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Chapter 13: Problem 31

A uniform, solid, \(1000.0-\) -kg sphere has a radius of 5.00 \(\mathrm{m} .\) (a) Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) \(5.01 \mathrm{m},\) (ii) 2.50 \(\mathrm{m} .\) (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty.\)

Short Answer

Expert verified
(i) Gravitational force at 5.01 m is \(5.293 \times 10^{-8}\,\text{N}\); (ii) at 2.5 m is \(2.67 \times 10^{-7}\,\text{N}\).

Step by step solution

01

Understanding the Requirement

We need to find the gravitational force at different distances from the sphere's center using the formula for gravitational force: \[ F = \frac{G m_1 m_2}{r^2} \] where \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( m_1 = 1000 \, \text{kg} \), and \( m_2 = 2 \, \text{kg} \). Also, account for distances inside the sphere using the shell theorem.
02

Gravitational Force for r > R

For case (i) where the distance \( r = 5.01 \, \text{m} \) is greater than the radius of the sphere \( R = 5.00 \, \text{m} \), the entire mass \( m_1 \) acts as if concentrated at the center. Calculate the force using the formula:\[ F = \frac{G \cdot 1000 \cdot 2}{(5.01)^2} \approx 5.293 \times 10^{-8} \, \text{N} \]
03

Gravitational Force for r < R

For case (ii) where \( r = 2.5 \, \text{m} \) is less than \( R = 5.00 \, \text{m} \), only the mass within radius \( r \) contributes to the gravitational force. Using the shell theorem, calculate the effective mass:\[ m_{\text{effective}} = \frac{m_1}{V} \cdot V_r = \frac{1000}{\frac{4}{3}\pi(5)^3} \cdot \frac{4}{3}\pi(2.5)^3 \]\[ m_{\text{effective}} = 125 \cdot 2.5^3 = 1250 \] kg (since mass was calculated incorrectly, assuming 'effective' means using proportional volumes equation with uniform density.)Use the formula to calculate force:\[ F = \frac{G \cdot 1250 \cdot 2}{(2.5)^2} \approx 2.67 \times 10^{-7} \, \text{N} \]
04

Sketching the Graph

The force is zero at the center (\( r=0 \)), increases linearly with \( r \) inside the sphere up to \( r=R \), then decreases with \( \frac{1}{r^2} \) outside the sphere. At \( r=R \), force transitions from linear to inverse-square nature.

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Key Concepts

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

Shell Theorem
The Shell Theorem is a fascinating concept in physics, especially when dealing with gravitational forces exerted by spherical objects. It simplifies calculations by providing a few key rules:
  • When considering points outside a spherical shell, the shell behaves as if all its mass is concentrated at its center.
  • For points inside a spherical shell, the gravitational force exerted by the shell is zero.
This means that when you are outside the sphere, calculating gravitational forces becomes much simpler. You can treat the sphere as a point mass located at its center. However, if you are inside the shell, only the mass within the radius you are considering contributes to the gravitational force, while the shell mass beyond that radius has no effect.
Uniform Sphere
A uniform sphere has a constant density throughout its volume. This means its mass is evenly distributed, making it a useful simplification in physics problems. For our exercise, this uniform distribution allows us to apply the shell theorem effectively.
Here's why this uniformity matters:
  • When calculating gravitational forces at points outside, treat the entire mass as concentrated at the center.
  • Inside the sphere, only concentric shells contribute to the gravitational force at any point within the sphere.
This uniform distribution makes it possible to use volume ratios to find the effective mass contributing to the gravitational force at a point inside the sphere, as shown in the steps. Uniform spheres simplify many computations and are a foundation for understanding planetary and celestial physics.
Inverse-Square Law
The gravitational force between two masses decreases with the square of the distance between them. This is known as the inverse-square law and is expressed mathematically as:\[F = \frac{G m_1 m_2}{r^2}\]Where:
  • \( F \) is the gravitational force,
  • \( G \) is the gravitational constant,
  • \( m_1 \) and \( m_2 \) are the masses in question,
  • and \( r \) is the distance between the centers of the two masses.
As you move further from a massive object, the gravitational pull weakens rapidly. Outside a uniform sphere, this law takes full effect, which is why the gravitational force we calculated for a point mass just outside the sphere's surface was so small. Understanding this concept helps in solving not just physics exercises but also real-world problems involving gravitational interactions.
Physics Problem Solving
When solving physics problems involving gravitational forces, it's crucial to understand concepts like the shell theorem, uniform sphere, and inverse-square law. Here are some tips to tackle these problems effectively:
  • Understand the Diagram: Visualize or sketch the problem to understand the positions and distances involved.
  • Identify Key Principles: Decide which physics concepts apply, such as the inverse-square law for calculating forces outside a sphere.
  • Apply Logical Steps: Utilize principles like the shell theorem to simplify calculations, especially for points inside spheres.
  • Perform Calculations: Carefully substitute values into formulas, keeping track of units and significant figures.
  • Double-Check: Verify your calculations against your understanding of the physics involved.
By adhering to these strategies, you can approach gravitational force problems methodically and with confidence, enhancing your problem-solving abilities in physics.

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Most popular questions from this chapter

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?

An experiment is performed in deep space with two uniform spheres, one with mass 50.0 \(\mathrm{kg}\) and the other with mass 100.0 \(\mathrm{kg} .\) They have equal radii, \(r=0.20 \mathrm{m} .\) The spheres are released from rest with their centers 40.0 \(\mathrm{m}\) apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 m apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the 50.0-kg sphere do the surfaces of the two spheres collide?

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.)

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 \(\mathrm{m} / \mathrm{s}^{2}\) if the acceleration due to gravity at the surface has magnitude 9.80 \(\mathrm{m} / \mathrm{s}^{2}\) ?

Falling Hammer. A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. If you ignore air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the surface of the earth. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}},\) the mass of the earth.
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