Approximately: 282201568539770880000 cubic feet
282 Quintillion (rounded)
weighing
approximately 11288062741590835200 pounds! :-)
11.3 Quintillion (rounded)
I first determined the "approximate" weight of the atmosphere by calculating the average surface area of the earth: surface area = 4 x pi x r^2 = 4 x 3.1415 x (3900^2) = 191128860^2 miles
to square inches
191128860 x 4014489600 (square inches in a square mile) = 767284820729856000 sq/in of earth's surface
x 14.7 lbs (weight of atmosphere per square inch.) = 11279086864728883200 pounds
/ .04 pounds per cubic foot (to get the cubic feet of the atmosphere) *This is an averaged weight from sea level to vacuum of space = 281977171618222080000 cubic feet.
These numbers are slightly different than the ones above due to my use of pi.
2006-07-27
11:03:06
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2 answers
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asked by
TommyTrouble
4
in
Science & Mathematics
➔ Physics
injanier
Thank you for your response. I was looking to "Approximate" the volume of the Earth's atmosphere not sure "Anyone" could calculate the "exact" volume. As you noted, there are many variables to consider. I also "Approximated" the earths surface. I'm sure if I used a Terascale Computing System or similar I could get a little closer to the "exact" figure. But, I do believe that what this demonstrates, is that the Earths Atmosphere is quite large by anyone’s standards.
2006-07-27
15:02:22 ·
update #1
injanier,
THANK YOU FOR THE GREAT LINK PAL 8-)
Looking at what the National Center for Atmospheric Research estimates, "The total mean mass of the atmosphere is 5.1480 x 1018 kg
If we multiply 2.2 (pounds per kilogram) times 5.148 Quintillion kg - we get 11.3256 Quintillion pounds WOOOOHOOO! HIGH FIVE!!!!!!!!!!!!!!
2006-07-27
15:11:36 ·
update #2
Should read: 5.1480 x 10^18th
2006-07-27
15:13:14 ·
update #3