lecture dynamics rotation

20
Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014 Earth reference frame Geoid Rotation definitions Centrifugal force Coriolis force Inertial motion READING: DPO: Chapter 7.2.3, 7.5.1 Stewart chapter 7.6, 9.1 Tomczak and Godfrey chapter 3, pages 29-35 alley SIO 210 (2014)

Upload: thoi-hoa-do

Post on 21-Nov-2015

13 views

Category:

Documents


2 download

DESCRIPTION

rotation

TRANSCRIPT

PowerPoint Presentation

Dynamics III: Earth rotationL. Talley SIO 210 Fall, 2014Earth reference frameGeoidRotation definitionsCentrifugal forceCoriolis forceInertial motion

READING:DPO: Chapter 7.2.3, 7.5.1 Stewart chapter 7.6, 9.1Tomczak and Godfrey chapter 3, pages 29-35

Talley SIO 210 (2014)1The Earth as a reference frameCoordinate system for most oceanography:

Local Cartesian (x,y,z), ignoring sphericityWest-east (zonal) x direction (positive eastwards)South-north (meridional) y-direction (positive northwards)Down-up (vertical) z-direction (positive upwards)Talley SIO 210 (2014)Add diagram2An aside on the Earths reference frame for pressure: the geoidWhen we calculate horizontal pressure gradients, they are relative to a level surface along which the pressure gradient vanishes (no force, no motion)

Earths mass is not distributed evenly AND Earth is not a perfect ellipsoid. That is, the level surface is not locally flat.

The surface of constant pressure, i.e. along which the gravitational force has no changes, is the geoid

Talley SIO 210 (2014)Good geoid site http://www.ngs.noaa.gov/GEOID/geoid_def.html

3GeoidEarths mass is not distributed evenly AND Earth is not a perfect ellipsoid

http://www.ngs.noaa.gov/GEOID/

Geoid map using EGM96 data, from http://cddis.gsfc.nasa.gov/926/egm96/egm96.htmlTalley SIO 210 (2014)200 meter variation4GeoidEarths mass is not distributed evenly AND Earth is not a perfect ellipsoid

http://www.ngs.noaa.gov/GEOID/

Talley SIO 210 (2014)5Rotating coordinatesThe Earth is rotating. We measure things relative to this rotating reference frame.

Quantity that tells how fast something is rotating:Angular speed or angular velocity = angle/second360 is the whole circle, but express angle in radians (2 radians = 360)For Earth: 2 / 1 day = 2 / 86,400 sec = 0.707 x 10-4 /sec

Also can show = v/R where v is the measured velocity and R is the radius to the axis of rotation (therefore v = R)

RAt home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.) Talley SIO 210 (2014)6At home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.)Rotating coordinatesVector that expresses direction of rotation and how fast it is rotating:vector pointing in direction of thumb using right-hand rule, curling fingers in direction of rotation

R

Talley SIO 210 (2014)7Centripetal and Centrifugal forces(now looking straight down on the rotating plane)

Centripetal force is the actual force that keeps the ball tethered (here it is the string, but it can be gravitational force)Centrifugal force is the pseudo-force (apparent force) that one feels due to lack of awareness that the coordinate system is rotating or curvingcentrifugal acceleration = 2R (outward) (Units of m/sec2)Talley SIO 210 (2014)8Effect of centrifugal force on ocean and earth Centrifugal force acts on the ocean and earth. It is pointed outward away from the rotation axis. Therefore it is maximum at the equator (maximum radius from axis) and minimum at the poles (0 radius). = 0.707 x 10-4 /sec At the equator, R ~ 6380 km so 2R = .032 m/s2Compare with gravity = 9.8 m/s2Centrifugal force should cause the equator to be deflected (0.032/9.8) x 6380 km = 21 km outward compared with the poles. (i.e. about 0.3%)

Talley SIO 210 (2014)9A = 6000 km; omega = 2pi/86400 sec = 7.292e10-4/secOmega2 a cos(latitude) = 3.17 cm/sec2 x cos(latitude)Compare 3.17 with gravity of 980 cm/sec2Earth surface should be 0.3% farther out at equator = 20 kmEffect of centrifugal force on ocean and earth Radius:Equatorial 6,378.135 kmPolar 6,356.750 kmMean 6,372.795 km(From wikipedia entry on Earth)The ocean is not 20 km deeper at the equator, rather the earth itself is deformed! We bury the centrifugal force term in the gravity term (which we can call effective gravity), and ignore it henceforth. Calculations that require a precise gravity term should use subroutines that account for its latitudinal dependence.

Talley SIO 210 (2014)10Return here need to state that at the equator, centrigufal force reduces the effective gravity the most Coriolis effect

Inertial motion:motion in a straight line relative to the fixed stars

Coriolis effect: apparent deflection of that inertially moving body just due to the rotation of you, the observer.

Coriolis effect deflects bodies (water parcels, air parcels) to the right in the northern hemisphere and to the left in the southern hemisphere

Talley SIO 210 (2014)11http://www.physics.orst.edu/~mcintyre/coriolis/index.html add this reference and animationsCoriolis effect

look on youtube for countless animationsTalley SIO 210 (2014)12http://www.physics.orst.edu/~mcintyre/coriolis/index.html add this reference and animationsCoriolis forceAdditional terms in x, y momentum equations, at latitude horizontal motion is much greater than vertical)

x-momentum equation: -sinv = -f v

y-momentum equation: sinu = f u

f = sinis the Coriolis parameter. It depends on latitude (projection of total Earth rotation on local vertical)

Talley SIO 210 (2014)13Coriolis forcef = sinis the Coriolis parameter

= 1.414x10-4/sec

At equator (=0, sin=0): f = 0

At 30N (=30, sin=0.5): f = 0.707x10-4/sec

At north pole (=90, sin=1): f = 1.414x10-4/sec

At 30S (=-30, sin=-0.5): f = -0.707x10-4/sec

At north pole (=-90, sin=-1): f = -1.414x10-4/sec

Talley SIO 210 (2014)14Edited 11/3/14Complete force (momentum) balance with rotationThree equations:

Horizontal (x) (west-east)acceleration + advection + Coriolis = pressure gradient force + viscous term

Horizontal (y) (south-north)acceleration + advection + Coriolis = pressure gradient force + viscous term

Vertical (z) (down-up)acceleration +advection (+ neglected very small Coriolis) = pressure gradient force + effective gravity (including centrifugal force) + viscous termTalley SIO 210 (2014)15Full equations in wordsFinal equations of motion (momentum equations in Cartesian coordinates)x: u/t + u u/x + v u/y + w u/z - fv = - (1/)p/x + /x(AHu/x) + /y(AHu/y) +/z(AVu/z) (7.11a)

y: v/t + u v/x + v v/y + w v/z + fu = - (1/)p/y + /x(AHv/x) + /y(AHv/y) +/z(AVv/z) (7.11b)

z: w/t + u w/x + v w/y + w w/z (+ neglected small Coriolis) = - (1/)p/z - g + /x(AHw/x) + /y(AHw/y) +/z(AVw/z) (7.11c)(where g contains both actual g and effect of centrifugal force)

Talley SIO 210 (2014)16Full equations for the physics-based students in class. The class is not responsible for knowing these full equations.The g term includes centrifugal force. The neglect of Coriolis in the z-momentum equation is called thetraditional approximation..Coriolis in action in the ocean: Observations of Inertial CurrentsSurface drifters in the Gulf of Alaska during and after a storm.Note the corkscrews - drifters start off with clockwise motion, which gets weaker as the motion is damped (friction)DAsaro et al. (1995)

DPO Fig. 7.4Talley SIO 210 (2014)17Inertial currentsBalance of Coriolis and acceleration terms: push the water and it turns to the right (NH), in circles.

DPO Fig. S7.8aTalley SIO 210 (2014)18What is the inertial period?Equator => infinite since f = 0Poles 2pi/f = Inertial currents: periodsTalley SIO 210 (2014)

sin() = latitudeCoriolis parameterf = 2sin()Inertial periodT = 2/fInertial currents: force balanceThree APPROXIMATE equations (all other terms are much smaller):

Horizontal (x) (west-east)acceleration + Coriolis = 0

Horizontal (y) (south-north)acceleration + Coriolis = 0

Vertical (z) (down-up) (hydrostatic) (not important for this solution)0 = pressure gradient force + effective gravity

That is:x: u/t - fv = 0y: v/t + fu = 0Solution: solve y-eqn for u and substitute in x eqn (or vice versa). Solutions are velocity circles:u = Uosin(ft) and v = Uocos(ft) where Uo is an arbitrary amplitudeTalley SIO 210 (2014)20Equations (bottom) not required for class.Inertial currents: relation to internal waves

Energy in inertial oscillations (which are wind-forced) as measured by surface drifters (Elipot et al., JGR 2010)Talley SIO 210 (2014)