Question

Ear "popping" is an unpleasant phenomenon sometimes experienced when a change in pressure occurs, for example,in an airplane. If you are in a two-seater airplane at 3000 m and a descentof 100 m causes your ear to "pop, what is the pressure change that your ear pops at in millimeters of mercury? If the airplane now rises to 8000 m and again begins descending, how far will the airplane descend before your ears "pop" again?

Answer 1

Answer:

H = 0.00058m = 0.58mm of Mercury

h = 160m for popping at 8000m

Explanation:

Air density decreases with altitude (increasing height).

At sea level air density = 1.22kg/m^3

while at 3000m above it is approximately 0.8 kg/m^3

ASSUMPTION:

we neglect the change in density across 100m vertically

density of mercury = 13600kg/m ^3

we equate the pressure for mercury and air

AIR(pgh) = Mercury(pgh)

0.8 * g * 100 = 13600 * g * H

H = 0.00058m = 0.58mm of Mercury

At even farther heights, the air density further drops

at 8000m above it is 0.52kg/ m^3  (source:https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html)

***for the ear to pop again, it has to experience same change in pressure as of at 3000m

Change in P for AIR at 3000m = Change in P for AIR at 8000m

ASSUMPTION : gravitational acceleration does not change with altitude

pgh = pgh

0.8 * g * 100 = .5 * g *h

h = 160m