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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.