On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept?

Question

I am playing around with data from exoplanets.org, and am interested in the plot of surface gravity to planet mass. I reproduced this plot:

after downloading their data and performed a non-linear regression model to fit the curve. Unfortunately,the covariance matrix of my fit has infinite values, so I am now trying out a linear fit on a log-log plot, shown below:

My fit, for $y=ax+b$, is $a=0.9511$, $b=0.8631$.

I am now thinking about what I am plotting. I suspect there may not be anything interesting in $\log \frac{GM}{R^2}$ vs. $\log M$, but, regardless, I am trying to understand if there is any meaning in the y-intercept.

What are possible explanations for the value $b$?

Answer 1

I think what you have established here is just that $\rho$ tends to increase linearly with mass for giant planets, as a consequence of their radii being held almost constant by electron degeneracy pressure.

Let density depend on mass as $\rho = \rho_0 (M/M_{earth})^{\alpha}$, so that $M = (4/3)\pi R^{3} \rho_0 (M/M_{earth})^{\alpha}$

Then $$g = \frac{GM}{R^2} = \frac{4\pi G}{3} R \rho$$

Replace $R$ with $(3M/4\pi \rho)^{1/3}$ so that $$ g = \frac{4\pi G}{3} \left(\frac{3M}{4\pi \rho}\right)^{1/3} \rho$$ $$ g = \left(\frac{4\pi}{3}\right)^{2/3} G M^{1/3} \rho_0^{2/3} (M/M_{earth})^{2\alpha/3}$$ $$ g = \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3} (M/M_{earth})^{(2\alpha+1)/3}$$

So, bar a (highly possible) algebraic slip, if you plot $\log g$ vs $\log M$, the gradient is $(2\alpha+1)/3$, which from your plot, gives $\alpha \simeq 0.92$ - i.e. the average planet density increases almost linearly with mass.

The intercept then is $$ b = \log \left[ \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3}\right],$$ which yields $\rho_0 \simeq 3.5$ kg/m$^3$ (NB: I subtracted 2 from your $b$ to make it SI; giving a density of around 814 kg/m$^3$ at a Jupiter mass).

The fact that density is almost proportional to mass can be found from the same dataset. e.g. see below. Of course, the value of $\rho_0$ above is nothing like the density of Earth and that is because below 0.1 Jupiter masses, the relationship breaks down, though in actual fact very few of the densities for such planets are very accurately measured (since it requires a radius from a transit), but it works well enough in the range you have plotted. The physics here is that gas giants are governed by a partially (electron) degenerate equation of state that results in them all having a similar radius from about a tenth of a Jupiter mass to about 50 Jupiter masses (albeit with considerable and largely unexplained scatter). Thus the density is proportional to mass. This relationship does not work for small rocky planets, where the radius decreases for smaller masses. Thus your nice line in log space does not continue to lower masses (see plot at bottom - NB: some of the low-mass points have huge uncertainties).

Answer 2

For planets of constant mean density you have: $$ M=\rho \times 4\pi r^3 $$ and the surface value of $g$ is: $$ g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r $$ So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ as $r \to 0$

In terms of mass: $$ g(M)={G(4\pi \rho)^{2/3}M^{1/3}} $$ So as $M\to 0$ we have $g(M)\to 0$ and again $\log(g(M))\to -\infty$ and the intercept of the $\log-\log$ plot as $M$ goes to zero alows you to calculate the mean density.