Question

What is the appropriate mass of the atmosphere of earth? Assume the radius of earth to be $6370\,km$.

Answer 1

In order to solve this question we need to understand atmospheric pressure. So pressure on a surface is defined as the force exerted on it per unit surface area.So atmospheric pressure is defined as pressure exerted by the atmosphere on earth, and the atmosphere exerts pressure because the force of gravity is pulling it downward and hence it exerts an upward force to be in equilibrium.

Complete step by step answer:
To find the mass of the atmosphere we need to consider the force of gravity and force exerted by the atmosphere in upward direction to remain stable above earth.Let the mass of the atmosphere be, $m$.Since the acceleration due to gravity is, $g = 10\,m{\sec ^{ - 2}}$.
So the force of gravity acting downward is given by, ${F_1} = mg$
Since the radius of earth be, $R = 6370 \times {10^3}m$
So area of earth surface is given by, $A = 4\pi {R^2}$
Since we know pressure exerted by atmosphere is,
$P = 1\,atm$
$\Rightarrow P = 1.03125 \times {10^5}\,N{m^{ - 2}}$
$\Rightarrow P = 103125\,N{m^{ - 2}}$

So the force exerted by atmosphere in above direction is,
${F_2} = A \times P$
$\Rightarrow {F_2} = 4\pi {R^2}P$
Equating both the force as they are equal and opposite to each other,
${F_1} = {F_2}$
$\Rightarrow mg = 4\pi {R^2}P$
$\Rightarrow m = \dfrac{{4\pi {R^2}P}}{g}$
Putting values we get,
$m = \dfrac{{4 \times \pi \times {{(6370 \times {{10}^3})}^2} \times (103125)}}{{10}}$
$\therefore m = 5.258 \times {10^{18}}\,kg$

So the mass of the atmosphere is $5.258 \times {10^{18}}\,kg$.

Note: It should be remembered that while calculation we have assumed that the earth’s atmosphere exerts uniform pressure at all points on the earth, also we have assumed that earth is a perfect sphere to find its radius and we also considered that we are measuring mass of atmosphere from the ground level. However this is not the real situation and that is why we have appropriate mass.