# Homework 3: The Coriolis Parameter and the Geostrophic Wind

Due May 10.

(a) Discuss why Coriolis Effect is maximum at the Poles and minimum at the Equator. Your answer should include reasoning based upon the equation in the handout, but should also have a section giving descriptive reasoning. (20 pts)

Coriolis acceleration can be viewed as being dependent on the fact that different points on the earth's surface at different distances from the axis of rotation are moving at different linear speeds relative to a point in space. This is true despite the fact that the earth is in constant angular rotation of 15 degrees longitude counterclockwise.

In order for there to be a Coriolis acceleration, therefore, the earth's surface must be perpendicular to the axis of rotation. This is true at the Poles and not true at the Equator, where the earth's surface is tangent to the axis of rotation.

The relationship of the earth's surface at any latitude to the axis of rotation can be resolved into two "components", one perpendicular to the earth's axis and one parallel to it. The degree to which one of those "components" is perpendicular to the axis of rotation is measured by the factor sin phi (where phi is the latitude).

Since sin 0 is 0 and sin 90 is 1, there is no Coriolis acceleration at the equator, and Coriolis acceleration at the poles is maximum. This is consistent with the qualitative reasoning given above.

(b) Calculate the value of the Coriolis acceleration at our latitude for a wind speed of 10 m s^{-1}. The angular velocity of the earth is 7.292 X 10^{-5} s^{-1} and our latitude is approximately 40^{0} N. 40^{0} is approximately 0.6891 radians. (30 pts)

The Coriolis acceleration is given by 2 omega V sin latitude.

V= 10 m s^{-1}, omega = 7.292 X 10^{-5}s^{-1}, sin 40^{0}= .6428

Inserting each into the equation gives

Coriolis Acceleration = 9 X 10^{-4}m s^{-2}

(c) You are provided the 500 mb analysis (without data
plots) for 12 UTC 9 February 2005. Note the station in southeastern Oklahoma (indicated by the
square). The line segment shown is
a portion of the "n" axis (normal to the flow) at that location, with
the segment length shown there (delta n) = 300 km.

^{-2}, f is the Coriolis parameter, numerically equal to approximately 10 X 10

^{-5}s

^{-1}at around the latitude of the station shown, delta z is the difference in 500 mb heights normal to the flow measured along the increment delta n (remember, to estimate a gradient, take the value at location 2, furthest along the positive directon of the given axis, and subtract from it the value at location 1.)

Estimate the value of the geostrophic wind in meters per second, kilometers per hour and knots at the location shown and plot the value with conventional weather map symbols for wind directon and wind speed (knots) on the map.

Caution, Caution:the challenge here will be to keep units consistent initially. Please remember that when you solve for the geostrophic wind speed using the above expression, your result will initially be in meters per second. You will need to convert that to kilometers per hour and knots. Remember that a nautical mile is 6040 feet.1. (30 points)

g = 9.8 m s^{-2}

f= 10 X 10^{-5}s^{-1}

∆z/∆n= -(5580m-5700m)/3.0 X 10^{5}m = - 4.0 X 10^{-3}

V_{g}= - [( 9.8 m s^{-2})/10 X 10^{-5}s^{-1}] [- 4.0 X 10^{-3}] = 39.1 m s^{-1}=141.1 km h^{-1}

(39.1 m s^{-1}) X 100 cm/m X 1 inch/2.54 cm X 1 foot/12 in X 1 nautical mile/6040 ft X 3600 s/h = 77.1 knots

*
*

2. With a long
arrow/streamline drawn right on the chart, locate the position of the fastest
current (the polar jet stream).

Fig. 1: 500 mb chart
for 1200 UTC 9 February 2005