Question

Atmospheric Pressure Earth's atmospheric pressure \(p\) is often modeled by assuming that the rate \(d p / d h\) at which \(p\) changes with the altititude \(h\) above sea level is proportional to \(p\) . Suppose that the pressure at sea level is 1013 millibars (about 14.7 lb/in \(^{2} )\) and that the pressure at an altitude of 20 \(\mathrm{km}\) is 90 millibars. (a) Solve the initial value problem $$\begin{array}{ll}{\text { Differential equation: }} & {\frac{d p}{d h}=k p} \\\ {\text { Initial condition: }} & {p=p_{0} \text { when } h=0}\end{array}$$ to express \(p\) in terms of \(h .\) Determine the values of \(p_{0}\) and \(k\) from the given altitude-pressure data. (b) What is the atmospheric pressure at \(h=50 \mathrm{km} ?\) (c) At what altitude does the pressure equal 900 millibars?

Short answer

The atmospheric pressure at an altitude of 50 km is approximately 30.789 millibars. The altitude at which the pressure equals 900 millibars is approximately 116.3 km.

Step-by-step solution

Finding the general solution of the differential equation

Our differential equation is given by \( \frac{dp}{dh} = kp \). This is a first-order linear differential equation. Separating the variables and integrating, we obtain the general solution: \( p(h) = p_0 e^{kh} \). Where \( p_0 \) is the initial pressure (pressure at sea level), \( k \) is a constant, and \( h \) is altitude.

Calculating the constant k

To get the value for \(k\), we use the provided atmospheric pressure at the altitude of 20 km. We solve the equations \( 90 = p_0 * e^{20k} \) and \( 1013 = p_0 \), which gives \( k = \frac{1}{20} \ln \left( \frac{90}{1013} \right) \). Now we have our full model \( p(h) = 1013 * e^{ \frac{h}{20} \ln \left( \frac{90}{1013} \right) } \).

Finding the pressure at an altitude of 50 km

We can directly substitute \( h = 50 \) km into our model equation to give: \( p(50) = 1013 * e^{ \frac{50}{20} \ln \left( \frac{90}{1013} \right) } \).

Determining the altitude at which the pressure equals 900 millibars

We find the altitude \( h \) for when \( p = 900 \) mb by setting our model equal to 900 and solving for \( h \): \( 900 = 1013 * e^{ \frac{h}{20} \ln \left( \frac{90}{1013} \right) } \). Then, isolate \( h \) to give the altitude: \( h = 20 \frac{\ln \left(\frac{900}{1013}\right)}{\ln \left(\frac{90}{1013}\right)} \).