Fundamental Forces in the Atmosphere Questions and Answers

Answer: C

Solving the gradient wind equation for anticyclonic flow gives a real root only while the pressure gradient stays below a limiting value proportional to \( f^{2}R/4 \), which is why anticyclones are always broad and gentle while cyclones can be small and intense. The wind around a high is supergeostrophic rather than subgeostrophic, and the curvature is finite. This asymmetry between highs and lows is one of the most striking results obtainable from a simple force balance.

Answer: D

The mathematical limit on the anticyclonic pressure gradient has no cyclonic counterpart, so there is no dynamical ceiling on how deep a low may become. Latent heat release does help intensify lows and is a genuine second reason, but the balance argument alone suffices and applies even to dry systems. Friction acts in both, and the Coriolis parameter depends on latitude alone and does not change sign with the sense of the curvature.

Answer: A

Sigma is pressure divided by surface pressure, so it equals one at the ground whatever the orography and zero at the top, which removes the problem of pressure surfaces intersecting mountains. The price is that the pressure gradient term becomes a difference of two large quantities over steep terrain. Isentropic and log pressure coordinates serve other purposes, and plain pressure coordinates are exactly the ones whose intersection with terrain sigma was invented to avoid.

Answer: B

On a steeply sloping sigma surface the horizontal pressure gradient is computed as the difference between two terms each much larger than their difference, so truncation error can produce a spurious wind. This is a well known source of error in models of mountainous regions and has motivated hybrid and step-terrain coordinates. Mass conservation and the diagnosis of vertical motion are unaffected, and \( f \) depends only on latitude.

Answer: C

Any body moving over the rotating Earth is subject to the same Coriolis deflection as an air parcel, so a navigator must correct for it exactly as a gunner does. The magnitude for an aircraft at 250 metres per second in mid-latitudes is about 0.025 metres per second squared, which accumulates to a significant lateral error over an hour. The other options describe effects that do not follow from rotation in this way.

Answer: D

Naval gunnery at ranges of twenty kilometres and more required Coriolis corrections from the late nineteenth century, and the failure to reverse the sign for the southern hemisphere is a well known episode of the First World War. Aviation came later and shipping speeds are too low for the effect to matter. Radio propagation is unaffected, since the frame plays no part in the behaviour of electromagnetic waves.

Answer: A

A parcel at rest in the rotating frame has no acceleration in that frame and no apparent forces acting, since both Coriolis and centrifugal terms vanish for zero velocity, so the real forces must sum to zero. That means the upward pressure gradient force exactly balances the weight, which is hydrostatic balance. Recognising still air as the trivial case of the full equation is a useful check on any formulation that has been written out at length.