/*! 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 83 An object in the shape of a thin... [FREE SOLUTION] | 91Ó°ÊÓ

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An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (Fig. 12.35 ). What is the gravitational force that the sphere exerts on the ring- shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\) .

Short Answer

Expert verified
The gravitational force is \( F = \frac{G M m}{x^2 + a^2} \) and approaches \( \frac{G M m}{x^2} \) when \( x \gg a \).

Step by step solution

01

Understanding the System

We have a thin ring of radius \(a\) and mass \(M\), and a uniform sphere with mass \(m\) placed at a distance \(x\) from the ring's center along a line perpendicular to its plane. We need to find the gravitational force the sphere exerts on the ring.
02

Gravitational Force Contribution from a Ring Element

Consider a small element of mass \(dm\) on the ring. The gravitational force \(dF\) on this element due to the sphere is given by Newton's law of universal gravitation:\[ dF = \frac{G \cdot dm \cdot m}{r^2} \]where \(G\) is the gravitational constant, and \(r\) is the distance between the sphere and the ring element, \(r = \sqrt{x^2 + a^2}\).
03

Expression for Total Gravitational Force

Integrate the expression for \(dF\) over the entire ring. Since the gravitational force along any direction that is not aligned with the line joining the center of the ring and the sphere cancels out due to symmetry:\[ F = \int dF = \int \frac{G \cdot dm \cdot m}{(x^2 + a^2)} = \frac{G \cdot m}{(x^2 + a^2)} \int dm \]Since \(\int dm = M\) (the total mass of the ring), obtain:\[ F = \frac{G \cdot M \cdot m}{x^2 + a^2} \]
04

Large Distance Simplification

When \(x\) is much larger than \(a\), we can approximate \(x^2 + a^2 \approx x^2\), which simplifies the force expression to:\[ F \approx \frac{G \cdot M \cdot m}{x^2} \]This is the standard gravitational force equation for two point masses, valid when the distance between them is much larger than their sizes.

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Key Concepts

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

Newton's Law of Universal Gravitation
Sir Isaac Newton introduced the Law of Universal Gravitation in the late 17th century. This groundbreaking concept describes how every mass attracts every other mass in the universe. The force of attraction is dependent on two primary factors:
  • The masses of the two objects involved.
  • The distance between the centers of the two objects.
Mathematically, this force is described as: \[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]where F is the gravitational force, \(m_1\) and \(m_2\) are the masses of the objects, \(r\) is the distance between their centers, and G is the gravitational constant. This law is pivotal in our understanding of gravity and is applied universally to a range of scenarios, from falling apples to the orbits of planets. In our exercise, we applied this law to assess the gravitational pull between a sphere and a ring.
Integration in Physics
Integration is a powerful mathematical tool used in physics to accumulate small quantities over a certain range to find a total effect. In our problem, we dealt with the total gravitational force exerted by a sphere on a ring. Here, integration helped us sum up the infinitesimal gravitational forces acting on each small mass element of the ring, denoted as \(dm\). Given the symmetry and continuous distribution of mass in the ring, the integration simplifies to:
  • The gravitational force contribution from a single element: \[dF = \frac{G \cdot dm \cdot m}{r^2}\]
  • Since each small force needs to be added up, the integral is \[F = \int dF = \int \frac{G \cdot dm \cdot m}{x^2 + a^2} = \frac{G \cdot m}{x^2 + a^2} \int dm\]
Integration in such problems ensures that we consider the entire geometry of the mass distribution, yielding an accurate result. It is particularly important in physics to obtain a comprehensive, net effect from continuous distributions.
Symmetry in Physics
Symmetry plays a vital role in simplifying complex physical problems. In physics, symmetry often allows us to make assumptions that streamline calculations and reveal deeper insights into the underlying principles. In the context of our ring and sphere problem, the symmetry of the ring ensures that the gravitational components perpendicular to the axis through the center cancel each other out.
This means we only need to consider forces acting along the line joining the sphere's center and the center of the ring. This reduces the complexity of our calculations. Key aspects of symmetry include:
  • Reduction in computational complexity by narrowing down directions where net forces are non-zero.
  • Providing a visual and conceptual understanding of how forces distribute and cancel out.
Recognizing symmetry helps in breaking down many physics problems, as it allows focusing on the essential dynamics without getting bogged down in unnecessary details.

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Most popular questions from this chapter

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.

One of the brightest comets of the 20 th century was Comet Hyakutake, which passed close to the sun in early 1996 . The orbital period of this comet is estimated to be about \(30,000\) years. Find the semi-major axis of this comet's orbit. Compare it to the average sun-Pluto distance and to the distance to Alpha Centauri, the nearest star to the sun, which is 4.3 light-years distant.

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{\mathrm{E}},\) radius \(R_{\mathrm{E}} )\) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\) . (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{km},\) much less than \(R_{\mathrm{E}}=6380 \mathrm{km} .\) You can ignore the kimetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement: "In terms of energy, low earth orbit is halfway to the edge of the universe."

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{100}\) 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 than the density of rock, which is one piece of evidence that Titania is made primarily of ice. \()\)

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 \(\mathrm{km}\) above the earth's surface; at the high point, or apogee, it is 4000 \(\mathrm{km}\) above the earth's surface. (a) What is the period of the spacecraft's orbit the earth's surface. (a) What is the period of the spacecraft's orbit? (b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?

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