# Homework 2

(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.

*The expression for Coriolis Acceleration is*

**∆V/∆t = 2 Ω V sin Ø**

**where Ω is the angular velocity of the earth and Ø is latitude. The geographic equator is at 0 ^{0 }latitude and at the poles, 90^{0}. Since sine 90^{0} of is 1 and sine 0^{0} is 0, then there is no Coriolis at the Equator. At latitudes between the Equator and the Poles, sine Ø gradually increases to a maximum value of 1. Thus, Coriolis acceleration (or force) is maximum at the Poles and minimum at the Equator.**

**In addition, the Coriolis acceleration is dependent upon a frictionless object moving relative to a plane perpendicular to the axis of rotation. The earth's surface is perpendicular to the earth's axis at the Poles, but parallel to it at the Equtor. Hence, there can be no Coriolis acceleration at the Equator. **

**In intermediate latitudes, a portion of the earth's surface is at angle with respect to the axis. To that extent, there is an increasing amount of deflection due to Coriolis as one moves poleward from the Equator.**

(b) Caculate 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.

*The expression for Coriolis Acceleration is*

**∆V/∆t = 2 Ω V sin Ø **

**sine 40 ^{0} = 0.6428; Ω = 7.292 X 10^{-5} s^{-1}; and V = 10 m s^{-1}. Substitute these values into the expression for Coriolis Acceleration. This gives:**

**∆V/∆t = 2 * (7.292 X 10 ^{-5} s^{-1}) * (10 m s^{-1}) * 0.6428 = 9.37 X 10^{-4} m s^{-2}**