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I am hoping to create a planet with a greater rotational speed, making it oblong, thus causing differential gravitaion from the equator to the poles. I am hoping for the gravitarion to be lighter than the moon (possibly just heavy enough to prevent a person from jumping off the planet [the tangential speed would factor into this]) at the equator, and heavier than Earth's gravity (possibly by a lot) at the poles. I am hoping the difference can be relatively substantial, but I'm not sure how that would work, exactly. If the shape of a planet based on the various factors of rotational speed, mass/gravity, etc. can be obtained (a graph, like y = x^2), that would be most helpful.

I am also hoping to have the planet be "always day", such that there is always a relatively large amount of light impacting the entire surface. I considered having just the rotational speed account for this, but that wouldn't work with the necessary mass of the planet. The rotational speed could help factor into it, though. It was suggested to me that the atmosphere be denser, thus refracting the light from the star, which could possibly allow a person to see the sunrise and sunset at the same time (nifty). Plus, with the change in density, there would be a change in pressure, thus creating differences in planetary chemistry that would be interesting to explore. And as buoyancy is determined by density (I believe), it could result in certain elements floating a ways up in the atmosphere (a literal layer of water floating in the air sounds intriguing). As a last resort, multiple stars can be used. I'm hoping to simply "manipulate" the light, but a multi-star system could work if it would be the best possible option.

After the proposed rotational speed, mass, etc. has been presented, any additional effects they would cause on the planet would be very interesting, and greatly appreciated.

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    $\begingroup$ See Mesklin. $\endgroup$
    – JDługosz
    Jan 17, 2017 at 22:33
  • $\begingroup$ @JDlugosz Yeah, I saw that. Very intriguing. Thanks for the link. $\endgroup$
    – Iter
    Jan 18, 2017 at 0:21
  • $\begingroup$ See also the older Q A planet with changing gravity for some ideas that may apply. $\endgroup$
    – JDługosz
    Jan 18, 2017 at 0:57
  • $\begingroup$ There are several Qs about always daylight, and one in particular (I can’t find right now) I recall where the impossibility of having stars on both sides of the planet is discussed. $\endgroup$
    – JDługosz
    Jan 18, 2017 at 1:08
  • $\begingroup$ You should read up on Mesklin en.wikipedia.org/wiki/Mesklin $\endgroup$
    – Tim B
    Feb 1, 2017 at 12:22

2 Answers 2

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It is certainly possible to have a planet with Moon-like gravity at the equator and heavier-than-Earth at the poles, if it rotates fast enough.

Unfortunately, figuring out exactly what shape a planet of a given mass would have at various rotational velocities, and what the gravitational force curve looks like across the surface, is rather complicated. If you're not too afraid of math, you can check out this lecture on calculating the equilibrium shape of the Earth for an example. It particular, it depends on the internal structure of the planet, and above certain critical values of angular momentum, there are actually multiple solutions, with rapidly-spinning bodies potentially taking on some pretty funky multi-lobed shapes. Fortunately, however, a simple oblate ellipsoid is always one of the potential solutions (as long as the rate of spin is low enough that the planet doesn't fly apart completely, anyway), and there are ways to simplify the problem. Mesklin, for example (already mentioned in the question comments) is assumed to have a degenerate-matter core, so the majority of its mass is concentrated near the center, and you can approximate the correct solution by ignoring the effects of deforming the distribution of the rest of the planet's mass away from spherical.

If you do something similar with your world, then we can figure out its shape given a certain mass and rotational period, and the gravitational force as a function of latitude, by finding the equipotential surface for the combination of centrifugal and gravitational potentials.

The effective potential is $U(r,\theta) = -\frac{1}{2}(\omega r \cos\theta)^2 - \frac{GM}{r}$, where $r$ is radius, $\theta$ is latitude, and $\omega$ is angular velocity. Setting $U$ to a constant value and solving for $r$ is kinda gross (it's a cubic equation), so you'll probably want to use a graphing program to figure out the exact shape numerically, but once you've got that, it's easy to calculate the gravity at the poles vs. the equator, and you can work backwards from there guessing-and-checking until you get a combination of polar gravity, equatorial gravity, and rotation rate that you like.

The "always day" bit is trickier, and hinges on just what counts as "always" and "day". Is it sufficient, for example, if "always" means "for a few hundred years at a time in one hemisphere" (plenty of time and space for a story to take place), and for "day" to mean "consistently illuminated at least to the level of a well-lit interior room, but with variations in brightness and heat above that point permitted"? In that case, I'd just put this planet in a system with a distant companion star on a long-period orbit highly inclined to the plane of the planet's orbit around its primary sun; that way, the companion star can provide constant illumination over one whole hemisphere, so there's never a real "night", on top of the seasonal and daily variations in illumination at any given point due to the primary sun.

Another option might be to make the world a rogue planet near an active galactic nucleus (quasar), which could provide habitable levels of insolation at several light years distance, with millenia-long orbits. It's the same general idea, just replacing the companion star with something bright enough to do the whole job of warming the planet all on its own, eliminating the need for another sun with its own superimposed illumination cycles, while still maintaining a sufficient distance to make the period of constant summer daylight seem like forever. Quasars, of course, have the downside of turning on and off rapidly and somewhat unpredictably, and they tend not to live in very nice neighborhoods, but the purposes of a good story I'd be willing to believe that, somewhere in the universe, there's a quasar that just happens to have had sufficiently steady output for sufficiently long to make this one unique world possible.

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  • $\begingroup$ I love your description of the planet's shape with differential gravity. I have some basic, college-level knowledge of Calculus, but sadly my physics knowledge is not yet great enough that I easily understand all of it. Nevertheless, I'm attempting to manipulate the equation you provided, using Desmos Graphing Calculator, to see if I can get it to work and fully understand it in the process. I fairly certain that the G is gravitational constant and M is mass, and you explained everything else. But why is the equation set to U of r and theta? Overall, excellent answer. As for "constant day" $\endgroup$
    – Iter
    Aug 4, 2017 at 1:18
  • $\begingroup$ I was considering having the planet sit in a Lagrange Point in a twin-star solar system. Though, depending on the rotation to-be-decided, a system such as one you described may be a better alternative. If I can figure out how to properly manipulate that equation, I think this may be the answer to my question. As an added note, I may forgo trying to create a constant day via sun/star light, and opt for bioluminescence that gives light throughout a good bit of the planet. While every answer is helpful to pull ideas from, this has definitely been very helpful. $\endgroup$
    – Iter
    Aug 4, 2017 at 1:26
  • $\begingroup$ G is the indeed the gravitational constant, and M is the mass of the planet. The equation is in terms of $U(r,\theta)$ because the potential and gravitational force for a non-spherically symmetric world varies with both the radius and the latitude. Which is why, if you set $U$ to a constant value, you find that $r$ varies with $\theta$- which is what describes the non-spherical shape of the world. As far as sitting at a Lagrange point goes, only L1 would give you constant daylight, and that's extremely dynamically unstable. No planet could form there without active technological support. $\endgroup$ Aug 4, 2017 at 2:27
  • $\begingroup$ I was going to edit my answer, but it would have made it a poor copy of yours :-D ... so I'm saving time and recommending yours to be the accepted one. $\endgroup$
    – LSerni
    Aug 4, 2017 at 16:08
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Without getting into the math, carving the planet out of a cube shape, taking out close to half an ellipse from the side facing its star, with the shorter axis perpendicular to the star direction, would yield the desired properties.

It would be thicker at the poles, thus there would be more mass to attract smaller objects like humans standing there; atmosphere could fill most of the carved sphere, thicker at the poles again.

As long as it is close enough to the star to be tidally locked, probably an M star so the planet has to be quite close to be in the habitable zone, having constant daylight is natural.

I will try to draw and make the calculations later, but hopefully you get the idea.

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  • $\begingroup$ Why do so many people just toss out probably be tidally locked? A spin-orbit resonance of 3:2 is what we would expect. $\endgroup$
    – JDługosz
    Jan 17, 2017 at 22:36
  • $\begingroup$ Well, they may toss out the suggestion for other questions. For my proposal, I don't think spin-orbit would work, as the planet is not your regular spheroid, but a 3D negative of one; so it seems to me that with your suggestion, the poles of the planet would eclipse its own middle latitudes often enough... $\endgroup$ Jan 17, 2017 at 22:47
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    $\begingroup$ With a less-than-perfect spherical mass (it’s lopsided) then the slightest eccentricity will disuade a locked rotation and encourage a half odd multiple. $\endgroup$
    – JDługosz
    Jan 17, 2017 at 22:49
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    $\begingroup$ I think I can visualize that. Maybe not perfectly, but I think I have at least a partial understand of what you're saying. Interesting using geometry like that, but could that really exist very long in an orbit? It would seem that it couldn't support a moon, so there's no satellite to protect it from meteors. It would have to rely on its closeness to the star to burn the debris, but at that point, could you properly inhabit the planet? $\endgroup$
    – Iter
    Jan 18, 2017 at 0:36
  • $\begingroup$ Well, the specific geometry suggested would probably need to be artificial; similar to RingWorld, it would need some stabilization every so often, and defenses for the debris. That being said, M star systems can be older and clearer from debris for that reason. $\endgroup$ Jan 18, 2017 at 17:00

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