/*! 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 47 The earth orbits the sun once a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The earth orbits the sun once a year \(\left(3.16 \times 10^{7}\right.\) s) in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{m} .\) With respect to the sun, determine (a) the angular speed of the earth, (b) the tangential speed of the earth, and (c) the magnitude and direction of the earth's centripetal acceleration.

Short Answer

Expert verified
(a) \(1.99 \times 10^{-7}\) rad/s; (b) \(2.98 \times 10^4\) m/s; (c) \(0.00594\) m/s² towards the Sun.

Step by step solution

01

Calculate the Angular Speed

To find the angular speed \( \omega \) of the Earth, use the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of the Earth's orbit. The period \( T \) is given as \( 3.16 \times 10^7 \) seconds. Thus, \( \omega = \frac{2\pi}{3.16 \times 10^7} \approx 1.99 \times 10^{-7} \text{ rad/s}. \)
02

Calculate the Tangential Speed

The tangential speed \( v \) can be found using the formula \( v = \omega r \), where \( r = 1.50 \times 10^{11} \text{ m} \) is the radius of the orbit. Given \( \omega \approx 1.99 \times 10^{-7} \text{ rad/s} \), \( v = 1.99 \times 10^{-7} \times 1.50 \times 10^{11} \approx 2.98 \times 10^4 \text{ m/s}. \)
03

Calculate the Centripetal Acceleration

The centripetal acceleration \( a_c \) is given by \( a_c = \frac{v^2}{r} \). Substitute \( v = 2.98 \times 10^4 \text{ m/s} \) and \( r = 1.50 \times 10^{11} \text{ m} \) into the expression: \( a_c = \frac{(2.98 \times 10^4)^2}{1.50 \times 10^{11}} \approx 0.00594 \text{ m/s}^2 \). This acceleration is directed towards the center of the circular orbit, which is the Sun.

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.

Angular Speed
Angular speed, often symbolized by \( \omega \), is a measure of how quickly an object rotates or revolves relative to a point or axis. In the context of our planet's orbit around the sun, angular speed describes how fast the Earth travels through its orbit in radians per unit time. To calculate it, you need to know the full circular orbit time, which for Earth is its orbital period around the sun. This is typically one year, or approximately \( 3.16 \times 10^7 \) seconds.
The formula used to find angular speed is:
  • \( \omega = \frac{2\pi}{T} \)
Here, \( T \) is the orbital period. The \( 2\pi \) term comes from the fact that a full orbit represents a complete circle, which measures \( 2\pi \) radians. Plugging in Earth's period, the angular speed turns out to be around \( 1.99 \times 10^{-7} \text{ rad/s} \). This number gives a sense of how much of its orbit Earth covers in a short amount of time.
Tangential Speed
Tangential speed is essentially the linear speed along the edge of a circular path. It measures how fast the Earth is moving in its orbit as if you were moving along the orbit’s perimeter. This speed can be envisioned as how quickly you would be traveling if you were riding on the edge of Earth as it revolved around the sun.
To calculate tangential speed (
  • \( v \)
), you use the formula:
  • \( v = \omega r \)
Where \( \omega \) is the angular speed, and \( r \) is the radius of Earth's orbit, given as \( 1.50 \times 10^{11} \text{ m} \). By substituting the values, we find that Earth's tangential speed is approximately \( 2.98 \times 10^4 \text{ m/s} \). This immense speed is what allows Earth to maintain its stable orbit around the sun and not veer off into space.
Centripetal Acceleration
Centripetal acceleration is the acceleration that occurs when an object moves along a circular path. It is directed towards the center around which the object is moving. For Earth, this center is the Sun. Regardless of the consistent speed Earth maintains, acceleration is always present due to the continuous change in direction of Earth's travel path.
The formula for calculating centripetal acceleration \( a_c \) is:
  • \( a_c = \frac{v^2}{r} \)
Here, \( v \) is the tangential speed, and \( r \) is the radius of the orbit. By substituting Earth's tangential speed (\( 2.98 \times 10^4 \text{ m/s} \)) and the radius of orbit (\( 1.50 \times 10^{11} \text{ m} \)), the centripetal acceleration turns out to be approximately \( 0.00594 \text{ m/s}^2 \).
This acceleration is not felt as 'pushing' because it is a fundamental aspect of circular motion. It keeps Earth in orbit, always pulling it inward toward the Sun to balance the outward linear momentum of Earth's travel in space.

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

The initial angular velocity and the angular acceleration of four rotating objects at the same instant in time are listed in the table that follows. For each of the objects (a), (b), (c), and (d), determine the fi nal angular speed after an elapsed time of 2.0 s. $$ \begin{array}{lcc} & \begin{array}{c} \text { Initial angular } \\ \text { velocity } \omega_{0} \end{array} & \begin{array}{c} \text { Angular } \\ \text { acceleration } \alpha \end{array} \\ \text { (a) } & +12 \mathrm{rad} / \mathrm{s} & +3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (b) } & +12 \mathrm{rad} / \mathrm{s} & -3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (c) } & -12 \mathrm{rad} / \mathrm{s} & +3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (d) } & -12 \mathrm{rad} / \mathrm{s} & -3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \end{array} $$

The drawing shows a device that can be used to measure the speed of a bullet. The device consists of two rotating disks, separated by a distance of \(d=0.850 \mathrm{m},\) and rotating with an angular speed of \(95.0 \mathrm{rad} / \mathrm{s} .\) The bullet first passes through the left disk and then through the right disk. It is found that the angular displacement between the two bullet holes is \(\theta=0.240\) rad. From these data, determine the speed of the bullet.

Energy of a Bullet Dissipated by Plywood. As part of a criminal investigation, you need to determine how much of a bullet's energy is dissipated by a 0.500 -inch piece of plywood. You construct a device that consists of three disks that are separated by a distance \(d=0.950 \mathrm{m}\) and rotate on a common axis. The bullet is fired through the first disk (a few inches above its center), which is composed of a light plastic that has a negligible effect on the speed of the bullet. The bullet then passes through the second disk, which is composed of 0.500-inch plywood. Finally, the bullet strikes the third disk, where it becomes embedded. The disks rotate with an angular velocity of \(\omega=92.0 \mathrm{rad} / \mathrm{s} .\) The angular displacement between holes in the first and second disks is \(\Delta \theta_{12}=0.255\) rad, and the angular displacement between the holes in the second and third disks is \(\Delta \theta_{23}=0.273\) rad. If the mass of the bullet is \(15.0 \mathrm{g}\), find \((\mathrm{a})\) the initial speed of the bullet and \((\mathrm{b})\) the energy dissipated by the 0.50-inch plywood.

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of \(1.20 \mathrm{m} / \mathrm{s}\) on its circular path. The rope holding the bucket unwinds without slipping on the barrel of the crank. Find the linear speed with which the bucket moves down the well.

A motorcycle accelerates uniformly from rest and reaches a linear speed of \(22.0 \mathrm{m} / \mathrm{s}\) in a time of \(9.00 \mathrm{s}\). The radius of each tire is \(0.280 \mathrm{m}\) What is the magnitude of the angular acceleration of each tire?

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.