Chapter 20, Problem 62Q

(a)

To determine

The brightness of the star Betelgeuse in terms of a fraction of the Sun’s brightness. It is given that Betelgeuse is transformed from a red supergiant to a Type II supernova at the distance of 425 ly from Earth.

Answer to Problem 62Q

Solution:

$7.1\times {10}^{-7}{\text{b}}_{\odot }$

Explanation of Solution

Given data:

The distance of the star from Earth is $425\text{ly}$.

Formula used:

The expression for apparent magnitude of a supernova is,

$m=M+5\log d-5$

Here, $m$ is the apparent magnitude, $M$ is the absolute magnitude of Type II supernova and d is the distance of the star from Earth in parsecs.

The expression for ratio of brightness of two objects is,

$\frac{b_1}{b_2}={10}^{\frac{(m_1-m_2)}{2.5}}$

Here, $b_1\text{and}{b}_2$ are the brightness and $m_1\text{and}{m}_2$ are the apparent magnitudes of the objects.

Explanation:

Convert the distance from light years to parsec as follows:

$1\text{ly}\text{=}\text{0}\text{.3}\text{pc}$

Therefore, the provided distance of the star from Earth in parsecs is,

$\begin{array}{c}d=425\text{ ly}\left(\frac{0.3\text{ pc}}{1\text{ly}}\right)\\ =130\text{ pc}\end{array}$

Write the formula for apparent magnitude of Type II supernova.

$m=M+5\log d-5$

The absolute magnitude for Type II supernova is $-17$. Substitute $-17$ for $m$ and 130 for $d$.

$\begin{array}{c}m=-17+5\log (130)-5\\ =-11.4\end{array}$

The apparent magnitude of the Sun is $-26.8$.

Write the expression for the brightness ratio of Betelgeuse and Sun.

$\frac{b}{b_S}={10}^{\frac{(m-m_S)}{2.5}}$

Here, the subscript S refers to the corresponding quantities for the Sun and b is the brightness of Betelgeuse.

Substitute $-26.8$ for $m_S$ and $-11.4$ for m.

$\begin{array}{c}\frac{b_B}{b_S}={10}^{\frac{-11.4-\left(-26.8\right)}{2.5}}\\ ={10}^{\frac{-15.4}{2.5}}\\ b_B=7.1\times {10}^{-7}{\text{b}}_{\odot }\end{array}$

Conclusion:

So, the supernova is $7.1\times {10}^{-7}$ times brighter than the Sun.

(b)

To determine

The comparison between the brightness of the supernova and that of Venus. It is given that it is transformed from a red supergiant to Type II supernova at the distance of 425 ly from Earth and the brightness of Venus is ${10}^{-9}{\text{b}}_{\odot }$.

Answer to Problem 62Q

Solution:

The ratio of the brightness of the supernova to that of Venus is 710.

Explanation of Solution

Given data:

The brightness of Venus is ${10}^{-9}{\text{b}}_{\odot }$.

The distance of the star from Earth is $425\text{ly}$.

Formula used:

The expression for apparent magnitude of a supernova is,

$m=M+5\log d-5$

Here, $m$ is the apparent magnitude, $M$ is the absolute magnitude of Type II supernova and d is the distance of the star from Earth in parsecs.

The expression for the ratio of brightness of two objects is,

$\frac{b_1}{b_2}={10}^{\frac{(m_1-m_2)}{2.5}}$

Here, $b_1\text{and}{b}_2$ are the brightness and $m_1\text{and}{m}_2$ are the apparent magnitudes of the objects.

Explanation:

Refer to part (a). The brightness of the star with respect to that of the Sun is $7.1\times {10}^{-7}{\text{b}}_{\odot }$.

In order to compare the brightness of the star with that of Venus, determine the ratio of their respective brightness (relative to the Sun), that is,

$ratio=\frac{b}{b_V}$

Substitute $7.1\times {10}^{-7}{\text{b}}_{\odot }$ for $b$ and ${10}^{-9}{\text{b}}_{\odot }$ for $b_V$.

$\begin{array}{c}ratio=\frac{7.1\times {10}^{-7}{\text{b}}_{\odot }}{{10}^{-9}{\text{b}}_{\odot }}\\ =710\end{array}$

Conclusion:

So, the supernova is 710 times brighter than Venus.