Problem 83 - University Physics with Modern Physics

Question

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 \(a\).

Step-by-Step Solution

Answer

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 1: 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 \(a\).

Step 2: 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.

Step 3: 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 \(\sqrt{x^2 + a^2}\).

Step 4: 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.

Step 5: 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 \(F = G Mm \frac{x}{(x^2 + a^2)^{3/2}}\).

Step 6: 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 \(x\).

Key Concepts

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.