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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 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 $b$?

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  • $\begingroup$ Considering the values for your plot, I would think that the error with determining the slope and the lack of points between 0.5 and 2 Earth masses is responsible for the large deviation of an object with almost no mass having a significant surface gravity. It would be best to plot the surface gravity for objects between 0.5 and 2 Earth masses to see if there is a statistical error. $\endgroup$ – LDC3 May 3 '15 at 0:16
  • $\begingroup$ BTW, where did the data come from? $\endgroup$ – LDC3 May 3 '15 at 0:18
  • $\begingroup$ Thanks for your comments. The data are from exoplanets.org. Now that I am thinking about it, though, log(1)=0, so the y-intercept represents the surface gravity of a 1 $M_{J}$ planet, no? $\endgroup$ – user5341 May 3 '15 at 0:25
  • $\begingroup$ You're right. I guess I should have double checked my calculation. So, the surface gravity of a planet with $log (M_O)=0$ is $log(a)=0.8631$, or $a=7.3cm/s^2$. Are you sure you have the right units? Since $9.98m/s^2=998cm/s^2$ and $log(998cm/s^2)=2.999$. $\endgroup$ – LDC3 May 3 '15 at 0:34
  • $\begingroup$ Yes this is confusing... Right now, I think that since surface gravity also depends on radius, this line is giving information about the radius. But nothing particularly useful, since $\log \frac{GM}{R^2}$ vs $\log M$ goes like $1 + \frac{k \log R}{\log M}$, after using some logarithm rules, I think. Please let me know if this does not sound correct to you, I am very new to this type of thing! $\endgroup$ – user5341 May 3 '15 at 1:52
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I think what you have established here is just that $\rho$ tends to increase with mass. The density of planets isn't constant.

Let $\rho = \rho_0 (M/M_{earth})^{\alpha}$, 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. Below 0.1 Jupiter masses, the relationship appears to break 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).

Planet density versus mass

Surface gravity versus mass

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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.

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  • $\begingroup$ Shouldn't that be $g = (\tfrac{2}{3}\sqrt[3]{6\pi^2})(G\sqrt[3]{\rho^2M})$? Note that the OP's question is actually mostly about $\log M\to 0$, but otherwise, I agree, the most straightforward physical thing it corresponds to is mean density, fit to the data. $\endgroup$ – Stan Liou May 3 '15 at 10:22
  • $\begingroup$ Isn't the problem with this idea that $\log g = (1/3)\log M + const$, which is not what the OP gets? $\endgroup$ – ProfRob May 3 '15 at 10:40
  • $\begingroup$ @RobJeffries The OP is far-extrapolating gas giants, so there $\rho$ is also strongly dependent on $M$--but that shouldn't change the correspondence to mean density for any particular fixed $M$, which was actually OP's question as stated. Really, though, the OP's extrapolation is physically inappropriate, and for light exoplanets ($M\lesssim 4$ in units of $R_\oplus = M_\oplus = 1$), Seager et al. (2007) $$\renewcommand{\lg}{\log_{10}} \lg R = k_1 + \frac{1}{3}\lg{M}-k_2M^{k_3}\text{,}$$ and so $\lg g = \frac{1}{3}\lg M - 2k_1 + 4k_2M^{k_3}$. $\endgroup$ – Stan Liou May 3 '15 at 11:17
  • $\begingroup$ That $4k_2$ should be $2k_2$ in my comment (typo). $\endgroup$ – Stan Liou May 3 '15 at 20:08

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