I was thinking about the physics behind a hypothetical scenario where a planet the size and the mass of the Earth is orbiting so close to a very hot star and what the long-term fate of such a planet would be. There are of course so many variables to consider if I expect this question to be answered, so I will try to make the most important assumptions.

Assumptions are as following:

1- The planet is an Earth-like planet.

2- The star is an O-type star with effective surface temperature of 40,000 °K and a radius of 15 solar radii.

3- The Earth-like planet is orbiting at 20 solar radii from the center of the star, or 5 solar radii from its surface.

I have calculated that an Earth-sized planet at this distance from this star would intercept ~ $5\times10^{-8}$ of the total radiation emitted by this star, or about $10^{25}$ W.

Let's now assume that it would take an amount of energy equal to the graviational potential energy of the Earth which is $2.5\times10^{32}$ J to disassemble all the planet's mass. It would actually take a little more energy than that in order to raise the temperature to the boiling point of rocks, but this energy should be less than the gravitational energy and so we can ignore it for simplicity.

To accumlate this amount of energy at the rate our hypothetical planet would be intercepting, it would take $25\times10^{6}$ seconds, or almost a year. This rate is about 60 million times the current rate at which the Earth is intercepting radiation from the Sun. And since radiant flux is proportional to the fourth power of temperature, then the temperature on this planet should reach about 25,000 °K, which is way higher than the boiling point of any known compound on Earth.

This is actually the part where my question lies. The heating of the planet can't be 100% efficient, and the planet will be re-radiating energy at a very high rate plus the fact that the already vaporized material will be shielding the material beneath it. So, would this planet ever get completely vaporized after a long time? Or would it heat to a point where its radiation flux is equal to its absorbed flux and so nothing happens?

Also one other thing I thought of is the thermal conductivity of rock. Since rock has an average thermal conductivity of about 2 W/m.K which is not that high, it would take so long for layers beneath the surface to actually heat up. But at the same time, the surface should be vaporizing, so would this accelerate the process and save the time needed for thermal conduction?

So if any mass-loss rate can be estimated within an order of magnitude level of accuracy, that would definitely answer my question.

  • $\begingroup$ Imo, and I hope you get a better answer later from someone that I can learn from too, is that the material of the planet will dissipate the material of the planet anyway. In other words, the vaporised material "shielding" the remainder of the planet would need have to be constantly replenished, as it would eventually drift or be blown away, and the planet is doomed in the long run. $\endgroup$
    – user81619
    Commented Aug 19, 2015 at 1:57
  • $\begingroup$ I am assuming, that the shield would get so hot it would not last. By this I mean, and my apologies for my badly worded statement, that as you say in your comment below regarding Maxwell distribution effects, the Maxwell velocity distribution effect would leave only colder particles, which would then heat up and repeat the process, assuming the sun lasts much longer than the planet, I assume the planet material would continue in this cycle until it was gone. $\endgroup$
    – user81619
    Commented Aug 19, 2015 at 1:57
  • $\begingroup$ I might give this a shot later, but my flippant answer is that the planet will be plowing through so much stellar wind that its orbit will "rapidly" decay and send it plunging into the star. I think. Would have to run some numbers to be sure. $\endgroup$
    – Kyle Oman
    Commented Aug 19, 2015 at 2:37
  • $\begingroup$ @KyleOman, I remember I once calculated this too. I didn't do it to figure out if the planet's orbit will decay, but rather to see how much mass the planet would be intercepting from the stellar wind. I remember it was negligible compared to planet's mass, which make sense comparing the sizes of an Earth-sized planet and a 15 solar radii star. $\endgroup$ Commented Aug 19, 2015 at 8:26

2 Answers 2


I think there is a better way to think about it. To a small factor, you are essentially saying the planet is on the photosphere of the star. The star will fill almost half the solid angle. Therefore the flux (power per unit area) from the star at the substellar point on the planet is $$ f \simeq \pi \int B_{\nu}\ d\nu =\sigma T^{4} = 1.5\times 10^{11}\ W/m^2.$$ There would be a small factor reduction due to projection effects over the rest of the illuminated surface. Let's round down the absorbed power to $$P = 10^{11}\times 2\pi R^2\ W.$$

The equilibrium temperature of the illuminated surface would be a little lower than 40,000 K (as you say), but it could never get this hot before the surface was vapourised, so I would conclude that radiative losses from the planet will be negligible. Heat conduction is inefficient so little of the energy is conducted into the planet.So to first order I would say that the energy goes into (a) vaporising the rock, (b) liberating it from the planet.

The specific heat capacity of generic rock or magma is about 1000 J/(K kg). The latent heat of fusion is around $3\times 10^{5}$ J/kg, but the latent heat of vaporisation at a few thousand K is likely an order of magnitude larger than this. I will assume it takes $5\times 10^6$ J to evaporate a kg.

To liberate material from the surface of an Earth-like planet requires of order $GM/R=6\times 10^{7}$ J/kg so appears to be considerably more important.

So the mass loss rate is around $$\dot{M} = \frac{PR}{GM} \simeq \frac{2\times 10^{21}}{\rho} \ kg/s, $$ where $\rho$ is the average density (in this model, as the planet gets smaller the mass gets smaller, but the lower potential energy requirements are balanced by intercepting less stellar flux).

This gives an evaporation timescale for an Earth-like planet of $1$ year! So I am agreeing with your calculation.

Perez-Becker & Chiang study a similar scenario (rocky, evaporating planet very close to a star) in some detail. They point out the following inadequacies of the treatment above. First, at the substellar point, the gravity of the star helps to pull the material away from the planet. Second, as the gaseous atmosphere expands, it cools and dust can condense. Third, the latent heat can reheat the wind, but the dust can also shield the planet from radiation. Fourth, the wind from the star can interact with the wind from the planet, significantly altering mass-loss rates. It turns out that these are extremely important. In the examples considered (see section 4.1 of the paper), they say that the simple mass loss rates calculated here would need to be multiplied by between $10^{-4}$ and $10^{-8}$!

However, these calculations are for the modest situation of planets close to a fairly dim K-type star. My gut feeling is that in the scenario you propose the wind would not cool sufficiently to form dust and the evaporation timescale would still be extremely short, and/or the wind from the star would blast the evaporating material away, preventing obscuration.

If the evaporating material reached 10,000 K, even vaporised iron could escape the Earth's gravity. A spherically symmetric wind would have $$\dot{M} = 4\pi R^2 \rho_w v$$ If the expansion velocity were of order 10 km/s, then for the mass-loss rate discussed above wind density would be $$ \rho_w \simeq 0.08 \left(\frac{R}{R_E}\right)^{-2} \left(\frac{v}{10\ km/s}\right)^{-1}\ kg/m^3.$$ The (ram) pressure (for $v=10$ km/s would only be $8\times10^{6}$ Pa. Typical O-star winds might have a velocity of 1000 km/s and a mass-loss rate of $10^{24}$ kg/s. At a radius of $20R_{\odot}$ this suggests a wind density of $4\times10^{-4}$ kg/m$^3$, but a ram pressure of $4\times 10^{8}$ Pa. On the basis of this calculation I would suggest that the O-star wind simply blasts away the evaporating material. Thus no dust obscuration and thus the original evaporation timescale is ok.

  • 2
    $\begingroup$ Where have you been all this time? We've all been waiting for an answer from a real astrophysicist :) $\endgroup$ Commented Sep 26, 2015 at 10:04
  • $\begingroup$ But in such a situation would a planet ever have been possible? $\endgroup$ Commented Sep 26, 2015 at 15:57
  • $\begingroup$ @ThorbjornRavnAndersen Rocky planets have been found very close to stars, but not O stars. The core of an evaporated hot Jupiter might also resemble a rocky planet. $\endgroup$
    – ProfRob
    Commented Sep 26, 2015 at 16:14
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    $\begingroup$ @RobJeffries, sorry for keeping coming back to this question although it is already answered. But I have noticed that in many paper, the evaporation models of planets seem to take only the ionizing luminosity (UV and X-rays) of the star into consideration. Here's an example: arxiv.org/pdf/1504.07170v1.pdf. So is there any explanation for why not the entire luminosity of the star is used ? $\endgroup$ Commented Oct 17, 2015 at 12:07
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    $\begingroup$ @AbanobEbrahim I think that is correct. Most power is above 10.2 eV. $\endgroup$
    – ProfRob
    Commented Oct 18, 2015 at 7:34

The precise happenings cannot be predicted. However, let's try using the basics of radiation heat transfer to suppose what will happen.

As we know, the radiation emitted by a body is directly proportional to $T^4$, where $T$ represents the temperature of the body. Thus, the increase in emission of radiation increases very rapidly with an increase in temperature. A very important part of our discussion here.

By current information, we know that approximately $35\%$ of the incoming radiation is reflected back into space, by the Earth's atmosphere. Approximately $17\%$ is absorbed at various levels of the atmosphere. That leaves us with $48\%$ of the incoming radiation to be dealt with by the ground. Let's consider that this entire radiation is absorbed by the ground, for simplicity's sake.

We know that: $$Q_\text{emitted} = \epsilon \sigma A T^4$$

Let's consider for Earth to be a perfect emitter ($\epsilon = 1$). The temperature needed for emission of all energy incident ($10^{25} \times 0.5 = 5 \times 10^{24}$) Upon calculation, the temperature comes out to be approximately $20200~\text{K}.$

As a large part of the Earth is composed of water, that will evaporate quickly, increase the density of water vapour in the air. This will further increase the amount of radiation reflected back by the atmosphere. Moreover, it will also increase Earth's surface area.

Thus, I believe that we would reach a point where almost all the incoming radiation will be equal to the outgoing radiation, and no change in temperature will take place. So no, Earth will not vaporize.

  • $\begingroup$ Your calc should give an equilibrium temperature of 20,300 K. $\endgroup$
    – ProfRob
    Commented Sep 26, 2015 at 12:57
  • $\begingroup$ @RobJeffries Aha.... Maybe I made a calculation mistake? Are you sure? And obviously considering the fact that you have posted your answer, don't think mine holds much value anyways. $\endgroup$ Commented Sep 26, 2015 at 15:25

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