# Calculating pressure in high altitude?

dmk90

## Homework Statement

What is the change in pressure on a special-op soldier who must scuba dive at a depth of 20 m in seawater one day and parachute at an altitude of 7.6 km the next day? Assume that the average air density within the altitude
range is 0.87 kg/m3.

p=p0 + ρgh

## The Attempt at a Solution

I understand how to calculate pressure using the above formula for something at the surface, but how do I apply it to a high altitude? In the solution manual, they just ignore atmospheric pressure, so I am confused. At that high altitude, doesn't it mean the person is not affected by the usual atmospheric pressure of 1 atm?

## Answers and Replies

Homework Helper
Pressure is less at the top of the sea than at the bottom of the sea ...
we live on the bottom of a "sea of air" that's only about 8km deep.
Pressure is zero at the "top" of the atmosphere, the edge of space.

Mentor
In the solution manual, they just ignore atmospheric pressure, so I am confused. At that high altitude, doesn't it mean the person is not affected by the usual atmospheric pressure of 1 atm?

I doubt that the solution manual ignored the atmospheric pressure at the surface of the earth. Please show us how the manual solution got the pressure at 7.6 km.

Chet

dmk90
This is from the solution manual:

The gauge pressure at a depth of 20 m in seawater is
p1swgd = 2.00 x 105 Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p2airgh=6.48 x 104 Pa
Therefore, the change in pressure is
Δp = p1 - p2 = 1.4 x 105 Pa

Mentor
This is from the solution manual:

The gauge pressure at a depth of 20 m in seawater is
p1swgd = 2.00 x 105 Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p2airgh=6.48 x 104 Pa
Therefore, the change in pressure is
Δp = p1 - p2 = 1.4 x 105 Pa
The result for 7.6 km from the solution manual is incorrect. They have the sign wrong. It should be p2= - ρairgh= - 6.48 x 104 Pa. The absolute pressure at 7.6 should be 1. x 105 - 6.48 x 104 Pa. This is where the atmospheric pressure at the surface comes in.

Of course, this error makes the final answer incorrect also.

Chet

dmk90
Isn't 105 only applies to things at roughly sea level? Wouldn't something at 7.6 km up not be subjected to this pressure?

Mentor
Isn't 105 only applies to things at roughly sea level? Wouldn't something at 7.6 km up not be subjected to this pressure?
Sure. It''s less. That's what I calculated.

Chet