Hydrostatic Equilibrium Questions and Answers

Answer: A

With no overlying air there is no weight to support and the pressure is zero, which follows immediately from the hydrostatic interpretation. The solar wind pressure is many orders of magnitude below anything a barometer could detect. Gravity acts on the instrument but produces no reading on a properly constructed barometer.

Answer: B

Martian gravity is only about 3.7 metres per second squared against 9.8 on Earth, and although the temperature is lower and the molecular mass higher, the reduced gravity dominates and gives a scale height near 11 kilometres. The total amount of atmosphere does not enter the scale height at all, which is the point the question tests. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

Scale height is \( RT/g \), and the specific gas constant depends on the mean molecular mass, so those three quantities alone determine it. The total amount of atmosphere sets the surface pressure but not the rate at which it falls with height. Rotation does modify effective gravity slightly but the effect is negligible for this purpose. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

Integrating the hydrostatic equation from the level of the instrument to the top of the atmosphere gives exactly that result. The reason therefore explains the assertion completely. It follows that surface pressure changes reflect redistributions of mass, which is why the surface pressure tendency equation involves the vertically integrated divergence rather than any local thermal effect.

Answer: A

The reason states the condition correctly, and a convective updraught a few kilometres wide and ten deep violates it, so vertical accelerations of order a metre per second squared occur. The assertion is therefore false. This is precisely why convection permitting models must solve the non-hydrostatic equations, and it is the clearest practical consequence of the scaling argument.

Answer: B

Both statements follow directly from the hypsometric relation, since thickness is proportional to mean virtual temperature. They are the same result stated from two directions. Together they explain why upper level troughs correspond to cold air and why thickness charts and upper air charts show such similar patterns.

Answer: C

The first two statements are correct, but the balance does not require the atmosphere to be at rest, only that the vertical acceleration be negligible compared with gravity. Horizontal motion of any speed is perfectly compatible with it. Confusing hydrostatic with static is a natural slip that the etymology encourages.

Answer: D

The 300 hectopascal surface lies near 9 kilometres, not 5, so the last pairing is wrong. The other three are correct for the standard atmosphere. Carrying this ladder of approximate heights makes any upper air chart immediately interpretable without reference to a table, which is worth the small effort of memorising it.

Answer: A

The geostrophic wind arises from the horizontal balance between the pressure gradient and Coriolis forces and has nothing to do with the vertical balance. The hypsometric equation, the barometric formula and the use of pressure as a vertical coordinate all follow directly from hydrostatic balance. Keeping the horizontal and vertical balances separate is essential when the results are combined, as they are in the thermal wind relation.

Answer: B

Reducing through 2000 metres of air that does not exist requires an assumed temperature for that column, and a plausible range of assumptions shifts the reduced value by several hectopascals. This is why analyses over the Rockies, the Andes and the Tibetan Plateau are drawn on the 850 or 700 hectopascal surface instead. The instrument works perfectly well and the gravity correction is small. Most errors here come from carrying a result across from a neighbouring case where it does not hold.