/*! 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 1 Find the acceleration of the moo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon \(=3 \cdot 85 \times 10^{5} \mathrm{~km}\) and the time taken by the moon to complete one revolution around the earth \(=27 \cdot 3\) days.

Short Answer

Expert verified
The acceleration of the Moon with respect to the Earth is approximately \(2.75 \times 10^{-3} \text{ m/s}^2\).

Step by step solution

01

Convert distances and time

First, convert the distance between the Earth and Moon from kilometers to meters: \[ d = 3.85 \times 10^{5} \text{ km} \times 10^{3} \text{ m/km} \] \[ d = 3.85 \times 10^{8} \text{ m} \]Next, convert the period from days to seconds:\[ T = 27.3 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \] \[ T = 27.3 \times 86400 \text{ s} \] \[ T = 2.36 \times 10^{6} \text{ s} \]
02

Calculate the circumference of the Moon鈥檚 orbit

The Moon travels in a nearly circular orbit, so calculate the circumference:\[ C = 2\pi \times d \] \[ C = 2\pi \times 3.85 \times 10^8 \text{ m} \] \[ C \approx 2.42 \times 10^9 \text{ m} \]
03

Determine the Moon鈥檚 orbital velocity

The orbital velocity can be found by dividing the circumference by the period:\[ v = \frac{C}{T} \] \[ v = \frac{2.42 \times 10^9 \text{ m}}{2.36 \times 10^6 \text{ s}} \] \[ v \approx 1.03 \times 10^3 \text{ m/s} \]
04

Calculate centripetal acceleration

The centripetal acceleration of the Moon can be calculated using:\[ a = \frac{v^2}{d} \] \[ a = \frac{(1.03 \times 10^3 \text{ m/s})^2}{3.85 \times 10^8 \text{ m}} \] \[ a \approx 2.75 \times 10^{-3} \text{ m/s}^2 \]

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 91影视!

Key Concepts

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

Orbital Velocity
Orbital velocity is a critical term that describes the speed at which an object travels while orbiting another body under the influence of gravity. In simpler terms, it's the speed necessary to keep an object in motion along its orbit without falling into or escaping from the gravitational pull of the primary body it orbits.

To find the orbital velocity, you divide the total distance traveled in one complete orbit by the time it takes to complete that orbit. For example, in the Earth-Moon system, the Moon's orbital velocity is determined by its circular path length, or circumference, divided by the time for a full orbit.

This value reflects how fast the Moon needs to go to maintain its path around the Earth, which is approximately 1,030 meters per second as found in the original solution.
Distance Conversion
Converting distances is essential to ensure units are compatible for calculations. In scientific contexts, distances are often transformed into meters, the standard unit of distance.

For the Earth-Moon system exercise, the distance between them was given in kilometers. To convert kilometers to meters, multiply the distance by 1,000 (since 1 km = 1,000 m). This conversion is crucial when calculating the Moon鈥檚 orbital velocity and centripetal acceleration. In this example, the original distance of 385,000 km converts to 385 million meters.

Performing such conversions helps avoid errors and ensures that all components of the formulae align, providing accurate results.
Time Conversion
Just like distance, converting time into a consistent unit is vital, particularly when carrying out physics calculations. For the Moon's orbit calculations, time was originally provided in days and needed conversion to seconds.

The conversion process involves several steps: converting days to hours, hours to minutes, and finally, minutes to seconds. The equations used are:
  • 1 day = 24 hours
  • 1 hour = 60 minutes
  • 1 minute = 60 seconds
For the Moon's orbital period of 27.3 days, these conversions lead to a total of approximately 2.36 million seconds. This step ensures that the time component in calculations like orbital velocity and centripetal acceleration is correctly aligned with distance and speed measurements.
Earth-Moon System
The Earth-Moon system showcases a fascinating celestial dance as the Moon orbits Earth due to gravitational forces. This system illustrates several key physics concepts: orbital motion, gravitational interaction, and centripetal forces.

The distance between Earth and the Moon defines the scale of this orbit. This vast distance ensures that calculations done to determine speeds and accelerations are accurate, whether for scientific purposes or academic exercises.

The calculations from this system also exemplify the conversion of real-world astronomical data into mathematical models, making them more manageable for mathematical analysis and enhancing understanding of celestial mechanics.

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 wire of length \(2 \cdot 00 \mathrm{~m}\) is stretched to a tension of \(160 \mathrm{~N}\). If the fundamental frequency of vibration is \(100 \mathrm{~Hz}\), find its linear mass density.

A stone is fastened to one end of a string and is whirled in a vertical circle of radius \(R\). Find the minimum speed the stone can have at the highest point of the circle.

The equation of a standing wave, produced on a string fixed at both ends, is $$ y=(0 \cdot 4 \mathrm{~cm}) \sin \left[\left(0-314 \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(600 \pi \mathrm{s}^{-1}\right) t\right] $$ What could be the smallest length of the string?

A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius \(R\) (figure 7-E5). A smooth pulley of small radius is fastened to the table. Two masses \(m\) and \(2 m\) placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

A tuning fork of frequency \(440 \mathrm{~Hz}\) is attached to a long string of linear mass density \(0 \cdot 01 \mathrm{~kg} \mathrm{~m}^{-1}\) kept under a tension of \(49 \mathrm{~N}\). The fork produces transverse waves of amplitude \(0-50 \mathrm{~mm}\) on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Quantum Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Force

Read Explanation

Physical Quantities And Units

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.