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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 (see Fig. E13.33). 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 \(F = G Mm \frac{x}{(x^2 + a^2)^{3/2}}\). When \(x \gg a\), this simplifies to the usual point mass force \(F \approx G \frac{Mm}{x^2}\).

Step by step solution

01

Understand the Problem

We need to calculate the gravitational force that a uniform sphere exerts on a thin ring. The system comprises a ring of mass \(M\) and radius \(a\), and a sphere of mass \(m\) and radius \(R\). The sphere's center is at a distance \(x\) from the ring's center, along a line perpendicular to the plane of the ring. We are given that \(x\) is larger than \(a\).
02

Use the Law of Gravitation

The gravitational force between two point masses \(m_1\) and \(m_2\) is given by Newton's law as \(F = G \frac{m_1 m_2}{r^2}\), where \(G\) is the gravitational constant and \(r\) is the distance between the masses. Since the ring is symmetric, the force between a small element of the ring and the sphere can be integrated.
03

Setup the Integration

Consider a small mass element \(dm\) on the ring. The ring can be divided into infinitesimally small mass segments \(dm = \frac{M}{2\pi a} \cdot d\theta\), where \(d\theta\) is a small angle segment on the ring. The distance from this element to the center of the sphere is \(\sqrt{x^2 + a^2}\).
04

Calculate the Gravitational Force

The differential gravitational force \(dF\) on \(dm\) due to the sphere is given by \(dF = G \frac{dm \cdot m}{x^2 + a^2}\). The components of \(dF\) perpendicular to \(x\) will cancel out due to symmetry, leaving only the components along the line joining the centers, which will be summed up.
05

Integrate to Find Total Force

Integrate \(dF_x = dF \cdot \frac{x}{\sqrt{x^2 + a^2}}\) over the entire ring (from \(0\) to \(2\pi\)), using symmetry to simplify. The integral simplifies to \(F = G Mm \frac{x}{(x^2 + a^2)^{3/2}}\).
06

Analyze the Limiting Case

When \(x\) is much larger than \(a\), \(x^2 + a^2 \approx x^2\). Substituting this into our expression for force gives \(F \approx G \frac{Mm}{x^2}\), the standard gravitational force between point masses \(M\) and \(m\) separated by distance \(x\).

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Key Concepts

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

Newton's law of gravitation
Newton's law of gravitation is a fundamental principle in physics. It states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula for gravitational force is:\[ F = G \frac{m_1 m_2}{r^2} \]where:- \( F \) is the gravitational force between two masses,- \( G \) is the gravitational constant, approximately equal to \( 6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2 \),- \( m_1 \) and \( m_2 \) are the masses of the objects,- \( r \) is the distance separating the centers of the two masses.This law helps us understand how objects influence each other gravitationally. In the context of the exercise, we apply this law by calculating the gravitational force a uniform sphere exerts on a thin ring. When the ring is symmetrical and the masses are distributed evenly, integrating these small gravitational forces of ring elements can help calculate the total force on the ring.
symmetry in physics
Symmetry in physics is a powerful concept that helps simplify problems and understand fundamental interactions. It refers to the idea that certain physical aspects of a system remain unchanged, even when transformations such as rotation or translation are applied. For the problem of the gravitational force between a thin ring and a sphere, symmetry plays a crucial role. We consider: - The ring is circular and symmetric about its center. - Any horizontal force component from one side of the ring is balanced out by an equal and opposite component from the other side. Consequently, only the components of the gravitational force that act along the line joining the sphere and the center of the ring do not cancel out. This aspect simplifies the integration process, as it allows us to consider only the perpendicular distance from the ring to the sphere's center.
integration in physics
Integration in physics is a tool used to add up infinite, infinitesimally small quantities to calculate quantities like area, volume, and total force. In our exercise, integration allows us to determine the total gravitational force a sphere exerts on a ring by summing up small force contributions from each segment of the ring.Here's how it works in our context:- The ring is considered as composed of tiny mass elements.- For each element, the gravitational force is calculated using Newton's law.- Due to symmetry, the net gravitational force contribution along certain axes cancels out, simplifying our calculations.- The integration is performed over the entire ring, which means summing the contributions of these small forces around a full circle.The formula to integrate over the ring is:\[ F = \int dF_x = \int G \frac{dm \cdot m}{x^2 + a^2} \cdot \frac{x}{\sqrt{x^2 + a^2}} \]This integration takes advantage of constants and symmetry to simplify into a manageable form. This work also helps in analyzing the limiting case when distance \(x\) is much larger than radius \(a\), showing the force behaves as it would between point masses.

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Most popular questions from this chapter

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2} .\) Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a 5.0 -kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

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 \(r \rightarrow \infty.\)

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?

A particle of mass 3\(m\) is located 1.00 \(\mathrm{m}\) from a particle of mass \(m\) (a) Where should you put a third mass \(M\) so that the net gravitational force on \(M\) due to the two masses is exactly zero? (b) Is the equilibrium of \(M\) at this point stable or unstable (i) for points along the line connecting \(m\) and \(3 m,\) and (ii) for points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and 3\(m ?\)

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{km}\) and the other at \(64,000 \mathrm{km} .\) Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

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