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Density and pressure in Earth's atmosphere are proportional: \(\rho=p / h_{0} g,\) where \(h_{0}=8.2 \mathrm{~km}\) is a constant called the scale height and \(g\) is the gravitational acceleration. (a) Integrate Equation 15.2 for this case to show that atmospheric pressure as a function of height \(h\) above the surface is given by \(p=p_{0} e^{-h / h_{0}}\), where \(p_{0}\) is the surface pressure. (b) At what height will the pressure have dropped to half its surface value?

Short Answer

Expert verified
a) The atmospheric pressure as a function of height above the surface is given by \(p = p_0 e^{-h/h_0}\). b) The pressure will have dropped to half its surface value at a height of \(h = h_0 \ln(2)\).

Step by step solution

01

Transform the Proportional Relation into a Differential Equation

Start by transforming the given proportional relationship between density and pressure into a differential equation. We know that \(\rho = \frac{p}{h_0g}\), and that the density \(\rho\) can also be represented as \(\frac{dp}{dh}\). This yields a differential equation: \(\frac{dp}{dh} = -\frac{p}{h_0g}\)
02

Solve the Differential Equation

Next, solve the differential equation using the method of separation of variables. To do this, collect all terms involving p on one side and all terms involving h on the other, giving \(\frac{dp}{p} = -\frac{dh}{h_0g}\). This differential equation can be solved by integrating both sides.
03

Integrate Both Sides

Performing the integration, we get \(\int \frac{1}{p} dp = -\int \frac{1}{h_0 g} dh\). The left integral will be the natural logarithm of the absolute value of p, and the right integral will become -h/h_0g. Plus, we should add an integration constant C. So, \(\ln|p| = -\frac{h}{h_0g} + C\).
04

Determine the Constant of Integration

Next, to determine the constant of integration, apply the boundary condition that at the surface of the earth (h = 0), the pressure is the surface pressure \(p_0\). So, directly substitute these values into the equation and find that \(C= \ln|p_0|\). So, the equation becomes \(\ln|p| = -\frac{h}{h_0g} + \ln|p_0|\).
05

Transform The Result

By taking the exponential of both sides and simplifying, we get \(p = p_0 e^{-h/h_0}\).
06

Solve for Half-Pressure Height

For part (b), set the pressure p equal to half the surface pressure \(p_0/2\), and solve for h in the equation we just obtained. This leads to a solution of \(h = -h_0 \ln(0.5)\), which can be simplified to \(h = h_0 \ln(2)\) since the natural logarithm of 0.5 is negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density in Earth's Atmosphere
In the Earth's atmosphere, density is an essential concept. It measures how much mass is contained in a unit volume of air. The formula connecting density (\( \rho \)), pressure (\( p \)), scale height (\( h_0 \)), and gravitational acceleration (\( g \)) is given by \( \rho = \frac{p}{h_0 g} \).
This relationship indicates that the density of the atmosphere depends on the pressure and varies with height.
As we move higher above the surface, the pressure decreases, and thus the density decreases as well. - **Key Characteristics**: - Density decreases with height. - Directly proportional to pressure in the atmosphere.- **Practical Implications**: - Understanding density is crucial for predicting weather patterns and understanding how airplanes fly through different layers of the atmosphere.
Differential Equations
Differential equations are equations involving derivatives, representing rates of change. In the context of Earth's atmosphere, it helps us understand the change in pressure with respect to height. The differential equation derived from the relationship \( \rho = \frac{p}{h_0g} \) is \( \frac{dp}{dh} = -\frac{p}{h_0g} \).This equation helps us express the relationship between the change in atmospheric pressure (\( dp \)) with a change in height (\( dh \)). - **Solving the Equation**: - We use separation of variables to isolate terms involving \( p \) and \( h \) on different sides. - This leads to \( \frac{dp}{p} = -\frac{dh}{h_0g} \).- **Integration**: - By integrating both sides, we find how pressure varies with height.The solution to this provides us with an exponential function revealing how atmospheric pressure decreases exponentially with height.
Scale Height
Scale height (\( h_0 \)) is a critical concept that defines the atmospheric structure. It is the height over which the pressure decreases by a factor of \( e \) (approximately 2.718). For Earth's atmosphere, \( h_0 \) is approximately 8.2 km.- **Importance**: - It provides a measure of the "thickness" of the atmosphere. - Helps in understanding how pressure decreases with altitude.- **Mathematical Expression**: - The exponential form \( p = p_0 e^{-h/h_0} \) illustrates how pressure reduces exponentially with height.A deeper understanding of the scale height helps us predict how quickly atmospheric pressure will reduce with increased altitude, impacting everything from weather forecasting to aviation.
Gravitational Acceleration
Gravitational acceleration (\( g \)) plays a pivotal role in atmospheric dynamics. It is the acceleration due to Earth's gravity, approximately \( 9.81 \text{ m/s}^2 \).- **Role in Atmospheric Pressure**: - It's a key factor in the proportional relationship \( \rho = \frac{p}{h_0g} \). - Influences how quickly the pressure changes with altitude.- **Physical Significance**: - It keeps the atmosphere bound to Earth. - Affects air density and pressure, contributing to weather patterns and flight dynamics.Understanding gravitational acceleration allows us to comprehend its influence on atmospheric pressure and density, showing how gravity assists in maintaining our atmosphere's structure.

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