<p>The formula appears to compare the squares of distances, not taking into account the difference in extinction (which is small for small zenith distances, and for different altitudes provided they're high enough), this means <span class="math-container">$M - m = 5log(\frac{h}{cos(Z)})-5log(h_1)=-5log(cos z) - 5log(\frac{h_1}{100})$</span>. Also, I believe the formula should say -5 cos z ... the meteor from the picture provided is farther away from the one at zenith and at the same altitude and should have higher (numerical value) magnitude.</p>

CC BY-SA 4.0
2020-07-03T13:48:42.560Z
2020-07-03T13:54:15.377Z
2020-07-03T13:54:15.377Z
Tosic
<p>I have a question regarding the apparent and absolute magnitude of space debris objects or meteors reentering the earth's atmosphere. I already understand the difference between apparent and absolute magnitude. The absolute magnitude is the brightness of the object if the object is at a distance of 100km and at zenith. And the apparent magnitude is how bright the objects appears to the observer.</p>
<p>But how can I determine the apparent magnitude of an objects observed from an observer? The objects has the distance <span class="math-container">$d$</span>, the zenith distance <span class="math-container">$Z$</span> and the height <span class="math-container">$h$</span>. On the internet I found the following equation:</p>
<p><span class="math-container">$$
M = m +5  \log(\cos(Z))
$$</span>
<span class="math-container">$M$</span> = absolute magnitude, and <span class="math-container">$m$</span> = apparent magnitude. I posted a picture below.</p>
<p>But how is this equation derived? I do not understand the equation. What can I do, if the height <span class="math-container">$h$</span> of the objects does not equal 100 km?</p>
<p>The book where i found this is &quot;The Observer's guide to Astronomy&quot; (Volume 2) pages 659 to 660!</p>
<p><a href="https://books.google.de/books?id=TXc54LfKsSQC&amp;pg=PA659&amp;lpg=PA659&amp;dq=zenithal+magnitude&amp;source=bl&amp;ots=fPeMVD1sOq&amp;sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&amp;hl=de&amp;sa=X&amp;ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&amp;q=zenithal%20magnitude&amp;f=false" rel="nofollow noreferrer">https://books.google.de/books?id=TXc54LfKsSQC&amp;pg=PA659&amp;lpg=PA659&amp;dq=zenithal+magnitude&amp;source=bl&amp;ots=fPeMVD1sOq&amp;sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&amp;hl=de&amp;sa=X&amp;ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&amp;q=zenithal%20magnitude&amp;f=false</a></p>
<p><a href="https://i.sstatic.net/4jbji.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4jbji.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2020-07-03T12:13:09.973Z
Hello and welcome ! It might help potential answerers to have the reference of where you found that equation.
usernumber
CC BY-SA 4.0
2020-07-03T12:19:44.763Z
Thank you for your answer. I have found this in the book "The Observer's guide to astronomy" (volume 2) on page 659-660!
https://books.google.de/books?id=TXc54LfKsSQC&pg=PA659&lpg=PA659&dq=zenithal+magnitude&source=bl&ots=fPeMVD1sOq&sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&hl=de&sa=X&ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&q=zenithal%20magnitude&f=false
Alex
<meteor><apparent-magnitude>
apparent and absolute magnitude of meteors and space debris
Alex
# apparent and absolute magnitude of meteors and space debris

(Asked by: username_0 on 2020-07-03)

I have a question regarding the apparent and absolute magnitude of space debris objects or meteors reentering the earth's atmosphere. I already understand the difference between apparent and absolute magnitude. The absolute magnitude is the brightness of the object if the object is at a distance of 100km and at zenith. And the apparent magnitude is how bright the objects appears to the observer.

But how can I determine the apparent magnitude of an objects observed from an observer? The objects has the distance $d$, the zenith distance $Z$ and the height $h$. On the internet I found the following equation:

$$
M = m +5 \log(\cos(Z))
$$
$M$ = absolute magnitude, and $m$ = apparent magnitude. I posted a picture below.

But how is this equation derived? I do not understand the equation. What can I do, if the height $h$ of the objects does not equal 100 km?

The book where i found this is "The Observer's guide to Astronomy" (Volume 2) pages 659 to 660!

<https://books.google.de/books?id=TXc54LfKsSQC&pg=PA659&lpg=PA659&dq=zenithal+magnitude&source=bl&ots=fPeMVD1sOq&sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&hl=de&sa=X&ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&q=zenithal%20magnitude&f=false>

[![enter image description here](https://i.sstatic.net/4jbji.jpg)](https://i.sstatic.net/4jbji.jpg)

**Tags:** meteor, apparent-magnitude
> **Comments:**
> **username_1**: Hello and welcome ! It might help potential answerers to have the reference of where you found that equation.  
> **username_0**: Thank you for your answer. I have found this in the book "The Observer's guide to astronomy" (volume 2) on page 659-660! https://books.google.de/books?id=TXc54LfKsSQC&pg=PA659&lpg=PA659&dq=zenithal+magnitude&source=bl&ots=fPeMVD1sOq&sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&hl=de&sa=X&ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&q=zenithal%20magnitude&f=false  

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## 1 answers

### Answer by username_2 on 2020-07-03. Score: 1

The formula appears to compare the squares of distances, not taking into account the difference in extinction (which is small for small zenith distances, and for different altitudes provided they're high enough), this means $M - m = 5log(\frac{h}{cos(Z)})-5log(h\_1)=-5log(cos z) - 5log(\frac{h\_1}{100})$. Also, I believe the formula should say -5 cos z ... the meteor from the picture provided is farther away from the one at zenith and at the same altitude and should have higher (numerical value) magnitude.


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