<p>If you're just looking for an equivalent expression, but relative to the sun, the modification is pretty simple.  Let's start with the original expression:
<span class="math-container">$$B=\frac{A \cdot L_S \cdot R^2}{D^2 \cdot d^2}$$</span></p>
<p>where</p>
<p><span class="math-container">$B$</span> = Brightness of observed object in watts<br/>
<span class="math-container">$A$</span> = Albedo of observed object<br/>
<span class="math-container">$L_S$</span> = Luminosity of the Sun in watts<br/>
<span class="math-container">$R$</span> = Radius of the observed object in meters<br/>
<span class="math-container">$D$</span> = Distance between the Sun and the observed object in meters<br/>
<span class="math-container">$d$</span> = Distance between the Earth and the observed object in meters<br/></p>
<p>To make this relative to the sun we're like an expression for <span class="math-container">$B^* = B / L_S$</span> so we divide both side of the original expression by <span class="math-container">$L_S$</span>:
<span class="math-container">$$B^* = \frac{B}{L_S} = \frac{A \cdot R^2}{D^2 \cdot d^2}$$</span></p>
<p>where</p>
<p><span class="math-container">$B^*$</span> = Brightness of the observed object relative to the brightness of the sun</p>
<p>This is of course in linear units. To convert to relative <a href="https://en.wikipedia.org/w/index.php?title=Magnitude_(astronomy)&amp;oldid=1064062972" rel="nofollow noreferrer">magnitude</a>, which might be more helpful:
<span class="math-container">$$\Delta M = -2.5\log_{10}(B^*)$$</span></p>

CC BY-SA 4.0
2022-01-22T12:23:53.160Z
2022-01-22T12:23:53.160Z
GrapefruitIsAwesome
<p>I know that we can determine the apparent brightness of an object in our solar system using the following formula:</p>
<p><span class="math-container">$$B=\frac{A\cdot L_S\cdot R^2}{D^2\cdot d^2}$$</span></p>
<p><span class="math-container">$B$</span> = Brightness of observed object in watts<br />
<span class="math-container">$A$</span> = Albedo of observed object<br />
<span class="math-container">$L_S$</span> = Luminosity of the Sun in watts<br />
<span class="math-container">$R$</span> = Radius of the observed object in meters<br />
<span class="math-container">$D$</span> = Distance between the Sun and the observed object in meters<br />
<span class="math-container">$d$</span> = Distance between the Earth and the observed object in meters</p>
<p>However, I find this isn't helpful in visualizing the brightness in your mind. I was wondering what modifications would be needed to this formula to make the result relative to the sun.</p>

CC BY-SA 4.0
2021-02-27T10:20:37.930Z
Hi! As you have seen, I have edited your question using MathJax. For future questions, you might find this helpful: [MathJax basic tutorial and quick reference](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)
jng224
CC BY-SA 4.0
2021-02-27T10:22:20.493Z
Could you maybe give the source of your formula?
User123
CC BY-SA 4.0
2021-02-27T10:31:49.917Z
What do you mean with: I find this isn't helpful in visualizing the brightness in your mind?
User123
CC BY-SA 4.0
2021-02-27T10:40:31.457Z
You could simply divide by the apparent brightness of the sun which would give you a result in percent of the sun's brightness.
jng224
<mathematics><brightness>
How to calculate apparent brightness in relation to the sun
TrEs-2b
# How to calculate apparent brightness in relation to the sun

(Asked by: username_0 on 2021-02-27)

I know that we can determine the apparent brightness of an object in our solar system using the following formula:

$$B=\frac{A\cdot L\_S\cdot R^2}{D^2\cdot d^2}$$

$B$ = Brightness of observed object in watts  
$A$ = Albedo of observed object  
$L\_S$ = Luminosity of the Sun in watts  
$R$ = Radius of the observed object in meters  
$D$ = Distance between the Sun and the observed object in meters  
$d$ = Distance between the Earth and the observed object in meters

However, I find this isn't helpful in visualizing the brightness in your mind. I was wondering what modifications would be needed to this formula to make the result relative to the sun.

**Tags:** mathematics, brightness
> **Comments:**
> **username_1**: Hi! As you have seen, I have edited your question using MathJax. For future questions, you might find this helpful: [MathJax basic tutorial and quick reference](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)  
> **username_2**: Could you maybe give the source of your formula?  
> **username_2**: What do you mean with: I find this isn't helpful in visualizing the brightness in your mind?  
> **username_1**: You could simply divide by the apparent brightness of the sun which would give you a result in percent of the sun's brightness.  

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## 1 answers

### Answer by username_3 on 2022-01-22. Score: 1

If you're just looking for an equivalent expression, but relative to the sun, the modification is pretty simple. Let's start with the original expression:
$$B=\frac{A \cdot L\_S \cdot R^2}{D^2 \cdot d^2}$$

where

$B$ = Brightness of observed object in watts  
$A$ = Albedo of observed object  
$L\_S$ = Luminosity of the Sun in watts  
$R$ = Radius of the observed object in meters  
$D$ = Distance between the Sun and the observed object in meters  
$d$ = Distance between the Earth and the observed object in meters

To make this relative to the sun we're like an expression for $B^\* = B / L\_S$ so we divide both side of the original expression by $L\_S$:
$$B^\* = \frac{B}{L\_S} = \frac{A \cdot R^2}{D^2 \cdot d^2}$$

where

$B^\*$ = Brightness of the observed object relative to the brightness of the sun

This is of course in linear units. To convert to relative [magnitude](https://en.wikipedia.org/w/index.php?title=Magnitude_(astronomy)&oldid=1064062972), which might be more helpful:
$$\Delta M = -2.5\log\_{10}(B^\*)$$


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