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 *10^{r} kg.
Provide your reasoning as well.
We shall jointly evaluate the accuracy later.
(In reply to
My solution by Jer)
I'm sure it doesn't matter that the weight is being concentrated into a slightly smaller surface than at the top of the atmosphere; any weight is felt, and no more, at the surface.
I had basically the same solution, but limited the precision to the four significant figures that the surface area has, especially as the surface area begins with a 1 in the units given and the weight begins with a 5, in its chosen units.
As for the portion displaced by land, the Wikipedia article on the earth's atmosphere (legitimate after having done the puzzle) states that the portion displaced by land is actually 2.5%:
The average atmospheric pressure at sea level is defined by the International Standard Atmosphere as 101325 pascals (760.00 Torr; 14.6959 psi; 760.00 mmHg). This is sometimes referred to as a unit of standard atmospheres (atm). Total atmospheric mass is 5.1480×10^{18} kg (1.135×10^{19} lb),^{[25]} about 2.5% less than would be inferred from the average sea level pressure and Earth's area of 51007.2 megahectares, this portion being displaced by Earth's mountainous terrain. Atmospheric pressure is the total weight of the air above unit area at the point where the pressure is measured. Thus air pressure varies with location and weather.

Posted by Charlie
on 20160927 16:02:30 