Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about $8.0 \times 10^{11}$ solar masses. A star orbiting on the galaxy’s periphery is about $6.0 \times 10^{4}$ light-years from its center.
(a) What should the orbital period of that star be?
(b) If its period is $6.0 \times 10^{7}$ instead, what is the mass of the galaxy? Such calculations are used to imply the existence of “dark matter” in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.

The orbital period of the star is ${8.45×10}^{15} s$ .
The mass of the galaxy for the given orbital period is ${3.04×10}^{43} kg$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of Galaxy

A system having millions or billions of stars, as well as gas and dust, that are gravitationally bound together.

Step 2: Calculating the star's orbital period

The connection that can be used to calculate a parent body's mass M from its satellites' orbits:

$\frac{r^{3}}{T^{2}} = \frac{G}{4 π^{2}} M$ ....................... ( 1 )

If a satellite's r and T are known, the parent's mass M can be estimated, or if r and M are known, T can be determined.

(a)

A light-year is the speed of light multiplied by a year so that it can be written as, $1 Lightyear = 3 \times 10^{8} \frac{m}{s} \times \frac{3600 s}{1 hour} \times \frac{24 hour}{1 day} \times \frac{365 days}{1 year} = 9.46 \times 10^{15} \frac{m}{year}$ .

The Orbital radius of the star $r = 6.00 \times 10^{4} Lightyears = 6.00 \times 10^{4} \times 9.46 \times 10^{15} m = 5.68 \times 10^{20} m$ .

The mass of the Milky Way galaxy or parent body is, $M = 8.0 \times 10^{11} solar mass = 8.0 \times 10^{11} \times 1.9 \times 10^{30} kg = 1.52 \times 10^{42} kg$ .

Putting these values in the equation

$\begin{matrix} \frac{{(5.68 \times 10^{20})}^{3}}{T^{2}} = \frac{6.673 \times 10^{- 11}}{4 π^{2}} \times (1.52 \times 10^{42}) \\ T^{2} = \frac{6.673 \times 10^{- 11}}{4 π^{2}} \times \frac{(1.52 \times 10^{42})}{{(5.68 \times 10^{20})}^{3}} \\ T = \sqrt{\frac{6.673 \times 10^{- 11}}{4 π^{2}} \times \frac{(1.52 \times 10^{42})}{{(5.68 \times 10^{20})}^{3}}} \\ T = 8.45 \times 10^{15} s \end{matrix}$

Hence, the orbital period of the star is ${8.45×10}^{15} s$ .

Step 3: Determining the mass of the galaxy for the given orbital period

(b)

The Orbital radius of the star, $r = 6.00 \times 10^{4} Lightyears = 6.00 \times 10^{4} \times 9.46 \times 10^{15} m = 5.68 \times 10^{20} m$ .
The orbital period of the star is, $T = 6.0 \times 10^{7} y = 6.0 \times 10^{7} y \times \frac{365 d}{1 y} \times \frac{24 hr}{1 d} \times \frac{3600 s}{1 hr} = 1.89 \times 10^{15} s$ .

Putting these values in the equation ( 1 ) and we get,

$\begin{matrix} \frac{{(5.68 \times 10^{20})}^{3}}{{(1.89 \times 10^{15})}^{2}} = \frac{6.673 \times 10^{- 11}}{4 π^{2}} \times M \\ M = \frac{{(5.68 \times 10^{20})}^{3}}{{(1.89 \times 10^{15})}^{2}} \times \frac{4 π^{2}}{6.673 \times 10^{- 11}} \\ M = 3.04 \times 10^{43} kg \end{matrix}$

Hence, the mass of the galaxy for the given orbital period is ${3.04×10}^{43} kg$ .

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Definition of Galaxy

Step 2: Calculating the star's orbital period

Step 3: Determining the mass of the galaxy for the given orbital period

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Definition of Galaxy

Step 2: Calculating the star's orbital period

Step 3: Determining the mass of the galaxy for the given orbital period

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