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Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is \(2.2 \times 10^{20} \mathrm{~m},\) and the angular speed of the sun is \(1.2 \times 10^{-15} \mathrm{rad} / \mathrm{s}\). (a) What is the tangential speed of the sun? (b) How long (in years) does it take for the sun to make one revolution around the center?

Short Answer

Expert verified
(a) Tangential speed is \(2.64 \times 10^{5} \, \text{m/s}\). (b) It takes about \(1.66 \times 10^{8} \) years for one revolution.

Step by step solution

01

Identify given values

We are given the radius of the sun's orbit as \( r = 2.2 \times 10^{20} \, \text{m} \) and its angular speed \( \omega = 1.2 \times 10^{-15} \, \text{rad/s} \).
02

Find the tangential speed of the sun

The formula to find the tangential speed \( v_t \) is \( v_t = \omega \times r \). Substitute the given values:\[ v_t = (1.2 \times 10^{-15} \, \text{rad/s})(2.2 \times 10^{20} \, \text{m}) = 2.64 \times 10^{5} \, \text{m/s}. \]
03

Find the period of one revolution

The period \( T \) of one revolution is given by the relation \( T = \frac{2\pi}{\omega} \). Substitute the angular speed:\[ T = \frac{2\pi}{1.2 \times 10^{-15} \, \text{rad/s}} = 5.24 \times 10^{15} \, \text{s}. \]
04

Convert period from seconds to years

There are \( 60 \times 60 \times 24 \times 365 \) seconds in a year. Divide the period by the number of seconds in a year to convert:\[ T_{\text{years}} = \frac{5.24 \times 10^{15} \, \text{s}}{60 \times 60 \times 24 \times 365} \approx 1.66 \times 10^{8} \, \text{years}. \]

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Key Concepts

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

Tangential Speed
When we discuss the sun's motion around the Milky Way, one of the key terms is tangential speed. Tangential speed gives us an idea of how fast an object is moving along a circular path. In essence, it is the linear speed of an object moving along a circular trajectory. To calculate the tangential speed (\(v_t\)), we rely on the relation between the radius (\(r\)) of the path and the angular speed (\(\omega\)) of the object. The simple formula to determine this speed is:- \(v_t = \omega \times r\).For our sun, we substitute its given angular speed and radius into the formula. Doing so gives us the sun's tangential speed as approximately 264,000 meters per second (\(2.64 \times 10^5 \text{ m/s}\)). This means that as the sun orbits the galaxy, it travels an astonishing distance every second due to its massive circular path.
Angular Speed
Angular speed is a measure of how quickly an object moves around a circular path. It tells us how much angle—in radians—an object covers per second. In our solar context, the angular speed of the sun is given as:- \(\omega = 1.2 \times 10^{-15} \text{ rad/s}\).This might seem extremely small, but remember that the sun is moving around an enormous path. Angular speed helps us understand the rotational dynamic, providing insight into how large objects like stars, planets, or even galaxies perform their circular journeys.In these calculations:- A smaller angular speed indicates a slower rotation,- whereas a larger value indicates rapid rotation around the center.This angular speed, when combined with the orbit's radius, helps in finding that all-important tangential speed.
Revolution Period
The revolution period refers to the time it takes for an object to make one complete orbit around another object or point. For the sun, it indicates how long it takes to move once around the center of the Milky Way galaxy.To find the revolution period (\(T\)), we use the formula:- \(T = \frac{2\pi}{\omega}\).Using the sun's angular speed, we find the period of one full orbit to be around 5.24 trillion seconds (\(5.24 \times 10^{15} \text{ s}\)).Since this number is quite large, converting it into years gives us a more comprehensible figure—approximately 166 million years (\(1.66 \times 10^8 \text{ years}\)).In essence:- The revolution period helps understand long-term astronomical dynamics,- It highlights the vast scales of time involved when dealing with celestial movements.This further underscores the grandeur of cosmic cycles and the patience of celestial mechanics.

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Most popular questions from this chapter

Refer to Interactive Solution \(\underline{8.49}\) at in preparation for this problem. A car is traveling with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{~m} .\) If the car speeds up with a linear acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s},\) find the angular displacement of each wheel during this period.

Interactive LearningWare 8.1 at reviews the approach that is necessary for solving problems such as this one. A motorcyclist is traveling along a road and accelerates for \(4.50 \mathrm{~s}\) to pass another cyclist. The angular acceleration of each wheel is \(+6.70 \mathrm{rad} / \mathrm{s}^{2}\) and, just after passing, the angular velocity of each is \(+74.5 \mathrm{rad} / \mathrm{s}\), where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

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 of diameter \(4.50 \times 10^{-2} \mathrm{~m}\) is just passing by one of the rotating blades. What must be the minimum speed of the ball so that it will not be hit by the next blade?

A pitcher throws a curveball that reaches the catcher in \(0.60 \mathrm{~s}\). The ball curves because it is spinning at an average angular velocity of 330 rev \(/ \mathrm{min}\) (assumed constant) on its way to the catcher's mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher?

An auto race is held on a circular track. A car completes one lap in a time of \(18.9 \mathrm{~s},\) with an average tangential speed of \(42.6 \mathrm{~m} / \mathrm{s}\). Find (a) the average angular speed and (b) the radius of the track.

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