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

\hat{z}

f = 2Ω sin ϕ

0 = - \frac{\nabla _{h} p}{ρ} - f \hat{z} \times u_{g}, u_{g} = \frac{z ^ \times \nabla _{h} p}{ρ f}, u_{g} = - \frac{p _{y}}{ρ f}, v_{g} = \frac{p _{x}}{ρ f} .

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 Pa / (2.00 \times 1 0^{5} m) = 1 0^{- 3} Pa m^{- 1}

, while

p_{x} = 0

. Using the external nominal Earth's rotation rate

Ω = 7.292115 \times 1 0^{- 5} s^{- 1}

u_{g} (3 0^{\circ} N) = - 11.4 m s^{- 1}, u_{g} (6 0^{\circ} N) = - 6.60 m s^{- 1}, v_{g} = 0.

Both zonal velocities are westward; their signs reverse if the supplied gradient is interpreted as a northward decrease. ^[ref]

U / (∣ f ∣ L) ≪ 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.

- r u

with

r > 0

. The steady horizontal momentum equation gives

(r f - f r) (v u) = - \frac{1}{ρ} (p _{y} p _{x}), u = - \frac{f p _{y}}{ρ ( f ^{2} + r ^{2} )}, v = - \frac{r p _{y}}{ρ ( f ^{2} + r ^{2} )} (p_{x} = 0) .

Drag 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