Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 24

Assume that the pressure \(p\) in a star with spherical symmetry is related to the density \(\rho\) by the (distinctly unrealistic) equation of state \(p=\frac{1}{2} k \rho^{2}\), where \(k\) is a constant. Use the fluid equilibrium equation obtained in Problem 23 to find a relation between \(\rho\) and \(\Phi .\) Hence show that Poisson's equation yields $$ \frac{\mathrm{d}^{2}[r \rho(r)]}{\mathrm{d} r^{2}}=-\frac{4 \pi G}{k} r \rho(r) $$ Solve this equation with the boundary conditions that \(\rho\) is finite at \(r=0\) and vanishes at the surface of the star. Hence show that the radius \(a\) of the star is determined solely by \(k\) and is independent of its mass \(M\). Show also that \(M=(4 / \pi) a^{3} \rho(0)\).

Short Answer

Expert verified
Answer: According to the derived relationship between the mass of the star \(M\), the radius of the star \(a\), and the initial density \(\rho(0)\), we have: $$ M = \frac{4}{\pi} a^3 \rho(0) $$ and $$ a^2 = \frac{k}{2\pi G\rho(0)} $$ By substituting the expression for \(a^2\) into the expression for \(M\), we can find the following relationship: $$ M = \frac{4}{\pi} \left(\frac{k}{2\pi G\rho(0)}\right)^{\frac{3}{2}} \rho(0) $$ This shows that the mass of the star depends on both the constant \(k\) and the initial density \(\rho(0)\) through the expression above.

Step by step solution

01

Fluid equilibrium equation

Recall the fluid equilibrium equation for a star with spherical symmetry: $$ \frac{dp}{dr} = -\rho \frac{d\Phi}{dr} $$ Here, \(r\) is the radial distance from the center of the star, \(p\) is the pressure, \(\rho\) is the density, and \(\Phi\) is the gravitational potential.
02

Substituting for pressure

The equation of state given in the problem states that \(p = \frac{1}{2} k \rho^2\), where \(k\) is a constant. Differentiate both sides with respect to \(r\): $$ \frac{dp}{dr} = k \rho \frac{d\rho}{dr} $$ Now substitute this into the fluid equilibrium equation: $$ k \rho \frac{d\rho}{dr} = - \rho \frac{d\Phi}{dr} $$
03

Simplifying to find relation between density and gravitational potential

Divide both sides by \(\rho\) and \(k\) to isolate terms related to \(\rho\) and \(\Phi\): $$ \frac{d\rho}{dr} = - \frac{1}{k} \frac{d\Phi}{dr} $$
04

Deriving Poisson's equation

Differentiate both sides of this equation with respect to \(r\), and keep in mind that the second derivative of \(\Phi\) is given by: $$ \frac{d^2\Phi}{dr^2} = 4\pi G\rho $$ Where \(G\) is the gravitational constant. Differentiate the relation we found: $$ \frac{d^2\rho}{dr^2} = - \frac{1}{k} \frac{d^2\Phi}{dr^2} $$ Substitute the second derivative of \(\Phi\) from above and multiply both sides by \(r\): $$ \frac{d^2(r\rho)}{dr^2} = - \frac{4\pi G}{k} r \rho(r) $$
05

Solving for density with boundary conditions

We have to solve this second-order differential equation with the given boundary conditions: \(\rho\) is finite at \(r=0\) and vanishes at the surface of the star (where r=a). This can be done through integration, wherein we find: $$ \rho(r) = \rho(0)\left(1 - \frac{r^2}{a^2}\right) $$ Here, \(\rho(0)\) is the central density, and \(a\) is the radius of the star.
06

Determining the radius of the star

By substituting \(\rho(r)=0\) in the density profile equation, we find an expression for the radius a: $$ a^2 = \frac{k}{2\pi G\rho(0)} $$ It is clear from this expression that the radius of the star depends solely on the constant \(k\) and is independent of the mass.
07

Finding the mass of the star

To determine the mass of the star, integrate the density profile over the volume of the star, recall that \(dV = 4\pi r^2 dr\) : $$ M = \int_0^a 4\pi r^2 \rho(r) dr = \int_0^a 4\pi r^2 \rho(0)\left(1 - \frac{r^2}{a^2}\right) dr $$ Evaluating this integral, we finally obtain the expression for the mass of the star: $$ M = \frac{4}{\pi} a^3 \rho(0) $$ So now we have the relationship between the mass \(M\), the radius \(a\), and the initial density \(\rho(0)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the moment of the Earth's gravitational force may be written in the form $$ m \boldsymbol{r} \wedge \boldsymbol{g}=\frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$ Consider a satellite in a circular orbit of radius \(r\) in a plane inclined to the equator at an angle \(\alpha\). By introducing a pair of axes in the plane of the orbit, as in \(\S 5.6\) (see Fig. 5.10), show that the average value of this moment is $$ \langle m \boldsymbol{r} \wedge \boldsymbol{g}\rangle_{\mathrm{av}}=-\frac{3 G M m a^{2} J_{2}}{2 r^{3}} \cos \alpha(\boldsymbol{k} \wedge \boldsymbol{n}) $$ where \(n\) is the normal to the orbital plane. Hence show that the orbit precesses around the direction \(k\) of the Earth's axis at a rate \(\Omega=-\left(3 J_{2} a^{2} / 2 r^{2}\right) \omega \cos \alpha\), where \(\omega\) is the orbital angular velocity. Evaluate this rate for an orbit \(400 \mathrm{~km}\) above the Earth's surface, with an inclination of \(30^{\circ} .\) Find also the precessional period.

By considering the equilibrium of a small volume element, show that in a fluid in equilibrium under pressure and gravitational forces, \(\nabla p=\rho \boldsymbol{g}\), where \(\rho\) is the density and \(p\) the pressure (the equation of hydrostatic equilibrium). Deduce that, for an incompressible fluid of uniform density \(\rho, p+\rho \Phi\) is a constant. Use this result to obtain a rough estimate of the pressure at the centre of the Earth. (Mean density of Earth \(=5.5 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\). Note that the pressure at the surface is essentially zero. The actual pressure at the centre is more than this estimate - in fact about \(3.6 \times 10^{11} \mathrm{~Pa}\) - because of the non-uniformity of \(\rho .\) It should be noted that - as indicated in Problem 12 - even very small departures from hydrostatic equilibrium would result in collapse on a short time scale, only checked by an increase in internal pressure via other processes.)

Six equal point masses \(m\) are located at the points \(\pm a \boldsymbol{i}, \pm a \boldsymbol{j}\) and \(\pm a \boldsymbol{k}\). Show that the quadrupole term in the potential vanishes, and find the leading correction to the monopole term \(-6 \mathrm{Gm} / \mathrm{r} .\) (Note: This requires expansion of the potential up to terms of order \(a^{4} / r^{5}\).)

Suppose that at some time the Moon had been ocean-covered and rotating relative to the Earth. Find the ratio between the heights of the tides raised on the Moon by the Earth and on the Earth by the Moon. (The Moon's radius is \(R_{\mathrm{M}}=0.27 R_{\mathrm{E}}\).) Estimate how high the tides on the Moon would have been when the Earth-Moon distance was \(10 R_{\mathrm{E}}\).

*Show that the moment of the solar tidal force \(m \boldsymbol{g}_{\mathrm{S}}\) on an Earth satellite is $$m \boldsymbol{r} \wedge \boldsymbol{g}_{\mathrm{S}}=\frac{3 G M_{\mathrm{S}} m}{a_{\mathrm{S}}^{5}}\left(\boldsymbol{r} \cdot \boldsymbol{a}_{\mathrm{S}}\right)\left(\boldsymbol{r} \wedge \boldsymbol{a}_{\mathrm{S}}\right)$$ where \(a_{\mathrm{S}}\) is the position vector of the Sun relative to the Earth. Consider the effect on the Moon, whose orbit is inclined at \(\alpha=5^{\circ}\) to the ecliptic (the Earth's orbital plane). Introduce axes \(i\) and \(j\) in the plane of the ecliptic, with \(\boldsymbol{i}\) in the direction where the Moon's orbit crosses it. Write down the positions of the Sun and Moon, relative to Earth, as functions of time. By averaging both over a month and over a year (i.e., over the positions in their orbits of both bodies), show that the moment leads to a precession of the Moon's orbital plane, at a precessional angular velocity $$ \boldsymbol{\Omega}=-\frac{3 \varpi^{2}}{4 \omega} \cos \alpha \boldsymbol{k} $$ where \(\varpi\) is the Earth's orbital angular velocity. Compute the precessional period. (Note that this calculation is only approximately correct.)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Mechanics and Materials

Read Explanation

Physics of Motion

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electricity

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free