Chapter 22, Problem 46Q

(a)

To determine

The volume of the disk in cubic parsecs if the diameter of the disk is 50 kpc and thickness is 600 pc.

Answer to Problem 46Q

Solution:

The volume is $1.2\times {10}^{12}p{c}^3$.

Explanation of Solution

Given data:

The diameter of the disk is 50 kpc and its thickness is 600 pc.

Formula used:

The expression for volume of a disk is:

$V_G=\pi {\left(\frac{D}{2}\right)}^{2}T$

Here $V_G$ denotes the volume of disk, $D$ denotes the diameter of Sun and $T$ denotes the thickness of the disk.

Explanation:

Recall the formula for volume of a disk.

$V_G=\pi {\left(\frac{D}{2}\right)}^{2}T$

Substitute 50,000 pc for $D$ and 600 pc for $T$.

$\begin{array}{c}{V}_G=\pi {\left(\frac{50,000}{2}\text{ pc}\right)}^2\left(600\text{ pc}\right)\\ =1.2\times {10}^{12}{\text{ pc}}^3\end{array}$

Conclusion:

The volume of the disk is $1.2\times {10}^{12}{\text{ pc}}^3$.

(b)

To determine

The volume of a sphere of radius 300 pc orbiting the Sun.

Answer to Problem 46Q

Solution:

The volume is $1.1\times {10}^8{\text{ pc}}^3$.

Explanation of Solution

Given data:

The radius of the sphere is 300 pc.

Formula used:

The expression for volume of a sphere is:

$V_s=\frac{4}{3}\pi R^3$

Here $V_s$ denotes the sphere volume and $R$ denotes the sphere radius.

Explanation:

Recall the expression for volume of a sphere.

$V_s=\frac{4}{3}\pi R^3$

Substitute 300 pc for $R$.

$\begin{array}{c}{V}_s=\frac{4}{3}\pi {\left(300\text{ pc}\right)}^3\\ =1.1\times {10}^8{\text{ pc}}^3\end{array}$

Conclusion:

The volume of sphere is $1.1\times {10}^8{\text{ pc}}^3$.

(c)

To determine

The probability of a supernova within the radius of the Sun, that is, 300 pc, if supernovae occur randomly throughout the volume of the galaxy. Also, calculate the time for these explosions if supernovae occur three times in a century.

Answer to Problem 46Q

Solution:

The probability is $9.2\times {10}^{-5}$ and the time of occurrence of a supernova is 370,000 y.

Explanation of Solution

Given data:

The radius of the sun is $300\text{ pc}$.

Formula used:

The expression to find the number of 300 pc spheres is:

$N=\frac{V_G}{V_S}$

Here $N$ denotes the number of spheres, $V_G$ denotes the volume of disk and $V_s$ denotes the volume of sphere.

The expression for probability is:

$P=\frac{1}{N}$

Here $P$ denotes the probability.

Explanation:

The volume of the sphere of radius $300\text{ pc}$ from sub part (b) is equal to $1.1\times {10}^8{\text{ pc}}^3$ and the volume of disk from the sub part (a) is equal to $1.2\times {10}^{12}{\text{ pc}}^3$.

Recall the expression for number of spheres.

$N=\frac{V_G}{V_S}$

Substitute $1.2\times {10}^{12}p{c}^3$ for $V_G$ and $1.1\times {10}^{8}p{c}^3$ for $V_s$.

$\begin{array}{c}N=\frac{1.2\times {10}^{12}{\text{pc}}^3}{1.1\times {10}^8{\text{pc}}^3}\\ =1.09\times {10}^4\\ \simeq 11,000\end{array}$

Recall the expression for probability.

$P=\frac{1}{N}$

Substitute 11,000 for $N$.

$\begin{array}{c}P=\frac{1}{11,000}\\ =9.2\times {10}^{-5}\end{array}$

If there are three supernovae then their occurrence will be:

$\begin{array}{l}=\frac{11,000}{3/1000\text{ y}}\\ =3.67\times {10}^5\text{ y}\\ \text{=3,67,000 y}\end{array}$

Conclusion:

The probability of a supernova within 300 pc is $9.2\times {10}^{-5}$ and the expected time of occurrence of a supernova is about once in 370,000 y.