It is my understanding that a planet's equatorial bulge is caused by the centrifugal force produced by its rotation, and the faster a planet rotates the bigger the bulge and flattening ratio of a planet is.

Jupiter's rotational period is 9.925 hours while Saturn's is 10.56 hours. However, Saturn not only has a higher flattening ratio than Jupiter, but also has a bigger equatorial bulge (11,801 km vs. 9,276 km) despite rotating slower and being smaller. Am I missing something?


1 Answer 1


Consider the dimensionless parameter $k$ for some planet, given by $$k \equiv \frac{\Omega^2 a^3}{G M}$$ where

  • $\Omega$ is the planet's equatorial sidereal rotation rate,
  • $a$ is the planet's equatorial radius,
  • $G$ is the universal gravitational constant, and
  • $M$ is the planet's mass.

The values for this dimensionless parameter are

The factor $k$ is essentially the ratio of centrifugal effects to gravitational effects at the "surface" at the equator of the planet, surface in scare quotes because giant planets don't quite have a surface. Ignoring higher order terms, the flattening of a uniform density planet in hydrostatic equilibrium can be shown to be $\frac54k$. The mathematics that go into this derivation are a bit hairy, even for a planet of uniform density, so for now I'll just state this as a fact. (But see Rotational Flattening, for example. Note well: The derivation at this site involves a bit of hand-waving. Overcoming the hand-waving involves even more mathematics.)

The assumption of uniform density is not that good. Even if one does allow for density variations as a function of radial distance from the center of the planet, the parameter $k$ defined above plays a significant role in the computation of what the flattening should be. The observed flattening of Saturn versus that of Jupiter is consistent with the fact that the value of $k$ for Saturn is 1.76 times that of the value for Jupiter.

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    $\begingroup$ But in order to calculate k, we'd have to know the value for a planet's equatorial radius, which is the radius if it was a perfect sphere + the equatorial bulge. Is it not possible to calculate the equatorial bulge on its own? $\endgroup$
    – user177107
    May 23, 2020 at 16:03
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    $\begingroup$ @user177107 - $a$ is the observed equatorial radius. Calculations that involve non-uniform density and higher-order spherical harmonics terms have been done, but they all involve the ratio outlined in my answer. The observed value of $k$ for a given planet and the observed size of the planet's equatorial bulge give a clue regarding density variations within a planet, and also give a clue regarding how far away the planet is from hydrostatic equilibrium. $\endgroup$ May 23, 2020 at 16:34
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    $\begingroup$ For example, the ongoing Juno mission to Jupiter has revealed knowledge of Jupiter's interior just by observing Jupiter's exterior shape and by observing how the spacecraft orbits Jupiter. The observations of the spacecraft's non-Keplerian orbit yield information on Jupiter's spherical harmonic gravity coefficients, some of which are intimately coupled with Jupiter's equatorial bulge. $\endgroup$ May 23, 2020 at 16:34

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