during ascent, and especially during decent, volume changes of trapped air in the middle ear can cause ear discomfort until the middle-ear pressure and exterior pressure are equalized. If the rapid descent rate of 7.0 m/s or faster commonly causes ear discomfort, what is the maximum rate of atmospheric pressure that is tolerable to most people?

Question

dboome · Answer

Here's my reasoning, see what you think:

We know that the pressure P at a height h in the atmosphere is given by$$P(h)=p_0e ^{-h/h_0}$$ where:
p = atmospheric pressure
(measured in bars)
h = height (altitude)
p0 = is pressure at height h = 0 (surface pressure)
h0 = scale height (which we know is about 7000m)

Taking the derivative with respect to time of that, you get $$dP/dt = p_0*d(e ^{-h/h_0})/dt$$$$dP/dt = p_0\frac{ -\frac{ dh }{ dt} }{ h_0 }e ^{-h/h_0}$$ However, with dh/dt given at -7m/s, h0 as 7km, h as 0 (is that reasonable?)  and p0 = 1 bar, i arrive at the answer 1000bar/seconds, which doesn't seem plausible. Spot my errors(s)?

anonymous · Answer

Thank you so much! I do believe it requires a differential equation and that fits the bill.