According to the NASA press release, the Trappist planets are close enough (only a few million kilometers) that "the tidal forces between the planets are not negligible". The speaker says that this could cause ocean tides on the planets. Are they close enough that the tidal forces could heat the interior of the planets?


2 Answers 2


This is a complicated question that would really require a full physics simulation and better knowledge of the system to accurately answer. But let's try a few back of the envelope calculations to see what we get.

Calculating tidal forces from TRAPPIST-1c on TRAPPIST-1b

I'm going to calculate the tidal effects of TRAPPIST-1c on TRAPPIST-1b (simply because, a priori, this seems likely to be where the strongest tidal heating will be induced). See the figure below which describes the parameters.

Diagram of setup

The tidal force of 1c on 1b is defined as the differential force of gravity across 1b, that is, the difference of the force of gravity on the side of 1b facing towards 1c and the force of gravity on the side of 1b facing away from 1c. Mathematically, we get.

$$F_{tide,c-b} = F_{g,-R_b} - F_{g,+R_b} = \frac{GM_bM_c}{(D-R_b)^2} - \frac{GM_bM_c}{(D+R_b)^2} = 4GM_bM_c\frac{DR_b}{(D^2-R_b^2)^2}$$

We can presume that $R_b << D$ (for this case $R_b/D = 1\%$) and reduce this to

$$F_{tide,c-b}(D) \approx 4GM_bM_c\frac{R_b}{D^3}$$

But this isn't enough to determine the amount of tidal heating that may occur. Tidal heating only occurs when the tidal force changes. It is this constantly changing tidal force which results in tidal flexing of the planet and thus creating heat through tidal friction. Fortunately, for these two planets, the tidal force will be changing since $D$ will be constantly changing. So let's calculate $F_{tide}$ for the two extremes where these planets are as close as possible and as far as possible and difference them.

$$\Delta F_{tide,c-b} = F_{tide}(0.004\:\mathrm{AU}) - F_{tide}(0.026\:\mathrm{AU})$$

If I plug in numbers to this, I find that

$$\Delta F_{tide,c-b} \approx 3.7\times10^{20}\:\mathrm{N}$$

Okay, but what do we do with this number? It's somehow a metric of the change in tidal forcing that 1c imparts on 1b, but is it negligible? To determine this, we have to compare it to something. Let's compare this to the tidal forcing that TRAPPIST-1b would receive from the star.

Calculating tidal forces from TRAPPIST-1 on TRAPPIST-1b

I've already set up the math, so we don't need to go over that again. But first, let me discuss where this tidal forcing is actually coming from. Quoting an article from space.com, the paper's author, Gillon, states:

Because the seven alien worlds orbit so tightly, they're probably all tidally locked, Gillon said. That is, they likely always show the same face to their host star, just as Earth's moon only shows the "near side" to us.

As I said above, the only way to produce tidal heating is to have changing tidal forces. These planets are likely synchronized and always present the same side to the star. It's marginally possible these planets aren't perfectly tidally locked, but rather have some higher spin-orbit resonance. That is, their spin orbit resonance might not be 1:1 (as it would be if they were tidally locked) but instead could be something like 3:2 (which is what Mercury has). I'll ignore that complication though and just assume 1:1 resonance. So if they're tidally locked, they can't experience differing tidal forces through their own rotation. Instead, the differential tidal forcing comes from the orbit's ellipticity. Sometimes the planet will be closer and sometimes it will be farther, causing a differential tidal force on TRAPPIST-1b from the star as it orbits. This is exactly what occurs in the tidal heating of Io. Let's calculate $\Delta F_{tide,*-b}$ by using the varying distances TRAPPIST-1b will have from the star. I found that TRAPPIST-1b will orbit between $0.0101\:\mathrm{AU}$ and $0.0119\:\mathrm{AU}$1. This means the differential tidal force is:

$$\Delta F_{tide,*-b} \approx 4GM_bM_*R_b\left(\frac{1}{(0.0119\:\mathrm{AU})^3} - \frac{1}{(0.0101\:\mathrm{AU})^3}\right) = 1.8\times10^{23}\:\mathrm{N}$$

Is planetary tidal heating non-negligible?

The back of the envelope calculations show that the differential tidal force on TRAPPIST-1b from TRAPPIST-1c is about $0.2\%$ of the differential tidal force due to the star. Whether you consider this negligible or not is up to you. I'd personally consider it a pretty small effect and say that most of the tidal heating these planets experience comes from the star itself.

Could the inter-planetary tidal heating still contribute to the tidal heating of the planets enough to heat the interior?

This is a remarkably hard question to answer and I can't even really do a back of the envelope calculation without making wild, unjustifiable assumptions. The calculations above simply determined the maximum tidal force variation over time. That doesn't tell us anything about how much tidal heating this may induce though. That requires knowing more about the planet itself, in particular the planet's Love numbers which define the rigidity of the body and thus how easy it is to stretch via differential tidal forces. You can vary your tidal forcing as much as you want, but if your planet is pure iron (and thus very rigid) you're unlikely to have as much of an effect as if it were primarily silicate (and thus much less rigid). The paper produces the plot below which defines the potential constituents of each planet. This would be a first step in determining the planet rigidities, but as you can see from the error bars, it would be highly uncertain.

enter image description here

Overall, and this is entirely opinion based and from my calculations above, but I'd say the chances of the inter-planetary tidal heating having significant effects on the interior heat of of these planets is negligible. More than likely the biggest contributing factor is radioactive decay, followed by tidal heating from the star (but this is amplified by the eccentric orbits induced by planetary gravitational perturbations)

1 Note that this calculation involves using the eccentricity and the paper only provides an upper limit. These distances then represent an upper limit as well and the final answer will also be an upper limit. It may be less.

Values used in calculations:

  • $G = 6.67\times10^{-11}\:\mathrm{m^3 kg^{-1} s^{-2}}$
  • $M_b = 5.075\times10^{24}\:\mathrm{kg}$
  • $M_c = 8.239\times10^{24}\:\mathrm{kg}$
  • $M_* = 1.604\times10^{29}\:\mathrm{kg}$
  • $R_b = 7.34\times10^{6}\:\mathrm{m}$
  • $\begingroup$ While long, this answer is not quite correct. Rather than downvote, I'll post an alternative. $\endgroup$ Mar 6, 2017 at 2:52
  • $\begingroup$ @DavidHammen What about this is not quite correct? I believe I explained everything in my answer that you put in yours. Particularly the point about the eccentric orbit being induced by other planets, causing tidal heating. $\endgroup$
    – zephyr
    Mar 6, 2017 at 13:38
  • $\begingroup$ It ignores orbital resonances, which are a primary driver in the tidal heating of the Jovian moons, and it ignores that a tidally locked object in a perfectly circular orbit about the primary would undergo zero tidal heating from the primary. $\endgroup$ Mar 6, 2017 at 13:42
  • $\begingroup$ @DavidHammen I don't specifically mention orbital resonances, but I do have a whole paragraph describing the effects of the planets on each others ellipticities (and in fact mention and link to the case for Io). I also explicitly say "if they're tidally locked, they can't experience differing tidal forces" and "the differential tidal forcing comes from the orbit's ellipticity". This is equivalent to saying no tidal heating occurs for circular orbits. I don't believe I said anything incorrect as you suggest, I just didn't mention all the exact points you might have. $\endgroup$
    – zephyr
    Mar 6, 2017 at 13:46

What drove the comment that "the tidal forces between the planets are not negligible" was the Jovian moons. The three innermost of the Galilean moons of Jupiter, Io, Europa, and Ganymede, are in a 4:2:1 orbital resonance. Io would not exhibit any tidal heating if its orbit was circular.

Io's orbit is not circular, thanks to those orbital resonances. One of the consequences of these resonances is that Europa and Ganymede act to pull Io's orbit out-of-round; i.e., more elliptical. The elliptical nature of Io's orbit results in time-varying tidal stresses on Io, which makes Io geologically active. Those tidal stresses by Jupiter in turn act to circularize Io's orbit.

The tidal stresses become less severe as Io's orbit becomes closer to circular. Io cools down, resulting in an increase in its tidal qualify factor Q. This makes it less susceptible to further circularization. The competing forces from Europa and Ganymede can then make Io's orbit more eccentric. Tidal stresses eventually come into play again, warming up Io and decrasing its tidal quality factor. Now Jupiter is the driver. This makes for a rather nice hysteresis loop.

What prompted that comment that the TRAPPIST-1 planets may be subject to tidal heating is that some of those planets appear to be in an orbital resonance, with periods being very close to small integer multiples of one another.

  • 1
    $\begingroup$ I understand how tidal heating works, especially with the Jovian system. My question was more about if the planets of the Trappist system were close enough to their star and to each other to be tidally heated. $\endgroup$
    – Phiteros
    Mar 6, 2017 at 4:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .