All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
No data - just a weighing problem (Posted on 2016-09-27) Difficulty: 4 of 5
You are free to look up the physical dimensions of the planet Earth, math formulas, physical constants etc using books, tables, Google etc. You are precluded to ask (i.e. Google) a direct question:

What is the total weight of Earth's atmosphere?

prior to posting your answer.

Please submit your estimate in a format N.PQ *10r kg.
Provide your reasoning as well.
We shall jointly evaluate the accuracy later.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
My solution | Comment 1 of 2
I hope I didn't cheat.

https://en.wikipedia.org/wiki/Atmospheric_pressure
gives pressure at one atmosphere of 14.696 pounds/in^2

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=25&cad=rja&uact=8&ved=0ahUKEwj18b-XkLDPAhVIJCYKHcgvDecQmhMIvAEwGA&url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FEarth&usg=AFQjCNHYV-e8KqN_90wfUv8kboBKe55hSA
(wikipedia)
gives the surface area of the earth as 196.9 million square miles

Converting square miles to square inches and multiplying the two numbers gives
1.1616*10^19 pounds

which converts to 5.2689*10^18 kg.

I could probably improve this by subtracting the air that is displaced by land, but I doubt this matters much.

This also assumes a column of air has parallel sides, but they actually meet at the center of the earth.  I think this is probably negligible.

  Posted by Jer on 2016-09-27 13:30:12
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information