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Milky Way Our Sun is \(2.3 \times 10^{4}\) ly (light-years) from the center of our Milky Way galaxy and is moving in a circle around the center at a speed of \(250 \mathrm{~km} / \mathrm{s}\). (a) How long does it take the Sun to make one revolution about the galactic center? (b) How many revolutions has the Sun completed since it was formed about \(4.5 \times 10^{9}\) years ago?

Short Answer

Expert verified
One revolution takes approximately 1.734 × 10^8 years. The Sun has completed around 25.95 revolutions.

Step by step solution

01

- Convert the distance to kilometers

First, convert the distance from light-years to kilometers. 1 light-year is approximately equal to 9.46 × 10^{12} km. Thus, the distance is: \[ 2.3 \times 10^{4} \text{ ly} × 9.46 \times 10^{12} \text{ km/ly} = 2.1738 \times 10^{17} \text{ km} \]
02

- Calculate the circumference of the orbit

The circumference of a circle is given by: \[ C = 2πr \] Substitute the radius (which is the distance from step 1): \[ C = 2π (2.1738 \times 10^{17} \text{ km}) \] \[ C ≈ 1.3669 \times 10^{18} \text{ km} \]
03

- Calculate the time for one revolution

Using the speed of the Sun (given), we can find the time it takes for one revolution: \[ t = \frac{C}{v} \] where v is the speed of the Sun (250 km/s): \[ t = \frac{1.3669 \times 10^{18} \text{ km}}{250 \text{ km/s}} \] \[ t ≈ 5.4676 \times 10^{15} \text{ s} \]
04

- Convert the time from seconds to years

Convert seconds into years using the conversion factor: 1 year ≈ 3.154 \times 10^7 seconds. \[ t = \frac{5.4676 \times 10^{15} \text{ s}}{3.154 \times 10^7 \text{ s/year}} \] \[ t ≈ 1.734 \times 10^8 \text{ years} \]
05

- Calculate the number of revolutions

Given that the Sun is approximately 4.5 × 10^9 years old. The number of revolutions the Sun has completed is: \[ \text{Number of Revolutions} = \frac{4.5 \times 10^9 \text{ years}}{1.734 \times 10^8 \text{ years/revolution}} \] \[ \text{Number of Revolutions} ≈ 25.95 \]

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is the study of the motions of celestial objects under the influence of gravity. It involves understanding how planets, stars, and other bodies move in space. An important formula in orbital mechanics is the circumference of a circle: \[ C = 2\fibigr(}\fibigr) \]. This helps us determine the distance traveled in an orbit.
Light-Year Conversion
A light-year is the distance light travels in one year. This is about 9.46 × 10^{12} km. Converting light-years to kilometers helps in understanding astronomical distances. For example, if our Sun is 2.3 × 10^{4} light-years away from the center of the Milky Way, we convert this distance to kilometers for practical calculations by multiplying with 9.46 × 10^{12}.
Revolution of the Sun
The Sun revolves around the center of the Milky Way galaxy. This journey takes approximately 1.734 × 10^{8} years. We can calculate this duration by finding the orbit's circumference and dividing it by the Sun's speed. This is crucial to understanding how long celestial bodies take to complete their orbits in the galaxy.
Galactic Orbit Calculations
To determine the number of revolutions the Sun has completed since its formation, we divide the Sun's age by the time it takes to complete one orbit. Given that the Sun is about 4.5 × 10^{9} years old and it takes approximately 1.734 × 10^{8} years for one revolution, we find the Sun has completed roughly 26 revolutions. This calculation helps appreciate the dynamic nature of our galaxy.

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Most popular questions from this chapter

Vinyl Record A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) rev/min. (a) What is its rotational speed in radians per second? What is the translational speed of a point on the record at the needle when the needle is (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

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Uniform Spherical Shell A uniform spherical shell of mass \(M\) and radius \(R\) rotates about a vertical axis on frictionless bearings (Fig. 11-39). A massless cord passes around the equator of the shell. over a pulley of rotational inertia \(I\) and radius \(r\), and is attached to a small object of mass \(m\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it falls a distance \(h\) from rest? Use energy considerations.

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