I've seen statements to the effect of: "the gravity on the surface of Jupiter is about 2.5 times that of Earth". The problem with such a statement is that as essentially a ball of gas, Jupiter is not believed to have a solid surface. Behind the "2.5g" claim there must be some criteria applied to select a radius, which when combined with a figure for Jupiter's mass, will yield an acceleration value according to the classic gravitation formula.

The meat of my question pertains to the criteria which produce radius values for the gas planets which presumably are the basis of statements such as I mentioned. As we know, Earth's atmosphere extends quite far into space, just getting thinner and thinner. So as one approaches a gas giant, one would presumably first encounter an extremely thin atmosphere, getting progressively denser, eventually as dense as a liquid. Somewhere along the way, we crossed a point which represents what we deem to be the planet's radius. What is going on at that point? Is it the same for all gas planets, or do the conditions of each planet necessitate different choices? Are the same criteria applied regardless of endeavor (astronomy vs spacecraft engineering vs exometeorology, etc.)?

Also, for what its worth, how does the planet's deemed radius relate to its visible horizon i.e. the visible but indistinct boundary between planet and space?

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    $\begingroup$ Btw - one of the things Juno is trying to determine is whether there is a solid core in there or not. Some people think there is one, and it might mass 20x Earth. $\endgroup$
    – kim holder
    Jul 7, 2016 at 2:09
  • $\begingroup$ Interesting tidbit: size of gas giants like Jupiter is pretty consistent regardless of mass - the extra gravity compresses the gas leaving about the same radius of the planet. Only once the threshold of fusion is crossed (at about 20x Jupiter mass), and the planet becomes a Brown Dwarf star, the radius will start climbing with mass. $\endgroup$
    – SF.
    Jul 7, 2016 at 9:38
  • $\begingroup$ The same kind of question applies to stars as well, and the answer depends on the purpose for which you are asking the question. See physics.stackexchange.com/questions/33695/… on Physics (yet another place where this kind of question is on topic). $\endgroup$ Jul 7, 2016 at 14:29

3 Answers 3


The radius of Jupiter and the other gas giants is defined, somewhat arbitrarily, to be the radius at which the atmosphere has a pressure of 1 bar. As your question points out, they had to pick something. So that's what they picked. This convention is used for all of the gas giants in our solar system.

For a visible boundary definition, you would need to define the opacity threshold and the wavelength, since the opacity depends on wavelength. Even then you would have some ambiguity due to features with different opacity. Would Jupiter have a different radius at the Great Red Spot if your wavelength was red?

Though I suppose the pressure definition has a similar problem, as the Great Red Spot is probably a low pressure system. However the uncertainty would be small as the pressure changes very rapidly with altitude.

The radius of the Sun used to be defined by opacity (at an optical depth of 2/3). That photosphere radius has some measurement interpretation issues, and remains a subject of investigation. Apparently the IAU got tired of it, so in 2015 the "nominal" radius of the Sun was defined to be exactly 695,700 km. This allows the use of "solar radii" as a unit without everyone wondering what radius to use to convert that to distance units.

  • $\begingroup$ Not to be a smart ass, but where did this definition come from? Is it the same for the other gas giants? Thanks! $\endgroup$
    – Erik
    Jul 7, 2016 at 3:48
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    $\begingroup$ I have no idea who is responsible for starting that convention. The rationale is that the surface of the Earth is at 1 bar, so we might as well make the surface of the gas giants at the same pressure. Though if we applied that logic to Venus, it would suddenly get bigger. $\endgroup$
    – Mark Adler
    Jul 7, 2016 at 4:13
  • $\begingroup$ I found the equation for scale height, but stopped when I realized I don't know any of the other values besides acceleration. I'm guessing that the atmosphere climbs from 1 bar to say 100 bar for example pretty fast, relative to Jupiter's large size? $\endgroup$
    – uhoh
    Jul 7, 2016 at 7:54
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    $\begingroup$ @Gusdor: At least Earth has a fairly consistent sea level. What about Phobos, which isn't even a decent sphere? $\endgroup$
    – SF.
    Jul 7, 2016 at 9:35
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    $\begingroup$ Earth isn't a decent sphere either. Most planets' radii are given as two numbers: equatorial and polar. For Earth, those are 6378 km and 6357 km respectively. That ellipsoidal description is sufficient for most purposes, but the real equipotential surface is more complicated. Bodies like Phobos require a shape model, but a triaxial ellipsoidal model isn't too far off for Phobos. 13 x 11.4 x 9.1 km. $\endgroup$
    – Mark Adler
    Jul 7, 2016 at 14:46

As we don't seem to have come up with an official standard definition, I would have a tendency to simply consider the average density of gas per volume as a reference point. That would simply be the average number of gas molecules measured per cubic centimeter at any given time. After we have measured this, all we need to do is define a reference value stating at which measurement we do consider this statistically significant for defining our gas planet radius.


here is the definition for the earth, the Kármán line My two cents, generally I think for other planets it would be related to where the density of the atmosphere reduces such that it becomes a rarefied gas. That is where the separation between the fluid molecules becomes significant.

If you think of a body entering an atmosphere it will first encounter molecules spaced very far apart such that they have negligible resistance (that said, over many orbits even this rarefied fluid will contribute to orbital decay). As the fluid becomes more dense there would be a transition from flow through a rarefied fluid to that through a Newtonian fluid. I think that would be a good basis for a drawing a line.

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    $\begingroup$ This question is not talking about the perimeter of the atmosphere, but about the surface of the planetary body proper, corresponding (on Earth) to the ground. $\endgroup$ Jul 7, 2016 at 9:03

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