# A star has an apparent magnitude of 10 and an absolute magnitude of -5. How far away is it?

##### 1 Answer
Jan 23, 2017

The star is ${10}^{4}$ parsecs away.

#### Explanation:

We know that the apparent magnitude ${m}_{\text{app}}$ of an object tells us how bright that object will appear as observed from Earth.

And the absolute magnitude ${M}_{\text{abs}}$ of an object tells us how bright that object would appear when it is observed from a standard distance of 10 parsecs.

Difference between apparent and absolute magnitudes of an object,${m}_{\text{app" – M_"abs}}$, is called the distance modulus.

Magnitude system is based on the response of the human eye which shows a logarithmic response. Also the magnitude system, a logarithmic scale, assumes that a factor of $100$ in intensity corresponds exactly to a difference of $5$ magnitudes.
Therefore, we have this scale of base ${100}^{\frac{1}{5}} = 2.512$.

For two stars $A \mathmr{and} B$ if there magnitudes and intensities are denoted by $m \mathmr{and} I$ respectively, we have the expression connecting both as
${I}_{A} / {I}_{B} = {\left(2.512\right)}^{{m}_{B} - {m}_{A}}$

Taking the log of both sides and using ${\log}_{10} {M}^{p} = p {\log}_{10} M$ we get
${\log}_{10} \left({I}_{A} / {I}_{B}\right) = \left({m}_{B} - {m}_{A}\right) {\log}_{10} 2.512$

This is commonly expressed in the form
${m}_{B} - {m}_{A} = 2.5 {\log}_{10} \left({I}_{A} / {I}_{B}\right)$ ......(1)

We know that intensity of a light source follows inverse square law of distances. After comparing intensities and magnitudes of two different stars as in equation (1) let us compare the intensities and magnitudes of the same star at two different distances.
Substituting ${\left({d}_{B} / {d}_{A}\right)}^{2}$ for $\left({I}_{A} / {I}_{B}\right)$, (1) becomes

${m}_{B} - {m}_{A} = 2.5 {\log}_{10} {\left({d}_{B} / {d}_{A}\right)}^{2}$

$\implies {m}_{B} - {m}_{A} = 5 {\log}_{10} \left({d}_{B} / {d}_{A}\right)$

When ${d}_{A} = 10 \text{ pc}$, so that ${m}_{A} = {M}_{\text{abs}}$, and ${d}_{B} = d$ be specified in pc, above equation reduces to
${m}_{\text{app" - M_"abs}} = 5 {\log}_{10} \left[\frac{d}{10}\right]$

Above can be rewritten as
${m}_{\text{app" - M_"abs}} = - 5 + 5 {\log}_{10} d$
$\implies d = {10}^{\frac{{m}_{\text{app" - M_"abs}} + 5}{5}}$ .....(2)

Inserting given values in (2) above we get
$d = {10}^{\frac{10 - \left(- 5\right) + 5}{5}}$
$\implies d = {10}^{4} \text{ pc}$