¶ |
| = |
| = |
−m∂v
|
correponds to the deformation flow in Fig. 7.4a.
Integrating, we obtain
m = m0e−∂v∂y(t−t0)
Proof: From (7.3)
C ≡I Vsds
where ωr2is the angular momentum per unit mass.
7.13 Extend Fig. 7.8 by adding the positions of the marble at points
13—24.
(b) The marble is released from point r0 with initial clockwise
motionΩr0 in the fixed frame of reference. Show that in the rotating
frame of reference the marble remains stationary at the point of
release.
Analysis: In the rotating frame of reference, the motion is circular,
with radius r0 as in (a) but in the clockwise direction. Hence, relative
to the surface of the dish beneath it, the marble is moving
clockwise
Cabs = Crel + Ccoords (1)
From (7.3) it follows that if the loop is infinitesimally small, so
that the relative vorticity ζ can be considered to be uniform, Crel =
ζA. For the special case of motion close to the poles, where the Earth’s
surface is nor-mal to the axis of rotation, it follows from Exercise 7.1
that Cccords = 2ΩA. More generally, at latitude φ, the solid body
rotation of the Earth’s sur-face can be resolved into a component 2Ωsin
φ in the plane perpendicular to the Earth’s rotation and a component
2Ωcos φ in the plane of the axis of rotation. The vorticity and the
circulation reside in the perpendicular component. Hence, Ccoords =
2Ωsin φ = f. Substituting for Crel and Ccoords in (1) yields
Cabs = (ζ + f) A
(ΩRE + u)2/RE where u is the zonal velocity along the equator.
Substituting values, we obtain
(20)2
(6.37 × 106) = 0.627 × 10−4 m s−2
(b) Compute the apparent Coriolis force in the rotating coordinate
sys-tem.
Integrating (1) yields
w = w0 −gt (3)
Now let us consider the zonal velocity and displacement of the
projectile. Combining (2) and (3) yields
Solution: The Coriolis force is transverse to the train tracks and
directed toward the right of the motion of the train in the northern
hemisphere and toward the left in the southern hemisphere. Its magnitude
in units of N is given by mfV , where m is the mass of the locomotive, f
= 2Ωsin φ is the Coriolis parameter at the latitude φ of the locomotive,
and V is the velocity of the locomotive. Substituting values m = 2 ×
104kg, f = 10−4 s−1, and V = 40 m s−1yields 80 N. To put this force into
perspective, the ratio of this force to the weight of the train is
Since the isobars are oriented east-west, with lower pressure toward
the pole, the geostrophic wind is from the west.
7.20 (a) Derive an expression for the divergence of the geostrophic
wind (7.45) and give a physical interpretation of this result.
| = |
| = |
|
(2Ωsin φ)2 × 2Ω∂∂ysin φ
|
7.21 Two moving ships passed close to a fixed weather ship within a
few minutes of one another. The first ship was steaming eastward at a
rate of 5 m s−1 and the second was steaming northward at 10 m s−1.
During the 3-h period that the ships were in the same vicinity, the
first recorded a pressure rise of 3 hPa while the second recorded no
pressure change at all.
During the same 3-h period, the pressure rose 3 hPa at the location
of the weather ship (50◦N, 140◦W). On the basis of these data, calculate
the geostrophic wind speed and direction at the location of the weather
ship.
Applying (1) to the eastward-moving ship yields
from which it follows that
| = |
2.78 × 10−3
1.25 × 2 × 7.29 × 10−5sin 50◦19.9 m s−1
|
where Va ≡V −Vg is the ageostrophic component of the wind. Proof: In
the absence of friction, the horizontal equation of motion is
Proof: Gradient wind balance requires that
where a = V2/RT, b = f and c = Pn. Real solutions exist only if
b2−4ac > 0, i.e., if
f2> 4 |P| /RT
|
Proof: To obtain the first expression, we differentiate the
geostrophic wind equation (7.15a) with respect to pressure, which
yields
yields the thermal wind equation in (x, y, p) coordinates
Differentiating (2) with respect to height yields
Substituting for ∂p/∂z from the hydrostatic equation, we obtain
| ∂z |
|
|
| coordinates |
∂Vg |
|
|
(4) |
|
∂z |
fT×∇
|
∂zg
|
|
| GTA = Vg · ∇T ' −Vg · ∇T |
(2) |
Combining (1), (2), and (3) yields
It is also evident from the figure that the approximation (3) is
valid pro-vided that the thermal wind is fairly uniform within the
layer. The layer need not necessarily be thin.
(b) At a certain station located at 43◦N, the geostrophic wind at the
1000-hPa level is blowing from the southwest (230◦) at 15 m s−1while at
the 850-hPa level it is blowing from the west—northwest (300◦) at 30 m
s−1. Estimate the geostrophic temperature advection.
The geostrophic temperature advection is positive because the wind is
veering with height.
1Such large time rates of temperature are observed only in very
strong (and narrow) frontal zones. They are not sustained over intervals
of more than a few hours.
7.29 Prove that in the absence of friction, the circulation
[cs] ≡H c · ds
[cs] is the tangential velocity averaged over the length of the loop and
L = H accompanied by a proportionate decrease in the mean tangential
velocity along the loop and vice versa.
ds is the length of the loop. Hence, a lengthening of the loop must
be
14
7.33 Consider a sinusoidal wave along latitude φ with wavelength L
and am-plitude v in the meridional wind component. The wave is embedded
in a uniform westerly flow with speed U. (a) Show that the amplitude of
the geopotential height perturbations associated with the wave is
fvL/2πg where f is the Coriolis parameter and g is the gravitational
acceleration. (b) Show that the amplitude of the associated vorticity
perturbations is (2π/L)v. Show that the maximum values of the advection
of planetary and relative vorticity are βv and (2π/L)2Uv, respectively,
and that they are coincident and of opposing sign. (c) Show that the
advection terms exactly cancel for waves with wavelength
15
LS is referred to as the wavelength of a stationary Rossby wave. For
L < LS, the advection of relative vorticity is larger than the
advection of planetary vorticity and the waves propagate eastward and
for L > LS, the advection of planetary vorticity dominate and the
waves propagate westward. This behavior is actually observed when the
waves propagating along a prescribed latitude circle such as 50◦N are
decomposed into wave-numbers by harmonic analysis. Zonal wavenumber 1
tends to propagate westward and wavenumbers 4 and higher propagate
eastward.
| 1 |
dt(f + ζ) −1 H |
|
| (f + ζ) |
Making use of the identity dy/y = −dx/x, where y = 1/x, the above
expression can be rewritten as
|
| 1 |
d
dt(f + ζ) +
|
1 |
d(1/H) |
|
| (f + ζ) |
(1/H) |
dt |
|
7.36 Consider barotropic ocean eddies propagating meridionally along
a sloping continental shelf, with depth increasing toward the east, as
pictured in Fig. 7.25, conserving barotropic potential vorticity in
accordance with (7.27). There is no background flow. In which direction
will the eddies propagate?
At point A, the flow in the eddies is toward shallower water so the
depth H of a moving column of water is decreasing. Absolute vorticity (f
+ ζ) decrease by a proportionate amount so that barotropic potential
vorticity (f + ζ)/H remains constant. The planetary vorticity does not
change appreciably because the water column at A is not moving
meridionally. Hence there must be an anticyclinic tendency in relative
vorticity ζ. By the same line of reasoning it can be argued that there
must be a cyclonic vorticity tendency at pont B, where water columns are
being stretched as they move away from the shore.
7.37 During winter in middle latitudes, the meridional temperature
gradient is typically on the order of 1◦per degree of latitude, while
potential tem-perature increases with height at a rate of roughly 5◦C
km−1. What is a typical slope of the potential temperature surfaces in
the meridional plane? Compare this result with the slope of the 500-hPa
surface in Exercise 7.19.
In the meridional plane
Comparing with Exercise 7.19, we find that the isentropic in
extratropical latitudes surfaces slope about an order of magnitude more
steeply than the pressure surfaces
The vertical motion term is approximately equal to
From the equation of state
18
where V is the root mean squared velocity, p is the mean surface
pressure, and g is gravity. The potential energy (per unit area)
released by lowering the center of mass of the atmosphere by the
increment δz is given by
Substituting values, we obtain
δz = (17)2
2 × 9.8 = 14.7 m
If the air parcel moves poleward to latitude φ, ΩR2 Ewhile conserving
an-gular momentum,
(ΩRE cos φ + u) RE cos φ = ΩR2 E
7.45 On average over the globe, kinetic energy is being generated by
the cross-isobar flow at a rate of ∼2 W m−2. At this rate, how long
would it take to “spin up” the general circulation, starting from a
state of rest?
The “spin up” time is
T =KE (1) G
20
