Jupiter's Equatorial Rotation Period

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Jupiter's equatorial rotation period can be calculated by comparing centrifugal force to gravitational force, as the material will be torn apart when the former exceeds the latter. The equations for gravitational force (Fg) and centrifugal force (Fc) can be set up as an inequality to solve for the rotation period (P). Using Jupiter's radius (7.14 x 10^7 m) and mass (1.9 x 10^27 kg), the derived rotation period is approximately 7.39 minutes, significantly shorter than its current period of 590.5 minutes. The approach taken in the calculations appears correct, though confidence in the results can vary. This method effectively illustrates the relationship between rotation speed and gravitational stability.
hfenton
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How would I find Jupiter's equatorial rotation period when it begins to tear the planet apart?

Is it true that if the centrifugal force is greater than the gravitational force the material can be torn apart?

If this is right, can I just set up an inequality for the centrifugal force > gravitational force

Fg = GM/R^2 Fc = (4(pi^2)R) / P^2

and then just solve for P throughout the inequality...

P > sqrt[ (4(pi^2)R^3)/ GM) where R(Jupiter) = 7.14*10^7 and M (Jupiter) = 1.9*10^7

If I do this, I will get a rotation period of 7.39 minutes. I am wondering if the way I went about the problem is right. As far as Jupiter's current rotational period of 590.5 mins, the period at which it would be torn apart is nearly 80x shorter. Any other suggestions for doing this problem?
 
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Sounds reasonable, what's the problem?
 
I just am never very confident with my answers. Thanks for responding though!
 
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