Wikipedia's reference for Saturn's gravity gives $10.44 m/s^2$ at the equator, but this conflicts with Britannica, which gives $8.96 m/s^2$ at the equator and $12.14 m/s^2$ at the poles. All values are at whatever elevation has a pressure of $1 bar$.

I've noticed this has led to the funny situation of people Goggling the subject and erroneously citing that one would weigh more on the surface of Saturn... which is wrong. It's certainly an understandable error, because the Wikipedia number could mean one of several things:

  • "the literal gravity vector on the surface in an inertial reference frame"
  • the apparent surface acceleration in the reference frame of a point on the surface
  • EDIT: A false metric of gravity by just using the mass and equatorial radius

Not so clear is who lies at fault here. Both Wikipedia and Britannica call their numbers "gravity". Here's Britannica's very intentional wording:

The equatorial gravity of the planet, 896 cm (29.4 feet) per second per second, is only 74 percent of its polar gravity.

How correct is this wording? What is the best way to handle the subject?

  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented May 28, 2013 at 11:31

2 Answers 2


It seems that Britannica is more accurate.

The Britannica values of $g$ can be traced back to Table 2 of this paper: The atmosphere of Saturn - an analysis of the Voyager radio occultation measurements, Lindal et al (1985). As stated in the title, these values are derived from the Voyager measurements.

At first, I thought that the wiki/Nasa value was based on the more recent Casini measurements. However, it turns out that this value actually comes from a simplified (and inaccurate) calculation.

I haven't found an explicit recent value of $g$ in the literature, but I did find out how to calculate it: on page 3 of the paper Interior Models of Saturn: Including the Uncertainties in Shape and Rotation (Helled & Guillot, 2013), it is stated that the effective potential of a rotating planet is $$ U(r,\varphi) = \frac{GM}{r}\!\left(1 - \sum_{n=1}^\infty\left(\frac{r_\text{s}}{r}\right)^{2n}J_{2n}P_{2n}(\sin\varphi)\right) + \frac{1}{2}\omega^2r^2\cos^2\varphi, $$ where $M$ is the mass of the planet, $r_\text{s}$ is a reference equatorial radius, $\varphi$ is the latitude, $\omega=2\pi/P$ is the angular velocity (and $P$ is the rotation period), $P_{2n}(x)$ are Legendre polynomials and $J_{2n}$ are coefficients that express the deviation of the planet from a perfect sphere. The gravitational acceleration $\mathbf{g}$ is then a vector with components $$ \begin{align} g_r(r,\varphi) &= -\frac{\partial U}{\partial r},\\ g_\varphi(r,\varphi) &= -\frac{\partial U}{\partial \varphi}. \end{align} $$ Since $g_\varphi=0$ at the poles and the equator, let's focus on $g_r$ only. We find $$ g_r = \frac{GM}{r^2}\!\left(1 - \sum_{n=1}^\infty(2n+1)\left(\frac{r_\text{s}}{r}\right)^{2n}J_{2n}P_{2n}(\sin\varphi)\right) - \omega^2r\cos^2\varphi, $$ which is the same as the formula given in the Appendix of Lindal et al (1985). Now let's plug in some numbers: from Helled & Guillot we have $$ \begin{align} r_\text{s} &= 60,330\;\text{km},\\ J_2 &= 16,290.71\times 10^{-6}, \end{align} $$ ignoring the smaller higher-order coefficients $J_4$, $J_6$,...

In Table 2 of Anderson & Schubert (2006) we find: $$ \begin{align} a &= 60,357.3\;\text{km},\\ q &= 0.158904,\\ P &= 10^\text{h}32^\text{m}35^\text{s} = 37,955\;\text{s}, \end{align} $$ where $a$ is the equatorial radius and $q$ the so-called smallness parameter. From this, we obtain $$ \begin{align} \omega &= 2\pi/P = 1.65543\times 10^{-4}\;\text{rad}\,\text{s}^{-1},\\ GM &= \omega^2a^3/q = 3.79207\times 10^{16}\;\text{m}^3\,\text{s}^{-2}. \end{align} $$ Finally, we need the equatorial and polar radius at 1 bar. According to the IAU report on cartographic coordinates and rotational elements (2006), pag. 173, we find $$ \begin{align} r_\text{eq} &= 60,268\;\text{km},\\ r_\text{pol} &= 54,364\;\text{km}. \end{align} $$ Thus, at the equator we have (ignoring higher-order terms) $$ \begin{align} g_r(r_\text{eq},0^\circ) &= g_1 + g_2 + g_3\\ &= \frac{GM}{r_\text{eq}^2} + \frac{GM}{r_\text{eq}^2}\frac{3}{2}\!\frac{r_\text{s}^2}{r_\text{eq}^2}\!J_2 - \omega^2r_\text{eq}, \end{align} $$ so, the gravitational acceleration comes from three contributions: the gravity of a spherical planet, a correction due to non-sphericity and a centrifugal influence due to the rotation. We obtain $$ \begin{align} g_1 &= 10.44\;\text{m}\,\text{s}^{-2},\\ g_2 &= 0.256\;\text{m}\,\text{s}^{-2},\\ g_3 &= -1.652\;\text{m}\,\text{s}^{-2}, \end{align} $$ so that the acceleration in a frame rotating with the planet is $$ g_r(r_\text{eq},0^\circ) = 9.04\;\text{m}\,\text{s}^{-2}. $$ Contributions of the $J_4$, $J_6$ terms may change the last digit.

Now we see where the 10.44 value comes from: it is only the spherical term, incorrectly ignoring the effects of non-sphericity and rotation. At the poles, we have $$ \begin{align} g_r(r_\text{pol},90^\circ) &= g_1 + g_2\\ &= \frac{GM}{r_\text{pol}^2} - \frac{GM}{r_\text{pol}^2}\!3\!\frac{r_\text{s}^2}{r_\text{pol}^2}\!J_2, \end{align} $$ which results in $$ \begin{align} g_1 &= 12.83\;\text{m}\,\text{s}^{-2},\\ g_2 &= -0.772\;\text{m}\,\text{s}^{-2},\\ g_r(r_\text{pol},90^\circ) &= 12.06\;\text{m}\,\text{s}^{-2}. \end{align} $$ So the equatorial acceleration is indeed about 75% of the acceleration at the poles. Conclusion: Britannica is more accurate, although the values are not up-to-date.

  • 1
    $\begingroup$ That 10.44 number is... significantly more wrong than what I expected. It actually comes from just $GM/r_{eq}^2$. Even if the equatorial "gravity" is meant to not include rotational acceleration, it seems that should be 10.44+0.256= 10.696 $\endgroup$ Commented May 28, 2013 at 12:50
  • 1
    $\begingroup$ @AlanSE Yes, it surprised me as well. I didn't expect such sloppiness from Nasa. $\endgroup$
    – Pulsar
    Commented May 28, 2013 at 12:58

Although somewhat tangential, I want to show a separate approach to get valid numbers for Saturn's gravity. For this, I will use Gauss' Law for gravitation and rotation to get the planet's average gravity. I summarized some methods here, so I will not go into them now.

For the average apparent gravity (that is, including rotational acceleration), we can divide the total apparent gravitational flux by the surface area. That gravitational flux (or perhaps more accurately, "acceleration flux") is composed of two terms, one from the matter and one from the rotation. The apparent gravity is always normal to the "surface", and by surface the 1 bar isobaric surface. Thus, this average is the area-averaged downward acceleration.

$$ g_{avg} = \frac{ 4 \pi G M - 2 \omega^2 V}{ SA } $$

The merit of this approach may be open for discussion, because if you used local gravity to derive the mass or density, then you might just be back-calculating the measured quantity you obtained in the first place. There is still hope, however, that the mass of Saturn was derived from large-scale orbital dynamics (likely), and that surface area and volume might have been obtained from purely observational/geometric evidence (also likely). Going by the Wikipedia set of data (which is questionable, as we've shown), I obtain this result:

$$ g_{avg} = \frac{ 4 \pi G \left( 5.6846 \times 10^{26} kg \right) - 2 \left( \frac{2 \pi}{10.57 hr} \right)^2 \left(8.27 \times 10^{14} km^3 \right) }{ 4.27 \times 10^{10} km^2 } = 10.109 \frac{m}{s^2} $$

Compare to known correct figures which range around $9.0 m/s^2$ for the equator and $12.1 m/s^2$ for the pole. These are the nearest significant figure and approximate average for the numbers given by Britannica Encyclopedia and user Pulsar. The former may be preferable, but these are overall good consensus figures. All the numbers I'm citing here are what I call "apparent" gravity, which includes the subtraction due to rotation.

We find that my calculation for average apparent surface gravity is about $35.4 \text{%} $ between the two. That is, my average gravity calc is closer to the experienced equatorial gravity than the polar gravity. This sounds consistent with an area-averaged value. If rotation was the only factor (no matter redistribution), we could integrate $\sqrt{x^2+y^2}$ on the unit sphere to find the average value, and doing this I find that the average would be $21 \text{%}$ the difference between the equatorial and polar gravity values. I believe this shows consistency with the results from the "why is the Earth so fat" question where people consistently found that models that didn't include matter redistribution were off by a significant factor.

With confidence, I would report these gravity figures with rotation included, in $m/s^2$

  • Equator 9.0
  • average 10.1
  • Pole 12.1

To get the numbers excluding rotation, you can repeat my calculation excluding the $-2 \omega^2 V$ term. The polar number is the same for both because it has no rotational acceleration. Then the equator number is trickier, but others have found this for me.

  • average 11.2
  • Equator 10.7 (from Pulsar)

In a sense, these are all valid numbers for the gravity of Saturn for various stipulations. Not among these is the 10.44 number, which has no valid physical interpretation.


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