1
$\begingroup$

My planet is a gas giant, it is called Klarloth (Inspired by a character from "Star Mouse" by Fredric Brown).

Klarloth is a strange planet. For one thing, the planet is large in size, more larger than usual for a gas giant (Most gas giants above $1$ $M_J$ tend to get denser, not bigger, as they gain mass, mostly hovering at around $1-1.5$ $R_J$). Characteristics of Klarloth:

  • Radius: 285,000 km

  • Parent Star: None, Klarloth is a rogue planet.

  • Mass: 500 $M_J$

Yes, you heard that right. Klarloth is 500 times the mass of Jupiter.

Not even a star, not even a brown dwarf. Klarloth is a PLANET.

The reason is well (relatively) mundane, when you get to it.

The nebula, where the planet Klarloth formed was deficient in deuterium/lithium for some reason. I will figure out why in a later question. But for now, let's assume the fact that for some reason, the presence of deuterium/lithium in the nebula that formed planet Klarloth, was non-existent. Meaning that even if it wasn't a star, Klarloth couldn't effectively be a brown dwarf, as they was not enough deuterium in the nebula that it could fuse and generate energy. Even without deuterium, lithium wasn't present the nebula that formed Klarloth, so even if deuterium wasn't present, it couldn't resort to fusing lithium for energy. (How? and Why? is happened is out of the scope of the question, as I have handwaved the chances of such a nebula occuring, so I am primarily concerned with how- [REVEALED LATER IN QUESTION])

The next reason, however is a bit more exciting.

You see, Klarloth, although primarily made of hydrogen, has a core made of heavier elements. Although Klarloth has plenty of hydrogen, the reason why it cannot fuse it despite it being wayyyy more massive than $80$ $M_J$, is especially, there is no hydrogen in the core.

The situation is similar to a AGB Star (Red Giant, blue supergiant, red hypergiant etc etc). Let's take Betelguese for an analogy of Klarloth (I know that a red supergiant is a poor analogy for a supermassive planet, but I cannot find any other reference)

Betelguese still has more than half of it's hydrogen remaining. However, it is paradoxically fusing heavier elements. If current calculations are current, Betelguese is still fusing either helium or carbon.

The main reason is not because of lack of hydrogen. Betelguese still has plenty of hydrogen. But the hydrogen has been shunted out of the core. In other words, you have plenty of food at the table, but the food has been shunted out of reach, leaving only plates on the table. Meaning that, Betelguese's core is now littered with heavier elements like carbon/oxygen. The hydrogen has been shunted out to the outside of the core, where it is not hot enough to fuse.

My planet Klarloth, similarly also faces a similar situation, although in this case, there is NO fusion going on. The planet started out as a ball of rock which grew up in size and gained tons of hydrogen gas, which continued to grow up, till it reached $500$ $M_J$.

As of today, Klarloth's core is composed of the following chemicals in the given ratios-

  • 55% Iron, mostly in the form of oxides.
  • 16% Oxygen, bound to iron as oxides.
  • 29% Thorium sulfide

Basically, the planet Klarloth has such a large core of heavy metals, that at a certain radius, it should be able to displace hydrogen from the center. Meaning that hydrogen is now in an envelope surrounding the core of heavy metals.

The only thing that is left to calculate, is the radius of the core....

Basically, I want to find how big the core has to be, so that it can displace hydrogen from the center, preventing it from fusing.

So, if an electron-degenerate iron core about 2 times as wide as Germany, can halt silicon fusion in a red supergiant (before collapsing into a neutron star in a fraction of a second, of course, but that's completely unrelated to the question's scope), I figure that, the core of the planet can be small and still prevent fusion in the core. Sadly, my planet is not a red supergiant, so my previous assumptions in the previous sentence were proven wrong. I figure that the core of the planet has to be much larger to order to prevent fusion of hydrogen in the core. I simply want a supermassive planet. I have solved the deuterium/lithium deficit problem, now I just need to find out the radius of the core, at what size it can displace enough hydrogen from the center, to stop fusion.

My dear Worldbuilders, I present you with this question:-

  • How large (in radius) should the core of my supermassive planet be, to prevent fusion of hydrogen?

Bonus question (Not necessary to answer, I just added in, just because I thought it might be related)

  • How massive should the core of the planet be?

The reason I have added the tag is because, I need calculations and stuff to check the answer, the resultant required radius of Klarloth's core.

Remember that although ironically handwavium has been used in a question tagged , handwavium has only been used for parts of the story that are not the scope of the question (explaining how it formed, why it has a deficit of deuterium/lithium etc). The main problem here is to find the radius of the core of Klarloth.

Note about the Hydrogen envelope, surrounding the heavy-element core:-

The envelope (you can call it the "mantle" of Klarloth. Gas giants don't really have a "crust"), is composed of 90% hydrogen and 10% helium.

Note that the core of the gas giant isn't a "fuzzy core" like Jupiter. If that was the case, results would be disastrous, as the hydrogen could seep in and ignite fusion. Klarloth's core is a proper, well-defined core.

$\endgroup$
5
  • $\begingroup$ 13 hours after posting and no comments or answers. You might get more interaction if you script this out a little better, lay bare the relevant points. Reads almost like an internal monologue. $\endgroup$
    – BMF
    Commented Jun 18, 2023 at 22:08
  • $\begingroup$ For ex., when I answer Q's I don't exactly memorize all the relevant details. I work an answer piecewise and go back into the body of the Q to scrounge up the info I need next. I think it's an interesting idea, but being this verbose is not helping anyone. $\endgroup$
    – BMF
    Commented Jun 18, 2023 at 22:10
  • $\begingroup$ Brown dwarfs are massive enough to burn deuterium, but burning deuterium isn't necessary to make something a brown dwarf. As for lithium, its presence is a sign that something is likely a brown dwarf, but it is not required for something to be a brown dwarf. As for the thorium...thorium is not a common element and such a huge concentration would be extremely unnatural. It's even more unnatural for it to be present without its decay and transmutation products. The deuterium and lithium are not what you need to worry about explaining... $\endgroup$ Commented Jun 19, 2023 at 2:53
  • $\begingroup$ @ChristopherJamesHuff As I said, no need to worry about the immense presence of thorium, I have handwaved that away. For now, I just need to find the radius of the core of the Planet $\endgroup$
    – Alastor
    Commented Jun 19, 2023 at 10:50
  • $\begingroup$ You didn't say that. And nearly a third of the core's mass being a rare element that can sustain fission reactions seems...concerning. Even if it somehow has avoided neutrons all this time, if it's of similar age to Earth, ~20% of it should have decayed to lead...ignoring the decay products that should have come with it from its original source. $\endgroup$ Commented Jun 19, 2023 at 15:11

1 Answer 1

2
$\begingroup$

We must stay below the point of collapse

As you will see in a moment, our limit in mass and size comes entirely from the point at which our supermassive planetary body collapses into a black hole, and before that, there's a point at which matter starts to become degenerate matter. Still, we can make things easy if see can prevent fusion in the first place by choosing the composition of the body.

No Fusion beyond...

So before we elaborate on our size and mass, we figure out what the planet is made from. This is very simple if we know one thing: Iron is the last element created by fusion in a star, everything with more protons in the core is created in a supernova by bombarding matter with neutrons and protons.

So we set up our planet as a massive sphere of iron. This also makes one thing easy for us: the normal density of Iron is $\rho=\pu{7874 \frac{kg}{m^3}}$. Uniform material is also elegant in that we know at which mass our body starts to degenerate and convert into a star: that's the Chandrasekhar limit.

So our mass is known to be $M=\pu{2.765 \times 10^30 kg}$, or about 1.4 Solar Masses, as beyond that our planet needs to collapse into a star.

Simple Geometry now kicks in: We know the density and mass, we know the volume of a sphere is $V=\frac 4 3 \pi r^3$, and the connection between Volume and density is $\frac M V=\rho$. Reformatting to solve for r we get...

$$\frac {M}{\frac 4 3 \pi r^3}=\rho$$ $$\frac {3M} {4\rho \pi}=r^3$$

Plugging in the Numbers into Wolframalpha you get: $$\pu{8.383 \times 10^{25} m^3}=r^3$$ or solved for simple r $$\pu{4.376\times 10^8 m}=r$$

That is a hard-to-compare number, but again, easy to calculate: just divide that by the earth's radius at the equator and you learn that your iron planet has a radius of 68.6188 earth radii, its diameter is the same correlation to earth diameters.

More complex setups?

Well, the formula above is easy to adjust! You keep the mass, you alter the density to the average density in kg/m³ and the output will be your planet's absolute maximum radius. From there you can start to calculate all other factors.

$\endgroup$
1
  • $\begingroup$ I just want the radius of the planet's core, not the planet itself. Thanks for the calculations tho, =) $\endgroup$
    – Alastor
    Commented Jun 19, 2023 at 13:48

You must log in to answer this question.

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