Question

The density of the atmosphere at sea level is $1.29kg{m^{ - 2}}$. Assume that it does not change with altitude. How high would the atmosphere extend?

Answer 1

The density of the atmosphere is given to us and the pressure exerted by the atmosphere will be constant and the acceleration due to gravity is also constant hence all these three values are known. Now by applying a fluid pressure formula, the air present in the atmosphere is a fluid. Now we can solve it by putting the known value to find the height of the extension.

Complete step by step answer:
As per the given problem we know that the density of the atmosphere at sea level is $1.29kg{m^{ - 2}}$.
We need to calculate how high the atmosphere would extend.
Now we can use the fluid pressure formula we will get,
$P = \rho gh$
Where,
P is atmospheric pressure and it will remain the same as in the problem it is given that the density does not change with altitude the pressure also does not change.
Now atmospheric pressure at sea level, $P = 1.013 \times {10^5}N{m^{ - 2}}$
$\rho$ is the density of the atmosheper $= 1.29kg{m^{ - 2}}$
$g$ is the acceleration due to gravity $= 9.8m{s^{ - 2}}$
Now putting all the known value in the pressure formula we will get,
$1.013 \times {10^5}N{m^{ - 2}} = 1.29kg{m^{ - 2}} \times 9.8m{s^{ - 2}} \times h$
Rearranging the above equation we will get,
$\dfrac{{1.013 \times {{10}^5}N{m^{ - 2}}}}{{1.29kg{m^{ - 2}} \times 9.8m{s^{ - 2}}}} = h$
Now on solving this we will get,
$h = 8012.9km$ Which is nearly equal to $8013km$.

Note:
Always keep in mind that whenever we move higher and higher in the atmosphere, practically the density of the atmosphere increases and we know density and pressure are related, hence the pressure also increases. That's why in the problem it is already given that the density does not change with height. For constant density there is a constant pressure of the atmosphere.