The atmosphere effectively ends at an altitude of 120 km and produces a pressure of 1.01 x 10^5 Pa at the ground. Using g = 9.80 m/s^2, what is the average density of the whole atmosphere?
A0.0859 kg/m^3
B1.29 kg/m^3
C0.000859 kg/m^3
D8.59 kg/m^3
Answer & Solution
Correct answer: A. 0.0859 kg/m^3
1. Start from pressure = h rho g and solve for the average density: rho = pressure / (h g).
2. The pressure is the whole weight of the air column, 1.01 x 10^5 N/m^2.
3. The height of that column is 120 km, which is 1.20 x 10^5 m.
4. Denominator = (1.20 x 10^5 m)(9.80 m/s^2) = 1.176 x 10^6 m^2/s^2.
5. rho = (1.01 x 10^5 N/m^2) / (1.176 x 10^6 m^2/s^2) = 8.59 x 10^-2 kg/m^3, that is 0.0859 kg/m^3.
6. The value 1.29 kg/m^3 is the density of air at sea level, about fifteen times the average, and is the trap for anyone who quotes a table value instead of computing.
7. The values 8.59 kg/m^3 and 0.000859 kg/m^3 arise from leaving the height as 120 m or as 1.2 x 10^7 m.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.4 Variation of Pressure with Depth in a Fluid_
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