Problem 84 - Thomas Calculus

Question

Mass of planet's atmosphere \(A\) spherical planet of radius \(R\) has an atmosphere whose density is \(\mu=\mu_{0} e^{-c h},\) where \(h\) is the altitude above the surface of the planet, \(\mu_{0}\) is the density at sea level, and \(c\) is a positive constant. Find the mass of the planet's atmosphere.

Step-by-Step Solution

Answer

The mass of the planet's atmosphere is \( M = \frac{8\pi \mu_0}{c^3} e^{-cR} (1 + cR + c^2 R^2) \).

Step 1: Define the Problem

To find the mass of the planet's atmosphere, we need to integrate the density over the entire volume of the atmosphere. The density given by \( \mu = \mu_0 e^{-ch} \) changes with altitude \( h \). The goal is to determine \( M \), the total mass, by integrating this density from the surface to infinitely high in a spherical shell fashion.

Step 2: Express Volume Element

Consider a thin spherical shell at radius \( r = R + h \) from the center of the planet, where \( R \) is the radius of the planet. The volume element \( dV \) of this spherical shell can be expressed as \( 4\pi r^2 dr \).

Step 3: Define Mass Element

The mass element \( dm \) of a thin shell at altitude \( h \) is given by the product of the density and the volume element: \( dm = \mu \, dV = \mu_0 e^{-ch} \, 4\pi (R+h)^2 \, dh \).

Step 4: Set Up the Integral for Total Mass

The total mass \( M \) of the atmosphere is obtained by integrating \( dm \) from \( h = 0 \) to \( h = \infty \): \ \[ M = \int_{0}^{\infty} \mu_0 \, e^{-ch} \, 4\pi (R+h)^2 \, dh. \]

Step 5: Simplify the Integral

Simplify the integral by recognizing it can be expressed as: \ \[ M = 4\pi \mu_0 \int_{0}^{\infty} (R+h)^2 \, e^{-ch} \, dh. \] \.This integral can be solved by a substitution \( u = R + h \), \( du = dh \).

Step 6: Integrate Using Substitution

Perform the substitution \( u = R + h \), making \( h = u - R \); replace the limits accordingly. \ \[ M = 4\pi \mu_0 \int_{R}^{\infty} u^2 e^{-cu+ cR} \, du. \] \ Since it is from \( h = 0 \) originally, maintain \( h \to \infty \) directly implies \( u \to \infty \).

Step 7: Solve the Final Integral

Now, solve the integral: \ \[ M = 4\pi \mu_0 e^{cR} \int_{R}^{\infty} u^2 e^{-cu} \, du. \] \ This is a standard exponential integral and the solution involves using integration by parts or lookup of the result: \ \( M = \frac{8\pi \mu_0}{c^3} e^{-cR} (1 + cR + c^2 R^2) \, .\)

Step 8: Final Result Interpretation

The result gives the total mass of the atmosphere in terms of the initial parameters: \( \mu_0 \), \( c \), \( R \) and constants. Notice how the exponential decay \( e^{-cR} \) modulates the density across the spherical atmosphere.

Key Concepts

Density Function

The density function is a mathematical construct used to describe how a particular quantity, like mass, is distributed over a certain space. In this exercise, the density function of the planet's atmosphere is given by \[ \mu = \mu_0 e^{-ch}, \]where \(\mu_0\) is the initial density at the planet's surface (altitude \(h = 0\)), \(c\) is a positive constant that affects how quickly the density decreases with altitude \(h\) above the planet's surface.

• \(\mu_0\) represents the density of the air at sea level.
• The function \(e^{-ch}\) describes how the density decreases as you move away from the planet's surface.

The key point is that the density decreases exponentially with increasing altitude, meaning it never truly drops to zero but becomes negligibly small at very high altitudes.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. This kind of decay occurs in various natural processes, including the decline in atmospheric density with altitude in this exercise.When we express the density of the atmosphere as\[ \mu = \mu_0 e^{-ch}, \]we are using an exponential function where:

• \(e\) is the base of the natural logarithm, approximately equal to 2.718.
• \(-ch\) is the exponent which determines the rate of decay.

This expression implies that as altitude \(h\) increases, the term \(-ch\) becomes more negative, making the density \(\mu\) decrease towards zero. This reflects how the atmosphere becomes thinner the higher you go above the planet's surface, which is typical of most planetary atmospheres, including Earth's.

Integration by Parts

Integration by parts is a technique in calculus useful for solving integrals involving products of functions. It's based on the product rule for differentiation and is particularly handy for integrals where one function is easily integrable while the other is easily differentiable.For this problem, we approach an integral of the form:\[ \int (R+h)^2 e^{-ch} \, dh, \]which appears when calculating the total mass \(M\) of the atmosphere. Using substitution or integration by parts can simplify this integral. Integration by parts is defined as:\[ \int u \, dv = uv - \int v \, du, \]where you choose parts of the function as \(u\) and \(dv\), and differentiate/integrate these as needed. This method turns a complex integral into a simpler one that's easier to solve or look up.

Spherical Coordinates

Spherical coordinates are an extension of Cartesian coordinates, commonly used in problems involving symmetrical objects like spheres. This coordinate system uses three values: the radial distance \(r\), the polar angle \(\theta\), and the azimuthal angle \(\phi\).In this problem, we deal with spherical shells where:

• \(r = R + h\), the distance from the center of the planet.
• The symmetry of the problem allows us to simplify the volume element for integration: \(dV = 4\pi r^2 \, dr\).

The calculation uses the shape's symmetry to our advantage, converting a potentially complex 3D integral into a simpler form. Spherical coordinates expedite solving for mass distribution in spherical atmospheres by breaking down the volume into manageable spherical layers.