(a) Surface map showing areas of high and low pressure. The solid lines are
isobars drawn at 4-mb intervals. The arrows represent wind direction. Notice
that the surface winds blow across the isobars. (b) The 500-mb map for the
same day as the surface map. Solid gray lines on the map are contour lines in
meters above sea level. Dashed red lines are isotherms in °C. Arrows show
wind direction. Notice that the wind tends to blow parallel to the contour lines.
Question: Why do the winds blow more or less parallel to the contour lines?
Figure 8-16 p209
Newton’s Laws of Motion
• Newton’s 1st Law:
– An object at rest will remain at rest, and an
object in motion will remain in straight line
motion with constant velocity as long as no
force is exerted on the object
Newton’s Laws of Motion
• Newton’s 2nd Law:
– The force exerted on an object equals its
mass times the acceleration produced
Acceleration = rate of change of velocity
An alternative way of writing Newton’s 2nd Law:
Stull (9.3c)
Conservation Laws
• Conservation of Energy
– 1st Law of Thermodynamics
• Conservation of Momentum
– Newton’s 2nd Law of Motion
• Conservation of Mass
– Continuity equation (Stull pp. 196-197, will be
discussed later)
Forces
• Newton’s 2nd Law: acceleration = force/mass
(Or acceleration = force per unit mass)
• Since acceleration = rate of change of velocity,
to find how velocity changes, we need to know
what the forces are
• What are the forces giving rise to changes in
horizontal atmospheric motion?
– Note that “centripetal force” (Ahrens p.211) is not
actually a force on its own but represents the
imbalance between the other forces, hence should
not be classified with the other forces
Pressure-Gradient Force
• Pressure gradient force: force due to
pressure difference over a distance
– The larger the pressure difference, the stronger the
force
– The shorter the distance, the stronger the force
– PG = p/d (Ahrens p. 211)
– Pressure gradient force
Stull (9.8)
x: eastward, y: northward; U: eastward velocity, V: northward velocity
Numerical example
• Station B is located 200 km north of station A. The MSL
pressure at station A is 1010 hPa, while that at station B is
1013 hPa. What is the pressure gradient force per unit
mass in the north-south direction? Assume  = 1.2 kgm-3.
• For the two stations, y = 200 km = 200000 m, p = 3hPa
= 300 Pa
• Using Stull (9.3b), we get
– Hence Fy/m = -1/1.2 x 300/200000 = -1.25x10-3 ms-2
• See also example in Stull (p.183)
Coriolis force
• Coriolis force is an apparent force due to
the rotation of the earth
• Straight line motion (when observed in
space) appears curved in a rotating frame
– Note that Ahrens’ satellite illustration (p. 213,
left column) is wrong
• Properties of the force:
– Perpendicular to the velocity
• To the right in the Northern Hemisphere (to the left
in the Southern Hemisphere)
– The larger the velocity, the stronger the force
– Horizontal component largest at the pole, zero
at the equator
Coriolis Force
• Coriolis paremeter:
Stull (9.10)
– Where 2 = 1.458 x 10-4 s-1
– Hence in the mid-latitudes ( ~ 45 deg), f ~ 10-4 s-1
• Coriolis force:
Stull (9.11)
Note:  is angular velocity of the earth = 2/1day = 2/86400s
Numerical example
• Find the Coriolis parameter for Stony Brook
• If the wind at Stony Brook is 20 ms-1 from the south,
what is the magnitude and direction of the Coriolis force?
• From Stull (9.10) f = 2 sin()
– Latitude of Stony Brook is about 41 °N, hence
f = 1.458 x 10-4 x sin(41) = 9.6 x 10-5 s-1
• For the Coriolis force, wind 20 ms-1 from the south,
meaning
V = 20 ms-1, and U = 0.
– From Stull (9.11), Fy/m = -fU = 0;
– Fx/m = fV = 9.6 x 10-5 x 20 = 1.92x10-3 ms-2
– Hence the force is from west to east
See also example from Stull (p.184)