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I am curious to see if I have calculated this correctly:

A Type I delta-Cepheid star with a one month (30day) period is put at the distance of the star Vega.
a) Determine the apparent magnitude this star would have.
b) Determine how many times more luminous the delta-Cepheid star is compared to Vega.

My workings:

a) Absolute Magnitudes given for Type I delta-Cepheid star: $$M=-2.8log\left(P\right)-1.43$$

using the information given $P=30$

$$M=-2.8log(30)-1.43=-5.58$$

Absolute magnitude is given by $$m-M=5log\left(d\right)-5$$

using information for calculate value of $M$ and that vega = $8Pc$

$$m=5log\left(8\right)-5+5.58=5.10$$

b) Apparent magnitude is given by $$m_1-m_2=-2.51log(\frac{F_2}{F_1})$$

as the two start are at the same distance then I will be left with just luminosity.

so adapting the above equation and subbing in the required values

$$0-5.10=-2.51log\left(\frac{L_c}{Lv}\right)$$

Rearrange

$$10^{\frac{5.10}{2.51}}=\frac{L_c}{L_v}$$

$$107.61=\frac{L_c}{L_v}$$

so therefore

$$L_c=108L_v$$

where subscript c is delta-Cepheid and subscript v is vega

Is this calculation correct?

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  • $\begingroup$ This looks like your homework. Such questions are rarely of any use to others unless they are made less specific and are of more general interest. Some people on this site are very much against answering such questions. $\endgroup$
    – ProfRob
    Jul 5, 2021 at 19:55

1 Answer 1

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Pay attention to your signs: Remember that the value of $M$ is negative so on the second equation when you move $M$ to the right hand side, is $ 5 \log(𝑑)−5- M$ giving you an apparent magnitude of $\sim -6$ which is acceptable as cepheids are the brightest stars in the galaxies

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  • $\begingroup$ Welcome to astronomy SE! I fixed the typesetting for you. $\endgroup$
    – B--rian
    Feb 5, 2021 at 16:59
  • $\begingroup$ The brightest star in the sky (that isn't the Sun) is a lot fainter than $m=-6$. $\endgroup$
    – ProfRob
    Mar 2 at 23:23

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