Explanation Formula used: The orbital velocity formula that relates the mass and the velocity of the orbiting object is as follows: M = r v 2 G Here, M be the mass of the Milky way, v be the orbital velocity, and r is the radius of the orbital in which the cloud is revolving, and G is the gravitational constant. Now, substitute 1000 km/s for v , 20 light days for r, and 6.67 × 10 − 11 N ⋅ m 2 ⋅ kg -2 for G in the equation M = r v 2 G . M = ( 20 light days ) ( 1000 km/s ) 2 6.67 × 10 − 11 N ⋅ m 2 ⋅ kg -2 = ( 20 light days ) ( 1000 km/s ) 2 6

Essential Cosmic Perspective, The, Books a la Carte Edition & Modified MasteringAstronomy with Pearson eText -- ValuePack Access Card -- for The Essential Cosmic Perspective Package

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ISBN: 9780133879216

Author: Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit

Publisher: PEARSON

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We may see thousands of stars in the dark sky. Our universe consists of billions of stars. Stars may appear tiny to us but they are huge balls of gasses. Sun is a star of average size. Some stars are even a thousand times larger than the sun. The stars d…

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It is an astronomical phenomenon. In this phenomenon, increase in wavelength with corresponding decrease in photon energy and frequency of radiation of light. It is the displacement of spectrum of any kind of astronomical object to the longer wavelengths…

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Chapter 15, Problem 47EAP

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The mass of the mass of central black hole.

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Chapter 15, Problem 46EAP

Chapter 15, Problem 48EAP

Chapter 15 Solutions

Essential Cosmic Perspective, The, Books a la Carte Edition & Modified MasteringAstronomy with Pearson eText -- ValuePack Access Card -- for The Essential Cosmic Perspective Package

Ch. 15 - Prob. 1VSC Ch. 15 - Prob. 2VSC Ch. 15 - Prob. 3VSC Ch. 15 - Prob. 4VSC Ch. 15 - Prob. 5VSC Ch. 15 - Prob. 6VSC Ch. 15 - Prob. 1EAP Ch. 15 - Prob. 2EAP Ch. 15 - 3. Describe and contrast stellar orbits in the...Ch. 15 - Prob. 4EAP

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