Chapter 12, Problem 6P

To determine

The lifetime of main-sequences of Star A compared to the sun.

The lifetime of main-sequences of Star B compared to the sun.

The luminous of main-sequences of Star A compared to the sun.

The luminous of main-sequences of Star B compared to the sun.

Answer to Problem 6P

The lifetime of main-sequences of Star A compared to the sun is $4\times {10}^{-6}\text{solar}\text{lifetimes}$.

The lifetime of main-sequences of Star B compared to the sun is $600\text{solar}\text{lifetimes}$.

The luminous of main-sequences of Star A is $4\times {10}^7\text{times more}$ luminous than the sun.

The luminous of main-sequences of Star B is $1\times {10}^{-4}\text{times}$ luminous as the sun.

Explanation of Solution

Write the expression for the stellar life expectancies of star A.

    $t_{*\text{A}}=\frac{1}{M_{*\text{A}}^{2.5}}$        (I)

Here, $t_{*\text{A}}$ is the stellar life expectancy of star A, $M_{*\text{A}}$ is the luminosity of the main-sequence star A.

Rewrite the above expression for luminosity of star A.

    $L_{*\text{A}}=\frac{M_{*\text{A}}}{t_{*\text{A}}}$        (II)

Here, $L_{*\text{A}}$ is luminous of main-sequences of Star A.

Write the expression for the stellar life expectancies of star B.

    $t_{*\text{B}}=\frac{1}{M_{*\text{B}}^{2.5}}$        (III)

Here, $t_{*\text{B}}$ is the stellar life expectancy of star B, $M_{*\text{B}}$ is the luminosity of the main-sequence star B.

Rewrite the above expression for luminosity of star B.

    $L_{*\text{B}}=\frac{M_{*\text{B}}}{t_{*\text{B}}}$        (IV)

Here, $L_{*\text{B}}$ is luminous of main-sequences of Star B.

Conclusion:

Substitute $150$ for $M_{*\text{A}}$ in (I) to find $t_{*\text{A}}$

    $\begin{array}{c}{t}_{*\text{A}}=\frac{1}{{\left(150\right)}^{2.5}}\\ =3.63\times {10}^{-6}\\ \simeq 4\times {10}^{-6}\text{solar}\text{lifetime}\end{array}$

Substitute $150$ for $M_{*\text{A}}$, $4\times {10}^{-6}\text{solar}\text{lifetime}$ for $t_{*\text{A}}$ in (II) to find $L_{*\text{A}}$.

    $\begin{array}{c}{L}_{*\text{A}}=\frac{150}{4\times {10}^{-6}}\\ =3.75\times {10}^7\\ \simeq 4\times {10}^7\text{times}\end{array}$

Substitute $0.08$ for $M_{*\text{B}}$ in (III) to find $t_{*\text{B}}$

    $\begin{array}{c}{t}_{*\text{B}}=\frac{1}{{\left(0.08\right)}^{2.5}}\\ =552.43\\ \simeq 600\text{solar}\text{lifetime}\end{array}$

Substitute $0.08$ for $M_{*\text{B}}$, $600\text{solar}\text{lifetime}$ for $t_{*\text{B}}$ in (IV) to find $L_{*\text{B}}$.

    $\begin{array}{c}{L}_{*\text{B}}=\frac{0.08}{600}\\ =1.33\times {10}^{-4}\\ \simeq 1\times {10}^{-4}\text{times}\end{array}$

Therefore, the lifetime of main-sequences of Star A compared to the sun is $4\times {10}^{-6}\text{solar}\text{lifetimes}$. The lifetime of main-sequences of Star A compared to the sun is $600\text{solar}\text{lifetimes}$. The luminous of main-sequences of Star A is $4\times {10}^7\text{times more}$ luminous than the sun.  The luminous of main-sequences of Star B is $1\times {10}^{-4}\text{times}$ luminous as the sun.