/*! 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 13 Does the Earth do any work on th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 13

Does the Earth do any work on the Moon as the Moon moves in its orbit?

Short Answer

Expert verified
Answer: No, the Earth does not do any work on the Moon as it moves in its orbit. This is because the gravitational force vector and the Moon's displacement vector are always perpendicular to each other, resulting in zero work done.

Step by step solution

01

Understand the concept of work

Work is done by a force when it displaces an object in the direction of the force. Mathematically, work (W) is calculated as the dot product of force (F) and displacement (d): W = F 鈰� d = |F| |d| cos(胃), where 胃 is the angle between the force vector and the displacement vector.
02

Determine the gravitational force between Earth and Moon

The gravitational force between the Earth and the Moon can be determined using Newton's law of universal gravitation: F = G * (m_E * m_M) / r^2, where F is the gravitational force, G is the gravitational constant (6.67430 x 10^{-11} m^3 kg^{-1} s^{-2}), m_E is the mass of the Earth (5.972 脳 10^{24} kg), m_M is the mass of the Moon (7.342 脳 10^{22} kg), and r is the average distance between the Earth and the Moon (3.844 脳 10^8 m).
03

Analyze the direction of the gravitational force and the Moon's orbit

The gravitational force between the Earth and the Moon always acts along the line joining their centers. The direction of this force is radially inward, pointing towards the center of the Earth. The Moon's orbit is approximately circular, and its motion is perpendicular to the line joining the Earth and the Moon. Therefore, the displacement vector of the Moon is always perpendicular to the gravitational force vector acting on it.
04

Calculate the angle between the gravitational force and the Moon's displacement

Since the gravitational force is always directed towards the center of the Earth and the Moon's displacement is perpendicular to it, the angle between the force vector and the displacement vector (胃) is 90掳.
05

Calculate the work done by the Earth on the Moon

Now we can calculate the work done (W) by the gravitational force on the Moon using the formula: W = F 鈰� d = |F| |d| cos(胃), Considering the angle between the force vector and the displacement vector is 90掳 and cos(90掳) = 0, the work done becomes: W = |F| |d| cos(90掳) = 0
06

Conclusion

Based on our calculations, the Earth does not do any work on the Moon as it moves in its orbit. This is because the gravitational force vector and the displacement vector are always perpendicular to each other (胃 = 90掳), resulting in zero work done.

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 91影视!

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 car of mass \(m=1250 \mathrm{~kg}\) is traveling at a speed of \(v_{0}=105 \mathrm{~km} / \mathrm{h}(29.2 \mathrm{~m} / \mathrm{s}) .\) Calculate the work that must be done by the brakes to completely stop the car.

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by \(5.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy, Calculate the original speeds of the two cars.

A force of \(5.00 \mathrm{~N}\) acts through a distance of \(12.0 \mathrm{~m}\) in the direction of the force. Find the work done.

A father pulls his son, whose mass is \(25.0 \mathrm{~kg}\) and who is sitting on a swing with ropes of length \(3.00 \mathrm{~m}\), backward until the ropes make an angle of \(33.6^{\circ}\) with respect to the vertical. He then releases his son from rest. What is the speed of the son at the bottom of the swinging motion?

How much work is done against gravity in lifting a \(6.00-\mathrm{kg}\) weight through a distance of \(20.0 \mathrm{~cm} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Space Physics

Read Explanation

Electromagnetism

Read Explanation

Wave Optics

Read Explanation

Energy Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation
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.