Density and pressure in Earth's atmosphere are proportional: \(\rho=p / h_{0} g,\) where \(h_{0}=8.2 \mathrm{km}\) is a constant 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 \(p=p_{0} e^{-h / l h_{0}},\) where \(p_{0}\) is the surface pressure. (b) At what height will the pressure have dropped to half its surface value?

Understanding the density of Earth's atmosphere is crucial for many scientific and engineering applications, including weather prediction, climate modeling, and aeronautics. Atmospheric density, often denoted as \(\rho\), is the mass per unit volume of air molecules and varies with altitude due to gravitational effects. At sea level, the atmosphere is denser because the weight of the air above compresses the air below. As altitude increases, atmospheric pressure and density decrease because there is less air above exerting force downwards.

Using the proportional relationship \(\rho = \frac{p}{h_{0}g}\), where \(p\) represents pressure, \(h_{0} = 8.2 \text{km}\) is a constant scale height, and \(g\) is the acceleration due to gravity, we can see that as pressure decreases with altitude, so does density. Understanding this relationship is essential for calculating changes in pressure and density with altitude, which is necessary for designing high-altitude aircraft or forecasting weather patterns.

Gravitational acceleration, usually symbolized as \(g\), is the rate at which an object accelerates towards the center of the Earth due to Earth's gravity. It's approximately \(9.81 \text{m/s}^2\) at the Earth's surface but decreases slightly with altitude. This fundamental force not only keeps us anchored to the planet but also governs the behavior of the atmosphere.

In the context of atmospheric pressure and density, gravitational acceleration is a key factor in the weight of the air, which contributes to the pressure exerted on the surface. The formula \(\rho = \frac{p}{h_{0}g}\) included in the exercise reflects how the density \(\rho\) varies inversely with the acceleration due to gravity \(g\). As we move upwards from the Earth's surface, not only does the pressure drop due to less air being above, but the gravitational acceleration also decreases slightly, further influencing the pressure and density of air.

The atmosphere exhibits an exponential decay in pressure with increasing altitude, which stems from the nature of air as a compressible fluid subjected to the force of gravity. Exponential decay is a way to express that a quantity decreases at a rate proportional to its current value, leading to a rapid decline initially that slows down as the value approaches zero.

The integration of the atmospheric pressure equation in the exercise demonstrates this decay: \(p = p_{0} e^{-h / l h_{0}}\), where \(p_0\) is the surface pressure and the exponential term \(e^{-h / l h_{0}}\) describes how pressure changes with height \(h\). As you move higher, the term \(e^{-h / l h_{0}}\) becomes smaller, and so does the pressure. To find the height at which pressure is halved, we can use this equation as seen in the solution to the exercise. By setting \(p = 0.5p_{0}\), and solving for \(h\), we get a quantitative grasp of how quickly pressure falls off with height, an important consideration for activities ranging from mountaineering to the design of pressurized cabins in aircraft.