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.

Homework Equations

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?

 

[lightgrav]

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.

 

[Chestermiller]

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
p₁=ρ_swgd = 2.00 x 10⁵ Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p₂=ρ_airgh=6.48 x 10⁴ Pa
Therefore, the change in pressure is
Δp = p₁ - p₂ = 1.4 x 10⁵ Pa

 

[Chestermiller]

This is from the solution manual:

The gauge pressure at a depth of 20 m in seawater is
p₁=ρ_swgd = 2.00 x 10⁵ Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p₂=ρ_airgh=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= - ρ_airgh= - 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?

 

[Chestermiller]

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