Question

What would be the height of the atmosphere if the air density (a) were uniform and (b) What would be the height of the atmosphere if the air density were decreased linearly to zero with height? Assume that at sea level the air pressure is$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{atm}$and the air density is $\mathbf{\text{1.3}}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$.

Short answer

a) The height of atmosphere when density is uniform is$7.9\mathrm{km}$

b) The height of atmosphere when density changes linearly to zero is$16\mathrm{km}$

Step-by-step solution

The given data

(i) Density of air, $\rho =1.3\mathrm{kg}/{\mathrm{m}}^3$

(ii) Air pressure at sea level, $p=1.0\mathrm{atm}$

(iii) Acceleration due to gravity,$g=9.8\mathrm{m}/{\mathrm{s}}^2$

Understanding the concept of pressure dependency on density

The atmospheric pressure depends on the density of the air. Therefore, if the density is constant, we can find the atmospheric pressure using the formula for pressure in terms of density, acceleration due to gravity, and height. From this, we can calculate the height.

If the density is changing linearly with height, then we have to define a function, which would give us this change in density with height, and at a certain height, the density will be zero. Using this function, we can find the height for the second case.

Formula:

Pressure applied on a body, $\mathrm{p}=\mathrm{\rho gh}$

a) Calculation of height when density d uniform

Using equation (i) and the given values, the height can be found as:

$1.01\times {10}^5=1.3\times 9.8\times \mathrm{h}$

$\mathrm{h}=7927.79\mathrm{m}$

$=7.9\mathrm{km}$

Hence, the value of height at uniform density is $7.9\mathrm{km}$

b) Calculation of height when the atmospheric pressure changes linearly to zero.

Now, density changes uniformly with atmosphere, so we can write the density as

$\mathrm{\rho }-{\mathrm{\gamma }}_0={\mathrm{\rho }}_0\frac{\mathrm{y}}{\mathrm{h}}$

${\mathrm{\rho }}_0$is the density of the air at the earth’s surface,‘y’ is the height at which the density is measured, and h is the height at which the density is zero.

We can simplify this as:

$\mathrm{\rho }={\mathrm{\rho }}_0\left(1-\frac{\mathrm{y}}{\mathrm{h}}\right)$

Now, pressure as a function of height can be written as:

$\mathrm{p}={\int }_0^{\mathrm{h}} \mathrm{\rho gdy}.$

Putting value of equation (1) in equation (2), we get

$\mathrm{p}={\int }_0^{\mathrm{h}} {\mathrm{\rho }}_0\mathrm{g}\left(1-\frac{\mathrm{y}}{\mathrm{h}}\right)\mathrm{dy}$

$\mathrm{p}={\left[{\mathrm{\rho }}_0\mathrm{gy}\right]}_0^{\mathrm{h}}-{\left[\frac{{\mathrm{\rho }}_0{\mathrm{gy}}^2}{2\mathrm{h}}\right]}_0^{\mathrm{h}}$

$\mathrm{p}={\mathrm{\rho }}_0\mathrm{gh}-\frac{{\mathrm{\rho }}_0\mathrm{gh}}{2}$

$\mathrm{p}=\frac{{\mathrm{\rho }}_0\mathrm{gh}}{2}$

$\mathrm{h}=\frac{2\mathrm{p}}{{\mathrm{\rho }}_0\mathrm{g}}$ (equation for height)

Now, substituting the given values, we have

$\mathrm{h}=\frac{2\times 1.01\times {10}^5}{1.3\times 9.8}$

$=15855.573\mathrm{m}$

$=16\mathrm{km}$

Hence, the value of height when the atmospheric pressure changes linearly to zero isrole="math" localid="1657536102648" $16\mathrm{km}.$