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Earth and Space Sciences · Glossary

What is Geostrophic balance, geostrophic wind?

Also known as: geostrophic balance · geostrophic wind

Definition 28.5 University Earth and Space Sciences — Year 1 · Chapter 28 — The General Circulation of the Atmosphere

A horizontal flow is in geostrophic balance when its pressure-gradient force and its Coriolis force cancel. The geostrophic wind v⃗g\vec v_g is the wind that achieves this balance for a given pressure field: −(1/ρ)∇p−f k⃗×v⃗g=0⃗-(1/\rho)\nabla p - f\,\vec k\times\vec v_g = \vec 0.

Forces on a parcel of air in the Northern Hemisphere, with low pressure to the north. Left: in the free atmosphere the pressure-gradient force (red) and the Coriolis force (blue) cancel, and the wind blows along the isobars with the low on its left. Right: near the ground friction (green) acts against the wind; the three forces balance only if the wind is slower and turned by an angle  towards the low (drawn for = 25 and the drag of ).
Forces on a parcel of air in the Northern Hemisphere, with low pressure to the north. Left: in the free atmosphere the pressure-gradient force (red) and the Coriolis force (blue) cancel, and the wind blows along the isobars with the low on its left. Right: near the ground friction (green) acts against the wind; the three forces balance only if the wind is slower and turned by an angle α\alpha towards the low (drawn for α=25∘\alpha = 25{}^{\circ} and the drag of Proposition 28.8).

Examples

Example 28.7 (A wind read from the isobars)

At 50∘50{}^{\circ} N, f=2×7.292×10−5×sin⁡50∘=1.117×10−4 s−1f = 2 \times 7.292 \times 10^{-5} \times \sin50{}^{\circ} = 1.117 \times 10^{-4}\,\mathrm{s}^{-1}. Isobars 4 hPa4\,\mathrm{hPa} apart and 400 km400\,\mathrm{km} apart give ∣∇p∣=400/4×105=1.0×10−3 Pa m−1|\nabla p| = 400/4 \times 10^{5} = 1.0 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{m}^{-1} and, with ρ=1.225 kg m−3\rho = 1.225\,\mathrm{kg}\,\mathrm{m}^{-3}, a geostrophic wind of 1.0×10−3/(1.225×1.117×10−4)=7.3 m/s1.0 \times 10^{-3}/(1.225 \times 1.117 \times 10^{-4}) = 7.3\,\mathrm{m}/\mathrm{s}. The same spacing at 20∘20{}^{\circ} N, where ff is less than half as large, would give 16.4 m/s16.4\,\mathrm{m}/\mathrm{s}.

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