# 2.5: Problems


Exercise $$\PageIndex{1}$$ International Standard Atmosphere

After the launch of a spatial probe into a planetary atmosphere, data about the temperature of the atmosphere have been collected. Its variation with altitude ($$h$$) can be approximated as follows:

$T = \dfrac{A}{1 + e^{\tfrac{h}{B}}},\label{eq2.5.1}$

where $$A$$ and $$B$$ are constants to be determined.

Assuming the gas behaves as a perfect gas and the atmosphere is at rest, using the following data:

• Temperature at $$h = 1000$$, $$T_{1000} = 250\ K$$;
• $$p_0 = 100000 \dfrac{N}{m^2}$$;
• $$\rho_0 = 1 \dfrac{Kg}{m^3}$$;
• $$T_0 = 300\ K$$;
• $$g = 10 \dfrac{m}{s^2}$$.

determine:

1. The values of $$A$$ and $$B$$, including their unities.
2. Variation law of density and pressure with altitude, respectively $$\rho (h)$$ and $$p (h)$$ (do not substitute any value).
3. The value of density and pressure at $$h = 1000 m$$.

We assume the following hypotheses:

(a) The gas is a perfect gas.

(b) It fulfills the fluidostatic equation.

Based on hypothesis (a):

$P = \rho RT.\label{eq2.5.2}$

Based on hypothesis (b):

$dP = -\rho gdh.\label{eq2.5.3}$

Based on the data given in the statement, and using Equation ($$\ref{eq2.5.2}$$):

$R = \dfrac{P_0}{\rho_0 T_0} = 333.3 \dfrac{J}{(Kg \cdot K)}$

1. The values of $$A$$ and $$B$$:
Using the given temperature at an altitude $$h = 0$$ $$(T_0 = 300\ K)$$, and Equation ($$\ref{eq2.5.1}$$):
$300 = \dfrac{A}{1 + e^0} = \dfrac{A}{2} \to A = 600 \ K.$
Using the given temperature at an altitude $$h = 1000$$ ($$T_{1000} = 250\ K$$), and Equation ($$\ref{eq2.5.1}$$):
$250 = \dfrac{A}{1 + e^{\tfrac{1000}{B}}} = \dfrac{600}{1 + e^{\tfrac{1000}{B}}} \to B = 2972\ m.$
2. Variation law of density and pressure with altitude:
Using Equation ($$\ref{eq2.5.2}$$) and Equation ($$\ref{eq2.5.3}$$):
$dP = -\dfrac{P}{RT} gdh.\label{eq2.5.7}$
Integrating the differential Equation ($$\ref{eq2.5.7}$$) between $$P(h = 0)$$ and $$P, h = 0$$ and $$h$$:
$\int_{P_0}^{P} \dfrac{dP}{P} = \int_{h = 0}^{h} -\dfrac{g}{RT} dh.\label{eq2.5.8}$
Introducing Equation ($$\ref{eq2.5.1}$$) in Equation ($$\ref{eq2.5.8}$$):
$\int_{P_0}^{P} \dfrac{dP}{P} = \int_{h = 0}^{h} -\dfrac{g(1 + e^{\tfrac{h}{B}})}{RA} dh.\label{eq2.5.9}$
Integrating Equation ($$\ref{eq2.5.9}$$):
$Ln \dfrac{P}{P_0} = -\dfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B) \to P = P_0 e^{-\tfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B)}.\label{eq2.5.10}$
Using Equation ($$\ref{eq2.5.2}$$), Equation ($$\ref{eq2.5.1}$$), and Equation ($$\ref{eq2.5.10}$$):
$\rho = \dfrac{P}{RT} = \dfrac{P_0 e^{-\tfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B)}}{R \tfrac{A}{1 + e^{\tfrac{h}{B}}}}\label{eq2.5.11}$
3. Pressure and density at an altitude of 1000 m:
Using Equation ($$\ref{eq2.5.10}$$) and Equation ($$\ref{eq2.5.11}$$), the given data for $$P_0$$ and $$g$$, and the values obtained for $$R, A$$, and $$B$$:
• $$\rho (h = 1000) = 1.0756 \tfrac{kg}{m^3}.$$
• $$P(h = 1000) = 89632.5 Pa.$$

2.5: Problems is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Manuel Soler Arnedo via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.