The weight of the atmosphere above 1 m^2 of Earth's surface is about 100,000 N. Density, of course, becomes less with altitude. But suppose the density of air were a constant 1.2 kg/m/s^3. Calculate where the top of the atmosphere would be.

Answer 1

With a uniform density and height H,l

(density)*g*H = 10^5 N/m^2
H = 10^5/9.8*1.2 = 8500 m.

That is approximately the height of Mt. Everest, the highest mountain.