<p>I think that should be <span class="math-container">$+ V_{Sun}$</span> in your second equation, not <span class="math-container">$- V_{Sun}$</span>.  Otherwise, a fainter Sun (larger magnitude) would give you a brighter Jupiter (smaller magnitude).   Those need to go in the same direction.</p>
<p>Applying that change (assuming the rest of the calculation is right) would increase your answer by <span class="math-container">$2 \times 4.83 = 9.66$</span>, which gets you much closer to the right magnitude.</p>
<p>Your estimate will now be a bit on the faint side - not sure why - but at least a lot closer than before.</p>

CC BY-SA 4.0
2020-08-08T12:48:56.050Z
Yes, I think so too. But a >2 magnitude difference is a tiny bit too much. I wonder if there's anything else wrong.
ChristieToWin
CC BY-SA 4.0
2020-08-08T12:50:19.607Z
If you look at the answer after the one with that formula, it mentions other factors that determine the apparent magnitude of a planet.
Eric Jensen
CC BY-SA 4.0
2020-08-08T12:20:41.720Z
2020-08-08T12:20:41.720Z
Eric Jensen
<p>I was searching for a formula to calculate the apparent magnitude of a planet knowing its physical characteristics and stumbled upon <a href="https://astronomy.stackexchange.com/a/5983/34121">this answer</a>.</p>
<p>The final formula given is:</p>
<p><span class="math-container">$$m_{planet} = V_{planet} + 5 \log_{10}\left( d_{e-p} \right) - 5$$</span></p>
<p>In which,</p>
<p><span class="math-container">$$ V_{planet}=-2.5 \log_{10}\left(a_p \frac{ r_p^2 }{ 4 d_s^2 } \right) - V_{sun}$$</span></p>
<p>where:</p>
<ul>
<li><p><span class="math-container">$m_{planet}$</span> is the planet in question's apparent magnitude,</p>
</li>
<li><p><span class="math-container">$V_{planet}$</span> is the planet's absolute magnitude,</p>
</li>
<li><p><span class="math-container">$a_{p}$</span> is the bond albedo of the planet,</p>
</li>
<li><p><span class="math-container">$r_{p}$</span> is the radius of the planet,</p>
</li>
<li><p><span class="math-container">$d_{s}$</span> is the distance of the planet from its star,</p>
</li>
<li><p><span class="math-container">$d_{e-p}$</span> is the distance between the planet and the observer (here on Earth) in parsecs, and,</p>
</li>
<li><p><span class="math-container">$V_{sun}$</span> is the star's absolute magnitude.</p>
</li>
</ul>
<hr />
<p>To test it out I tried to calculate the apparent magnitude of Jupiter as seen from Earth at an average opposition. The values then are:</p>
<ul>
<li><p><span class="math-container">$a_{p} = 0.343$</span>,</p>
</li>
<li><p><span class="math-container">$r_{p} = 69,911,000$</span> m,</p>
</li>
<li><p><span class="math-container">$d_{s} = 778.57 \times 10^9$</span> m (5.204 AU),</p>
</li>
<li><p><span class="math-container">$d_{e-p} = 628.97 \times 10^9$</span> m <span class="math-container">$= 0.0000203835294968$</span> parsecs (4.204 AU)</p>
</li>
<li><p><span class="math-container">$V_{sun} = 4.83$</span></p>
</li>
</ul>
<p>All the above values were taken from <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html" rel="nofollow noreferrer">NASA's Jupiter fact sheet</a>.</p>
<p>Plugging in the numbers and I get <span class="math-container">$m_{planet} = -10.38$</span>, which is much brighter than the actual apparent magnitude of Jupiter at average opposition of <span class="math-container">$-2.x$</span>. Where did I go wrong?</p>

<apparent-magnitude>
What is wrong with my calculations of Jupiter's apparent magnitude?
ChristieToWin
# What is wrong with my calculations of Jupiter's apparent magnitude?

(Asked by: username_0 on 2020-08-08)

I was searching for a formula to calculate the apparent magnitude of a planet knowing its physical characteristics and stumbled upon [this answer](https://astronomy.stackexchange.com/a/5983/34121).

The final formula given is:

$$m\_{planet} = V\_{planet} + 5 \log\_{10}\left( d\_{e-p} \right) - 5$$

In which,

$$ V\_{planet}=-2.5 \log\_{10}\left(a\_p \frac{ r\_p^2 }{ 4 d\_s^2 } \right) - V\_{sun}$$

where:

* $m\_{planet}$ is the planet in question's apparent magnitude,
* $V\_{planet}$ is the planet's absolute magnitude,
* $a\_{p}$ is the bond albedo of the planet,
* $r\_{p}$ is the radius of the planet,
* $d\_{s}$ is the distance of the planet from its star,
* $d\_{e-p}$ is the distance between the planet and the observer (here on Earth) in parsecs, and,
* $V\_{sun}$ is the star's absolute magnitude.

---

To test it out I tried to calculate the apparent magnitude of Jupiter as seen from Earth at an average opposition. The values then are:

* $a\_{p} = 0.343$,
* $r\_{p} = 69,911,000$ m,
* $d\_{s} = 778.57 \times 10^9$ m (5.204 AU),
* $d\_{e-p} = 628.97 \times 10^9$ m $= 0.0000203835294968$ parsecs (4.204 AU)
* $V\_{sun} = 4.83$

All the above values were taken from [NASA's Jupiter fact sheet](https://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html).

Plugging in the numbers and I get $m\_{planet} = -10.38$, which is much brighter than the actual apparent magnitude of Jupiter at average opposition of $-2.x$. Where did I go wrong?

**Tags:** apparent-magnitude


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## 1 answers

### Answer by username_1 on 2020-08-08. Score: 1

I think that should be $+ V\_{Sun}$ in your second equation, not $- V\_{Sun}$. Otherwise, a fainter Sun (larger magnitude) would give you a brighter Jupiter (smaller magnitude). Those need to go in the same direction.

Applying that change (assuming the rest of the calculation is right) would increase your answer by $2 \times 4.83 = 9.66$, which gets you much closer to the right magnitude.

Your estimate will now be a bit on the faint side - not sure why - but at least a lot closer than before.

> **Comments:**
> **username_0**: Yes, I think so too. But a >2 magnitude difference is a tiny bit too much. I wonder if there's anything else wrong.  
> **username_1**: If you look at the answer after the one with that formula, it mentions other factors that determine the apparent magnitude of a planet.  

---

