# How is the diameter of a gas giant calculated?

As we know atmospheres of celestial bodies don't just stop at a given distance. They gradually become less dense as you move away from the center.

I understand that the diameter of stars is typically given to be that of their photosphere (i.e. the boundary where a laser shined from behind the star toward an observer would just barely be visable).

I assumed that a similar method using optical depth also applied to gas giants. However wikipedia currently has the somewhat confusing explanation:

As Jupiter has no surface, the base of its atmosphere is usually considered to be the point at which atmospheric pressure is equal to 1 MPa (10 bar), or ten times surface pressure on Earth.

Which isn't quite the same thing as diameter, but it got me thinking: When wikipedia or a textbook or whatever lists a diameter for a gas giant how is that figure arrived at?

• Related: Why is Jupiter so sharply defined?. Jan 4, 2015 at 14:46
• That Wikipedia statement is crazy. They're talking about the "base" of the atmosphere (whatever that means), but the size is based on the "surface." Giant planets of course don't have a surface, so the "1-bar surface" is used as an arbitrary convention to define the size. Jan 30, 2022 at 0:54

I won't argue with the wikipedia definition (although the NASA Jupiter fact sheet lists it as the radius at 1 bar), but just to point out that the scale height of the atmosphere of Jupiter is given by, $$h \sim kT/m g$$, where $$T$$ is the temperature, $$m$$ is the mean molecular mass and $$g$$ is local gravity. Putting in some numbers: $$g \simeq 24.8$$ m/s$$^2$$, $$m \simeq 2.2 m_u$$ (atomic mass units), $$T \simeq 165$$ K, gives $$h \simeq 25$$ km. Thus the pressure changes extremely rapidly compared with the actual radius of a gas giant and is very small compared to the difference, for instance, between the polar and equatorial radius for a gas giant (like Jupiter) with significant rotation.
So, whether you give a planet's radius at 1 bar or 10 bar isn't going to make a lot of difference ($$<100$$ km).