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Expert-verified Found in: Page 224 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Astronomical observations of our Milky Way galaxy indicate that it has a mass of about ${\mathbf{8}}{.}{\mathbf{0}}{×}{{\mathbf{10}}}^{11}$ solar masses. A star orbiting on the galaxy’s periphery is about ${\mathbf{6}}{.}{\mathbf{0}}{×}{{\mathbf{10}}}^{4}$ light-years from its center.(a) What should the orbital period of that star be? (b) If its period is ${\mathbf{6}}{.}{\mathbf{0}}{×}{{\mathbf{10}}}^{7}$ 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.

1. The orbital period of the star is ${\text{8.45×10}}^{\text{15}}\text{ s}$.
2. The mass of the galaxy for the given orbital period is ${\text{3.04×10}}^{\text{43}}\text{ kg}$.
See the step by step solution

## 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:

$\frac{{r}^{3}}{{T}^{2}}=\frac{G}{4{\pi }^{2}}M$ ....................... ( 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, $1\text{ Lightyear}=3×{10}^{8}\text{ }\frac{\text{m}}{\text{s}}×\frac{3600\text{ s}}{1\text{ hour}}×\frac{24\text{ hour}}{1\text{ day}}×\frac{365\text{ days}}{1\text{ year}}=9.46×{10}^{15}\text{ }\frac{\text{m}}{\text{year}}$.

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

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

Putting these values in the equation

$\begin{array}{c}\frac{{\left(5.68×{10}^{20}\right)}^{3}}{{T}^{2}}=\frac{6.673×{10}^{-11}}{4{\pi }^{2}}×\left(1.52×{10}^{42}\right)\\ {T}^{2}=\frac{6.673×{10}^{-11}}{4{\pi }^{2}}×\frac{\left(1.52×{10}^{42}\right)}{{\left(5.68×{10}^{20}\right)}^{3}}\\ T=\sqrt{\frac{6.673×{10}^{-11}}{4{\pi }^{2}}×\frac{\left(1.52×{10}^{42}\right)}{{\left(5.68×{10}^{20}\right)}^{3}}}\\ T=8.45×{10}^{15}\text{ s}\end{array}$

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

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

(b)

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

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

$\begin{array}{c}\frac{{\left(5.68×{10}^{20}\right)}^{3}}{{\left(1.89×{10}^{15}\right)}^{2}}=\frac{6.673×{10}^{-11}}{4{\pi }^{2}}×M\\ M=\frac{{\left(5.68×{10}^{20}\right)}^{3}}{{\left(1.89×{10}^{15}\right)}^{2}}×\frac{4{\pi }^{2}}{6.673×{10}^{-11}}\\ M=3.04×{10}^{43}\text{ kg}\end{array}$

Hence, the mass of the galaxy for the given orbital period is ${\text{3.04×10}}^{\text{43}}\text{ kg}$. ### Want to see more solutions like these? 