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The book Mission of Gravity describes a planet with a mass of 16 times Jupiter spinning at a rate of one revolution every 17.75 minutes. This causes it to stretch out into an oblate spheroid and has an effective gravity of ~3g at the equator but a gravity at the poles that varies depending on how you calculate it from either 275g right the way through to 700g.

Leaving aside the improbability of the planet forming without gathering enough hydrogen to become a star, and without losing that enormous spin, and managing to cool down enough...

What I don't understand is the surface configuration. The force of rotation will indeed counteract gravitational pull so you get an effective 3g gravitational pull at the equator. However that will also cause the mass to move towards the equator, reducing the gravity at the poles.

Modelling this planet at my head then anywhere that effective gravity increases in strength is effectively lower potential energy "down" from anywhere that it doesn't. I would expect this to cause the planet to deform and reshape so that the effective gravitation was equal at all points on the surface.

In other words it would be 3g everywhere, because at the poles it would be much thinner than at the equator, but not have the spin throwing you away.

Is this mental model correct? Or have I missed something? Is the fact that the spin is throwing you away from the poles the missing factor?

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  • $\begingroup$ Well, on Earth $g$ is not the same everywhere. It's larger at the poles than the equator. No answer, because I can't comment on a planet of such extreme dimensions, but see en.wikipedia.org/wiki/Gravity_of_Earth $\endgroup$
    – zeldredge
    Commented Apr 23, 2015 at 20:05
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    $\begingroup$ @zeldredge Yes, but that's a variation over the entire surface of 0.7% - which I'd expect to be within the strength of the various materials making up the planet to allow it to hold that shape. $\endgroup$
    – Tim B
    Commented Apr 23, 2015 at 20:24
  • $\begingroup$ Ah, right. I wasn't sure from your question if you were objecting to the existence of a variation in general. $\endgroup$
    – zeldredge
    Commented Apr 23, 2015 at 20:25
  • $\begingroup$ Yeah, it was a good point. I should have mentioned that I can see small variations making sense. But variations that dramatic I can't picture in my mind how it would work and not have everything move out from the center until you had effective 3g everywhere. Planetary crust isn't that strong. $\endgroup$
    – Tim B
    Commented Apr 23, 2015 at 20:28
  • $\begingroup$ You might want to read this other question and its answers. It's for a non-rotating oblate spheroid but you can relatively easily take into account rotation by subtracting a contribution that depends on the latitude. $\endgroup$
    – wltrup
    Commented Apr 23, 2015 at 20:40

3 Answers 3

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Even though you were wrong, you nailed the critical question at the heart of the necessary understanding. Bold is my emphasis:

Modelling this planet at my head then anywhere that effective gravity increases in strength is effectively lower potential energy "down" from anywhere that it doesn't. I would expect this to cause the planet to deform and reshape so that the effective gravitation was equal at all points on the surface.

A planet will deform and reshape so that the gravitational potential along its surface is constant. This is not the same thing as gravitational field, and you can follow this back to the physics 101 rigor of separating the notions of fields and potentials. On large scales, you should be able to walk along the planet's surface without doing any work. Our brains want to extrapolate this statement to also imply that gravity is constant over the surface. Actually, it only dictates the direction of gravity is constant, and not its strength. In a pseudo 2D world, the basic idea works as follows:

curved arrows gravity

(EDIT: I have updated the image to be more accurate. The initial model had arrows going straight down, but this was problematic because it creates non-conservative fields, and I don't want to imply that is the case here. In the updated picture, the blue line could be reasonably imagined to be equipotential although my art skills are not great.)

In this weird world, you could take a train over to you neighbor's house and find the gravity to be noticeably different. This story does not violate the necessary physical criteria.

You might notice other strange things. For instance, if planes flew at an altitude where the air's density was a certain value, then we would find that they fly higher in the equatorial regions than in the polar regions. In fact, this effect is intensified further by the shape of the rotational false field. Since Mesklin rotates so fast, the height of a space elevator from the equator might only be a few 100 km or so. If you flew a plane high enough, the equations tell us you might make it all the way into space! In actuality, this tells us something different. If this were possible, the air would escape and leak out into space and Mesklin would be left as an airless world, even at its poles.

However, on the basic gravitational physics, the book had it right. The surface gravity varies as you travel along it. This does not mean that you walk up a slope to get to different regions with different values of gravity (that's what I hope the illustration helps with). Instead of walking "up" or "down" to a new gravity field, you walk "sideways" into it.

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  • $\begingroup$ So what you are saying is that at the surface your potential energy summing the two fields is constant, but that doesn't mean the gravitational strength at that point is also constant? $\endgroup$
    – Tim B
    Commented Apr 24, 2015 at 16:48
  • $\begingroup$ @TimB Potential energy is relative, so that's a little difficult to answer. Potential is constant along the ground, and it's also constant along the blue line. 1 arrow's length always represents the same change in potential (this is the relationship between fields and potential). $\endgroup$
    – Alan Rominger
    Commented Apr 24, 2015 at 17:23
  • $\begingroup$ @TimB Ugh, I just realized one of those nasty more complicated details. The field I drew is not conservative, which means you could make energy from it. It also means that the blue line isn't truly equipotential. However, for the rotating planet, the field is conservative by some geometry tricks. We would need to imagine curved lines for the gravity vector, and that the lines are perpendicular to the blue line as well. That's not quite what I drew, and I'm sorry about that. $\endgroup$
    – Alan Rominger
    Commented Apr 24, 2015 at 18:43
  • $\begingroup$ I notice that the field has a nonzero component parallel to the ground at finite heights. Presumably this would mean that pushing a trolley would take more work if it was taller than if it was really low? At an extreme, crawling along the ground would require no work but walking would. $\endgroup$
    – Emilio Pisanty
    Commented Apr 27, 2015 at 19:52
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    $\begingroup$ @EmilioPisanty You are, of course, strictly correct. But for that matter, when you start walking, you'll kick in dynamic components like the Coriolis force. Thus, we can't conclusively accept your narrative from a map of the field by itself (or without other assumptions, like walking "slowly"). For Mesklin itself, it would be like a cross-continent train ride raises your head less than 60 feet. For Saturn, it would be like 2 inches. $\endgroup$
    – Alan Rominger
    Commented Apr 27, 2015 at 21:10
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Your intuition is correct. At that scale, the planet behaves essentially like a liquid, and it's shape will equalize into an ellipsoid where:

  1. The local gravity vector is everywhere perpendicular to the surface, and
  2. The downward force felt by a person standing on the surface is the same everywhere.

But I don't think you could get an eccentricity as sever as that without the planet tearing itself apart.

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  • $\begingroup$ I feel Alan's argument, with a different conclusion, is stronger. $\endgroup$
    – Emilio Pisanty
    Commented Apr 24, 2015 at 16:19
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    $\begingroup$ "The downward force felt by a person standing on the surface is the same everywhere." Is incorrect. The potential is the same everywhere on the surface, but there is no requirement that the local gravitation have the same strength. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 24, 2015 at 18:33
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    $\begingroup$ Okay, my bad. I've now read a couple of papers on the subject. It turns out that, although a person standing at the equator has all the mass of the planet pulling in roughly the same direction, and the person at a pole has most of the mass pulling sideways rather than "down", that doesn't quite compensate for the increased average distance from the planet's mass distribution. So the person at the pole would feel a stronger effective gravity than the person at the equator. Rominger got it right--bummer. $\endgroup$
    – David Rose
    Commented Apr 27, 2015 at 17:10
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What is the shape of a self-gravitating rotating body?
I didn't read through all answers but skimmed them instead. So far as I've read, are missing a piece.

As @Alan Rominger pointed out the gravitational acceleration at any point on a self-gravitating sphere varies with latitude, like it does on the Earth.

But the shape the body is complex and does not equalize the acceleration due to gravity over the surface of the body. Isostacy attempts to equalize the pressure/force on materials (materials under higher pressures squirt into regions with lower pressure). The pressure any particular region possess is product of mass of materials above that location multiplied by their acceleration due to gravity. It turns out to be an integration to calculate this value because both variables are functions of $ r $.

Actually figuring out the body's shape is probably best left to a Finite Element Method solver.

Maximum rotation rate
It looks like simulations show the maximum rotation rate for a self-gravitating body is between 2 & 3 hours per revolution.

The video also mentions at rotational periods around this value, the body's shape is an oblate spheroid with 100% oblateness ($ r_{equator} = 2 \times r_{pole} $). Since they did use a simulation (I don't know if they used a FEM solver) to determine these values, I trust their numbers on this.

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