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For those of you who have read mission of gravity, or researched Saturn’s equatorial bulge, you will know that the faster a planet rotates, the greater the “flattening” effect it’s rotation has. This means that planets (like Saturn) which rotate very quickly have a “bulge” at the equator, reminiscent of a hole-less donut or a football that’s been stepped on. And the greater this “bulge” is, the greater the difference is between the surface gravity at the poles and the gravity at the equator.

Now, I have a terrestrial world which, like Saturn, has a tidal bulge. This is a terrestrial planet 1.6 times the mass of earth, with a density of around 5.5 g/cm3. It’s axial tilt is 24 degrees.

I would like this planet to have a gravitational attraction of about 40% earth-g at the equator, and 120% earth-g at the poles. How fast, then, would it have to rotate in order to create this difference?

If there is any information needed to work this out that I have neglected, please tell me! The above information is all I have on this planet so far, so additional data can easily be added.

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    $\begingroup$ You don't even need to have a bulge. The difference in angular speed between the equator and pole is sufficient to reduce the apparent gravity. $\endgroup$
    – Andrew
    Jun 10 at 19:35
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    $\begingroup$ With the gravity force at the pole three times as large, the equatorial bulge is going to be non-negligible. The simple calculations provided in the two answers posted now are going to be wrong by a significant amount. I'm busy trying to set up the calculation but it's evading me. $\endgroup$
    – Boba Fit
    Jun 11 at 1:13
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    $\begingroup$ @Andrew, you don't need a bulge, but you also can't avoid a bulge in the real world. $\endgroup$
    – Logan R. Kearsley
    Jun 11 at 2:05
  • $\begingroup$ A football that's been stepped on? You must be referring to a non-American football... :-D $\endgroup$
    – Michael
    Jun 12 at 1:24
  • $\begingroup$ Coriolis effects are going to be interesting. Variable with lattitude, and evere. $\endgroup$
    – Russell McMahon
    Jun 12 at 4:12

5 Answers 5

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Referring to the formula for how much bulging occurs for a given speed (up to on the first order but it's good enough), we have that the difference in polar and equatorial radii relative to the mean radius $a = \frac{a_e + a_e + a_p}{3}$ is that $$\frac{a_e - a_p}{a} = \frac{5}{4} \frac{\omega^2 a^3}{GM}$$. The only other constraint on $a_e, a_p$ we have is the volume of the planet derived from the mass and density, $\rho \cdot \frac{4 \pi}{3} a_e^2 a_p = M$. And what we want from the equatorial acceleration and polar acceleration is that $$3 g_e = 3 \left( \frac{GM}{a_e^2} - a_e \omega^2 \right) = q_p = \frac{GM}{a^2_p}$$ ignoring the non-spherical components of gravity. We can't decide the exact values as

We know from the volume constraint that $a_p = \frac{a_V^3}{a_e^2}$, where $a_V$ is the radius of the spherical planet. This turns the other two into an equation on $a_e$ and $\omega$. For the gravity magnitude relation, we get a quadratic on $a_e^3$, so $$a_e = \sqrt[3]{\frac{-3\omega^2 + \sqrt{9\omega^4 + 12 \frac{G^2M^2}{a^6_V}}}{2 \frac{GM}{a^6_V}}} \text{ or } \omega^2 = \frac{GM}{a_e^3} \left( 1 - \frac{a_e^6}{3a_V^6}\right)$$

Let $a_e = f a_V$ and we get the final monster formula when subbing into the bulge relation, $$\frac{3f^3 - 3}{2 f^3 + 1} = \left(1 - \frac{f^3}{3} \right) \frac{5}{4} \frac{\left(\frac{2}{3}f^3 + \frac{1}{3}\right)^3}{f^9}$$

This is an absolute monster of a polynomial, a quintic of $f^3$ and definitely not solvable by analytical methods. By graph, the answer is roughly $1.16017$. We could get more decimals, but we have enough error from our approximations. graph of the functions

We thus get our variables as $$ \begin{align} a_v &= \sqrt[3]{\frac{3M}{4 \pi \rho} } = 7458 \text{ km} \\ a_e &= f \cdot a_V = 8652 \text{ km} \\ a_p &= f^{-2} \cdot a_V = 5541 \text{ km} \\ \omega &= \sqrt{\frac{4 \pi \rho G}{3 f^3} \left( 1 - \frac{1}{3} f^6\right) } = 4.293 \cdot 10^{-4} \text{Hz} \\ \implies T &= \frac{2 \pi }{\omega} = 4 \text{ hours } 3 \text{ minutes } 56 \text{ seconds} \end{align}$$

Yeah that's fast. Definitely beyond the range of the approximations taken here.

Edit : BTW the gravitational acceleration at equator is 0.707 G and at the poles is 2.12G. If you scale your radius by $s$ while keeping density fixed, mass changes by $s^3$, while acceleration changes by $s$, but the time period remains fixed. So to get 0.4G at equator and 1.2 at poles at the same density, the planet must have a mass of $0.3 M_\oplus$, radii of $a_V = 4262 \text{ km}, a_e = 3166 \text{ km}, a_p = 4944 \text{ km}$.

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    $\begingroup$ As I get it, you cannot ignore that the planet is aspherical when calculating gravity on the pole, let alone in the center. So while your answer could be good enough for the first approximation, there had to be more to do with calculating exact gravity to an ellipsoid. (I don't remember if an inverse square function if integrated over an ellipsoid would return some pretty looking value, and if not, you'r in a world of too high maths for the entire Stack Exchange) $\endgroup$
    – Vesper
    Jun 12 at 8:32
  • $\begingroup$ I think it's worth pointing out that getting an exact answer for this is sufficiently complex, and dependent on compositional detials of the specific planet in question, that even Hal Clement got the answer wrong by a factor of about 3 when writing Mission of Gravity. $\endgroup$
    – Logan R. Kearsley
    Jun 12 at 13:19
  • $\begingroup$ @Vesper There was a decomposition till the first order in terms of Legendre Polynomials on cos(polar angle), the standard multi-pole expansion, but I really didn't feel up to calculating that. $\endgroup$
    – Varad Mahashabde
    Jun 14 at 1:44
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    $\begingroup$ @LoganR.Kearsley Of course, all possible calculations are dead in the water if you don't consider the planet to be uniform. AFAIK the core being slightly denser than the crust and mantle, which is the part that gets ellipsoidal, is a decently large factor in proper calculation, but at the end of the day they introduce a coefficient on the order of unity in our answer. $\endgroup$
    – Varad Mahashabde
    Jun 14 at 1:46
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    $\begingroup$ @VaradMahashabde Not all possible calculations; it is possible, after all, to do a finite-element numerical simulation to produce arbitrarily accurate results with arbitrary variations in density--you just have to be willing to throw a large enough amount of human effort and computer time at it! $\endgroup$
    – Logan R. Kearsley
    Jun 14 at 3:14
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At the poles the centrifugal force due to rotation is null, thus your question can be reworded as:

given a planet with surface gravity of 12 $m/s^2$, what is the rotation speed which would give a net gravity at the equator of 4 $m/s^2$?

The net gravity at the equator can be calculated as $g_{net}=g-r\omega^2$.

On a first approximation you can use the same radius for the pole and the equator. A more refined model would require accounting for the bulge and use different radia.

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    $\begingroup$ That first approximation will be extremely bad in this case, as there will be a significant departure from sphericality, and you have to account for the different polar and equatorial radii. $\endgroup$
    – Logan R. Kearsley
    Jun 11 at 2:07
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The acceleration due to gravity is given by $g = \frac{GM}{r^2}$. The apparent centrifugal acceleration on the surface is given by $c = \omega^2 r$. You want $g-c$ to be one-third of $g$, so $c=\frac{2}{3}g$:

$\omega^2 r = \frac23 \left(\frac{GM}{r^2}\right)$

$\implies \omega = \sqrt{\dfrac{2GM}{3r^3}}$

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    $\begingroup$ This is incorrect because the planet's radius at the equator is different than at the poles. Normally, this difference is so small it can be ignored for first-order approximations, but in this case I doubt it. $\endgroup$
    – E Tam
    Jun 11 at 1:35
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would like this planet to have a gravitational attraction of about 40% earth-g at the equator, and 120% earth-g at the poles. How fast, then, would it have to rotate in order to create this difference?

You can't.

The shape of the planet will be one in which the surface gravity along the surface is equal at the poles and at the equator.

If you spin it faster, the planet will just bulge further.

The material strength of a reasonably large planet required to have not differ significantly from this is simply insane.

(by "gravity", I mean the sum of the gravitational attraction of the planet plus the inertial effects of rotation).

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    $\begingroup$ Wrong for Earth at the very least, Earth has freefall acceleration difference of about 0.065 m/s^2 at sea level for poles vs equator. So, spinning a planet up could produce the required effect. After all, if a planet is about to get torn apart by its own spin, would it not have zero effective freefall acceleration on its equator? And on the pole, it's nonzero by construction. $\endgroup$
    – Vesper
    Jun 12 at 8:27
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    $\begingroup$ @Vesper your comment is nonsense. the definition of a planet is it forms its own shape under the force of gravity. Now rocks aren't perfectly liquid, and there are other forces, so there can be an imbalance of a few percent, which the answer has not mentioned. If there were any significant imbalance, the body fails to meet the definition of a planet $\endgroup$
    – camelccc
    Jun 12 at 11:39
  • $\begingroup$ @camelccc The surface of a planet must follow an approximate surface of equal potential for its own self-gravity. That is very much not the same thing as a surface of equal force. $\endgroup$
    – Logan R. Kearsley
    Jun 12 at 13:18
  • $\begingroup$ @camelccc No, I think Vesper is right. Take a gravitational+rotational force field. At equipotential a fluid can be at rest in this field. But the force of gravity at each spot is the derivative of this field; that derivative need not be equal over the equipotential surface. The derivative is even normal to the equipotential surface; hence, if you froze the liquid, a ball bearing wouldn't roll on the surface. No need for the body to be rigid at all. $\endgroup$
    – Yakk
    Jun 12 at 14:27
  • $\begingroup$ @yakk this is nonsense, if the force is not equal, so is the pressure of the liquid under the surface not equal, and if the body cant be considered a liquid, its not a planet. Obviously the planets surface must also be an equipotential. which implies no lateral forces so no pressure gradients else the liquid will move. In reality, any such body with this level of spin will fly apart, likely recombining to give a planet with rings $\endgroup$
    – camelccc
    Jun 13 at 14:57
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Stephen H. Dole, in Habitable Planets For Man (1964):

https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf

Discusses the requirements necessary for world to be habitable for human beings (and also for multicellular land animals with the same environmental requirements as humans).

Other scientific discussions of planetary habitability are for liquid water using life in general, and not for humans and lifeforms with the same environmental requirements in particular.

On pages 41 to 46 the cases of rapidly rotating planets which become oblate in shape.

And on pages 58 to 61 Dole discussed limits on the planetary rotation rate of human habitable planets. Dole believed that a planet would be rotating too fast when the surface gravity at the equator dropped to zero and matter was lost from the planet, or when the shape of the surface became unstable and axial symmetry was lost. Dole believed that for planets in he size range of Human habitable worlds, a rotation so fast that a single rotation took 2 to 3 hours would result in an unstable planetary surface.

So you should check whether a rotation rate fast enough for to there to be a surface gravity at the poles 3 times as high as at the equator would be safe and stable for the planet.

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