Chapter 22, Problem 46Q

To determine

(a)

The volume of the disk in cubic parsecs if the diameter of the disk is about $50\text{ kpc}$ and the thickness is about $600\text{ pc}$.

Answer to Problem 46Q

Volume of the disk in cubic parsecs = $1.178625\times {10}^{12}{\text{ pc}}^3$

Explanation of Solution

Given data:

Diameter of the disk = $50\text{ kpc}$

Thickness of the disk = $600\text{ pc}$

Formula used:

$\text{Volume of a disk}=\pi r^{2}h$

Calculation:

$\begin{array}{c}\text{Volume of the galactic disk}=\pi r^{2}h\\ =\frac{22}{7}\times {\left(\frac{50}{2}\times {10}^3\right)}^2\times 600\\ =3.141\times {\left(25\times {10}^3\right)}^2\times 600\\ =3.141\times 625\times {10}^6\times 600\\ =3.143\times 625\times {10}^6\times 600\\ =1178625\times {10}^6{\text{ pc}}^3\\ =1.178625\times {10}^{12}{\text{ pc}}^3\end{array}$

Conclusion:

The volume of the disk in cubic parsecs = $1.178625\times {10}^{12}{\text{ pc}}^3$

To determine

(b)

The volume in cubic parsecs of a sphere with a radius of $300\text{ pc}$ centered on the sun.

Answer to Problem 46Q

Volume in cubic parsecs of a sphere with a radius of $300\text{ pc}$ = $1.141428571\times {10}^8{\text{ pc}}^3$

Explanation of Solution

Given data:

Radius of the sphere centered on the sun = $300\text{ pc}$

Formula used:

$\text{Volume of a sphere}=\frac{4}{3}\pi r^3$

Calculation:

$\begin{array}{c}\text{Volume of a sphere}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\times \frac{22}{7}\times {\left(300\right)}^3\\ =\frac{88}{21}\times 27000000\\ =114142857.1{\text{ pc}}^3\\ =1.141428571\times {10}^8{\text{ pc}}^3\end{array}$

Conclusion:

Volume in cubic parsecs of a sphere with a radius of $300\text{ pc}$ = $1.141428571\times {10}^8{\text{ pc}}^3$

To determine

(c)

The probability of a supernova occurring within $300\text{ pc}$ of the sun if supernovae occur randomly throughout the volume of the galaxy.

The probability for one should expect to see supernovae within 300 pc of the Sun given that there are about 3 supernovae each century in our galaxy

Answer to Problem 46Q

Probability of a supernova occurring within $300\text{ pc}$ of the sun = $\text{9}\text{.684}\times {10}^{-3}\%$

Time on average per century one should expect to see supernovae within 300 pc of the sun = $3.228\times {10}^{-3}\%$

Explanation of Solution

Given data:

The radius of the sphere centered on the sun = $300\text{ pc}$

Number of supernovas occurring for a single century = $3$

Formula used:

$\text{Probability of a event occurring in an given area}=\frac{\text{ given area}}{\text{total area}}\times 100\%$

Calculation:

The probability that a supernova occurs within 300 pc of the Sun

$\begin{array}{l}\text{= }\frac{1.141428571\times {10}^8{\text{ pc}}^3}{1.178625\times {10}^{12}{\text{ pc}}^3}\times 100\%\\ \text{= 9}\text{.684}\times {10}^{-5}\times 100\%\\ \text{= 9}\text{.684}\times {10}^{-3}\%\end{array}$

Time on average per century, one should expect to see a supernova within 300 pc of the Sun $\text{= }\frac{1}{3}\times \text{9}\text{.684}\times {10}^{-3}\%\text{= }3.228\times {10}^{-3}\%$

Conclusion:

Probability of a supernova occurring within $300\text{ pc}$ of the sun = $\text{9}\text{.684}\times {10}^{-3}\%$

Time on average per century one should expect to see a supernova within 300 pc of the sun = $3.228\times {10}^{-3}\%$