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I had a vision of a planet twice the size of earth sliced in half like an apple by an invisible force, one half drifting away from the other into eternity.

Unlike the planet Earth, it did no longer have a molten core, but a solid one, as if the surface was all of equal solidity.

Could these ex-planets shaped like hemispheres be stable?

Stable as in "do they stay the way they physically are" stable.

Do let me know of your thoughts on this predicament.

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    $\begingroup$ @knowads I imagine a planet living a happy life when one day it is simply split in half and each half drifts away. According to my understanding of gravity, there are now two centers of gravity and each rotates faster now that there is less mass. Is that right? It certainly isn't a gas giant, but it might have a small atmosphere. $\endgroup$ Commented May 16, 2019 at 18:33
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    $\begingroup$ @ALambentEye you're ignoring a whole bunch of physics here, mostly around gravity. It's not the earth's spin that keeps it from collapsing, it's the mass of the stuff underneath it being in hydrostatic equilibrium. Cutting the planet neatly in half removes that, and both halves would immediately collapse into smaller spheres. The only way this would work is if you're starting with something much smaller than we normally think of as a planet. The only astronomical bodies that are able to remain stable in a non-spherical shape are things less than ~50km across. $\endgroup$ Commented May 16, 2019 at 18:37
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    $\begingroup$ @MorrisTheCat I am not ignoring it, I am uninformed concerning it, which is why I write these ignorant questions. Do feel free to write an answer and elaborate on it as much as you'd like. I will be thankful. $\endgroup$ Commented May 16, 2019 at 18:38
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    $\begingroup$ Related: Could you build a non-spherical structure that's >1000km long? $\endgroup$
    – Theraot
    Commented May 16, 2019 at 18:50
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    $\begingroup$ A simple slice won't separate them, either. Cut Earth in half, and the two halves will smack back together. You'll need to do something to put some significant relative velocity ( at least enough lateral velocity to put them in orbit round each other) between the two halves and do it gently enough that your halves aren't torn to pieces by the acceleration. $\endgroup$
    – notovny
    Commented May 16, 2019 at 19:10

3 Answers 3

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It depends on the size of the planet. A key concern is whether the body is in hydrostatic equilibrium. An object in hydrostatic equilibrium is approximately spherical, although it may become oblate due to rapid rotation. The question, then, is whether the division of the planet places it below the critical size required for hydrostatic equilibrium.

It turns out that there's no straightforward formula for this, but as a rule of thumb, a body with dimensions of 1500 km or more will become rounded, and a body with dimensions of 400 km or less will not (although this is composition-dependent!). In between the two is a transition zone. It seems likely that if the remnant of the planet is less than 400 km across, it will be stable; if not, it's unlikely to keep its shape.

I'll go out on a limb and say that if the body was already stable, it will remain so; if it was already in hydrostatic equilibrium, on the other hand, it will probably remain in hydrostatic equilibrium and be pulled into a sphere. The reason for this is that the precise threshold is currently not well known. I cited 400 km as a lower limit, but some rocky bodies may instead become spherical at 600 km while icy bodies become spherical at 400 km. Some authors propose even lower thresholds like 200 km. Essentially, only objects in the transition region (say, 600 km to 1500 km) are likely to shift from hydrostatic equilibrium out of it, and that's a fairly narrow range.

Of course, now that you've specified that the planet was originally twice the size of Earth, it seems clear that the fragments will, indeed, also be stable planets, as they are easily massive enough to be rounded by hydrostatic equilibrium - and thus, as per the IAU's definition, they can be planets (assuming they clear their respective orbits, of course). They will certainly be planetary-mass objects.

Rotation

It's worth talking about the spin of the fragments before and after the split. Initially, the planet is spinning at some angular velocity $\omega_o$ about an axis. It has a moment of inertia about that axis $I_p=\frac{2}{5}MR^2$, where $M$ and $R$ are its mass and radius. Then the angular momentum is $$L_o=I_p\omega_o=\frac{2}{5}MR^2\omega_o$$ Angular momentum should be conserved after the collision. Say the split happens along the planet's equator (this is a simple case, really, but it preserves axial symmetry). Then each fragment also happens to have moment of inertia $I_f=\frac{2}{5}mR^2$, where $m$ is the mass of the fragment. By symmetry, the fragments should have the same angular speed $\omega_f$ and total angular momentum $$L_f=I_f\omega_f+I_f\omega_f=2I_f\omega_f=\frac{4}{5}mR^2\omega_f$$ However, $m=M/2$, so $$L_f=\frac{2}{5}MR^2\omega_f$$ and we can see that as $L_f=L_o$, $\omega_f=\omega_o$; that is, the rotation does not change.

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  • $\begingroup$ Thank you for your informed answer! How would the spin of the original planet effect the spin of the ex-planets? Would they appear to drift apart and then start spinning along their own center of gravity with the axis paralell to the original spin? Or does this work entirely differently? $\endgroup$ Commented May 16, 2019 at 18:42
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    $\begingroup$ @ALambentEye I've edited in a simple case. For different ways of slicing the planets, you may get different results, but it's unlikely that the angular speeds would change very much in any scenario. $\endgroup$
    – HDE 226868
    Commented May 16, 2019 at 19:18
  • $\begingroup$ It hadn't, for some reason, occurred to me that the planet could be split along the equator. I was actually thinking of having it split from pole to pole. $\endgroup$ Commented May 16, 2019 at 19:21
  • $\begingroup$ If it's not large enough to be in hydrostatic equilibrium, it's by definition not a planet. $\endgroup$
    – Gene
    Commented May 16, 2019 at 20:16
  • $\begingroup$ 290 km is the limit is the limit I’m familiar with, adsabs.harvard.edu/full/1995MNRAS.277...99H $\endgroup$ Commented May 16, 2019 at 20:20
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This would be very hard to do, but you could make it work, maybe ... if you're not too strict.

Problems:

  1. Based on the IAU definition of a planet, your hemispherical planet no longer meets the classification of a planet due to its shape not being consistent with Hydrostatic equilibrium.
  2. Most planets are subject to the tendency toward hydrostatic equilibrium due to the fact that they either have molten cores, are gaseous, or that the solids making them up are easily crumbled.
  3. Planets would not naturally form this way.

Solutions:

  1. The IAU does not have to exist in your solar system, so their definition of a planet doesn't have to be the one that you use.
  2. The planet would have to be made of solid rock, and be very rigid. There would be a very steep precipice at the edge which would need to not cascade down the vertical slope.
  3. Planets are usually hot when they are formed and then cool off, so they would have been subject to hydrostatic equilibrium at one time. So you would have to actually have your planet form into a sphere, and then split into two after cooling in order for this to happen.

Other consequences.

  1. If some kid decides that it's fun to go throw things off the precipice, then whatever he throws will collect on the flat half of the hemisphere and eventually make a pile in the center making the planet not quite hemispherical. (Note that anything falling from the precipice has this problem, and that the kid in question isn't really necessary)
  2. The gravity on the planet would be uneven and the precipice would be similar to an unimaginably tall range of mountains in this regard.
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Planets are such because they are in hydrostatic equilibrium. This means that if you would cut it in half, it would crumble under gravity to a spherical shape again.

The orientation of the rotational axis has little to do with this.

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