# Given the mass and composition of a planet, can one determine what the radius should be?

I'm trying to auto-generate random solar systems, and I'm basically just allocating 2% of the total system mass to planets (and moons). It provides interesting results, I always have a few gas giants, I often have many Mars-and-Mercury-massed planets.

But to calculate surface gravity, I need a radius.

This largely depends on the planet's composition... which can be quite varied (and depends quite a bit on where it formed, how it formed, etc).

But if I have a planet with 0.7 Earth mass, and 35% of that is iron/nickel (or siderophile), and 50% lithophile, and so on, can a decent estimate of radius be determined?

Do I need the breakdown on composition to be per atomic element, or can this give decent ballpark numbers if I have the mass as the ratio of lithophile/siderophile/chalcophile/volatiles?

My understanding of physics in this arena is... inadequate. I do not believe it's enough to simply look up the density of these elements on Wikipedia and calculate backwards from volume. Certainly an iron core compresses a bit such that the density is quite higher than that of an iron ingot on the surface?

• You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly. – Joe Bloggs Sep 19 '18 at 21:42
• @JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version... – kingledion Sep 19 '18 at 23:08
• @kingledion: Apologies, I thought I’d jumped back to the normal version before link copying. – Joe Bloggs Sep 20 '18 at 5:38

# Mass, density and radius are related

Let $m$ be the mass of a planet.

For a given a density $\rho$, the relationship between mass and volume is \begin{align} V &= \frac{4}{3}\pi r^3\\ m &= \rho V = \frac{4}{3}\pi\rho r^3 \end{align}

The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:

\begin{align} m&=\frac{4}{3}\pi\rho r^3\\ r&=\sqrt{\frac{3m}{4\pi\rho}} \end{align}

Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.

# What should density be?

Since you need density in both these equations, what are some reasonable densities for a planet?

Object/Planet      Density (kg/m^3)
Earth's Inner Core            12800
Earth's Outer Core             9900
Earth                          5510
Mercury                        5430
Venus                          5240
Mars                           3930
Vesta (densest asteroid)       3420
Luna                           3340
Ceres (largest asteroid)       2080
Ganymede                       1940
Titan                          1880
Neptune                        1640
Jupiter                        1330
Uranus                         1270
A small chunk of water ice      934
Saturn                          690


A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.

• Earth's core is then higher than the density of iron (by quite a bit too). Is there so much uranium (and lead and such) that this boosts the density that high? Or is the pressure truly so great that it overcomes the general incompressibility of solids? – John O Sep 20 '18 at 4:46
• @JohnO The pressure is very high. Iron undergoes unusual phase changes at this pressure. – kingledion Sep 20 '18 at 10:42