If a planet has a core of super-dense material, how might that effect the geological composition of that planet?

Specifically: imagine a lunar-sized planet, with earth-like geology and density. If the center of that planet is a very small mass of super-dense unobtanium that cannot collapse in on itself (think neutron star density. This Super dense material is about 97.3% of earths mass compressed into the size of a shower stall), how would that effect the surrounding layers of this planet?

Would the normally earth-dense layers just flatten against the core and the rest of it become super-dense as well?

Would this planet be able to hold liquid water and an atmosphere?

Thanks for the feedback.


4 Answers 4


You'd have no atmosphere, and almost no liquid water.

Right, the atmosphere. The gravity of the planet will be quite strange, considering pretty much all of the gravitational material will be at the very centre. You'll have to ask someone that knows more about those things than me though. My concern would be that, without a large rotating core, you have no magnetic fields protecting the planet. That immediately puts you somewhere in the vicinity of a Mars like scenario, regardless of anything else. Solar wind will strip away your atmosphere, and simple mathematics will show that you don't have that much to lose. The volume of your planet is small so it won't take long, astronomically speaking, for you to lose your lighter elements. This will drop the temperature and pressure, losing any water that you have on the surface.

Secondly, you specified that you have a lunar sized planet, but 97.3% of the mass is at the centre, in a shower-stall sized block. Compared to even the Moon, that's absolutely tiny. Even if we assume a generous stall of maybe $3m^3$, that means that you have practically all of your mass in...wait for it...$1.37\times 10^{-19}$% of your planet. That's negligible. 3% of Earth's mass is about $1.8\times10^{23} kg$ which means that the density of everything other than your core is, on average, $7363kg/m^3$. That doesn't sound too bad; it's about the average density of cast iron, and it's more than Earth which is about $5515kg/m^3$. Consider that this planet is way smaller though. Again, I don't know what this would do to a planet but it would be weird; it obviously doesn't happen in nature. First thing that comes to mind is gravity: this planet would have, at the surface, a gravitational pull of $132m/s^2$. That's about $13.5g$ so any human, and most animals, would be dead at the surface. No large organisms can survive at that. Most plants on earth can't either. On the other hand, might help you keep an atmosphere for longer.

Lastly, you specified that the unobtanium magically doesn't collapse in, but you didn't say anything about the other way around. Neutron star matter is kept that way through truly enormous gravity, and our planet doesn't come close to that. Without supergravity to keep it in check, the core of the planet will explode. A teaspoon of neutron star matter will produce about $10^{27}J$ of energy as it decays; you have way more. You're planet will last maybe a few minutes before it's vapourised. With the amount of matter it'll produce the approximate energy output of the sun for 14 straight days. There'll be nothing left of your planet and anything surrounding it.

  • 1
    $\begingroup$ If you have a moon that stays between the planet and its sun (period of moon orbiting planet is same as planet orbiting star), the moon could have a magnetic field and atmosphere to shield the planet. If the planet has internal (geothermal) heat, then you might still have liquid water. $\endgroup$
    – SRM
    Jun 9, 2018 at 4:06
  • $\begingroup$ Maybe. I don't know how this would work as a moon though. Moons can get a bit bigger than Earth's, at least in our solar system, but an Earth-mass object might behave differenlty $\endgroup$
    – Serenical
    Jun 9, 2018 at 7:34
  • 1
    $\begingroup$ doesn't have to be massive, just needing to block solar wind. A really undense moon with a decent mag field. $\endgroup$
    – SRM
    Jun 9, 2018 at 8:13
  • $\begingroup$ No, this planet in the question would be massive. I'm not too well versed in astrophysics but I don't think a celestial body of Earth's weight is too common as a moon. Besides, this would produce almost no magnetic field, I think. Then there's the fact that it would vapourise after maybe 5 minutes $\endgroup$
    – Serenical
    Jun 9, 2018 at 11:20
  • 2
    $\begingroup$ Gravity doesn't just depend on overall mass: it depends on distance as well. A small object with the same mass as a large object will have a much higher surface gravity. If I recall, the formula for surface acceleration is $GM/{r^2}$ where G is the gravitational constant, M is the planetary mass, and r is the radius of the planet. So when r gets a lot smaller, the gravity goes up a lot. With most of the mass at the centre, the uneven distribution won't matter so much but he planet is weirdly dense. Earth is the densest in our solar system and you have that beaten by a lot. $\endgroup$
    – Serenical
    Jun 9, 2018 at 22:54

I loved Serenical's answer and upvoted it. You should too. In addition to it, think about this:

A lunar-sized object with earth-like gravity would experience:

  • Serious gravitic shear along both the curvature of its surface and radially from the center. In other words, anything bigger than an amoeba trying to live on it would likely be torn apart.

  • I don't think the non-unobtainium mass will collapse (I could be wrong about this, if I am it will be a glorious little star for a brief period of time), but it will be superheated magma. Do you remember what happens when ice skaters pull their arms in? Yup, they spin faster. The rotational speed of the lunar mass will be tremendous. With all the forces at play, I should think the non-unobtainium mass (all of it) would be a boiling liquid. No solid surface. In fact, it might vaporize into a gas dwarf planet.

The problem is that it's a whole lotta mass in a very small space. It's not condusive to stability.


Gravitic Shear:  Compared to the force of gravity, the radius of the earth is large enough that you can conceptualize the surface as a flat plane. As people walk along that plane, the force of gravity is statistically the same everywhere your feet, torso, and head pass.

enter image description here

But you're seriously reducing the radius. Now the force is concentrated closer to the ground than it is higher. I'd have to work the math to prove it (I apologize that I don't have the time), but walking would have minor effect, falling down and getting up would have a major effect. Buildings would be unable to withstand the shear (in my opinion).

enter image description here

By "shear" I mean substantial differences in gravitational force between two nearby regions.

Rotation: As an object of mass X grows closer to its center, it spins faster. Here's a passable example:

enter image description here

Think of a cylinder drawn around the skater. It begins with radius A (the outer extend of her leg). As she pulls her leg abover her head, the radius shrinks to radius B and she spins faster. This is an aspect of the conservation of momentum.

So, the lunar sized object (smaller radius than the Earth) has the same mass as the Earth. It must spin faster. A lot faster.

  • $\begingroup$ Thanks, JBH, can you give some more background on both parts of your answer? Specifically, how this object's rotation would be effected, and what you mean by gravitic shear. Thanks again for your time 👍 $\endgroup$
    – user49466
    Jun 9, 2018 at 13:26
  • $\begingroup$ I edited my answer. $\endgroup$
    – JBH
    Jun 13, 2018 at 0:26

Degenerate matter

Let's run some numbers.

Earth's mass compressed down to 3m^3 is roughly 2e24 kg/m^3. That's 10 million times more dense than a neutron star. This would be made up of degenerate quark matter.

A neutron star forms when a star that does not have enough mass to become a stellar black hole collapses. It becomes so dense its gravity overcomes the electromagnetic force keeping electrons and positrons apart. They fuse together to form neutrons. The neutrons can't be compressed further because the strong nuclear force becomes repulsive at 0.7 fm.

But you're 10 million times more dense than that. At your densities it's theorized that even neutrons would break down into quarks and gluons. It would also be very, very, very hot in the order of 1e12 Kelvins. With so little mass (and thus gravity) to hold itself together it would immediately blow itself apart.


Let's assume none of that happens, it isn't at some cosmologically high temperature, and it doesn't blow itself part. It's just magically really dense. There's still gravity to worry about. What's happening to the normal matter adjacent to this infernal shower stall?

Newtonian gravity is $F = Gm/r^2$.

  • $m$ is Earth's mass, 6e24kg
  • $r$ is the distance from the center of mass, about 1 meter
  • $G$ is the gravitational constant, 6.7e−11 $\frac{N m^2}{kg^2}$

Put that together and we get 402,000,000,000,000 N or 4e14 N. How much force is that?

It's a lot. Anything adjacent to the core will be immediately flattened onto the core. Anything above that will collapse. Your planet collapses into a thin layer of degenerate matter on the surface of the shower stall.

Tidal forces

It gets worse.

Gravity gets weaker with the square of the distance from the center of mass. At 2 meters from the worst shower stall in the universe, the gravitational attraction is 4 times less. At 4 meters it's 16 times less. At 8 meters it's 64 times less. This extreme force gradient is known as a tidal force and it tears things apart.

Imagine an 8 meter high rock 8 meters from the core. The end closest to the core is 8 meters away and will be experiencing $\frac{4e14N}{8}$ or 6e12 N. The further end is 16 meters away and will be experiencing $\frac{4e14N}{256}$ or 1.5e12 N. The rock will be torn apart by 4.5e12 N, the force of 100,000 Saturn V rockets.

It explodes anyway

It gets worse.

The force of all this normal matter falling into the dense unobtanium core and compressing will produce more energy than I care to calculate. Let's calculate it.

Let's say a hunk of rock falls towards the core in this intense gravitational field. Since we're dealing with such a crazy large gravity field we'll use special relativity. You can see the equation derived here.

$$ v = c \sqrt{1 - \left[ \frac{1}{\frac{GM}{c^2} \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) + 1} \right]^2}$$

  • $M$ is the mass of the shower stall, 6e24 kg.
  • $G$ is the gravitational constant, 6.7e−11 $\frac{N m^2}{kg^2}$
  • $r_\mathrm{initial}$ is the starting height above center mass.
  • $r_\mathrm{final}$ is the height above center mass when it lands, 1m.
  • $c$ is the speed of light, 3e8 m/s.
  • $v$ is its velocity when it hits the surface of the shower stall.

Running the numbers, falling from just 2 meters it's going 20,000,000 m/s or about 6.5% the speed of light. From 4 meters it's going 8%. From 10 meters it's going 8.9%. From 100 meters it's going 9.4% and it doesn't get much faster above that.

We can calculate its kinetic energy with the special relativity formulas...

$$K = (\gamma - 1) m c^2$$ $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

1 kg at 0.094c will impact the core with about 4e14 J or about 100 kilotons of TNT.

So the rock surrounding the core briefly pancakes onto the core before blowing the planet apart.


I've got a bunch of problems with what's come so far:

Neutron star material going boom: agree. However, it was only specified as unobtaininum of that density, it might not go boom.

The planet having Earth-like geology: nope, it's not going to happen. The problem is you don't have enough normal matter for radioactivity to keep the core molten. No plate tectonics. The planet erodes to basically flat and stays there.

Atmosphere: Nothing says unobtainium can't generate a magnetic field. However, it's not going to have an atmosphere because a world of that size can't hold onto one. It's not surface gravity that counts in holding an atmosphere, it's escape velocity. Escape velocity is determined both by surface gravity and how fast it drops off with distance. Your Luna-sized object has nowhere near the total gravity of Earth.

Gravitational sheer: I'm baffled here.

First, any creature that really lived in such an environment would be appropriately curved, the sheer would be built in and cause no abnormal force on them.

Second, while I can see there being such an effect if you were walking around on the core this is a Luna-sized world. The horizon is a lot closer than on Earth but it's still basically flat.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .