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Two concentric spherical shells with uniformly distributed masses \(M_{1}\) and \(M_{2}\) are situated as shown in Fig. \(13-41\). Find the magnitude of the net gravitational force on a particle of mass \(m\), due to the shells, when the particle is located at radial distance (a) \(a\), (b) \(b\), and (c) \(c\).

Short Answer

Expert verified
(a) 0, (b) \( \frac{G M_1 m}{b^2} \), (c) \( \frac{G (M_1 + M_2) m}{c^2} \)

Step by step solution

01

Understanding the Problem

Two concentric spherical shells are given, with masses \( M_1 \) and \( M_2 \). We need to find the gravitational force on a particle of mass \( m \) at different radial distances \( a \), \( b \), and \( c \) from the center. We can use the shell theorem, which states that a spherical shell exerts no net gravitational force on a particle located inside the shell.
02

Calculate the Force at Distance a (Inside Inner Shell)

Since the particle at distance \( a \) is inside both shells, the shell theorem tells us that there is no net gravitational force from either shell. Hence, the gravitational force on the particle at distance \( a \) is zero: \[ F(a) = 0 \]
03

Calculate the Force at Distance b (Between the Shells)

At distance \( b \), the particle is outside the inner shell and inside the outer shell. The shell theorem tells us the outer shell exerts no force. The gravitational force is only due to the inner shell. Using Newton's law of gravitation:\[ F(b) = \frac{G M_1 m}{b^2} \] where \( G \) is the gravitational constant.
04

Calculate the Force at Distance c (Outside Both Shells)

At distance \( c \), the particle is outside both shells. The net gravitational force on the particle is the sum of forces due to the masses of both shells:\[ F(c) = \frac{G (M_1 + M_2) m}{c^2} \]

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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 and practical concept in physics, introduced by Isaac Newton. It simplifies complex gravitational problems involving spherical shells.
Put simply, the shell theorem states that a spherical shell of uniform mass exerts no gravitational force on a particle located inside it. This is because the symmetrical spread of mass cancels out gravitational forces within the shell.
When the particle is outside the shell, the gravitational force it experiences can be calculated as if the entire mass of the shell were concentrated at its center. This greatly simplifies calculations, making it unnecessary to account for every point on the shell.
  • Inside the shell: No net force acts on the particle.
  • Outside the shell: The force is calculated as if all mass were at the center.
This theorem is instrumental when analyzing gravitational forces in systems like planets or stars, which can be modeled as spherical shells.
Spherical Shells
When addressing gravitational forces, especially in problems involving celestial bodies or hollow objects, spherical shells are a common aspect we encounter.
A spherical shell is essentially a hollow sphere with a uniform distribution of mass across its surface. This uniformity is key to understanding the effects of gravitational forces exerted by the shell.
With the use of the shell theorem, we can predict the gravitational behavior of these shells in different scenarios often more easily than with irregular shapes.
  • Uniform distribution simplifies gravitational calculations.
  • Important for understanding planets, moons, and other celestial bodies.

Gravitational effects from a spherical shell differ depending on the position of the object experiencing the force, whether it's inside, on the surface, or outside the shell. This distinct pattern of force application aids in numerous physics problems.
Newton's Law of Gravitation
Newton's law of gravitation is a cornerstone principle that describes how two masses in the universe attract each other. Discovered by Sir Isaac Newton, it states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula is expressed as: \[ F = \frac{G M_1 M_2}{r^2} \]where:
  • \( F \) is the gravitational force,
  • \( M_1 \) and \( M_2 \) are the masses,
  • \( r \) is the distance between their centers,
  • \( G \) is the gravitational constant.

This law explains why we stay anchored to the Earth and why planets orbit the Sun. It applies universally, from falling apples to the expansive motions of galaxies. Understanding this law helps in calculating gravitational interactions in systems ranging from everyday objects to large astronomical bodies.

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Most popular questions from this chapter

A very early, simple satellite consisted of an inflated spherical aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). Suppose a meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0\) \(\mathrm{m}\) of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth. What multiple of Earth's radius \(R_{E}\) gives the radial distance a projectile reaches if (a) its initial speed is \(0.500\) of the escape speed from Earth and (b) its initial kinetic energy is \(0.500\) of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Moon effect. Some people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth-Moon (center-to-center) distance is \(3.82 \times 10^{8} \mathrm{~m}\) and Earth's radius is \(6.37 \times 10^{6} \mathrm{~m}\).

Three dimensions. Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A}\) Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 m_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A} .\) A fourth particle \(D\), with mass \(4.00 m_{A}\), is to be placed near the other particles. In terms of distance \(d\), at what (a) \(x\), (b) \(y\), and (c) \(z\) coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C\), and \(D\) is zero?

An object lying on Earth's equator is accelerated (a) toward the center of Earth because Earth rotates, (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit, and (c) toward the center of our galaxy because the Sun moves around the galactic center. For the latter, the period is \(2.5 \times 10^{8} \mathrm{y}\) and the radius is \(2.2 \times 10^{20} \mathrm{~m} .\) Calculate these three accelerations as multiples of \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

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