Problem 24 - Thomas Calculus in SI Units

Question

Atmospheric pressure The earth's atmospheric pressure \(p\) is often modeled by assuming that the rate \(d p / d h\) at which \(p\) changes with the altitude \(h\) above sea level is proportional to \(p\) Suppose that the pressure at sea level is 1013 hectopascals and that the pressure at an altitude of \(20 \mathrm{km}\) is 90 hectopascals. a. Solve the initial value problem Differential equation: \(\quad d p / d h=k p \quad(k\) a constant) Initial condition: \(\quad p=p_{0} \quad\) when \(\quad h=0\) 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 hectopascals?

Step-by-Step Solution

Answer

Pressure at 50 km is approximately 17.13 hPa; for 900 hPa, the altitude is about 0.37 km.

Step 1: Set up the differential equation

We start with the given differential equation \( \frac{dp}{dh} = kp \). This means the rate of change of pressure with respect to altitude is proportional to the pressure itself.

Step 2: Solve the differential equation

The differential equation \( \frac{dp}{dh} = kp \) is separable. We separate the variables to get \( \frac{1}{p} dp = k dh \) and then integrate both sides: \( \int \frac{1}{p} dp = \int k dh \)

Step 3: Integrate and find the general solution

After integrating, we obtain \( \ln|p| = kh + C \), where \( C \) is the constant of integration. By exponentiating both sides, we get \( |p| = e^{kh + C} = e^{C}e^{kh} \). Setting \( e^{C} = C' \), we have \( p = C'e^{kh} \).

Step 4: Use the initial condition to solve for constants

Using the initial condition: \( p = p_0 \) when \( h = 0 \), we substitute into the equation \( p = C'e^{kh} \), giving \( p_0 = C'e^0 = C' \). So the equation simplifies to \( p = p_0 e^{kh} \).

Step 5: Solve for constant \( p_0 \)

We know that \( p_0 = 1013 \) hectopascals, which is the pressure at sea level (\( h = 0 \)). Thus, \( p_0 = 1013 \) hPa.

Step 6: Use additional data to determine \( k \)

We know the pressure at \( h = 20 \) km is \( 90 \) hPa. Substitute these values into the equation: \( 90 = 1013 e^{20k} \). Solving for \( k \), we get \( e^{20k} = \frac{90}{1013} \). Taking the natural log: \( 20k = \ln\left(\frac{90}{1013}\right) \), so \( k = \frac{1}{20}\ln\left(\frac{90}{1013}\right) \).

Step 7: Express \( p \/ in complete form

Now we have the complete formula for pressure \( p = 1013 e^{kh} \), where \( k = \frac{1}{20} \ln\left(\frac{90}{1013}\right) \).

Step 8: Calculate pressure at \( h = 50 \) km

Substitute \( h = 50 \) into the equation \( p = 1013 e^{50k} \). Compute \( 50k = \frac{50}{20} \ln\left(\frac{90}{1013}\right) = 2.5 \ln\left(\frac{90}{1013}\right) \). Then calculate \( p \) at 50 km.

Step 9: Solve for altitude with pressure 900 hPa

Set \( p = 900 \) in the equation: \( 900 = 1013 e^{kh} \). Solve for \( h \) by taking natural logs: \( e^{kh} = \frac{900}{1013} \), thus \( kh = \ln\left(\frac{900}{1013}\right) \). Finally, \( h = \frac{1}{k} \ln\left(\frac{900}{1013}\right)\).

Key Concepts

Atmospheric Pressure

Atmospheric pressure is a measure of the force that the earth's atmosphere exerts on objects. It varies with altitude due to the thinning of the air as one moves further from the earth's surface. At sea level, the atmospheric pressure is highest, typically around 1013 hectopascals (hPa). As altitude increases, atmospheric pressure decreases. This decrease occurs because the density of air molecules diminishes with height, leading to a reduction in force exerted on objects.

When modeling atmospheric pressure changes with altitude, we often use differential equations. In this particular case, the rate of change of pressure with respect to altitude is considered proportional to the current pressure. This means, as you go higher, the pressure drops at a rate dependent on its current value. Such modeling helps in accurately predicting pressures at various altitudes and is vital in fields like meteorology and aviation, ensuring efficient planning and operations.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a e^{bx}\), where \(e\) is approximately 2.718281828, known as Euler's number. These functions are fundamental in describing growth and decay processes, including changes in atmospheric pressure with altitude.

In our problem, after solving the differential equation \( \frac{dp}{dh} = kp \), we found the pressure to be represented as an exponential function: \( p = p_0 e^{kh} \). Here, \(p_0\) is the initial pressure at sea level, and \(k\) is a constant that dictates the rate of exponential decrease in pressure as altitude increases.

Exponential functions effectively model scenarios where changes happen at rates proportional to the current quantity. Just as pressure decreases at a rate proportional to its current value, exponential functions help reflect this continuous, relative change.

Initial Value Problems

An initial value problem involves finding a function based on a differential equation, given initial conditions. These kinds of problems are common when solving differential equations where specific starting information is provided.

In the exercise, the initial value problem is set by the conditions at sea level: the pressure \(p_0\) is 1013 hPa at altitude \(h = 0\). This initial condition allows us to determine the constants in our pressure function. By substituting the initial values into the general solution \(p = p_0 e^{kh}\), we can solve for the constants such as \(p_0\) and \(k\).

Initial value problems are crucial because they give context to a problem, allowing us to predict future behavior based on a known starting point. Using the pressure at another altitude, like at 20 km, helps determine the value of \(k\), which further provides complete knowledge of how pressure varies with altitude.