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I've plugged in the data listed for Jupiter into the equation for the Surface Gravity of a sphere: g = (6.67 x 10^-11 N m^2/kg^2)(1.8982 x 10^27 kg)/(6.9911 x 10^7 m)^2. But this equals 25.9 m/s^2. How come this website lists 24.79 m/s^2 for the surface gravity? It can't be because of centripetal acceleration. Using the website's number for v, a_cen = (12.6 x 10^3 m/s)^2/6.9911 x 10^7 m = 2.27 m/s^2. g - a_cen = 23.6 instead.Inkan1969 (talk) 20:50, 7 April 2025 (UTC)[reply]
OK. I see my mistake now. I didn't notice until later that equatorial radius is significantly larger than the average radius. So then the surface gravity listed is only if you use the equatorial radius. Inkan1969 (talk) 23:37, 8 April 2025 (UTC)[reply]
Thanks for the link. I saw this too, and it looked pretty speculative. But it is a peer-reviewed paper in Nature, so it's a reliably-sourced speculative computer model. Might be better to wait for other reactions, before adding this to the article. --Pete Tillman (talk) 18:03, 27 May 2025 (UTC)[reply]