Chapter 9, Problem 5P

Complete the following table:

To determine

Complete the given table with proper values.

Answer to Problem 5P

The required table is given below:

$m_v\left(\text{mag}\right)$	$M_V\left(\text{mag}\right)$	$d\left(\text{pc}\right)$	$p\left(\text{arc second}\right)$
7	7	10	0.1
11	1	1000	0.001
25	22	40	0.025
4	2	25	0.040

Explanation of Solution

Write the equation for the relation between absolute and apparent visual magnitude.

$m_v-M_V=-5+5\log \left(d\right)$        (I)

Here, $m_v$ is the apparent visual magnitude, $M_V$ is the absolute visual magnitude and $d$ is the distance in parsecs.

Write the expression between the relation of distance and parallax distance.

$d\left(\text{pc}\right)=\frac{1}{p\left(\text{arc}\text{sec}\right)}$        (II)

Here, $p$ is the parallax distance.

Conclusion:

First table:

Substitute, $7\text{mag}$ for $M_V$, $10\text{pc}$ for $d$ in equation (I) to find $m_v$.

$\begin{array}{l}{m}_v-\left(7\text{mag}\right)=-5+5\log \left(10\right)\\ m_v=7\text{mag}\end{array}$

Substitute, $10\text{pc}$ for $d$ in equation (II) to find $p$.

$\begin{array}{l}\left(\text{10 pc}\right)=\frac{1}{p\left(\text{arc}\text{sec}\right)}\\ p=0.1\text{arc}\text{sec}\end{array}$

Second table:

Substitute, $11\text{mag}$ for $m_v$, $1000\text{pc}$ for $d$ in equation (I) to find $M_V$.

$\begin{array}{l}\left(11\text{mag}\right)-M_V=-5+5\log \left(1000\right)\\ M_V=1\text{mag}\end{array}$

Substitute, $1000\text{pc}$ for $d$ in equation (II) to find $p$.

$\begin{array}{l}\left(\text{1000 pc}\right)=\frac{1}{p\left(\text{arc}\text{sec}\right)}\\ p=0.001\text{arc}\text{sec}\end{array}$

Third table:

Substitute, $0.025\text{arc sec}$ for $p$ in equation (II) to find $d$.

$\begin{array}{c}d=\frac{1}{\left(\text{0}\text{.025 arc}\text{sec}\right)}\\ =40\text{pc}\end{array}$

Substitute, $22\text{mag}$ for $M_V$, $40\text{pc}$ for $d$ in equation (I) to find $m_v$.

$\begin{array}{l}{m}_v-\left(22\text{mag}\right)=-5+5\log \left(40\right)\\ m_v=25\text{mag}\end{array}$

Fourth table:

Substitute, $0.040\text{arc sec}$ for $p$ in equation (II) to find $d$.

$\begin{array}{c}d=\frac{1}{\left(\text{0}\text{.040 arc}\text{sec}\right)}\\ =25\text{pc}\end{array}$

Substitute, $4\text{mag}$ for $m_v$, $25\text{pc}$ for $d$ in equation (I) to find $M_V$.

$\begin{array}{l}\left(4\text{mag}\right)-M_V=-5+5\log \left(25\right)\\ M_V=2\text{mag}\end{array}$

Thus, the required table is given below:

$m_v\left(\text{mag}\right)$	$M_V\left(\text{mag}\right)$	$d\left(\text{pc}\right)$	$p\left(\text{arc second}\right)$
7	7	10	0.1
11	1	1000	0.001
25	22	40	0.025
4	2	25	0.040