The calculation of atmospheric pressure at the summit of Mount Everest carried out in Example 13.3 used the model known as the isothermal atmosphere, in which gas pressure is proportional to density: \(p=\gamma \rho\), with \(\gamma\) constant. Consider a spherical cloud of gas supporting itself under its own gravitation and following this model. a) Write the equation of hydrostatic equilibrium for the cloud, in terms of the gas density as a function of radius, \(\rho(r) .\) b) Show that \(\rho(r)=A / r^{2}\) is a solution of this equation, for an appropriate choice of constant \(A\). Explain why this solution is not suitable as a model of a star.

Short Answer

Expert verified
Question: Prove that the gas density function 𝜌(𝑟)=𝐴/𝑟² is a solution to the hydrostatic equilibrium equation for a spherical cloud of gas with an isothermal atmosphere and explain why this solution is not suitable for modeling a star. Answer: The given gas density function 𝜌(𝑟)=𝐴/𝑟², when integrated into the hydrostatic equilibrium equation, does hold true for a suitable constant A. However, this solution is not suitable for modeling a star because the density goes to infinity as the radius approaches zero, which is not physically realizable for a star with finite and maximum density at its center.

Step by step solution

01

Write down the equation of hydrostatic equilibrium

The equation of hydrostatic equilibrium states that at any point within the cloud, the force due to pressure gradient is balanced by the force due to gravity. Mathematically, it is written as: $$\frac{dp}{dr}=-\rho g.$$ Where \(p\) is pressure, \(\rho\) is density function with respect to radius r, and \(g\) is the gravitational acceleration.
02

Substitute the isothermal atmosphere model

We are given the isothermal atmosphere model \(p = \gamma \rho\). Now substitute this into the hydrostatic equilibrium equation: $$\frac{d(\gamma \rho)}{dr}=-\rho g.$$
03

Gravitational force on the sphere gas cloud

Now we have to find the gravitational force on the gas cloud sphere of radius r and density \(\rho\). By Newton's Law of Gravitation: $$g=\frac{GM(r)}{r^2}.$$. Where \(G\) is the gravitational constant, \(M(r)\) is the mass of gas contained within a sphere of radius r. We know that mass can be written as a product of density and volume: $$M(r)=\int_0^r 4\pi r'^2 \rho(r') dr'.$$ Now, substitute this expression for \(g\) in the hydrostatic equation to get an expression for \(d(\gamma\rho)/dr\).
04

Solve the equation by substituting the given density function

We are given the gas density function \(\rho(r) = \frac{A}{r^2}\). Substitute the given density function into the hydrostatic equation: $$\frac{d(\gamma A / r^{2})}{dr}=-\frac{A}{r^2} \cdot \frac{GM(r)}{r^2}.$$. Now we need to integrate this equation and find a suitable constant A for which this equation holds true.
05

Explain why the given solution is not suitable for a star model

The given solution \(\rho(r) = \frac{A}{r^2}\) implies that the density decreases with the square of the radius and goes to infinity when the radius approaches zero (i.e., at the center of the cloud). This situation is not physically realizable for a star, as the density should be finite and maximum at the center. Hence, this solution is not suitable as a model of a star.

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Key Concepts

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

Isothermal Atmosphere Model
To begin understanding an isothermal atmosphere model, imagine a situation where the temperature within a planet's atmosphere remains constant with altitude. Although this is a simplification of real-life atmospheres, it serves as a helpful approximation for some calculations. In this model, the pressure and density of the atmosphere are directly proportional, leading to the equation \(p=\gamma \rho\), where \(\gamma\) is a constant.

This relationship implies that if you characterise the atmosphere's pressure, you can determine its density, and vice versa.

Hydrostatic Equilibrium in the Isothermal Model

In the context of a spherical gas cloud, as presented in the exercise, applying this model is used to derive the forces at play within the structure. The gas cloud is in hydrostatic equilibrium when the inward pull of gravity is exactly balanced by the outward pressure of the gas, ensuring that the cloud doesn't collapse inward or expand outward. The isothermal model aids us in simplifying the equations governing this balance, permitting easier analytical or numerical solutions.
Gravitational Force
Gravitational force is a fundamental concept not only in physics but also in understanding the stellar structure and the behavior of atmospheric models.

Newton's Law of Gravitation

Sir Isaac Newton postulated that every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this force can be expressed as \(F = G\frac{m_1m_2}{r^2}\), where \(G\) represents the gravitational constant, \(m_1\) and \(m_2\) are the masses in question, and \(r\) is the distance separating them. In the context of the spherical gas cloud described in the exercise, we use this law to infer the gravitational pull on every segment of the cloud, which is essential in establishing the condition for hydrostatic equilibrium.
Stellar Structure
Stellar structure refers to the composition and organization of stars, which are massive, spherical bodies composed primarily of hydrogen and helium, operating under principles of plasma physics. Stars maintain their structure through a delicate balance between the gravitational force pulling matter inward and the pressure of nuclear fusion reactions pushing outward.

Why the Isothermal Model Fails for Stars

The density solution \(\rho(r)=\frac{A}{r^2}\) from the isothermal model suggests that density unacceptably skyrockets as we move towards the star's core (\(r\) approaches zero) — fundamentally contradicting our understanding of physical density limitations and core conditions of real stars. In reality, stellar interiors have more complex relationships between pressure, temperature, and density, requiring sophisticated models that account for these interacting processes to describe their structure.

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Most popular questions from this chapter

A water-powered backup sump pump uses tap water at a pressure of \(3.00 \mathrm{~atm}\left(p_{1}=3 p_{\mathrm{atm}}=\right.\) \(3.03 \cdot 10^{5} \mathrm{~Pa}\) ) to pump water out of a well, as shown in the figure \(\left(p_{\text {well }}=p_{\text {ttm }}\right)\). This system allows water to be pumped out of a basement sump well when the electric pump stops working during an electrical power outage. Using water to pump water may sound strange at first, but these pumps are quite efficient, typically pumping out \(2.00 \mathrm{~L}\) of well water for every \(1.00 \mathrm{~L}\) of pressurized tap water. The supply water moves to the right in a large pipe with cross-sectional area \(A_{1}\) at a speed \(v_{1}=2.05 \mathrm{~m} / \mathrm{s}\). The water then flows into a pipe of smaller diameter with a cross-sectional area that is ten times smaller \(\left(A_{2}=A_{1} / 10\right)\). a) What is the speed \(v_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) b) What is the pressure \(p_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) c) The pump is designed so that the vertical pipe, with cross-sectional area \(A_{3}\), that leads to the well water also has a pressure of \(p_{2}\) at its top. What is the maximum height, \(h,\) of the column of water that the pump can support (and therefore act on ) in the vertical pipe?

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

Find the minimum diameter of a \(50.0-\mathrm{m}\) -long nylon string that will stretch no more than \(1.00 \mathrm{~cm}\) when a load of \(70.0 \mathrm{~kg}\) is suspended from its lower end. Assume that \(Y_{\text {nylon }}=3.51 \cdot 10^{8} \mathrm{~N} / \mathrm{m}^{2}\)

You are in a boat filled with large rocks in the middle of a small pond. You begin to drop the rocks into the water. What happens to the water level of the pond? a) It rises. d) It rises momentarily and then b) It falls. falls when the rocks hit bottom. c) It doesn't change. e) There is not enough information to say.

A tourist of mass \(60.0 \mathrm{~kg}\) notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he decides to dive for the chest. He inhales fully, thus setting his average body density to \(945 \mathrm{~kg} / \mathrm{m}^{3}\), jumps into the ocean (with saltwater density = \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) ), grabs the chain, and tries to pull the chest to the surface. Unfortunately, the chest is too heavy and will not move. Assume that the man does not touch the bottom. a) Draw the man's free-body diagram, and determine the tension on the chain. b) What mass (in kg) has a weight that is equivalent to the tension force in part (a)? c) After realizing he cannot free the chest, the tourist releases the chain. What is his upward acceleration (assuming that he simply allows the buoyant force to lift him up to the surface)?

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