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Centrifugal force effectively counteracts surface gravity making spinning worlds bulge and making gravity at the equator slightly less than at the poles. Hal Clement notably took this to an extreme with the planet Mesklin from Mission of Gravity, with 3g at the equator and around 700g at the poles... but what if we take it to an even greater extreme?

A neutron star spinning Sufficiently Fast should eventually end up with an equatorial bulge that has weak enough gravity for degenerate matter to start "re-inflating". Spin it even faster, and eventually you'll get a normal-matter rim that a human could stand on (while the sky is a series of fully-blurred circular streaks of stars).

But... what might that actually look like? What's the final shape? Would there actually be room to stand on that equator, or would it be too thin to inhabit? It would be relatively simple to work out the shape of an oblate spheroid of uniform density, or the equipotential surface around a point mass, but this thing is not really close to either ideal.

I realize a fully accurate answer would rely on information about degenerate nuclear matter physics that we just don't have yet, but I'd like to make this as accurate as possible for sci-fi. And if a precise answer is impractical to give, I would still appreciate references that might help me figure it out.

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  • $\begingroup$ I'm guessing you're looking for something different than "it's too hot", "gravity will crush you" and "the centrifugal forces will not allow anything to survive ine way or another"? $\endgroup$
    – Trioxidane
    Sep 8 '21 at 20:56
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    $\begingroup$ Neutron stars are much more extreme, their surface gravity is measured in many billions g. Which means that even if we are able to tune its spinning so that there will be 1g at the equator, the gradient would be too steep. 1 meter away from the surface the gravity will be in 1000s negative g. $\endgroup$
    – Alexander
    Sep 8 '21 at 21:01
  • $\begingroup$ @Trioxidane Yes. I do not care about the heat, and centrifugal forces balancing gravity so that it doesn't crush you is the whole point. $\endgroup$
    – Logan R. Kearsley
    Sep 8 '21 at 21:42
  • $\begingroup$ @Alexander Maybe so, but I'll cross that bridge when I come to it. Step is is figuring out what the shape, and he the precise gradient, actually is given that the equator itself is stand-on-able. $\endgroup$
    – Logan R. Kearsley
    Sep 8 '21 at 21:45
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    $\begingroup$ Figuring the exact shape would be an interesting challenge for Physics SE. "oblate spheroid of uniform density" - this will definitely won't be the case. Near the axle, it will be neutron degenerate matter, farther out - electron degenerate matter, and regular matter (iron?) on top of the equatorial bulge. $\endgroup$
    – Alexander
    Sep 8 '21 at 21:51
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While I still think it can't exist I realize my original analysis had a big problem.

The problem is an object that can't support itself (neutronium isn't a solid, it has no strength) will form an object with equal gravitational potential across it's surface. If it's 1g at the equator it's 1g at the poles, also. The star flattens out into a thin disk that I do not think has the pressure to remain a neutron star.

While I still think tides will be a big problem it's not so certain they are lethal.

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  • $\begingroup$ Can you support those assertions? It seems to me that the tidal gradient at the 1g surface will depend critically on the distance from the center and the mass distribution (since it won't be spherically symmetric), so you can't actually know that without first answering what the shape would be when spinning fast enough to produce a suitable surface gravity. $\endgroup$
    – Logan R. Kearsley
    Sep 9 '21 at 0:54
  • $\begingroup$ @LoganR.Kearsley: The lack of spherical symmetry shouldn’t matter since the star will still be radially symmetric. If you’re standing at the equator you can treat the mass as a point at distance r. Working out r is nontrivial though. $\endgroup$
    – Joe Bloggs
    Sep 9 '21 at 8:16
  • $\begingroup$ Equal gravitational potential does not mean equal gravity. $\endgroup$
    – Anton Sherwood
    Sep 13 '21 at 2:22
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A disk, or a toroid, or a binary. Or possibly a disk within a disk.

shapes

Equilibrium configurations of fluids and their stability in higher dimensions

As it spins it will flatten. Like a pizza crust. If it spins more it will thin out on the middle and eventually become a ring or toroid. Spinning more will yield a binary body.

I am irritated because I once found sweet 3d model images of all these shapes and now this is the best I can find.

But in the searching I also found a scholarly paper that predicted a torus with a central core, and also a torus within a torus as possible outcomes. Torus in a torus not depicted here - click thru to the original if you are digging it. No paywall!

[ Uniformly rotating axisymmetric fluid configurations bifurcating from highly flattened Maclaurin spheroids(https://arxiv.org/pdf/astro-ph/0208267.pdf)

more shapes

The math governing which of these shapes is the outcomes has to do (as far as I can tell) with the nature of the "fluid" and the forces holding it together. This math is heady stuff.

So Spinny the Star has got a number of fairly unstarlike conformations it can take on. The limit of course is spin so fast that it starts shedding mass. You want to not spin quite that fast.

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