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

CC BY-SA 4.0
2021-02-05T16:59:47.257Z
Welcome to astronomy SE! I fixed the typesetting for you.
B--rian
CC BY-SA 4.0
2022-03-02T23:23:33.123Z
The brightest star in the sky (that isn't the Sun) is a lot fainter than $m=-6$.
ProfRob
CC BY-SA 4.0
2021-02-05T16:56:27.343Z
2021-02-05T16:59:18.467Z
2021-02-05T16:59:18.467Z
v.n.s.
<p>I am curious to see if I have calculated this correctly: </p>

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

<p>My workings:</p>

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

<p>using the information given $P=30$</p>

<p>$$M=-2.8log(30)-1.43=-5.58$$</p>

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

<p>using information for calculate value of $M$ and that vega = $8Pc$</p>

<p>$$m=5log\left(8\right)-5+5.58=5.10$$</p>

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

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

<p>so adapting the above equation and subbing in the required values</p>

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

<p>Rearrange</p>

<p>$$10^{\frac{5.10}{2.51}}=\frac{L_c}{L_v}$$</p>

<p>$$107.61=\frac{L_c}{L_v}$$</p>

<p>so therefore </p>

<p>$$L_c=108L_v$$</p>

<p>where subscript c is delta-Cepheid  and subscript v is vega</p>

<p><strong>Is this calculation correct</strong>?</p>

CC BY-SA 4.0
2021-07-05T19:55:27.337Z
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.
ProfRob
<distances><cepheids>
Type I delta-Cepheid star, calculation of apparent maginute and luminosity
Jason Taylor
# Type I delta-Cepheid star, calculation of apparent maginute and luminosity

(Asked by: username_0 on 2018-01-26)

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**?

**Tags:** distances, cepheids

> **Comments:**
> **username_1**: 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.  

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## 1 answers

### Answer by username_2 on 2021-02-05. Score: -1

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()−5- M$ giving you an apparent magnitude of $\sim -6$ which is acceptable as cepheids are the brightest stars in the galaxies

> **Comments:**
> **username_3**: Welcome to astronomy SE! I fixed the typesetting for you.  
> **username_1**: The brightest star in the sky (that isn't the Sun) is a lot fainter than $m=-6$.  

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