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Chapter 8: Problem 47

The earth orbits the sun once a year \(\left(3.16 \times 10^{7} \mathrm{~s}\right)\) 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 \((\mathbf{c})\) the magnitude and direction of the earth's centripetal acceleration.

Short Answer

Expert verified
(a) \(1.99 \times 10^{-7}\text{ rad/s}\); (b) \(2.98 \times 10^4\text{ m/s}\); (c) \(5.93 \times 10^{-3}\text{ m/s}^2\) towards the sun.

Step by step solution

01

Understanding the Angular Speed

The angular speed \( \omega \) is defined as the angle divided by the time period for one complete revolution. Since one complete revolution around the sun equals \( 2\pi \) radians, and given the time period \( T = 3.16 \times 10^7 \text{ s}\), the angular speed can be calculated using:\[ \omega = \frac{2\pi}{T} \]Substitute \( T = 3.16 \times 10^7 \text{ s} \):\[ \omega = \frac{2\pi}{3.16 \times 10^7} \approx 1.99 \times 10^{-7} \text{ rad/s} \]
02

Calculating the Tangential Speed

The tangential speed \( v \) of an object in circular motion is related to the angular speed by the equation:\[ v = r \cdot \omega \]where \( r = 1.50 \times 10^{11} \text{ m} \) is the radius of the orbit. Substitute the values:\[ v = 1.50 \times 10^{11} \cdot 1.99 \times 10^{-7} \approx 2.98 \times 10^4 \text{ m/s} \]
03

Finding the Centripetal Acceleration

The centripetal acceleration \( a_c \) can be found using the formula:\[ a_c = \frac{v^2}{r} \]Substitute \( v = 2.98 \times 10^4 \text{ m/s} \) and \( r = 1.50 \times 10^{11} \text{ m} \):\[ a_c = \frac{(2.98 \times 10^4)^2}{1.50 \times 10^{11}} \approx 5.93 \times 10^{-3} \text{ m/s}^2 \]The direction of the centripetal acceleration is towards the center of the orbit, i.e., towards the sun.

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Key Concepts

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

Angular Speed
Angular speed refers to how fast an object rotates or revolves relative to another point, in this case, the Earth revolving around the Sun. This is a significant element in studying circular motion. Angular speed is often represented by the symbol \( \omega \). It is measured in radians per second. Angular speed provides insight into how much of an angle an object apparently 'sweeps' over a given time period.

To calculate angular speed, consider the entire circle made by Earth's orbit. One complete circle is equivalent to an angle of \( 2\pi \) radians. Since the Earth takes \( 3.16 \times 10^7 \) seconds to complete one revolution, we use the formula:
\[ \omega = \frac{2\pi}{T} \]
By substituting the given time period \( T \), we find the angular speed of Earth to be approximately \( 1.99 \times 10^{-7} \) radians per second.
  • This shows how efficiently we can track and define speeds, often making calculations easier.
  • Angular speed does not change unless there is a change in the orbit or revolution time, making it quite predictable and consistent for stable orbits.
Tangential Speed
Tangential speed is the linear speed of an object traveling along the edge of a circular path. It signifies how fast the object is moving along its path, not to be mistaken with angular speed which is concerned with the path's angle.

The formula to calculate tangential speed \( v \) is:
\[ v = r \cdot \omega \]
Where \( r \) is the radius of the circle formed by the path. For the Earth's orbit, it's given as \(1.50 \times 10^{11}\) meters and \( \omega \) is the angular speed we previously calculated. By plugging in these values, we deduce the Earth's tangential speed to be about \( 2.98 \times 10^4 \) meters per second.
  • Tangential speed provides insights into how quickly positions change over a time period.
  • This speed enables us to understand the actual velocity of motion along the circular path as opposed to the rotational perspective.
Centripetal Acceleration
Centripetal acceleration is essential in maintaining circular motion and describes how the object speeds up or slows down towards the path's center. The term 'centripetal' translates to 'center-seeking'. It ensures that, despite moving in circular paths, objects remain on course.

To determine centripetal acceleration \( a_c \), we utilize:
\[ a_c = \frac{v^2}{r} \]
This involves the tangential speed \( v \) and the circle鈥檚 radius \( r \). By substituting \( v = 2.98 \times 10^4 \) m/s and \( r = 1.50 \times 10^{11} \) m, we find that \( a_c \) is about \( 5.93 \times 10^{-3} \) m/s虏 towards the Sun.
  • Centripetal acceleration keeps the Earth in its orbit around the Sun. Without it, the Earth would drift away in a straight line.
  • It's the invisible force acting towards the center in every circular scenario, maintaining the constant movement.

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Most popular questions from this chapter

A wind turbine is initially spinning at a constant angular speed. As the wind鈥檚 strength gradually increases, the turbine experiences a constant angular acceleration of \(0.140 \mathrm{rad} / \mathrm{s}^{2}\). After making 2870 revolutions, its angular speed is \(137 \mathrm{rad} / \mathrm{s}\) (a) What is the initial angularvelocity of the turbine? (b) How much time elapses while the turbine is speeding up?

A dragster starts from rest and accelerates down a track. Each tire has a radius of 0.320 m and rolls without slipping. At a distance of 384 m, the angular speed of the wheels is 288 rad/s. Determine (a) the linear speed of the dragster and (b) the magnitude of the angular acceleration of its wheels.

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An electric fan is running on HIGH. After the LOW button is pressed, the angular speed of the fan decreases to \(83.8 \mathrm{rad} / \mathrm{s}\) in \(1.75 \mathrm{~s}\). The deceleration is \(42.0 \mathrm{rad} / \mathrm{s}^{2}\). Determine the initial angular speed of the fan.

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of \(1.25 \mathrm{rad} / \mathrm{s}\). The opening between successive blades is equal to the width of a blade.A golf ball (diameter \(4.50 \times 10^{-2} \mathrm{~m}\)) has just reached the edge of one of the rotating blades (see the drawing). Ignoring the thickness of the blades, find the \(minimum\) linear speed with which the ball moves along the ground, such that the ball will not be hit by the next blade.
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