After getting help from the members of this site with Worldbuilding an Earth-like world around an F-Type Star, I have another question. The Planet's radius is 90% of Earth's. The Planet's rotation is 11 hours and 50 Minutes. It's density is 120% of Earth. How do I figure out what elements are within the planet and at what percentages and within what layers?

It did not have a Mars sized body impact with the planet to form a moon, so the orbit is not stabilized and material was not added to it, which I have read is a major source of the Earth's layers.

I know it should have more heavy elements so such a density can compensate for the smaller radius, what I do not know is how the lack of a lunar event would effect its layers or how this difference in density affects it. How would I figure this out and what options do I have? This seems like two questions but I feel like I cannot have this question answered without at least mentioning the lack of a moon.

  • 1
    $\begingroup$ Can you link to your previous question and specify what "it" is for anyone who has not seen your last question? $\endgroup$
    – John Locke
    Sep 14, 2018 at 23:37
  • $\begingroup$ Edited it to explain further. $\endgroup$ Sep 14, 2018 at 23:40
  • $\begingroup$ "It's density is 120% of Earth." Is that the mean density (20% higher than Earth's 5.513 g/cm^3) or do you really mean that gravity is 20% higher than Earth's 9.8m/s^2? (The questions might or might not be related.) $\endgroup$
    – RonJohn
    Sep 15, 2018 at 1:12
  • $\begingroup$ Genuinely mean't the density is 120% of Earth, not gravity, I believe gravity is slightly lower than that of Earth's. The different radius effects cooling of the layers which might effect the size of them. Edit: I know I made a mistake referring to rotation of orbit sorry, I will make sure I am being careful with my wording from here on. $\endgroup$ Sep 15, 2018 at 3:59
  • $\begingroup$ What population is the star and/or what is it's metallicity? This will give us a few clues as to the elemental composition of the world(s) that orbit it. $\endgroup$
    – Ash
    Sep 15, 2018 at 13:54

1 Answer 1


Similar topics were discussed several times lately, and I recommend you read the questions, answers, and comments to have a better idea of how things work.

Geology on low density planets

Can A Theia-Like Object Make Earth Richer?

All the Radioactive Metals Inside Earth's Core--How Would They Affect Convection?

If Earth's Core had ALL Of the Heavy Metals

Leaving aside the numbers, what you want to do is basically compress Earth. For example, keep the same mass, but rearrange it so it's more tightly packed, thus making the radius smaller and density greater. The problem is that it is not possible to do without changing the chemical composition of the Earth (thus the mass), because radius results from density results from mineral composition results from chemical composition.

This means that we will have to change chemical composition a bit. The best way increase overall density would be to enlarge the metallic core, at the expense of the less dense rocky mantle. Conceptually, this is rather easy to do.

Let's approximate Earth with a chunk of solid iron (density 7.87) and a chunk of olivine (magnesium iron silicate: a solution of forsterite Mg2SiO4 and fayalite Fe2SiO4) of composition forsterite90fayalite10. This composition leads to a density of about 3.37. Having read some of the answers I linked to above, you should be aware that what we want to do is take iron from the olivine and put it in the core. This way, you're enlarging the dense core while shrinking the mantle. This is feasibly by removing oxygen, with the following chemical reaction:

10(Mg0.9Fe0.1)2SiO4 = 8Mg2SiO4 + 2MgSiO3 + 2Fe + O2

Note that enstatite (MgSiO3), another magnesium silicate is a product for this reaction.

What does this mean for our density and radius? The molecular weights for 1 mole of "forsterite" and "fayalite" are 140.69 and 203.78 g/cm3, respectively. So, 10(Mg0.9Fe0.1)2SiO4 will weigh 9×140.69 + 1×203.78 = 1470 g. The missing oxygen means you have lower mass: 1437.99, and 1326.30 g after putting the iron in the core.

Your "mantle" had 10 moles of a material with molar volume of 44.08 cm3/mol and you now have 8 moles of the same, and 2 moles of enstatite with a molar volume of 34.15 cm3/mol. Overall, your mantle occupies less volume.

So the results of this quick calculation lead to:

  1. Larger core (volume and mass) because we have more iron there.
  2. Smaller mantle (volume) because now the mineral composition has less molar volume.
  3. The density of the mantle remained the same because the densities of forsterite and enstatite are essentially identical.
  4. Lighter mantle (mass) because we removed some oxygen.

The combined effects of these is a smaller, denser "planet". Your g will slightly decrease because we lost mass in the form of oxygen, but you can always add some iron to the core compensate for that.

The percentage of change of density and radius (120% and 90% as requested by OP) is left as an exercise for the reader.

How does this apply to a real planet? Hard to tell. We don't exactly know the composition, density, proportions, etc of the mineral constituents of Earth's interior. For example, a paper published just one year ago (2017) in the journal Nature reported new significant findings (the specifics of which are not relevant at the moment). It is hard to know what to change and by how much to get to a certain outcome if we do not know what we are starting with.

Nonetheless, the basic idea is this: remove oxygen, which leads to a smaller lighter mantle and a larger core.

  • $\begingroup$ Thankfully you have already come up to a solution for the whole thing, for that I thank you. $\endgroup$ Sep 17, 2018 at 1:32
  • $\begingroup$ It seemed like it would be madness basing of of the first few answers. You seem to have solved most of the problem but I do still wonder what size the layers are as part of radius and percentage of mass. $\endgroup$ Sep 17, 2018 at 2:14
  • $\begingroup$ @BlindingLight well you have the numbers, now it's just a matter of plugging it into the equations of spheres and mass and density etc $\endgroup$
    – Gimelist
    Sep 17, 2018 at 2:41
  • $\begingroup$ This may take time, my knowledge of chemistry and units has disappeared. I need to figure out how to get moles into kilometers. So what is the volume of the planet's Core(the one made by your math?) Or where do I look in your equation to find it? Wait I think I see where the flaw in my knowledge and logic is sorry. $\endgroup$ Sep 17, 2018 at 4:26
  • $\begingroup$ @BlindingLight I might come up with an equation later when I have some time $\endgroup$
    – Gimelist
    Sep 17, 2018 at 5:11

You must log in to answer this question.

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