What's a Rayleigh number? Well, it doesn't have anything to do with Rayleigh scattering. Here's a Wikipedia page on this concept, but I believe Elvira Mulyukova's and David Bercovici's Mantle Convection in Terrestrial Planets put it better:

The competition between forcing by thermal buoyancy and damping by viscosity and thermal diffusion is characterized by a dimensionless ratio called the Rayleigh number.

In other words, a Rayleigh number represents the ratio between how much hot mass in a convection current rises and how much the viscosity and thermal conductivity of the mass stop it from rising and bleed away its heat, respectively. In this case, the mass in question is the mantle of a terrestrial planet, and the convection current in question is mantle convection.

Mantle Convection in Terrestrial Planets goes on to state the following:

$Ra$ [Rayleigh number] needs to exceed a certain value, called the critical Rayleigh number $Ra_c$, in order to excite convective flow. The value of $Ra_c$ is typically on the order of 1,000, with the exact value depending on the thermal and mechanical properties of the horizontal boundaries, (e.g., whether the boundary is rigid or open to the air or space, see Chandrasekhar, 1961).

Therefore, it can be concluded that, if, within a body's mantle, the ratio of "stuff-forcing-its-way-up" to "stuff-being-held-down" is ≤ ~1,000, said body won't experience convective flow in its mantle. As far as I know, convective flow in the mantle is necessary for a body to have plate tectonics; this is supported by the fact that the Earth and Venus, with the highest Rayleigh numbers calculated here, are fairly volcanically active, whereas Mercury, with the lowest one, is a dead rock, and Mars, with the second-lowest one, features stagnant-lid tectonics, if I recall correctly.

Continuing from Mantle Convection in Terrestrial Planets:

Assuming their material properties are similar to Earth’s, we can estimate the Rayleigh numbers for the mantles of other terrestrial planets: ${10}^4$ for Mercury, ${10}^7$ for Venus, and ${10}^6$ for Mars. With the exception of Mercury, whose $Ra$ is at most an order of magnitude above critical, the mantle of the rocky planets in the solar system appear to be cooling predominantly by convection.

Mulyukova and Bercovici calculated these Rayleigh numbers using the following values:

Property Mercury Venus Earth Mars
Density (kilograms per cubic meter) 3500 4000 4000 3500
Surface temperature (degrees Kelvin) 440 730 285 220
Core-mantle boundary temperature (degrees Kelvin) 3000 3500 3500 3000
Mantle thickness (kilometers) 400 2900 2900 2000
Gravity (meters per second squared) 3.7 10 10 3.5

The gravity values (if not others) are slightly inaccurate and could use slight refining (for instance, 9.80665 $m/s^2$ rather than 10 $m/s^2$ for the Earth's gravity), but that's irrelevant to this question.

For the purposes of this question, the equation for finding a Rayleigh number is: $Ra = \frac{ \rho g \alpha \Delta Td^3 }{\upsilon \kappa}$ where:

  • $\rho$ = density of mantle
  • g = planet's gravity
  • $\alpha$ = thermal expansivity of mantle
  • $\Delta T$ = difference in temperature between top and bottom boundaries of mantle
  • d = thickness of mantle
  • $\upsilon$ = viscosity of mantle
  • $\kappa$ = thermal diffusivity of mantle

My question is: what would the geology of a planet with a high (specifically, $Ra \geq 10^8$) Rayleigh number look like? The few planets for which I can find calculated Rayleigh numbers seem to fit a trend in which higher Rayleigh numbers correlate to more volcanism, more mountain-forming, more tectonic plates moving about, etc. Obviously, a sample size of 4 doesn't allow itself to accurate extrapolations, but that's why you're here: I'm asking this in the hope that someone more informed and/or qualified can provide further insight.

Let's use four hypothetical planets here:

  • $Ra = 10^8$
  • $Ra = 10^9$
  • $Ra = 10^{10}$
  • $Ra = 10^{11}$

If any of these are physically impossible, let me know.

Even before this is answered, I think I can draw several conclusions myself. Such a planet would cool relatively faster than planets with low Rayleigh numbers (how much more quickly, and what the relationship between Rayleigh number and heat loss is, I don't know) as the convection currents within its mantle would carry heat away from its core more quickly. I think that this means that, if the Rayleigh number-to-surface-area ratio (and, therefore, the Rayleigh number-to-planetary-radius ratio) is greater than that of planets with equal surface areas and ratios, the geothermal gradient would likely be steeper; more heat would be soaking out of the ground per unit of surface area. I also imagine such planets would see more volcanism, more mountain-forming, more tectonic plates moving about, etc.

I'm fairly sure this question fits Astronomy SE better than it fits Physics SE. If you think otherwise, let me know.

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    $\begingroup$ Kind of related: astronomy.stackexchange.com/q/40769/34513 $\endgroup$ Commented Jan 12, 2023 at 10:01
  • $\begingroup$ @Jean-MariePrival As a matter of fact, the answer to that post is the very thing which put me onto the concept of Rayleigh numbers in the first place. $\endgroup$
    Commented Jan 12, 2023 at 18:51

1 Answer 1


(This is more an answer to this part: "If any of these are physically impossible, let me know".)

As you mention volcanism, plate tectonics, mountains... I make the assumption that you consider rocky planets only. If so, it is pretty straightforward to calculate the theoretical maximum Rayleigh number of a rocky planet. The rocks being essentially the same as those on Earth/Mars/Venus, parameters like $\alpha$, $\upsilon$, $\kappa$ and $\rho$ will be similar. That leaves us $g$, $\Delta T$ and $d$ to work with. They all work the same way: the bigger the planet, the bigger they are, and the bigger $Ra$ gets.

Now, according to the following graph, the largest rocky super-Earth (near the "67 % Fe" dashed grey line) has a mass of about 7 $M_E$ and a radius of about 2 $R_E$ (a bit lower, but let's say 2 for the sake of simplicity):

enter image description here Scatter plot of super-Earth characteristics in context showing the mass, $m$, and radius, $r$, for transiting Super-Earths (black dots), public domain

This translates in:

  • $g = 17.26 \approx$ 2 times that of the Earth
  • $d^3 =$ 8 times that of the Earth
  • $\Delta T$ is trickier, but is fairly linear in Earth's mantle, so extrapolating the trend yields ~6000 K, about 2 times that of the Earth

This back-of-the-envelope calculation shows that the largest rocky super-Earth can have a $Ra$ of about $3 \times 10^8$, no more.

As to how such a vigorous mantle convection would translate in terms of surface processes, I agree with you: probably more volcanism and plate movements. The tectonic regime could also be different: some people think that in the Archean (when the Earth likely had a higher $Ra$) the dynamics was more vertical, as opposed to the modern horizontal tectonic regime.

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    $\begingroup$ I think this is a very good heuristic for finding the largest possible Rayleigh number. I'll note, however: larger rocky bodies have been hypothesized. For instance, Kepler-277b may have 87.4 M⊕ and 2.92 R⊕, and Kepler-145b may have 37.1 M⊕ and 2.65 R⊕. However, these figures equate to densities roughly comparable to depleted uranium and lead, respectively, which seem overly large, so your predictions likely hold true for bodies known to be rocky. Honestly, that's an entirely different question I should ask Astronomy SE. Also, thanks for the edit; these symbols can be tricky to work with. $\endgroup$
    Commented Jan 12, 2023 at 15:18
  • $\begingroup$ Did the math. If mass/radius figures are accurate, Kepler-277b has ~10.25⊕ g, 24.9⊕ d^3, and 2.92⊕ delta-T, for a Rayleigh number of ~7.45E9, and Kepler-145B has ~5.28⊕ g, 18.61⊕ d^3, and 2.65⊕ delta-T, for a Rayleigh number of ~2.604E9. Do you think such high numbers would equate to proportionately greater more volcanic and tectonic activity, though? At a certain point, it seems like it would be difficult for a planet to get more tectonically active and volcanic. $\endgroup$
    Commented Jan 12, 2023 at 18:48
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    $\begingroup$ Are you sure of your calculations? According to Wikipedia, Kepler-277b has a very large iron core, of 2.435 $R_{\oplus}$. That leaves only 0.485 $R_{\oplus}$ of rocky, convective layer (i.e., mantle), or ~3000 km, like Earth. So $d^3$ and $\Delta T$ would be similar, leaving just a factor of 10 to account for $g$, yielding a $Ra$ of $10^8$. $\endgroup$ Commented Jan 13, 2023 at 8:30
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    $\begingroup$ And I agree, there is probably a limit, at some point things would probably stop being solid, so no more plate tectonics or volcanism, just a big magma ocean world. $\endgroup$ Commented Jan 13, 2023 at 8:31

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