Explore pressure decrease and atmospheric scale height

Equal height steps remove equal fractions

At a fixed temperature, compare the pressure at 0, 5 and 10 km. The second five-kilometer interval removes fewer hectopascals than the first. Pressure does not decrease by a fixed subtraction per kilometer in this model. Instead, equal height increments multiply pressure by the same fraction.

The isothermal ideal-gas solution is P(z) = P_base × exp(−z/H). Here z is altitude above the reference base in meters, H is the scale height in meters, and exp means the exponential function with base e, approximately 2.718. The ratio z/H has no units, which is necessary for the exponent.

The scale height is H = RdT/g, with Rd = 287.05 J/(kg K), temperature T in kelvin and gravity g = 9.81 m/s². At 280 K it is about 8,193 m, or 8.19 km. At z = H, pressure is multiplied by exp(−1), approximately 0.368. Scale height is not the atmosphere’s total thickness. NASA discussion of an isothermal atmosphere and scale height.

Calculate a worked profile

For 5 km, first convert to 5,000 m. The ratio z/H is approximately 0.6103, so pressure is about 0.5432 times its base value, roughly 550.4 hPa. At 10 km, the exponent doubles and the retained fraction is the square of the five-kilometer fraction, about 0.2951.

On paper draw three bars using one shared pressure scale. Label the bars by altitude rather than treating them as consecutive moments in time. Calculate both the pressure differences and the ratios between adjacent bars. The differences vary, but the ratios match for equal increments in this idealized column.

The exponential behavior follows from hydrostatic balance combined with density that changes with pressure. It does not assume that the air has constant density. Isothermal means constant temperature with height; it does not mean every property is constant.

Find the height of a pressure level

Sometimes the question is reversed: at what altitude does pressure reach 500 hPa? Rearranging the profile gives z_500 = H × ln(P_base/50,000 Pa). The natural logarithm ln reverses the exponential function. Both pressures must use the same unit before taking their ratio.

The ratio P_base/50,000 is 2.0265, so the logarithmic factor is about 0.7063. At 280 K, the modeled 500 hPa level is therefore about 5.79 km. This is a calculated height for the selected column, not a universal altitude attached to 500 hPa.

Try 240 K and 320 K while keeping the same base pressure. A warmer column has a larger H, and the same fractional pressure decrease occurs over a greater height. The 500 hPa level moves upward. Capture both and explain the change using H rather than simply saying “heat rises.”

Hold the boundary condition in view

At positive altitude, the warmer selected column has higher pressure than the colder one with the same base pressure. Both still have the same total mass per area, P_base/g. Temperature changes how that mass is distributed vertically; it does not add air to the model.

A real atmospheric profile is not generally isothermal. Its temperature can vary with height, and models with layers or lapse rates use different assumptions. The fixed 280 K comparison line is another isothermal column, not a measured sounding or a standard-atmosphere implementation.

Transfer to half pressure

Using the default 280 K scale height, find the altitude where pressure is half the base value. Would that altitude equal one scale height? Show the inverse-function step before checking.

Check the pressure-altitude transfer

Set exp(−z/H) = 1/2, giving z = H ln(2). Since ln(2) is about 0.6931, the altitude is about 5.68 km. One scale height leaves about 36.8% of base pressure, so it is higher than the half-pressure level. This answer uses the constant-temperature model and its reference base.