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Chapter 13: Problem 97

An object of mass \(m\) is initially held in place at radial distance \(r=3 R_{E}\) from the center of Earth, where \(R_{E}\) is the radius of Earth. Let \(M_{E}\) be the mass of Earth. A force is applied to the object to move it to a radial distance \(r=4 R_{E}\), where it again is held in place. Calculate the work done by the applied force during the move by integrating the force magnitude.

Short Answer

Expert verified
The work done is \( \frac{G M_E m}{12R_E} \).

Step by step solution

01

Understanding Gravitational Force

The gravitational force between two masses, such as Earth and the object, is given by Newton's law of gravitation as \( F = \frac{G M_E m}{r^2} \), where \( G \) is the universal gravitational constant, \( M_E \) is the mass of Earth, \( m \) is the mass of the object, and \( r \) is the distance from the center of Earth.
02

Setting Up the Integral for Work Done

The work done by a force as it moves an object from one point to another is the integral of the force over the path of the object. Here, integrate the gravitational force as the object moves from \( r = 3R_E \) to \( r = 4R_E \). Thus, the work done, \( W \), is \( W = \int_{3R_E}^{4R_E} \frac{G M_E m}{r^2} \, dr \).
03

Calculating the Integral

To solve \( W = \int_{3R_E}^{4R_E} \frac{G M_E m}{r^2} \, dr \), integrate with respect to \( r \). The integral of \( \frac{1}{r^2} \) is \( -\frac{1}{r} \). Thus, \( W = G M_E m \left[-\frac{1}{r}\right]_{3R_E}^{4R_E} \).
04

Evaluating the Integral

Substitute the limits into the evaluated antiderivative: \( W = G M_E m \left( -\frac{1}{4R_E} + \frac{1}{3R_E} \right) \). Simplify this to \( W = G M_E m \left( \frac{1}{3R_E} - \frac{1}{4R_E} \right) \).
05

Simplify the Expression

Combine the fractions: \( \frac{1}{3R_E} - \frac{1}{4R_E} = \frac{4 - 3}{12R_E} = \frac{1}{12R_E} \). Hence, the work done is \( W = \frac{G M_E m}{12R_E} \).

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental interaction that occurs between two masses. Imagine it as a pulling force that both objects exert on each other due to their mass.
  • This force is why objects fall towards Earth, and why planets orbit stars.
  • It's described by the equation: \[ F = \frac{G M_E m}{r^2} \]
  • Here, \(G\) stands for the universal gravitational constant, \(M_E\) is the mass of Earth, \(m\) is the mass of the object experiencing the force, and \(r\) is the distance between the centers of the two masses.

Gravitational force decreases with the square of the distance between the objects. So, as the object moves further away from Earth, the gravitational pull weakens.
Integral Calculus
Integral calculus is a mathematical tool used to calculate quantities like area, volume, and work, that are accumulated over a continuous range. In this context, we're interested in finding the work done by a force as an object moves over a distance.
  • In our exercise, we needed to calculate the work done by integrating the gravitational force as the object moved from one position to another.
  • The formula to find work through integration is: \[ W = \int F(r) \, dr \] where the function \(F(r)\) represents the force depending on radius \(r\).

This process involves breaking down the path into infinitesimally small segments, calculating the force for each segment, and then summing these tiny amounts of work to find the total work done.
Newton's Law of Gravitation
Newton's law of gravitation explains how every mass in the universe attracts every other mass.
  • The law is mathematically expressed as: \[ F = \frac{G M_1 M_2}{r^2} \]
  • This states that the force \(F\) between two objects with masses \(M_1\) and \(M_2\) is directly proportional to the product of their masses and inversely proportional to the square of the distance \(r\) between their centers.

This law was a breakthrough as it unified celestial and terrestrial mechanics, explaining both the motion of planets and the falling of an apple in the same framework. It helps us understand how gravitational attraction influences the movement of heavenly bodies and everyday objects.
Universal Gravitational Constant
The universal gravitational constant, denoted as \(G\), is a key component of the gravitational force equation. It appears as a constant factor in Newton's law of gravitation.
  • Its value is approximately \(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\).
  • It represents the strength of gravity in the universe.

Without \(G\), it would be impossible to calculate gravitational forces. It ensures that the formula for gravitational force gives consistent and correct results across different scenarios and scales. It's a small value, which tells us that gravity, compared to other fundamental forces, is relatively weak. However, due to its long-range nature and universality, it becomes significant when dealing with astronomical bodies.

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Most popular questions from this chapter

(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?

The Sun, which is \(2.2 \times 10^{20} \mathrm{~m}\) from the center of the Milky Way galaxy, revolves around that center once every \(2.5 \times 10^{8}\) years. Assuming each star in the Galaxy has a mass equal to the Sun's mass of \(2.0 \times 10^{30} \mathrm{~kg}\), the stars are distributed uniformly in a sphere about the galactic center, and the Sun is at the edge of that sphere, estimate the number of stars in the Galaxy.

The orbit of Earth around the Sun is almost circular: The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times 10^{8} \mathrm{~km}\) respectively. Determine the corresponding variations in (a) total energy, (b) gravitational potential energy, (c) kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and conservation of angular momentum.)

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of \(20 \mathrm{~km}\), what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

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?
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