Chapter 17, Problem 25Q

(a)

To determine

The absolute magnitude of the star, HIP 72509, if the apparent magnitude of the star is $12.1$ and a parallax angle is $0.222\text{ arcsec}$.

Answer to Problem 25Q

Solution:

$13.831$

Explanation of Solution

Given data:

The apparent magnitude of the star, HIP 72509, is $12.1$.

A parallax angle of the star, HIP 72509, is $0.222\text{ arcsec}$.

Formula used:

The formula of the distance $\left(d\right)$ in terms of parallax angle $\left(p\right)$ is:

$d=\frac{1}{p}$

The relation between the star’s apparent magnitude and absolute magnitude is written as:

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

Here, $m$ is the apparent magnitude of a star, $M$ is the absolute magnitude of the star and $d$ is the distance from Earth to the star.

Explanation:

The formula of the distance $\left(d\right)$ in terms of parallax angle $\left(p\right)$ is:

$d=\frac{1}{p}$

The relation between the star’s apparent magnitude and absolute magnitude is written as:

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

Substitute $\frac{1}{p}$ for $d$.

$m-M=5\log \frac{1}{p}-5$

Substitute $12.1$ for $m$ and $0.222\text{ arcsec}$ for $p$.

$\begin{array}{c}12.1-M=5\log \frac{1}{0.222\text{ arcsec}}-5\\ M=12.1-5\log \frac{1}{0.222\text{ arcsec}}+5\\ M=13.831\end{array}$

Conclusion:

The absolute magnitude of the star, HIP 72509, is $13.831$.

(b)

To determine

The approximate ratio of luminosity of HIP 72509 to the Sun’s luminosity if the apparent magnitude of the star is $12.1$ and a parallax angle is $0.222\text{ arcsec}$.

Answer to Problem 25Q

Solution:

$5.1\times {10}^{-3}$

Explanation of Solution

Given data:

The apparent magnitude of the star, HIP 72509, is $12.1$.

A parallax angle of the star, HIP 72509, is $0.222\text{ arcsec}$.

Formula used:

The formula of the distance $\left(d\right)$ in terms of parallax angle $\left(p\right)$ is:

$d=\frac{1}{p}$

Magnitude difference related to the brightness ratio:

$m_2-m_1=2.5\log \left(\frac{b_1}{b_2}\right)$

Here, $m_1$ is the apparent magnitude of the star 1, $m_2$ is the apparent magnitude of the star 2, $b_1$ is the apparent brightness of the star 1 and $b_2$ is the apparent brightness of the star 2.

The relation between luminosity of the stars from inverse-square law is:

$\frac{L}{L_{\odot }}={\left(\frac{d}{d_{\odot }}\right)}^2\left(\frac{b}{b_{\odot }}\right)$

Here, $L$ is the luminosity of the star, $L_{\odot }$ is the luminosity of the Sun, $d$ is the distance of the star from Earth, $d_{\odot }$ the distance of the Sun from Earth, $b$ is the brightness of the star and $b_{\odot }$ is the brightness of the Sun.

Explanation:

The distance of star $\text{HIP 72509}$ is:

$d_{\text{HIP 72509}}=\frac{1}{p_{\text{HIP 72509}}}$

Recall the expression that relates the magnitude difference to the brightness ratio for star HIP 72509 to the Sun.

$m_{\odot }-m_{\text{HIP 72509}}=2.5\log \left(\frac{b_{\text{HIP 72509}}}{b_{\odot }}\right)$

The relation between luminosity of the star HIP 72509 and the Sun from inverse-square law is:

$\frac{L_{\text{HIP 72509}}}{L_{\odot }}={\left(\frac{d_{\text{HIP 72509}}}{d_{\odot }}\right)}^2\left(\frac{b_{\text{HIP 72509}}}{b_{\odot }}\right)$

Rearrange for brightness ratio.

$\frac{b_{\text{HIP 72509}}}{b_{\odot }}=\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(\frac{d_{\odot }}{d_{\text{HIP 72509}}}\right)}^2$

Combine the expression $m_{\odot }-m_{\text{HIP 72509}}=2.5\log \left(\frac{b_{\text{HIP 72509}}}{b_{\odot }}\right)$ and $\frac{b_{\text{HIP 72509}}}{b_{\odot }}=\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(\frac{d_{\odot }}{d_{\text{HIP 72509}}}\right)}^2$.

$m_{\odot }-m_{\text{HIP 72509}}=2.5\log \left(\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(\frac{d_{\odot }}{d_{\text{HIP 72509}}}\right)}^2\right)$

Substitute $\frac{1}{p_{\text{HIP 72509}}}$ for $d_{\text{HIP 72509}}$.

$m_{\odot }-m_{\text{HIP 72509}}=2.5\log \left(\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(p_{\text{HIP 72509}}{d}_{\odot }\right)}^2\right)$

The distance of the Sun from Earth is $1\text{au}$.

Substitute $13.8$ for $m_{\text{HIP 72509}}$ and $4.8$ for $m_{\odot }$, $1\text{ au}$ for $d_{\odot }$, and $0.222\text{ arcsec}$ for $p_{\text{HIP 72509}}$.

$\begin{array}{c}13.8-4.8=2.5\log \left(\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(\left(0.222\text{ arcsec}\right)\left(1\text{ au}\right)\right)}^2\right)\\ -9=2.5\log \left(\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(0.222\right)}^2\right)\\ -\frac{9}{2.5}=\log \left(\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(0.222\right)}^2\right)\end{array}$

Take anti log on both sides and solve as:

$\begin{array}{c}{10}^{-\frac{9}{2.5}}=\left(\frac{L_{\text{HIP 72509}}}{L_{\odot }}\right){\left(0.222\right)}^2\\ \frac{L_{\text{HIP 72509}}}{L_{\odot }}=\frac{{10}^{-\frac{9}{2.5}}}{{\left(0.222\right)}^2}\\ =5.1\times {10}^{-3}\end{array}$

Conclusion:

The luminosity ratio of the star to the Sun is $5.1\times {10}^{-3}$.