Hydrostatic Equilibrium Questions and Answers

Answer: A

Hydrostatic balance states that the upward force arising from the decrease of pressure with height exactly supports the weight of the air. It holds to better than one part in a hundred million for large scale motion, which is far tighter than the horizontal geostrophic balance. The other options describe geostrophic balance, the terminal velocity of a falling drop, and the combination that defines effective gravity, none of which is the vertical balance being asked about.

Answer: B

The relation \( dp/dz = -\rho g \) states that the pressure decreases upward at a rate equal to the weight per unit volume of the air. The minus sign is essential and reflects the fact that pressure falls as one climbs. Substituting the gas law for density converts it into an equation that can be integrated to give the pressure profile. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: C

Integrating the equation from a level to the top of the atmosphere shows that the pressure at any level equals the weight per unit area of everything above it. A barometer is therefore weighing the atmosphere. This interpretation makes surface pressure changes intelligible as redistributions of mass rather than as local thermal effects. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: D

Vertical accelerations in large scale flow are around \( 10^{-7} \) metres per second squared against a gravitational acceleration near 10, so the two vertical terms balance to about one part in \( 10^{8} \). That is why pressure can be used as a vertical coordinate with confidence. The horizontal geostrophic balance, by contrast, is good only to about ten per cent. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: A

The balance requires the horizontal scale of the motion to greatly exceed its vertical scale, and in a convective updraught only a few kilometres wide but ten deep that condition fails. Vertical accelerations of order a metre per second squared then occur. This is why convection permitting models must solve non-hydrostatic equations while global models traditionally did not. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: B

Scale analysis shows that the ratio of the vertical acceleration to gravity is of order the square of the aspect ratio, so a flat disturbance is hydrostatic and a tall narrow one is not. With a ratio of one in a hundred for synoptic motion the balance is extremely good. The Rossby number governs the horizontal balance instead and has no bearing on the vertical one. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

Substituting the gas law into the hydrostatic equation and integrating at constant temperature gives an exponential profile with scale height \( H = RT/g \). Pressure falls by a factor of e, about 2.718, over that distance. The half-pressure height is smaller by a factor of the natural logarithm of two, which is why 5.5 kilometres and 8 kilometres are both quoted for different definitions. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: D

With a mean temperature near 250 kelvin, \( RT/g \) works out at about 7.3 kilometres, and 8 kilometres is the value usually quoted for the lower atmosphere. The half-pressure height of 5.5 kilometres is a different quantity, being the scale height multiplied by the natural logarithm of two. Both figures are worth carrying but they answer different questions. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

Integrating the hydrostatic equation with constant temperature gives pressure proportional to the exponential of minus height over scale height. Real atmospheres are not isothermal but the exponential form is an excellent first approximation over a few kilometres. A linear decrease would be correct only for constant density, which the atmosphere emphatically is not. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Since \( H = RT/g \), a warmer layer has a larger scale height and the pressure falls more slowly through it. This is the physical basis of the thickness relation, which states that a warm layer is thicker between two given pressure surfaces. It also means that an aircraft altimeter reads too low in cold air, which matters for terrain clearance. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.