/*! 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 28 Assume a planet is a uniform sph... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{2} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

Short Answer

Expert verified
(a) \(R(\sqrt{2} - 1)\), (b) \(\frac{R}{2}\).

Step by step solution

01

Understand the given force at the surface

The gravitational force on the apple on the planet's surface is given by \[F_R = \frac{G M m}{R^2},\]where \(G\) is the gravitational constant, \(M\) is the planet's mass, \(m\) is the apple's mass, and \(R\) is the planet's radius.
02

Equation for gravitational force outside the planet

The gravitational force when the apple is at a distance \(r\) from the center of the planet (where \(r \geq R\)) is \[F = \frac{G M m}{r^2}.\] We want to find the distance \(d\) from the surface (where \(r = R + d\)) such that \[ \frac{G M m}{(R + d)^2} = \frac{1}{2} \cdot \frac{G M m}{R^2}. \]
03

Solve for distance from the surface (outside the planet)

Simplifying the equation from Step 2 \[\frac{R^2}{(R + d)^2} = \frac{1}{2}.\] Taking the square root of both sides gives \[\frac{R}{R + d} = \frac{1}{\sqrt{2}}.\] Solving for \(d\), we get \[R + d = R \sqrt{2},\]\[d = R(\sqrt{2} - 1).\]
04

Equation for gravitational force inside the planet

Inside the planet, using the shell theorem, the force is \[F = \frac{G M m r}{R^3}. \]We need to find \(d\) such that the force is half of \(F_R\), i.e., \[ \frac{G M m (R - d)}{R^3} = \frac{1}{2} \cdot \frac{G M m}{R^2}. \]
05

Solve for distance from the surface (inside the tunnel)

Simplifying the equation from Step 4, we have \[\frac{R - d}{R} = \frac{1}{2}.\]This simplifies to\[R - d = \frac{R}{2},\]so \[d = \frac{R}{2}.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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, proposed by Sir Isaac Newton. It describes the gravitational force exerted by a spherical shell of uniform density on a particle located outside or inside the shell. If a particle is positioned outside the shell, the shell acts as if all its mass were concentrated at its center, making calculations simpler. Conversely, if the particle is inside the shell, the gravitational forces cancel out, resulting in zero net force. This theorem aids significantly in gravitational calculations by reducing complex mass distributions into manageable mathematical expressions. Imagine it like simplifying a math problem using basic operations rather than tedious calculations.
Gravitational Constant
The gravitational constant, denoted as \(G\), is a key factor in Newton's Law of Universal Gravitation. It plays a crucial role in calculating the gravitational force between two masses. Measured as approximately \(6.674 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}\), this constant remains unchanged, ensuring that gravitational calculations are consistent and reliable. It acts as the linker in the equation \(F = \frac{G M m}{r^2}\), where \(F\) is the gravitational force, \(M\) and \(m\) are the masses, and \(r\) is the distance between the centers of the two masses. Knowing \(G\) is essential for solving physics problems related to gravitational interactions.
Uniform Sphere
A uniform sphere is a simple yet powerful concept in physics, particularly in gravitational studies. When a sphere is uniform, it means that its mass is evenly distributed throughout its volume. This assumption allows physicists to treat the sphere's mass as concentrated at its center for gravitational calculations regarding objects outside or far from the sphere. Understanding a uniform sphere is essential when applying the shell theorem, as it simplifies the mathematics involved in determining the gravitational force at various points in relation to the sphere. It plays a critical role in problems like the gravitational force inside and outside a planet, showing how a planet's mass distribution affects objects around it.
Physics Problem Solving
Solving physics problems, such as the one involving gravitational force on an apple, requires a systematic approach. Start by understanding the problem: what is given and what is required. Identify the relevant equations and concepts, such as those involving the gravitational constant or the shell theorem.
1. **Identify Key Concepts:** Recognize factors like gravitational force, distance, and mass.2. **Equation Formation:** Use fundamental equations, such as \(F = \frac{G M m}{r^2}\).3. **Apply Theorems:** Utilize helpful physics theorems, such as the shell theorem, to simplify calculations.4. **Solve Step-by-Step:** Simplify and solve the equations step-by-step to find solutions, such as the distance \(d\) where the force equals half of \(F_R\).5. **Verification:** Finally, always verify your answer to ensure it's reasonable in the context of the problem.This step-by-step approach helps in mastering concepts and applying them effectively to solve complex physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of \(160 \mathrm{~km}\) above Earth's surface? (b) What is the period of revolution?

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 \(6.37 \times 10^{6} \mathrm{~m}\).

(a) What is the escape speed on a spherical asteroid whose radius is \(500 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s} ?(\mathrm{c})\) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) ahove the surface?

A satellite is in elliptical orbit with a period of \(8.00 \times 10^{4} \mathrm{~s}\) about a planet of mass \(7.00 \times 10^{24} \mathrm{~kg} .\) At aphelion, at radius \(4.5 \times\) \(10^{7} \mathrm{~m}\), the satellite's angular speed is \(7.158 \times 10^{-5} \mathrm{rad} / \mathrm{s}\). What is its angular speed at perihelion?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.