12

I've just seen this Forbes article.

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?

Glorfindel
  • 4,790
  • 3
  • 25
  • 40
fadedbee
  • 223
  • 1
  • 5
  • 1
    Dupe of question on the Physics.SE site: https://physics.stackexchange.com/questions/26764/why-is-jupiter-so-sharply-defined – dotancohen Jun 12 '20 at 23:37
  • You might also take note of the mention of "impressive image processing skills" in that article. – jamesqf Jun 13 '20 at 03:15

1 Answers1

29

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 Jupiter

ProfRob
  • 151,483
  • 9
  • 359
  • 566
  • 2
    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 Jun 12 '20 at 20:33
  • 2
    @Kai Dividing by $R$ effectively does take viewing distance into account. – PM 2Ring Jun 13 '20 at 04:04
  • @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 Jun 13 '20 at 04:48
  • 2
    @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 Jun 13 '20 at 05:09
  • 3
    @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 Jun 13 '20 at 07:04
  • @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 Jun 13 '20 at 19:19