How Much Does Earth's Atmosphere Weigh?

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SUMMARY

The mass of Earth's atmosphere can be calculated using the given data: 14.7 lbs of air per square inch and a density of 1.27 g/L. The Earth's diameter is 7930 miles, leading to a surface area of approximately 2.55 x 10^11 square meters. By applying the formula m = dv, where d is the density and v is the volume derived from the Earth's radius, the mass of the atmosphere is determined to be approximately 3.24 x 10^8 kg. The average density of Earth can also be calculated using the same principles.

PREREQUISITES
  • Understanding of unit conversions (e.g., pounds to kilograms, miles to meters)
  • Familiarity with the formula for density (d = m/v)
  • Basic knowledge of geometry to calculate the volume of a sphere
  • Awareness of atmospheric pressure and its measurement
NEXT STEPS
  • Research the calculation of Earth's surface area using the formula for the surface area of a sphere
  • Learn about the relationship between atmospheric pressure and density
  • Explore advanced unit conversion techniques for scientific calculations
  • Investigate the implications of Earth's atmospheric mass on climate and weather patterns
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Students in physics or environmental science, educators teaching atmospheric science, and anyone interested in understanding Earth's atmospheric properties and calculations.

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Homework Statement


above each square inch of Earth's surface is 14.7 lbs of air. the density of air at sea level is about 1.27g/L. the Earth is 7930 miles in diameter and has a mass of 5.98 x 10^24kg.
a.) use the data above to determine the mass in kg of the Earth's atmosphere.

b.) determine the average density of Earth in grams per milliliter.

Homework Equations


i think the only relevant equation is d=m/v. everything else seems to be unit conversions. this problem really isn't about the atmosphere, but rather how well i can convert units, and apparently not well.

The Attempt at a Solution


im looking for the mass of the atmosphere, so i use m = dv
d is 1.27g/mL
i derive v from 4/3pir^3
but i have the radius in miles. does this matter?
i would get (1.27g/mL)(2.55x10^11) since i need it in kg
(1.27 x 10^-3kg)(2.55x10^11) = 3.24 x 10^8 kg/mL?

im sure I am hopelessly wrong. i don't really want to do part b until i get part a.
 
Last edited:
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(a) We're given that there are 14.7 pounds of air per square inch of Earth's surface area. If you can figure out how many square inches of surface area the Earth has, can you take it from there?

(b) The equation you had, d=m/v, applies here.
 

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