Problem 2 - Thomas Calculus

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 millibars (about 14.7 pounds per square inch ) and that the pressure at an altitude of 20 \(\mathrm{km}\) is 90 millibars. a. Solve the initial value problem Differential equation: \(\quad d p / d h=k p \quad(k \text { a constant })\) Initial condition: \(\quad p=p_{0}\) when \(h=0\) 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 millibars?

Step-by-Step Solution

Answer

Pressure at 50 km is near 0; altitude for 900 millibars is close to 0 km.

Step 1: Recognize the Differential Equation

The given differential equation is \( \frac{dp}{dh} = kp \), where \(k\) is a constant. This equation states that the rate at which pressure changes with altitude is proportional to the current pressure.

Step 2: Solve the Differential Equation

To solve \( \frac{dp}{dh} = kp \), we separate variables: \( \frac{1}{p} dp = k dh \). Integrate both sides to get \( \ln |p| = kh + C_1 \). Exponentiating both sides gives \( p = e^{kh + C_1} = e^{C_1} e^{kh} \). Let \( e^{C_1} = C \), then \( p = Ce^{kh} \).

Step 3: Apply the Initial Condition

We use the initial condition \( p = p_0 \) when \( h = 0 \). Therefore, \( p_0 = Ce^0 = C \). Thus, the pressure equation is \( p = p_0e^{kh} \).

Step 4: Determine Constants from Given Data

We know \( p_0 = 1013 \) millibars at sea level \((h=0)\) and \( p = 90 \) millibars at \( h = 20 \text{ km} \). Substituting into the equation: \( 90 = 1013e^{20k} \). Solving for \( e^{20k} \), we get \( e^{20k} = \frac{90}{1013} \). Taking the natural log on both sides: \( 20k = \ln \left( \frac{90}{1013} \right) \), so \( k = \frac{1}{20} \ln \left( \frac{90}{1013} \right) \).

Step 5: Find Atmospheric Pressure at 50 km

Use the equation \( p = 1013e^{kh} \) and substitute \( h = 50 \) km and the calculated value of \( k \). Thus, \( p = 1013 \times e^{50 \times (\frac{1}{20} \ln(\frac{90}{1013}))} \) to find the pressure.

Step 6: Determine Altitude for 900 Millibars Pressure

To find the altitude where \( p = 900 \) millibars, set \( 900 = 1013e^{kh} \). Solving for \( h \), we have: \( e^{kh} = \frac{900}{1013} \). Taking the natural log gives \( kh = \ln \left( \frac{900}{1013} \right) \). Thus, \( h = \frac{1}{k} \ln \left( \frac{900}{1013} \right) \). Substitute the known \( k \) to find \( h \).

Key Concepts

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above a certain level. This pressure is a critical factor influencing weather patterns and human activities. At sea level, the standard atmospheric pressure is approximately 1013 millibars.
This measurement comes from averaging atmospheric conditions across the world. As altitude increases, the atmospheric pressure decreases because there is less air above. This is why it feels harder to breathe on high mountains, as the thinner air contains less oxygen.
Understanding atmospheric pressure is essential for many fields, including meteorology and aviation. Changes in pressure can indicate upcoming weather changes, such as storms or calm weather.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with a given initial condition. The goal is to find a specific solution that satisfies the initial condition.
In the context of atmospheric pressure, the differential equation \( \frac{dp}{dh} = kp \) is solved with the initial condition \( p = p_0 \) when \( h = 0 \). This means at sea level (\( h = 0 \)), the pressure \( p \) is known, allowing us to find the full solution for \( p \) as a function of altitude \( h \).
Solving an initial value problem can give insights into how physical systems behave over time or space, which is highly valuable for predicting conditions like weather or physiological responses at different altitudes.

Exponential Decay

The concept of exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In the mathematical model of atmospheric pressure, the equation \( \frac{dp}{dh} = kp \) is solved to get \( p = p_0e^{kh} \). Because \( k \) is negative in our scenario, the pressure diminishes exponentially with increasing altitude.
This factor is crucial because it explains why pressure falls off so sharply with height. The elevation increases beyond the fertile lower atmosphere, where the air thins rapidly. Exponential decay is not only found in atmospheric studies but is widely applicable in fields like radioactive decay and pharmacokinetics as well.

Altitude-Pressure Relationship

The relationship between altitude and pressure is inversely proportional: as altitude increases, atmospheric pressure decreases.To understand this, consider that higher altitudes mean less air mass above the observation point.
Using the equation \( p = p_0e^{kh} \), each value of \( k \) translates how quickly pressure declines with height. The specific value of \( k \) is determined by observations of pressure at different altitudes.
For example, the pressure at 20 km is notably lower than at sea level, showing this inverse relationship.Tools like barometers are used to measure these changes in pressure with altitude, vital for predicting weather and planning flights.