16
$\begingroup$

Since gas giant consist of most gas components, where do we establish their "surface"?

My take is basically to take the limit in which all light is opaque. For example, in this photo:enter image description here

Photo of Jupiter. Credit to NASA, ESA, A. Simon, and M.H. Wong

The surface, then, will be the limit of the black blackground with the planet.

Any other way to formally define the "surface" of a gas giant?

$\endgroup$
  • 2
    $\begingroup$ Related question: astronomy.stackexchange.com/q/8410/24157 $\endgroup$ – antispinwards Jun 26 at 6:58
  • $\begingroup$ I won't give this as an answer because it's not formal, but informally one might consider the "surface" to be the point at which you can't descend any more. For a gas body this would probably be the point at which the density of the gas exceeded the object descending - as this boundary would be different depending on the object in question we would probably have to define an arbitrary point, much like we do with the "surface" of a body of water (which of course is much easier due to the sudden density increase at the boundary between air and water) $\endgroup$ – Michael Jun 26 at 20:40
  • $\begingroup$ @Michael given that the gas giants consist mostly of Hydrogen and Helium, I think you can dive quite a lot into it. (You will be squished to death before reaching this border.) $\endgroup$ – Paŭlo Ebermann Jun 27 at 20:25
  • $\begingroup$ @PaŭloEbermann Oh, definitely and not just squished probably fried too. I was reading somewhere that in the sun the depth at which you would "float" is halfway to the core. While not as hot I would imagine gas giants are quite a bit hotter at depth. But this isn't unique to gas giants - the surface of Venus is both hot and under a crushing atmosphere. $\endgroup$ – Michael Jun 27 at 21:27
22
$\begingroup$

There are two common definitions in use for the surface of gas planets:

  • The 1-bar surface: As pressure increases, the deeper in we go into the gas planet, we will hit a pressure of 1 bar at some altitude. Gas at this altitudes will usually sit deep enough in the gravitational well and be of a near-uniform density and temperature, as to not be influenced by exterior parameters, for example the solar wind. Therefore, the altitude of the 1-bar level will remain essentially constant, for short astronomical times.
  • The $\tau=2/3$-surface: This is the altitude, from which photons can escape freely into space. This happens at an average optical depth $\tau$ of 2/3. It is essentially what you see in your image as the limit of the black background. For the sun one end of the photosphere is the average $\tau=2/3$-surface, and for transiting exoplanets this is identical to the measured transit radius at that wavelength.

There is no hard relation between those two surfaces, but in general their altitude will not be different by more than a scale height, as at around 0.1-1 bar the gaseous atomic and molecular bands become enormously pressure broadened, which makes the atmosphere quickly opaque at most wavelengths, for the usual gas giant components.

| improve this answer | |
$\endgroup$
  • $\begingroup$ So, the optical depth has to be 2/3, when we consider all the wavelengths, right? $\endgroup$ – Carlos Vázquez Monzón Jun 25 at 23:09
  • $\begingroup$ Yes, I intentionally, for reasons of simplicity, omitted the discussion of which opacity to use when computing $\tau$. This is because of the line broadening argument: At about 1 bar most line forests are so broad that they form a single, grey opacity source. $\endgroup$ – AtmosphericPrisonEscape Jun 25 at 23:23
0
$\begingroup$

Any other way to formally define the "surface" of a gas giant?


Other than what Michael and AtmosphericPrisonEscape are proposing, you could also set the surface at the actual surface of Jupiter's metallic liquid core. The four gas giants (except Saturn perhaps which may be fully gaseous) have solid or liquid cores that have an actual surface. Jupiter's liquid core is about as big as the Earth and has about 10 Earth masses.

| improve this answer | |
$\endgroup$
  • $\begingroup$ This isn't a particularly useful definition though, and isn't commonly used. Jupiter's core seems to be somewhat "fuzzy" or "diluted", so there may not be a clear boundary. Thanks to a lack of orbiter missions, the interiors of the ice giants are not well constrained at all. $\endgroup$ – antispinwards Jul 21 at 16:44
  • $\begingroup$ @antispinwards Yeah one should send a probe(s) to Uranus and Neptune again. $\endgroup$ – Ioannes Jul 21 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.