/*! 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 81 Planets are not uniform inside. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 81

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be \(15.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the center and \(2.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the surface. What is the acceleration due to gravity at the surface of this planet?

Short Answer

Expert verified
The gravitational acceleration at the surface is approximately 9.44 m/s虏.

Step by step solution

01

Define the density function

Since the density decreases linearly from the center to the surface, we can define the density function as \( \rho(r) = \rho_c - k r \), where \( \rho_c = 15.0 \times 10^3 \text{ kg/m}^3 \) is the density at the center, and \( \rho_s = 2.0 \times 10^3 \text{ kg/m}^3 \) is the density at the surface. The planet's radius \( R = 6371 \text{ km} = 6.371 \times 10^6 \text{ m} \).
02

Find the value of k

The density function \( \rho(r) = \rho_c - k r \) gives us \( \rho(R) = \rho_s = 2.0 \times 10^3 \text{ kg/m}^3 \). Substituting, we find: \( 15.0 \times 10^3 - k \times 6.371 \times 10^6 = 2.0 \times 10^3 \). Solving for \( k \), we get \( k = \frac{15.0 \times 10^3 - 2.0 \times 10^3}{6.371 \times 10^6} \approx 2.040 \times 10^{-3} \text{ kg/m}^4 \).
03

Calculate mass of the planet

The mass \( M \) of a spherically symmetric object is given by integrating the density over its volume: \( M = \int_0^R \rho(r) 4\pi r^2 dr \). Substituting \( \rho(r) = 15.0 \times 10^3 - 2.040 \times 10^{-3} r \) into this integral, we get:\[ M = \int_0^{6.371 \times 10^6} \left( 15.0 \times 10^3 - 2.040 \times 10^{-3} r \right) 4 \pi r^2 \, dr \]This integral evaluates to approximately \( M \approx 6.12 \times 10^{24} \text{ kg} \).
04

Calculate gravity at the surface

The gravitational acceleration at the surface \( g \) is given by \( g = \frac{GM}{R^2} \), where \( G = 6.674 \times 10^{-11} \text{ N(m/kg)}^2 \). Substitute \( M = 6.12 \times 10^{24} \text{ kg} \) and \( R = 6.371 \times 10^6 \text{ m} \):\[ g = \frac{6.674 \times 10^{-11} \times 6.12 \times 10^{24}}{(6.371 \times 10^6)^2} \approx 9.44 \text{ m/s}^2 \].

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 Symmetry
Imagine a perfect ball, smooth and uniform in all directions; this is the principle of spherical symmetry. When dealing with cosmic bodies, such as planets, spherical symmetry implies that the planet鈥檚 properties only depend on the distance from its center, not on direction. This symmetry simplifies many calculations, as it assumes the planet has a smooth and consistent density variation, depending solely on the radial distance.
  • In our exercise, the density at the center is maximum and decreases gradually as we move toward the surface, representing this symmetry mathematically.
  • This concept allows us to describe the entire mass and gravitational fields of planets in straightforward equations, especially when applying Newton鈥檚 law of universal gravitation.
Through spherical symmetry, we can comprehend complex celestial phenomena in a more manageable way.
Density Function
The density function describes how matter is distributed within an object. In this problem, the planet's density decreases linearly from the center to its surface. This is represented as a linear function: \( \rho(r) = \rho_c - kr \), where
  • \( \rho_c \) is the density at the center, and
  • \( k \) is a constant derived from the change in density over the radius.
Understanding this function is key to modeling the planet's mass distribution.

Here鈥檚 why it鈥檚 significant:
  • Linear functions are simple, relying on basic arithmetic operations, making them easy to work with mathematically.
  • They provide a very realistic description of the physical situation within spherically symmetric bodies like planets.
The linearly decreasing density helps predict not just how much matter is at each point but also calculate overall mass and gravitational effects.
Integration
In the context of physics, integration helps us find the total amount of a quantity that is distributed throughout a space, such as mass in this case. To calculate the planet鈥檚 total mass, we need to integrate the density function over its entire volume.
The integral involved is: \[ M = \int_0^R \rho(r) 4\pi r^2 \, dr \] This equation calculates the planet's mass by summing up all tiny spherical shells of mass from the center to the radius \( R \).
  • \( 4\pi r^2 \) describes the surface area of each shell, while \( \rho(r) \) denotes how dense the shell is at a given radius \( r \).
  • This requires the concept of the integral as an 鈥渋nfinite sum鈥� of infinitesimal parts of the planet.
By understanding integration, we effectively combine the density and spherical symmetry concepts into a mathematical framework to model physical systems like planets.
Gravitational Acceleration
Gravitational acceleration signifies the rate at which an object accelerates due to gravity鈥檚 pull. At a planet鈥檚 surface, this attraction is described by the formula:\[ g = \frac{GM}{R^2} \] where:
  • \( G \) is the gravitational constant, a universal factor ensuring the right unit balance and strength of force.
  • \( M \) is the total mass calculated by integrating the density function.
  • \( R \) is the radius of the planet.
This acceleration is crucial in determining phenomena such as free-fall speeds and the weight of objects at the surface.
With this knowledge, one can understand how varying density profiles affect gravitational forces. It explains why different planets have different gravity strengths, which influences everything from atmosphere retention to potential human colonization viability.

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

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{1700}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is \(\rho(r)=A-B r,\) where \(A=12,700 \mathrm{kg} / \mathrm{m}^{3}\) and \(B=\) \(1.50 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{4}\) . Use \(R=6.37 \times 10^{6} \mathrm{m}\) for the radius of the earth approximated as a sphere. (a) Geological evidence indicates that the densities are \(13,100 \mathrm{kg} / \mathrm{m}^{3}\) and \(2,400 \mathrm{kg} / \mathrm{m}^{3}\) at the earth's center and surface, respectively. What values does the linear approximation model give for the densities at these two locations? (b) Imagine dividing the earth into concentric, spherical shells. Each shell has radius \(r\) , thickness \(d r\) , volume \(d V=4 \pi r^{2} d r,\) and mass \(d m=\rho(r) d V .\) By integrating from \(r=0\) to \(r=R,\) show that the mass of the earth in this model is \(M=\frac{4}{3} \pi R^{3}\left(A-\frac{3}{4} B R\right)\) (c) Show that the given values of \(A\) and \(B\) give the correct mass of the earth to within 0.4\(\%\) (d) We saw in Section 13.6 that a uniform spherical shell gives no contribution to \(g\) inside it. Show that \(g(r)=\frac{4}{3} \pi G r\left(A-\frac{3}{4} B r\right)\) inside the earth in this model. (e) Verify that the expression of part (d) gives \(g=0\) at the center of the earth and \(g=9.85 \mathrm{m} / \mathrm{s}^{2}\) at the surface. (f) Show that in this model \(g\) does not decrease uniformly with depth but rather has a maximum of \(4 \pi G A^{2} / 9 B=10.01 \mathrm{m} / \mathrm{s}^{2}\) at \(r=2 A / 3 B=5640 \mathrm{km} .\)

A uniform sphere with mass 60.0 \(\mathrm{kg}\) is held with its center at the origin, and a second uniform sphere with mass 80.0 \(\mathrm{kg}\) is held with its center at the point \(x=0, y=3.00 \mathrm{m} .\) (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 \(\mathrm{kg}\) placed at the point \(x=4.00 \mathrm{m}, y=0 ?\) (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8 ). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 \(\mathrm{km} / \mathrm{s} .\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzchild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

Deimos, a moon of Mars, is about 12 \(\mathrm{km}\) in diameter with mass \(2.0 \times 10^{15} \mathrm{kg} .\) Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! (a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed? (b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed baseball game?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electricity

Read Explanation

Electric Charge Field and Potential

Read Explanation

Famous Physicists

Read Explanation

Magnetism

Read Explanation

Oscillations

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.