/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Astronomers believe that the mas... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 3

Astronomers believe that the mass distribution (mass per unit volume) of some galaxies may be approximated, in spherical coordinates, by \(\rho=a e^{-b r}\), for \(0 \leq r \leq \infty\), where \(\rho\) is the density. Find the total mass.

Short Answer

Expert verified
The total mass of the galaxy is \(\frac{8\pi a}{b^3}\).

Step by step solution

01

Understanding the Problem

We are given a density function in spherical coordinates \[\rho = a e^{-b r}\]where \(a\) and \(b\) are constants, and \(r\) is the radial coordinate. We need to calculate the total mass of the galaxy using this density distribution.
02

Setting Up the Integral

In spherical coordinates, the differential volume element is given by \[dV = r^2 \sin(\theta) \, dr \, d\theta \, d\phi\]where \(r\) is the radial distance, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle. The total mass \(M\) can be found by integrating the density over the entire space:\[M = \int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} \rho \, r^2 \sin(\theta) \, dr \, d\theta \, d\phi\]
03

Substituting the Density Function

Substitute the given density function \(\rho = a e^{-b r}\) into the integral for mass:\[M = \int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} a e^{-b r} \cdot r^2 \sin(\theta) \, dr \, d\theta \, d\phi\]
04

Simplifying the Integration

The integral simplifies with separate limits:\[M = a \int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} e^{-b r} \, r^2 \sin(\theta) \, dr \, d\theta \, d\phi\]Recognize that the integral with respect to \(\theta\) and \(\phi\) can be computed independently from \(r\):\[\int_0^{2\pi} d\phi = 2\pi\]\[\int_0^{\pi} \sin(\theta) \, d\theta = 2\]
05

Solving the Radial Integral

Focus on the radial integral:\[\int_0^{\infty} e^{-b r} \, r^2 \, dr\]Use the substitution method by letting \(x = br\), then \(dx = b\,dr\) or \(dr = \frac{dx}{b}\). The integral then becomes:\[\int_0^{\infty} e^{-x} \left(\frac{x}{b}\right)^2 \frac{dx}{b} = \frac{1}{b^3} \int_0^{\infty} e^{-x} x^2 \ dx\]The integral \(\int_0^{\infty} e^{-x} x^2 dx\) is a known Gamma function result, equal to 2! = 2.
06

Calculating the Total Mass

Substituting back into the expression for mass:\[M = a \cdot 2\pi \cdot 2 \cdot \frac{2}{b^3} = \frac{8\pi a}{b^3}\]This is the total mass of the galaxy.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Spherical Coordinates
In the study of galaxies and their mass distribution, spherical coordinates can be an incredibly helpful mathematical tool. These coordinates extend our ability to understand three-dimensional space in a clear and structured manner. Unlike Cartesian coordinates, which represent points based on their distances along perpendicular axes, spherical coordinates use three different parameters: the radial distance, polar angle, and azimuthal angle.
  • The **radial distance** \( r \) is a measure of the direct linear distance from the origin to a given point in space.
  • The **polar angle** \( \theta \) is the angle between the positive z-axis and the line segment joining the origin and the point.
  • The **azimuthal angle** \( \phi \) is the angle in the x-y plane from the x-axis.
Spherical coordinates are crucial in fields like astronomy and physics because they often simplify complex systems, like the distribution of mass in galaxies, by aligning with the natural structures of these astronomical objects. In these cases, the complexity of traditional 3D graphs and functions can be reduced, making calculations more intuitive.
Density Function
A density function is a mathematical tool used to describe how much of a quantity, like mass, is present within a certain volume of space. In the realm of galaxies, density functions enable us to quantify how mass is spread throughout a galaxy's structure.
The given density function \( \rho = a e^{-b r} \) plays a key role in approximating the mass distribution in galaxies:
  • **\( a \)**: This constant scales the density function and can be related to the total mass concentration of the galaxy.
  • **\( b \)**: This constant determines how quickly the density decreases as you move radially outward from the center of the galaxy.
  • **\( e^{-b r} \)**: An exponential decay function that illustrates how the density diminishes with increasing distance \( r \) from the center, resembling many real-world galaxies where central regions tend to be more densely packed.
Understanding and utilizing density functions allows astronomers and astrophysicists to predict the total mass of galaxies and their dynamical properties.
Integral Calculus
Integral calculus is a branch of mathematics aimed at understanding the accumulation of quantities, such as areas under curves, and it is particularly powerful in evaluating total quantities when dealing with continuous distributions. In this exercise, integral calculus is applied to compute the total mass of a galaxy based on the provided density function.
The process involves setting up a triple integral in spherical coordinates to encompass:
  • The integration over \( \phi \), covering the entire azimuthal range from \( 0 \) to \( 2\pi \).
  • The integration over \( \theta \), encompassing the polar span from \( 0 \) to \( \pi \), reflecting the full 3D structure of the space.
  • The integration over \( r \), from \( 0 \) to infinity, containing the infinite spatial expansion of the galaxy.
In the calculation, recognizing separate integrals simplifications can enhance solving efficiency. Integral calculus, therefore, provides the necessary mathematical framework to derive significant physical insights from density distributions in astronomy.
Gamma Function
The Gamma function is a generalization of the factorial function to non-integer values, commonly appearing in the solutions of integrals involving exponential and polynomial expressions. It is an essential concept in advanced mathematics, often arising in problems involving distributions and probability.
For this exercise, the specific integral \( \int_0^{\infty} e^{-x} x^2 dx \) needed for determining the galaxy's total mass is recognized as a Gamma function result:
  • **Gamma Function Result**: This particular integral is equal to \( \Gamma(3) \), which simplifies to \( 2! = 2 \).
  • The method uses patterns of the function \( \Gamma(n+1) = n \times \Gamma(n) \) with \( \Gamma(1) = 1 \) to reach the result.
The utilization of the Gamma function showcases its elegance in simplifying otherwise intricate integrations and aids significantly in handling mathematical challenges encountered during mass distribution analyses in galaxies.

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

(a) In a series LC circuit driven by a DC voltage \((\omega=0)\), compare the energy stored in the inductor to the energy stored in the capacitor. (b) Carry out the same comparison for an LC circuit that is oscillating freely (without any driving voltage). (c) Now consider the general case of a series LC circuit driven by an oscillating voltage at an arbitrary frequency. Let \(\overline{U_{L}}\) and be the average energy stored in the inductor, and similarly for \(\overline{U_{C}}\). Define a quantity \(u=\overline{U_{C}} /\left(\overline{U_{L}}+\overline{U_{C}}\right)\), which can be interpreted as the capacitor's average share of the energy, while \(1-u\) is the inductor's average share. Find \(u\) in terms of \(L, C\), and \(\omega\), and sketch a graph of \(u\) and \(1-u\) versus \(\omega\). What happens at resonance? Make sure your result is consistent with your answer to part a.(answer check available at lightandmatter.com)

A hydrogen atom is electrically neutral, so at large distances, we expect that it will create essentially zero electric field. This is not true, however, near the atom or inside it. Very close to the proton, for example, the field is very strong. To see this, think of the electron as a spherically symmetric cloud that surrounds the proton, getting thinner and thinner as we get farther away from the proton. (Quantum mechanics tells us that this is a more correct picture than trying to imagine the electron orbiting the proton.) Near the center of the atom, the electron cloud's field cancels out by symmetry, but the proton's field is strong, so the total field is very strong. The voltage in and around the hydrogen atom can be approximated using an expression of the form \(V=r^{-1} e^{-r}\). (The units come out wrong, because I've left out some constants.) Find the electric field corresponding to this voltage, and comment on its behavior at very large and very small \(r\). (solution in the pdf version of the book)

Calculate the quantity \(i^{i}\) (i.e., find its real and imaginary parts).(answer check available at lightandmatter.com)

Find the capacitance of the surface of the earth, assuming there is an outer spherical "plate鈥� at infinity. (In reality, this outer plate would just represent some distant part of the universe to which we carried away some of the earth's charge in order to charge up the earth.)(answer check available at lightandmatter.com)

(a) Use complex number techniques to rewrite the function \(f(t)=4 \sin \omega t+3 \cos \omega t\) in the form \(A \sin (\omega t+\delta)\).(answer check available at lightandmatter.com) (b) Verify the result using the trigonometric identity \(\sin (\alpha+\beta)=\sin \alpha \cos \beta+\sin \beta \cos \alpha\).
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Thermodynamics

Read Explanation

Force

Read Explanation

Waves Physics

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.