<p>Consider the dimensionless parameter <span class="math-container">$k$</span> for some planet, given by
<span class="math-container">$$k \equiv \frac{\Omega^2 a^3}{G M}$$</span>
where</p>

<ul>
<li><span class="math-container">$\Omega$</span> is the planet's equatorial sidereal rotation rate,</li>
<li><span class="math-container">$a$</span> is the planet's equatorial radius,</li>
<li><span class="math-container">$G$</span> is the universal gravitational constant, and</li>
<li><span class="math-container">$M$</span> is the planet's mass.</li>
</ul>

<p>The values for this dimensionless parameter are</p>

<ul>
<li><a href="https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%28sidereal+day%29%29%5E2*%28Earth+equatorial+radius%29%5E3%29%2F%283.986004418e14+m%5E3%2Fs%5E2%29" rel="noreferrer">0.003461</a> for the Earth,</li>
<li><a href="https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%289.925+hours%29%29%5E2*%28Jupiter+equatorial+radius%29%5E3%29%2F%281.26686534e17+m%5E3%2Fs%5E2%29" rel="noreferrer">0.08919</a> for Jupiter, and</li>
<li><a href="https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%2810.56056+hours%29%29%5E2*%28Saturn+equatorial+radius%29%5E3%29%2F%283.7931187e16+m%5E3%2Fs%5E2%29" rel="noreferrer">0.1576</a> for Saturn.</li>
</ul>

<p>The factor <span class="math-container">$k$</span> 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 <span class="math-container">$\frac54k$</span>. 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 <a href="http://farside.ph.utexas.edu/teaching/336k/Newton/node109.html" rel="noreferrer">Rotational Flattening</a>, for example. Note well: The derivation at this site involves a bit of hand-waving. Overcoming the hand-waving involves even more mathematics.)</p>

<p>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 <span class="math-container">$k$</span> 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 <span class="math-container">$k$</span> for Saturn is 1.76 times that of the value for Jupiter.</p>

CC BY-SA 4.0
2020-05-23T16:03:11.643Z
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?
user177107
CC BY-SA 4.0
2020-05-23T16:34:38.130Z
@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.
David Hammen
CC BY-SA 4.0
2020-05-23T16:34:54.927Z
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.
David Hammen
CC BY-SA 4.0
2020-05-23T10:53:10.337Z
2020-05-23T11:09:23.703Z
2020-05-23T11:09:23.703Z
David Hammen
<p>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. </p>

<p>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 <em>vs.</em> 9,276 km) despite rotating slower <strong><em>and</em></strong> being smaller. Am I missing something? </p>

<rotation><jupiter><saturn>
Why does Saturn have a more prominent equatorial bulge and higher flattening ratio than Jupiter despite rotating slower?
user177107
# Why does Saturn have a more prominent equatorial bulge and higher flattening ratio than Jupiter despite rotating slower?

(Asked by: username_0 on 2020-05-22)

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?

**Tags:** rotation, jupiter, saturn

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## 1 answers

### Answer by username_1 on 2020-05-23. Score: 8 | Accepted answer

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

* [0.003461](https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%28sidereal+day%29%29%5E2*%28Earth+equatorial+radius%29%5E3%29%2F%283.986004418e14+m%5E3%2Fs%5E2%29) for the Earth,
* [0.08919](https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%289.925+hours%29%29%5E2*%28Jupiter+equatorial+radius%29%5E3%29%2F%281.26686534e17+m%5E3%2Fs%5E2%29) for Jupiter, and
* [0.1576](https://www.wolframalpha.com/input/?i=%28%28%282*pi%29%2F%2810.56056+hours%29%29%5E2*%28Saturn+equatorial+radius%29%5E3%29%2F%283.7931187e16+m%5E3%2Fs%5E2%29) for Saturn.

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](http://farside.ph.utexas.edu/teaching/336k/Newton/node109.html), 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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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.  
> **username_1**: 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.  

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