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)104 parsecs away.

Explanation:

We know that the apparent magnitude m_"app"mapp of an object tells us how bright that object will appear as observed from Earth.

And the absolute magnitude M_"abs"Mabs 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_"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^(1/5) = 2.512.

For two stars A and B if there magnitudes and intensities are denoted by m and I respectively, we have the expression connecting both as
I_A / I_B = (2.512)^(m_B - m_A)

Taking the log of both sides and using log_10 M^p = p log_10 M we get
log_10(I_A / I_B) = (m_B - m_A) log_10 2.512

This is commonly expressed in the form
m_B - m_A = 2.5 log_10 (I_A / I_B) ......(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 (d_B / d_A)^2 for (I_A / I_B), (1) becomes

m_B - m_A = 2.5 log_10 (d_B / d_A)^2

=>m_B - m_A = 5 log_10 (d_B / d_A)

When d_A = 10 " pc", so that m_A = M_"abs", and d_B=d be specified in pc, above equation reduces to
m_"app" - M_"abs" = 5 log_10 [ d / (10) ]

Above can be rewritten as
m_"app" - M_"abs" = -5 +5log_10 d
=>d=10^((m_"app" - M_"abs"+5)/5) .....(2)

Inserting given values in (2) above we get
d=10^((10 -(-5)+5)/5)
=>d=10^(4)" pc"