Chapter 20, Problem 34PQ

(a)

To determine

The fraction of the galaxy’s volume is occupied by stars.

Answer to Problem 34PQ

The fraction of the galaxy’s volume is occupied by stars is $\underset{\_}{1.2\times {10}^{-24}}$.

Explanation of Solution

Write the equation of fraction of galaxy’s volume is occupied by stars.

  $f=\frac{V_{star}}{V}$                                                                                                            (I)

Here, $f$ is the fraction, $V_{star}$ is the volume of star and $V$ is the total volume.

Write the expression for the volume of star.

  $V_{star}=N\left(\frac{4}{3}\pi {\left(0.63R_s\right)}^3\right)$                                                                                 (II)

Here, $N$ is the total number of star and $R_s$ is the radius of sun.

Rewrite the expression for the fraction.

  $f=\frac{N}{V}\left(\frac{4}{3}\pi {\left(0.63R_s\right)}^3\right)$                                                                                  (III)

Conclusion:

Substitute $0.098{\text{/pc}}^{\text{3}}$ for $\left(N/V\right)$ and $6.96\times {10}^8\text{m}$ for $R_s$ in equation (III) to find $f$.

  $\begin{array}{c}f=\left[\left(0.098{\text{/pc}}^{\text{3}}\right)\left(\frac{3.086\times {10}^{16}{\text{/m}}^{\text{3}}}{1{\text{/pc}}^{\text{3}}}\right)\right]\left(\frac{4}{3}\pi {\left(0.63\left(6.96\times {10}^8\text{m}\right)\right)}^3\right)\\ =\left(3.33\times {10}^{-51}{\text{m}}^{\text{-3}}\right)\left(\frac{4}{3}\pi {\left(0.63\left(6.96\times {10}^8\text{m}\right)\right)}^3\right)\\ =1.2\times {10}^{-24}\end{array}$

Thus, the fraction of the galaxy’s volume is occupied by stars is $\underset{\_}{1.2\times {10}^{-24}}$.

(b)

To determine

The mean free path of a star through galaxy.

Answer to Problem 34PQ

The mean free path of a star through galaxy is $\underset{\_}{8.81\times {10}^{31}\text{m}}$.

Explanation of Solution

Write the equation of mean free path.

  $\lambda =\frac{1}{\left(4\sqrt{2}\right)\pi {\left(0.63R_s\right)}^2\left(N/V\right)}$                                                                        (IV)

Here, $\lambda$ is the mean free path and $\left(N/V\right)$ is the number density.

Conclusion:

Substitute, $0.098{\text{/pc}}^{\text{3}}$ for $\left(N/V\right)$ and $6.96\times {10}^8\text{m}$ for $R_s$ in equation (IV) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{1}{\left(4\sqrt{2}\right)\pi {\left(0.63\left(6.96\times {10}^8\text{m}\right)\right)}^2\left[\left(0.098{\text{/pc}}^{\text{3}}\right)\left(\frac{3.086\times {10}^{16}{\text{/m}}^{\text{3}}}{1{\text{/pc}}^{\text{3}}}\right)\right]}\\ =\frac{1}{\left(4\sqrt{2}\right){\left(4.38\times {10}^8\text{m}\right)}^2\left(3.33\times {10}^{-51}{\text{m}}^{\text{-3}}\right)}\\ =8.81\times {10}^{31}\text{m}\end{array}$

Thus, the mean free path of a star through galaxy is $\underset{\_}{8.81\times {10}^{31}\text{m}}$.

(c)

To determine

The probability of collision.

Answer to Problem 34PQ

The probability of collision is $\underset{\_}{3.5\times {10}^{-13}}$.

Explanation of Solution

Write the equation for the probability.

  $p=\frac{{\lambda }_d}{\lambda }$                                                                                                              (V)

Here, $p$ is probability and ${\lambda }_d$ is the travelled distance.

Conclusion:

Substitute $1000\text{pc}$ for ${\lambda }_d$ and $8.81\times {10}^{31}\text{m}$ for $\lambda$ in equation (IV) to find $p$.

  $\begin{array}{c}p=\frac{\left(1000\text{pc}\right)\left(\frac{3.086\times {10}^{16}\text{m}}{1\text{pc}}\right)}{\left(8.81\times {10}^{31}\text{m}\right)}\\ =3.5\times {10}^{-13}\end{array}$

Therefore, the probability of collision is $\underset{\_}{3.5\times {10}^{-13}}$.