<p>In an isothermal atmosphere, the exponential scale height of the atmosphere is
<span class="math-container">$$ h \sim \frac{k_\mathrm B T}{\mu g},$$</span> 
where <span class="math-container">$g$</span> is the gravitational field, <span class="math-container">$\mu$</span> is the mean mass of a particle and <span class="math-container">$T$</span> is the temperature (in kelvin).</p>

<p>i.e. The pressure/density of the atmosphere falls exponentially, with an e-folding height given by the above expression.</p>

<p>I suppose what matters when you look at a photo, is how this height compares with the radius of the planet.
<span class="math-container">$$ \frac{h}{R} \sim \frac{k_\mathrm B T}{\mu g R}$$</span></p>

<p>Jupiter is half the temperature, 11 times the radius and with 3 times the gravity of Earth. However <span class="math-container">$\mu$</span> is about ten times smaller (hydrogen vs nitrogen/oxygen). Overall that means <span class="math-container">$h/R$</span> for Jupiter is of order 5–10 times smaller than for Earth and so it will appear "sharper".</p>

<p>EDIT: If you put some reasonable numbers in for Jupiter (<span class="math-container">$T \sim 130$</span> K, <span class="math-container">$\mu=2$</span>, <span class="math-container">$R=7\times 10^7$</span> m), then <span class="math-container">$h/R \sim 3 \times 10^{-4}$</span>. This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high.</p>

<p><a href="https://i.sstatic.net/0EBuU.gif" rel="noreferrer"><img src="https://i.sstatic.net/0EBuU.gif" alt="Earth"></a>
<a href="https://i.sstatic.net/mmdPW.jpg" rel="noreferrer"><img src="https://i.sstatic.net/mmdPW.jpg" alt="Jupiter"></a></p>

CC BY-SA 4.0
2020-06-12T20:33:12.227Z
I'm curious how much (relative) viewing distance plays into this as well, for example comparing Juno pictures of Jupiter with ISS pictures of Earth, both of which orbit at an altitude roughly 6% of the planets' radii.
Kai
CC BY-SA 4.0
2020-06-13T04:04:10.200Z
@Kai Dividing by $R$ effectively does take viewing distance into account.
PM 2Ring
CC BY-SA 4.0
2020-06-13T04:48:58.540Z
@PM2Ring It seems that the relevant scale to look at here is the angular size of the atmospheric "layer" from the viewpoint, which is not captured by comparing the atmospheric height to planetary radius
Kai
CC BY-SA 4.0
2020-06-13T05:09:07.493Z
@Kai Sure, but we're (approximately) scaling by the viewing distance anyway, when we make the planets have the same apparent radii in the photos. So if photos of Jupiter & the Earth have both planets with a diameter of 1000 pixels, the viewing distance to Jupiter is about 11 times the viewing distance to Earth.
PM 2Ring
CC BY-SA 4.0
2020-06-13T07:04:24.417Z
@Kai If you "fill" your picture with a planet, what matters is height of the atmosphere vs radius of the planet, which is what I have given.
ProfRob
CC BY-SA 4.0
2020-06-13T19:19:32.853Z
@RobJeffries this is a rare / singular case where I just can't agree with / don't understand what you're saying?  *"This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high."*.  If Earth fills a photo that is 3000 pixels, the edge is 10000% sharp.  If Jupiter fills a photo that is 3000 pixels, the edge is 10000% sharp.  The question is just nonsensical, wrong.
Fattie
CC BY-SA 4.0
2020-06-12T12:25:45.277Z
2020-06-13T21:55:31.187Z
2020-06-13T21:55:31.187Z
ProfRob
<p>I've just seen <a href="https://www.forbes.com/sites/jamiecartereurope/2020/06/11/nasas-spacecraft-at-jupiter-just-sent-back-some-jaw-dropping-photos/" rel="noreferrer">this Forbes article</a>.</p>

<p>Why do gas giants appear to have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?</p>

<p>Is it just a matter of scale?  Or is there some form of "surface-tension" for the hydrogen gas?</p>

CC BY-SA 4.0
2020-06-12T23:37:08.343Z
Dupe of question on the Physics.SE site: https://physics.stackexchange.com/questions/26764/why-is-jupiter-so-sharply-defined
dotancohen
CC BY-SA 4.0
2020-06-13T03:15:27.680Z
You might also take note of the mention of "impressive image processing skills" in that article.
jamesqf
<planetary-atmosphere><jupiter><gas-giants>
Why do gas giants have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?
fadedbee
# Why do gas giants have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?

(Asked by: username_0 on 2020-06-12)

I've just seen [this Forbes article](https://www.forbes.com/sites/jamiecartereurope/2020/06/11/nasas-spacecraft-at-jupiter-just-sent-back-some-jaw-dropping-photos/).

Why do gas giants appear to have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?

Is it just a matter of scale? Or is there some form of "surface-tension" for the hydrogen gas?

**Tags:** planetary-atmosphere, jupiter, gas-giants

> **Comments:**
> **username_1**: Dupe of question on the Physics.SE site: https://physics.stackexchange.com/questions/26764/why-is-jupiter-so-sharply-defined  
> **username_2**: You might also take note of the mention of "impressive image processing skills" in that article.  

---

## 1 answers

### Answer by username_3 on 2020-06-12. Score: 29 | Accepted answer

In an isothermal atmosphere, the exponential scale height of the atmosphere is
$$ h \sim \frac{k\_\mathrm B T}{\mu g},$$
where $g$ is the gravitational field, $\mu$ is the mean mass of a particle and $T$ is the temperature (in kelvin).

i.e. The pressure/density of the atmosphere falls exponentially, with an e-folding height given by the above expression.

I suppose what matters when you look at a photo, is how this height compares with the radius of the planet.
$$ \frac{h}{R} \sim \frac{k\_\mathrm B T}{\mu g R}$$

Jupiter is half the temperature, 11 times the radius and with 3 times the gravity of Earth. However $\mu$ is about ten times smaller (hydrogen vs nitrogen/oxygen). Overall that means $h/R$ for Jupiter is of order 5–10 times smaller than for Earth and so it will appear "sharper".

EDIT: If you put some reasonable numbers in for Jupiter ($T \sim 130$ K, $\mu=2$, $R=7\times 10^7$ m), then $h/R \sim 3 \times 10^{-4}$. This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high.

[![Earth](https://i.sstatic.net/0EBuU.gif)](https://i.sstatic.net/0EBuU.gif)
[![Jupiter](https://i.sstatic.net/mmdPW.jpg)](https://i.sstatic.net/mmdPW.jpg)

> **Comments:**
> **username_4**: I'm curious how much (relative) viewing distance plays into this as well, for example comparing Juno pictures of Jupiter with ISS pictures of Earth, both of which orbit at an altitude roughly 6% of the planets' radii.  
> **username_5**: @username_4 Dividing by $R$ effectively does take viewing distance into account.  
> **username_4**: @username_5 It seems that the relevant scale to look at here is the angular size of the atmospheric "layer" from the viewpoint, which is not captured by comparing the atmospheric height to planetary radius  
> **username_5**: @username_4 Sure, but we're (approximately) scaling by the viewing distance anyway, when we make the planets have the same apparent radii in the photos. So if photos of Jupiter & the Earth have both planets with a diameter of 1000 pixels, the viewing distance to Jupiter is about 11 times the viewing distance to Earth.  
> **username_3**: @username_4 If you "fill" your picture with a planet, what matters is height of the atmosphere vs radius of the planet, which is what I have given.  
> **username_6**: @RobJeffries this is a rare / singular case where I just can't agree with / don't understand what you're saying?  *"This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high."*.  If Earth fills a photo that is 3000 pixels, the edge is 10000% sharp.  If Jupiter fills a photo that is 3000 pixels, the edge is 10000% sharp.  The question is just nonsensical, wrong.  

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