GEF3450
Exercises for group session: Geostrophic Balance
Ada Gjermundsen
E-mail: ada.gjermundsen@geo.uio.no
November 2, 2015
Figure 1: Figure from UK Met Office
Let’s look at one of the most important force balances that drive the winds
and the ocean currents -it is a force balance that is a foundation of e.g. synoptic meteorology– the geostrophic balance.
Here’s how it works. We start with a pressure gradient–in figure 2, high
pressure is at the bottom and low pressure is at the top. Naturally the air
wants to flow from high pressure towards the low pressure, so it accelerates
in that direction. This force, that makes air move from high to low pressure,
is called the Pressure Gradient Force, or the “PGF” as it is often abbreviated.
The red arrow above represents the direction of this force–from high to low
pressure. The PGF only depends on the density of the air and the strength
of the pressure gradient–nothing else.
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Figure 2: Figure from http://lukemweather.blogspot.no/
However, there’s another “force” at work on this rotating planet of ours–the
Coriolis “force”. This isn’t really a true “force” but an apparent change in
the direction of the wind due to the fact that the earth is rotating out from
under it. This effect depends on the Coriolis parameter “f” and the velocity
of the wind (which is why I’ve abbreviated it as “fv” in the figure). The
faster the wind goes, the stronger this Coriolis turning effect becomes. In
the northern hemisphere, the Coriolis effect always acts to make the winds
appear to turn to the right of their direction of movement–always. Simply a
fact of the direction the earth rotates.
So as the air accelerates due to the pressure gradient force, the velocity increases and consequently the Coriolis effect (which depends on velocity) also
increases. This turns the winds more and more to the right the faster they
go. Finally, it gets to a point where the tendency of the Coriolis effect to
turn the winds to the right exactly balances the pressure gradient force, like
we see in figure above where the red PGF arrow is exactly balanced by the
blue Coriolis effect (fv) arrow. We see that the wind is no longer moving
from high to low pressure, but perpendicular to the pressure gradient.
From Luke’s weather blog: http://lukemweather.blogspot.no/
Note that the exact same applies for the ocean, just replace wind with current.
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The Geostrophic Equations:
ug = −
1 ∂p
f ρ ∂y
−vg = −
(1)
1 ∂p
f ρ ∂x
(2)
Exercises:
Exercises from “Fluid Mechanics” by Kundu and Cohen, “Atmosphere, Ocean and Climate Dynamics” by Marshall and Plumb, “Atmospheric Science” by Wallace and Hobbs,
“Dynamic Meteorology" by Holton and “Geophysical Fluid Mechanics” by Weber.
1 What is the pressure gradient required to maintain a geostrophic wind
at a speed of v = 10 ms−1 at 45◦ N? In the absence of a pressure
gradient show that air parcels flow around circles in an anti-cyclonic
sense of radius fv . What is the name of the motion?
2 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 ms−1 and the second northward at 10 ms−1 . During the 3 hr
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 hr 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 wind speed and direction at the location of the weather
ship. (Hint: assume geostrophic conditions)
3 Consider a low pressure system centered on 45◦ S, whose sea level pressure is described by
p = 1000hP a − ∆p · e−r
2 /R2
,
(3)
where r is the radial distance from the center. Determine the structure of the geostrophic wind around this system; find the maximum
geostrophic wind, and the radius of the maximum wind, if ∆p =
20 hPa, R = 500 km, and the density at sea level of interest is
1.3 kgm−3 . (Assume constant Coriolis parameter, appropriate to 45◦ S,
across the system.)
4 i) A typical hurricane at, say 30◦ latitude, may have low-level winds of
50 ms−1 at a radius of 50 km from its center: do you expect this flow
to be geostrophic?
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ii) Two weather stations near 45◦ N are 400 km apart, one exactly to
the northeast corner of the other. At both locations, the 500 hPa wind
is exactly southerly at 30 ms−1 . At the northeastern station, the height
of the 500 hPa surface is 5510 m; what is the height of this surface at
the other weather station?
(Hint: assume the wind to be geostrophic and parallel to lines of
constant height (Z) on an isobaric surface, then ug = − fg ∂Z
∂y and
g ∂Z
vg = f ∂x )
5 The Gulf stream flows northward along the coast of the United States
with a surface current of an average magnitude of 2 ms−1 . If the flow
is assumed to be in geostrophic balance, find the average slope of the
sea surface across the current at a latitude of 45◦ N.
6 Write down an equation for the balance of radial forces on a parcel of
fluid moving along in a horizontal circular path of radius r at constant
speed v (taken positive if the flow is in the same sense of rotation as
the Earth).
Solve for v as a function of r and the radial pressure gradient, and
hence show that:
i) if v > 0, the wind speed is less than its geostrophic value
ii) if |v| << f r, then the flow approaches its geostrophic value
iii) there is a limiting pressure gradient for the balanced motion when
v > −1/2f r.
(Hint: use the gradient wind balance equation)
7 Consider a circular tornado which rotates with a constant angular velocity ω around the vertical axes. In a distance r from the center of
the tornado , the tangential velocity is ωr.
i) Make a sketch of the forces acting on a fluid parcel at a distance
r from the center of the tornado. Assume the density is given by the
equation of state for dry air.
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ii) At a distance r0 from the center of the tornado, the pressure is p0 .
Show that the pressure at the center of the tornado is given by:
p = p0 exp(−
ω 2 r02
)
2RT
(4)
iii) If the temperature is 288 K, and pressure and wind speed at 100 m
from the center are 1000 hPa and 100 ms−1 , respectively, what is the
central pressure?
8 An aircraft flying a heading of 60◦ (i.e. 60◦ to the east of north)
at air speed 200 ms−1 moves relative to the ground due east (90◦ )
at 225 ms−1 . If the plane is flying at constant pressure, what is the
rate of change in altitude (in meters per kilometer horizontal distance)
assuming a steady pressure field, geostrophic winds and f = 10−4 s−1 ?
(Hint: geostrophic wind is parallel to lines of constant height (Z) on
g ∂Z
an isobaric surface (p=const), then ug = − fg ∂Z
∂y and vg = f ∂x )
9 An airplane pilot crossing the ocean at 45◦ N latitude has both a pressure altimeter and a radar altimeter, the latter measuring his absolute
height above sea. Flying an air speed of 100 ms−1 , he maintains altitude by referring to his pressure altimeter set for sea-level pressure of
1013 hPa. He holds an indicated 6000 m altitude. At the beginning of
one-hour period, he notes that his radar altimeter reads 5700 m, and
at the end of the hour he notes that it reads 5950 m. In what direction
and approximately how far has he drifted from his heading?
(Hint: assume the wind to be geostrophic and parallel to lines of
constant height (Z) on an isobaric surface, then ug = − fg ∂Z
∂y and
g ∂Z
vg = f ∂x )
Solutions:
1 We require a ∆p of 10 hPa in 1000 km. Inertial circles.
2 ug ' 20 ms−1
√
3 rmax = R/ 2 = 354 km. umax = −25.7 ms−1 .
4 a) Ro = fUL = 13.8, so no the flow is not expected to be in geostrophic
balance.
b) δz = ∂zδx + ∂zδy = ∂xδx = 89 m → Zw = 5421 m.
5 Eastward surface rise of 2.1 cm per km.
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7 iii) the central pressure is given by p ' p0 1 −
8 use ~uwind = ~uground − ~uair
−1 m/km
ω 2 r02
2RT
= 940 hPa.
→ vg = −100 ms−1 and ∂Z
∂x p =
f vg
g
'
9 The drift of the plane from its heading is D = vg t = 238 km. The drift
is to the left of the heading in the northern hemisphere.
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