Estimate the order of magnitude of the following quantities:The weight and the mass of Earth's atmosphere. $($Hint: At sea level the atmospheric

pressure is $1$atm.$)$

Solution: We know that the atmospheric pressure at sea level is $1$atm, which equals

${10}^5\frac{N}{m^2}.($The symbol $N$ here denotes the newton, which is the unit of force in the

International System of Units $($SI$).$ The newton is defined as the force that would give a

mass of one kilogram an acceleration of one meter per squared second.$)$ This means that

each squared meter on the surface of the Earth supports a column of air weighing ${10}^{5}N.$

We then need to calculate the surface area of the Earth. Taking the Earth as perfectly

spherical, with a radius $R$~~$6400km,$ we get

weight $of$ atmosphere ~~$4(6\mathrm{.}4*{10}^6(m){)}^2{10}^5\frac{N}{m^2}$~~$5*{10}^{19}N.$

The weight of an object on the Earth is a measure of the force with which the object is

attracted to the Earth and equals the product of the mass times the Earth's gravitational

acceleration $g$~~$9\mathrm{.}8\frac{m}{s^2}.$ Therefore the total mass of the atmosphere is given by

mass $of$ atmosphere $=\frac{\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}of\text{a}\text{t}\text{m}\text{o}\text{s}\text{p}\text{h}\text{e}\text{r}\text{e}}{g}$~~$\frac{5*{10}^{9}N\frac{m}{s^2}}{9\mathrm{.}8}$~~${10}^{18}kg$