Past exam of the geophysical fluid dynamics course of the University of Oxford / 2025 / 1 / b

Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward geostrophic wind. With Coriolis parameter $f$ independent of $z$, geostrophic balance, hydrostatic balance, and the ideal gas law giveHere $R_d$ is the specific gas constant for dry air. Taking a vertical derivative of the first equation and interchanging mixed partial derivatives givesIt matters that $\rho$ is not constant in the pressure gradient force. Substitutingcancels the two terms proportional to $gfu/(R_{d}T)$ and yields the height-coordinate thermal wind relationIf $|u{T}_z|\ll |T{u}_z|$, this reduces toThe discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\mathrm{km}$, $T=250\mathrm{K}$ and a atmospheric lapse rate of $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical temperature gradient; it is not an exact consequence of the two balances. If the temperature gradient vanishes vertically, the displayed approximation is exact within those balances.

The hydrostatic approximation is very good on the broad, shallow atmospheric scales considered. The geostrophic balance is normally justified by a small Rossby number and fails as a general approximation near the equator. At $30^{\circ }\mathrm{N}$, $f=7.292\times 10^{-5}{\mathrm{s}}^{-1}$. With the ideal $134\mathrm{m}{\mathrm{s}}^{-1}$ wind and a $1000$--$2000\mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq 0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure geostrophic balance, although a straight, steady zonal flow can satisfy it even when this scale estimate of the Rossby number is not small.

For the requested estimate, suppose the upper jet stream is $H=10\mathrm{km}$ above a weak surface wind, has temperature $T\simeq 250\mathrm{K}$, and spans a meridional width $L$. Integrating the approximate thermal wind relation in scale form givesThe northward temperature gradient is negative because the eastward wind strengthens upward. A $1000$--$2000\mathrm{km}$ jet width gives roughly $25$--$50\mathrm{K}$. No unique numerical difference follows without specifying the jet width, height and representative temperature; these are explicitly estimated quantities.