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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 · 2 answers · 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

2 answers

Your calculation of the weight of the atmosphere appears correct. Your volume calculation is based on an erroneous assumption of density as a linear function of altitude, and results in an atmosphere just 10 miles thick. I'm not sure what use the number is to you, but a more reasonable volume calculation would be to pick a number for the depth of the atmosphere and calculate the volume directly.

The exact height of the atmosphere is indeterminate, as there is no sharp demarcation between atmosphere and space. For spaceflight dynamics, atmospheric effects are considered significant below 75 miles. Other definitions have space beginning at 62 miles up.

2006-07-27 14:46:21 · answer #1 · answered by injanier 7 · 3 0

Look good to me. But, I have no idea what you're talking about.

2006-07-27 18:15:35 · answer #2 · answered by Jake W 3 · 0 0

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