Jupiter's Equatorial Rotation Period

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SUMMARY

The discussion centers on calculating Jupiter's equatorial rotation period at which the centrifugal force exceeds gravitational force, potentially leading to the planet's disintegration. The user proposes using the inequality Fg = GM/R^2 and Fc = (4(pi^2)R) / P^2 to derive the rotation period. By substituting Jupiter's radius (R = 7.14 x 10^7 m) and mass (M = 1.9 x 10^27 kg), the user calculates a critical rotation period of approximately 7.39 minutes, significantly shorter than Jupiter's current rotation period of 590.5 minutes. The approach is mathematically sound and highlights the dramatic difference in rotational dynamics.

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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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