5 ms·
How is the size defined for a gas planet? The gas density just keeps dropping, where do you draw the line (isosurface, rather)? Earth's radius is always the one
by est31 1y ago
How is the size defined for a gas planet? The gas density just keeps dropping, where do you draw the line (isosurface, rather)? Earth's radius is always the one without earth's atmosphere.
- Maxatar 1y agoIt's defined as the distance from the center of mass to the point where the pressure is equal to the pressure on Earth at sea level.
- amelius 1y agoIt's a matter of definitions, so we skip them and just choose something that makes sense to humans.
- philipov 1y agoThe most important thing about definitions is that we apply them consistently. A different definition might give different answers, but it's fine as long as it does so uniformly.
- JumpCrisscross 1y ago> most important thing about definitions is that we apply them consistently The most important consideration for a definition is its practical consequence. In this case, whether the line is drawn at 1 bar or an order of magnitude more or less doesn’t materially change that, on the same measure, Jupiter was 2x larger in the past. (Less than 1% in both cases.) In a different context, that difference may be meaningful and should thus be noted and tested for robustness.
- amelius 1y agoThe point is that there will be multiple definitions, so which one do you choose? From there your conclusion can be that we just use a loose definition that humans can easily grasp.
- jessriedel 1y agoAs others note, the definition of Jupiter’s radius is set by where the pressure is 1 bar. This is somewhat arbitrary, but the arbitrariness doesn’t matter much: the pressure drops to 1 microbar just 320 km higher, which is <0.5% of Jupiter’s ~70,000 km radius.
- gus_massa 1y agoFor comparison, extracting the numbers from the graphic in page 3 of https://projects.iq.harvard.edu/files/acmg/files/intro_atmo_chem_bookchap2.pdf https://projects.iq.harvard.edu/files/acmg/files/intro_atmo_... 1 microbar on Earth is like 50Km, that is 50/6400 ~= 0.8%
- queuebert 1y agoVenus would get a slight radius buff, too, if we applied that metric.
- hnuser123456 1y agoBut Venus has a solid surface.
- queuebert 1y agoYes, that's why I said buff. The radius as defined would increase, because surface pressure is 90 bar, so at 1 bar, you're pretty high in the atmosphere. I can see merit in such a definition because that is the level at which we wouldn't have to pressurize our space stations to be comfortable. (Really 1/3 bar is fine too.)
- hnuser123456 1y agoAh, I see what you're saying, didn't know the surface pressure was so high! I suppose Titan would also get an atmosphere radius buff!
- layer8 1y agoThe density falls off pretty steeply at the “edge”, so the exact definition only makes little difference for the radius: https://www.researchgate.net/figure/Density-vs-radius-for-a-realistic-Jupiter-model-solid-curve-a-toy-model-with-rc10-g_fig4_258247486 https://www.researchgate.net/figure/Density-vs-radius-for-a-...
- formerly_proven 1y agoThis is because of Newtonian gravity being inversely proportional to the square of the radius, right?
- skykooler 1y agoGravity changes little over that distance - it's more because of the compounding effect of atmospheric pressure (the deeper you go, the more air you have above you which raises the pressure, raising the density and meaning that pressure increases exponentially faster).
- vecter 1y agoWhat makes that curve exponential?
- westurner 1y agoNewtonian gravity (classical mechanics). Two-body gravitational attraction is observed to be an inverse square power law; gravitational attraction decreases with the square of the distance. g, the gravitational constant of Earth, is observed to be exponential; 9.8 m/s^2. Atmospheric pressure: https://en.wikipedia.org/wiki/Atmospheric_pressure#:~:text=Pressure%20(P)%2C%20mass%20(,atmospheric%20mass%20above%20that%20location. https://en.wikipedia.org/wiki/Atmospheric_pressure#:~:text=P... : > Pressure (P), mass (m), and acceleration due to gravity (g) are related by P = F/A = (m*g)/A, where A is the surface area. Atmospheric pressure is thus proportional to the weight per unit area of the atmospheric mass above that location.
- deleted 1y ago[deleted]