Consider a planet very similar to Earth, with the same radius and the same mass. In a very simplified model of its interior, 70 % of its mass is magnesium silicate and the remaining 30 % is iron. This latter, being denser than the first, it will sink towards the center, forming the planetary core.

Consider another planet with the same mass as the previous one, but now, iron represents 50 % of its mass and the other 50 % corresponds to magnesium silicate. It is logical to think that this planet will be smaller than the previous one, since iron is denser than magnesium silicate and, therefore, a certain amount of iron occupies a smaller volume than the same amount of magensium silicate. But the question is: how small would it be?

What i did was:

If it is known that the nucleus is a sphere of 100 % iron, its density would be the same as that of iron, and its mass, that is half the mass of the planet, is also known. Knowing the density and mass, it can be found the volume:


Solving for $v$:


I followed the same procedure to find the volume of the mantle. Then, by adding both volumes, the volume of the planet is found, and with him, the radius:


Solving for $r$:


By following this procedure, the radius of the planet turned out to be 45 % larger than the radius of the original planet, which is illogical. I would like to know: is this the procedure to follow in this case or am I doing something wrong?

  • $\begingroup$ When I solved for r, the result is r equals the cube root of three-quarters v divided by pi. How did you get your result? Are steps in your derivation of the equation that are missing? $\endgroup$
    – a4android
    Aug 19, 2019 at 6:38
  • $\begingroup$ @a4android I guess you mean the equation $r=\sqrt[3]{\frac{\frac{3v}{4}}{π}}$, if so, this equation and mine are equivalent, it is only expressed differently. $\endgroup$
    – URIZEN
    Aug 19, 2019 at 7:06
  • $\begingroup$ Yep I do. Your equation is expressing itself with an accent I don't recognize. :) $\endgroup$
    – a4android
    Aug 19, 2019 at 8:02
  • $\begingroup$ A note on symbols : normally capital $V$ would be used for volume and either "\rho" ($\rho$) or "\sigma" ($\sigma$) for density. $d$ and $v$ have other common uses in physics and mathematics so it's best to avoid using them (particularly $d$) for other purposes. $\endgroup$ Aug 19, 2019 at 9:17
  • 1
    $\begingroup$ "the radius of the [denser] planet turned out to be 45 % larger than the radius of the original planet, which is illogical". Correct, you got the wrong answer. But you haven't shown your work, so it's not possible to say what you did wrong when "following this procedure" and "turned out". $\endgroup$ Aug 19, 2019 at 14:56

3 Answers 3


Short version : the second planet (same total mass) will have a radius that is larger than the first unless the second planet's core is larger (in mass) than the first.

This makes perfect sense if you think about it. The smaller core has to have much more of the lighter exterior material to make up the total mass. The larger core needs less to make the same radius.

The hideous maths :-)

What you have is this :

$$M_1 = \frac 4 3 \pi \left[ \sigma_a r_{a1}^3 + \sigma_b(r_1^3-r_{1a}^3) \right]$$

$$M_2 = \frac 4 3 \pi \left[ \sigma_a r_{a2}^3 + \sigma_b(r_2^3-r_{2a}^3) \right]$$

Where the $a$ subscript refers to the Iron core and $b$ to the remaining material.

You know the masses are equal, so we can equate these two masses.

We also know the relative masses of the cores :

$$\frac{M_{1a}}{M_{2a}} = \frac{0.3}{0.5} = 0.6 = \frac {r_{1a}^3}{r_{2a}^3}$$

So we know the ratio of the radii of the cores, and I'm going to call that $x$ for convenience although it's actually $0.6^{\frac 1 3} = 0.8434$. I'm going to call $y=r_{1a}$ (to save my typing).

For simplicity we're going to use all radii relative to $r_1$ which we're setting to $1$.

$$\sigma_a y^3 + \sigma_b (1-y^3) = \sigma_a (xy)^3+\sigma_b(r_2^3-(xy)^3)$$

We know $x, \sigma_a$ and $\sigma_b$. We don't know the values of $y = r_{1a}$ and $r_2$.

So we get for the second planet's radius :

$$r_2^3 = \frac 1 {\sigma_b} \left[ \sigma_a y^3 + \sigma_b (1-y^3) - \sigma_a (xy)^3 + \sigma_b (xy)^3 \right]$$


$$r_2^3 = \frac 1 {\sigma_b} \left[ y^3 \left( \sigma_a - \sigma_b - \sigma_a x^3 + \sigma_b x^3 \right) +\sigma_b\right]$$

$$r_2^3 = \frac 1 {\sigma_b} \left[ y^3 (1-x^3) \left( \sigma_a - \sigma_b\right) +\sigma_b\right]$$

You asked :

It is logical to think that this planet will be smaller than the previous one

What this translates to is whether $r_2<1$ (because we set $r_1=1$ for convenience).

So this requires :

$$y^3 (1-x^3) ( \sigma_a - \sigma_b )<0$$

That is, it requires the radius of the first planet's core ($y=r_{1a}$) to be less than zero, which of course is impossible, which means that the second planet's radius is always larger !

To make the second planet smaller you need to have :

$$1-x^3 = 1 - \frac{{r_{1a}}^3}{{r_{1b}}^3} < 0$$

And that requires :

$$r_{1b} > r_{1a}$$

Or that the radius (and hence mass) of the second planet's core is larger than the first planet.

The planet with the smallest core has the largest radius if their total masses are equal.

You can also "fix" this if the core density ($\sigma_a$) was smaller than the exterior density ($\sigma_b$), but that's not physically realistic.

  • $\begingroup$ Can you really call any maths "hideous" if it doesn't include any integration? $\endgroup$ Aug 19, 2019 at 14:14
  • $\begingroup$ @StarfishPrime Integration is hideous? It's just taking an average... $\endgroup$
    – stix
    Aug 19, 2019 at 15:32
  • $\begingroup$ @stix xkcd.com/2117 $\endgroup$ Aug 19, 2019 at 15:35
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    $\begingroup$ @StarfishPrime Who tries to solve integrals analytically besides high school students? RK-4 it with a computer and be done. $\endgroup$
    – stix
    Aug 19, 2019 at 15:38
  • $\begingroup$ @stix duly noted ;-) $\endgroup$ Aug 19, 2019 at 16:36

Zeng et al. (2015) "Mass-Radius Relation for Rocky Planets based on PREM" might be useful for you.

They give the following equation:

$$\frac{R}{R_\oplus} = \left(1.07 - 0.21 \cdot \mathrm{CMF}\right) \cdot \left(\frac{M}{M_\oplus}\right)^{1/3.7}$$

Where $R$ and $M$ are the planetary radius and mass respectively, $R_\oplus$ and $M_\oplus$ are the radius and mass of the Earth respectively, and CMF is the core mass fraction (i.e. the percent by mass of iron). They state this equation is valid between ~1 and ~8 Earth masses and for CMF between 0 and 0.4.

If you want to go outside this range, you can interpolate the values in table 2 in the paper, which extends from 0.125 to 32 Earth masses, and iron fractions from 100% to 0%. The table also includes two-layer planets containing water and silicates.


Simpler Approach (no 𝜋 for you)

Assume the density of silica is 1, and the density of iron is 3:

  • The second planet's density will be 1*50% + 3*50% = 2.0
  • The first planet's density will be 1*70% + 3*30% = 1.6
  • The ratio of their densities will be 2.0 / 1.6 = 1.25
  • The ratio of their volumes will be 1 / 1.25 = .8
  • The ratio of their radii will be the cube root of .8, which is .928

I.e. the second, denser planet's radius will be 7.2% smaller than the first's.


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