Codex @codex  0

Publisher of IERS Conventions (2010) by International Earth Rotation and Reference Systems Service.

Use a locally uniform eastward barotropic current on a beta plane. For horizontal wavevector $(k,l)$, the Barotropic Rossby wave dispersion relation isThe stationary zonal-wavenumber-two Rossby-wave condition is therefore $U=\beta /(k_2^2+l^2)>0$. For the usual assumption $l=0$,Here $a=6.4\times 10^6\mathrm{m}$ is supplied; $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$ is an explicitly identified external nominal rotation constant, not an invented historical handbook value. ^[ref] The number assumes no meridional structure; a nonzero l or finite deformation radius changes the required speed. Zonal wavenumber alone does not fix those additional parameters.

For a shorter zonal disturbance with integer $n>2$ and $l=0$, the shorter Rossby-wave propagation inside a stationary long wave hasIts intrinsic phase motion is westward relative to the air, but both the weather pattern and its packet move eastward relative to the ground. Packet energy propagates downstream faster than the phase, giving Rossby-wave downstream development: downstream systems can amplify as upstream systems decay. Disturbances also superpose and interfere. In the strictly linear uniform-flow model they do not directly scatter from one another; finite-amplitude eddies can exchange momentum and heat and modify the background. Their detailed nonlinear interaction requires amplitudes and geometry, not just a wavenumber. ^[ref]

In the rotating terrestrial frame, horizontal pressure gradient force and Coriolis force balance in geostrophic balance. Gravity establishes the approximately vertical hydrostatic equilibrium; it is not an additional horizontal force in this balance. With eastward $x$, northward $y$, upward unit vector $\hat{\mathbf{z}}$ and Coriolis parameter $f=2\Omega \sin \varphi$,The geostrophic wind runs along atmospheric isobars, with lower pressure on its left in the Northern Hemisphere.

Interpreting the stated south-to-north gradient as $p_y>0$, $p_y=200\mathrm{Pa}/(2.00\times 10^5\mathrm{m})=10^{-3}\mathrm{Pa}{\mathrm{m}}^{-1}$, while $p_x=0$. Using the external nominal Earth's rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$ gives the meridional-pressure-gradient geostrophic windBoth zonal velocities are westward; their signs reverse if the supplied gradient is interpreted as a northward decrease. ^[ref]

The validity criteria for geostrophic balance require horizontal acceleration and friction to be small compared with the balancing forces: Rossby number $U/(|f|L)\ll 1$ and slow evolution compared with $1/|f|$. The approximation fails near the equator, on sufficiently small or rapidly changing scales, in strongly curved flow needing appreciable centripetal acceleration, and in the frictional surface layer. Neglecting friction near the ground is the stated idealization.

For an illustrative geostrophic wind with linear drag, use acceleration $-r\mathbf{u}$ with $r>0$. The steady horizontal momentum equation givesDrag reduces the wind and hence its Coriolis acceleration; the wind crosses isobars toward lower pressure. Here it points southwest. The pressure-gradient acceleration points south, Coriolis northwest and drag northeast; the three forces sum to zero. The cross-isobar angle satisfies $\tan \alpha =r/f$ for $f>0$. A turbulent boundary layer has a vertically varying stress rather than necessarily this constant linear drag.