The Sun has nowhere near enough mass to enter the branch of stellar evolution that would lead to a supernova, fortunately for us. However, there are planets that orbit stars that are destined to go supernova. These planets might not be habitable, because such massive stars live and die within short periods of time, but they could still be interesting.

Could a planet survive a supernova from the star it orbits? By "survive", I mean that the planet must have minimal orbital disruption and should remain in one piece and as undamaged as possible. The planet does not need to be in the habitable zone and survival of any life on the planet is not required.

I leave it to answers to choose the mass of the star and planet, the orbital radius, and other relevant parameters, because not all combinations of these will result in the planet surviving. A good answer should determine the boundary line between survivable and unsurvivable scenarios.

I'd still love to see an answer answer that discusses the effects of the supernova ejecta shell hitting the planet, and takes that into consideration when determining if the situation satisfies all the criteria for survivability.

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    $\begingroup$ Do you want just the planet to survive, or any life on it as well? The latter is considerably less likely. $\endgroup$
    – ArtOfCode
    Commented Jun 11, 2015 at 21:51
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    $\begingroup$ Answerers, do take note of the meta discussion linked in the opening sentence. hard-science may end up having strictly-enforced criteria for answers, which if an answer does not comply, may lead to deletion. $\endgroup$
    – ArtOfCode
    Commented Jun 11, 2015 at 21:55
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    $\begingroup$ And just so you know, I do plan to answer this, but because of the nature of the tag, it will take some time. $\endgroup$
    – ArtOfCode
    Commented Jun 11, 2015 at 22:15
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    $\begingroup$ Does the planet have to be in the goldilocks zone? $\endgroup$
    – Josiah
    Commented Jun 11, 2015 at 22:28
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    $\begingroup$ "[a physicist] told me his rule of thumb for estimating supernova-related numbers: However big you think supernovae are, they're bigger than that. Here's a question to give you a sense of scale: Which of the following would be brighter, in terms of the amount of energy delivered to your retina: 1.A supernova, seen from as far away as the Sun is from the Earth, or 2.The detonation of a hydrogen bomb pressed against your eyeball? Applying the physicist rule of thumb suggests that the supernova is brighter. And indeed, it is. by nine orders of magnitude. " - what-if.xkcd.com/73 $\endgroup$
    – Philipp
    Commented Jun 12, 2015 at 11:49

5 Answers 5


For big stars with the right supernova conditions, yes.

First a note. There are several types of supernova. In general, a Type I supernova doesn't leave much behind. Thus it is pointless to ask whether the planet exists with a minimal orbital disruption as there is nothing to orbit. A Type II supernova generally does leave something behind, like a neutron star or a black hole. This is what we're interested in here.

To date, four pulsar planets have been discovered. A pulsar planet is, of course, a planet orbiting a pulsar. A pulsar is a remnant of a supernova. So, clearly there are planets which orbit what's left of a supernova.

However, we have never observed a system which had planets, went supernova, and confirmed that the planets remained. So while we have observed planets orbiting stars which at some point went supernova, we do not know for certain whether they formed after the supernova or before.

It is also impossible for us to know how the orbits or structure of the planets were affected. One National Geographic article reports that one newer model suggests planets could remain through a supernova:

The new model also hints that—in very rare cases—some survivor planets may remain bound to the supernova remnants, finding new orbits around the neutron stars or black holes left behind by the explosions.

The paper itself says specifically:

Planets around > 20M⊙ black hole progenitors may easily survive or readily be ejected depending on the core collapse and superwind models applied

So really big stars that supernova might be able to hold on to their planets given the right conditions.

  • $\begingroup$ It is also impossible for us to know how the orbits or structure of the planets were affected. This could be calculated, though, right? $\endgroup$
    – HDE 226868
    Commented Jun 11, 2015 at 22:30
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    $\begingroup$ Orbital disruption is discussed in the paper. I can't extract any good one liners, but the gist is that the orbit will be significantly disrupted. I would argue that even a multiple AU orbital disruption is minimal compared to ejection, and thus satisfies the requirements for "surviving". $\endgroup$
    – Samuel
    Commented Jun 11, 2015 at 22:45

If we focus on the luminosity, ignoring shell impact, we can say that inner planets of the star-system do get destroyed whereas outer might survive (to some extent).

Brightness of a supernova

From this question on physics.SE, we get a rough estimate of the supernova shell/nebula peak brightness at about 60 days, with a variation of 3 magnitudes (a brightness ratio of 16:1). As a rough number, figure the average luminosity is 1/10 of peak for that period.

So how bright is a supernova? Consider SN2011fe, a Type 1a supernova which produced a peak brightness about 2.5 x $10^9$ that of the sun. To be conservative, let's figure on an average shell/nebula luminosity over 60 days of about $10^8$ suns.

Energy received by the Earth

Under ordinary conditions, the solar power intercepted by the Earth is about 174 x $10^{15}$ watts, and has an albedo of about 0.3. So the total absorbed power is on the order of 6 x $10^{26}$ watts. After the shell passes, the brightess will approximately double, and remain more or less constant for the next 60 days, since it is inside the nebula. Which leads to an absorbed power of 6 x $10^{26}$ watts for 5 x $10^6$ seconds, for a total energy of 3 x $10^{33}$ joules.

Energy withstood by the Earth and consequences

Modelling the earth as 6 x $10^{24}$ kg, this provides 5 x $10^8$ joules/kg.

Iron has a vaporization energy of 4.25 x $10^5$ J/mol, and with an atomic weight of about 56, that's about 18 mol/kg. So the energy required to vaporize iron is about 7.6 x $10^6$ J/kg. This is an upper limit, since the core of the earth is a good deal warmer than 20°C.

As a result, a rough estimate says that the earth will be vaporized after about 22 hours. Even if ablation shields the unvaporized portions of the planet by 98%, the earth is completely vaporized after 46 days.

It's tough to recover from that.

Case of outer planets

Now, about Jupiter. Jupiter's orbit is a bit over 5 AU. Its diameter is about 10 times earth and its mass is about 318 times earth.

So the energy power intercepted by it will be roughly $10^2$/$5^2$, or 4 times as great as earth. It has 318 times the mass, but it's all hydrogen, and I'm not sure of the energy required to blow it apart. As a guess, let's use the gravitational binding energy. For the 4 gas giants, the binding energies are (from "Gravitational Potential Energy of the Major Planets", Bursa & Hovorkova):

Jupiter - 2.6 x $10^{36}$ J
Saturn - 3.6 x $10^{35}$ J
Uranus - 1.6 x $10^{34}$ J
Neptune - 2.2 x $10^{34}$ J

The total energy received for each planet will be (approximately)

Jupiter - 1.2 x $10^{34}$ J
Saturn - 2.5 x $10^{33}$ J
Uranus - 1.2 x $10^{32}$ J
Neptune - 5 x $10^{31}$ J

In all cases the binding energy of the planet is at least 2 orders of magnitude greater than the received energy, so by this measure they ought to survive, although the inner ones, especially Jupiter, should expect to lose significant mass.

By contrast, the binding energy of Earth is 2.5 x $10^{32}$ J, rather less than the 3 x $10^{33}$ J of energy it will receive, so it should expect to be destroyed in about 6 days, which seems to be in pretty good agreement with the vaporization argument.


So basically, a rough estimate says that the inner planets get vaporized, while the outer planets ought to survive.

  • $\begingroup$ The heat of vaporization is the energy required to bring it from liquid at boiling temperature to gas at boiling temperature, and doesn't take into account the energy required to bring it to that temperature in the first place. The actual energy is 4.25e5 J/mol or 7.61e6 J/kg to bring iron from solid @ 20C to gas @ 3134K. Also, you wrote the unit wrong. $\endgroup$
    – evankh
    Commented Jun 11, 2015 at 23:48
  • $\begingroup$ What about Jupiter? And Neptune? $\endgroup$ Commented Jun 11, 2015 at 23:59
  • $\begingroup$ @knave - Oops. Thanks. Forgot the runup. I've edited. But note that only the surface is at 20 C, the interior is warmer so the energy requirements go down. And I'm drawing a blank on the bad unit. $\endgroup$ Commented Jun 12, 2015 at 0:01
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    $\begingroup$ It seems a little pointless to analyze a Type 1a supernova, there won't be anything left for a planet to orbit. It already fails the "minimal orbital disruption" requirement. $\endgroup$
    – Samuel
    Commented Jun 12, 2015 at 16:21
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    $\begingroup$ the earth is completely vaporized after 46 days. It's tough to recover from that. $\endgroup$
    – Michael
    Commented Feb 2, 2016 at 1:57


Wikipedia has an entry on Pulsar Planets. This entry indicates that only four confirmed pulsar planets have been found and a fifth is a candidate. These planets seemed to have formed from three different mechanisms.

These planet formation mechanisms are:

PSR_B1257+12's Planets

Condensed from supernova debris

The planets are believed to be the result of a second round of planetary system formation[4] resulting from unusual supernova remnants or a quark-nova.

PSR B1620-26 b

Captured after supernova

At some point during the 10 billion years, the neutron star is thought to have encountered and captured the host star of the planet into a tight orbit, probably losing a previous companion star in the process. About half a billion years ago, the newly captured star began to expand into a red giant

PSR J1719-1438 b

Core remnant left by a "cooked" companion white dwarf

We show that it is in a binary system with an orbital period of 2.2 h. Its companion’s mass is near that of Jupiter, but its minimum density of 23 g cm−3 suggests that it may be an ultra-low mass carbon white dwarf. This system may thus have once been an Ultra Compact Low-Mass X-ray Binary, where the companion narrowly avoided complete destruction.

(emphasis mine)

If a star barely survived a supernova explosion then no planet would be able to withstand that blast (unless it was sufficiently far away).


Nearly all known exoplanets, including pulsar planets, are quite close to their primaries with correspondingly short orbital periods, since we need several orbits worth of data to detect them. But for this question we want to go to the other extreme and consider planets as far away from their primary as possible, about which we don’t know very much since we’ve only detected a handful by direct optical imaging. It’s perfectly possible and indeed likely that a type II supernova will have gas giant planets at a distance of half a light year or more, and they will easily survive the supernova, but there’s no way we could detect them without millennia of data. We couldn’t even detect a gas giant planet of our own sun at that distance, and one or more may well exist.


I might note the planets orbiting the remnants of a supernova explosion ("pulsar planets") are sometimes thought to be "new" planets formed by condensing the materials vaporized by the initial supernova explosion.

Given the amount of energy being received by the outer gas and ice giant planets during the supernova explosion, I would expect that even far off Neptune analogues are reduced to the exposed core, with the atmosphere stripped away. Since the cores are thought to be about Earth sized, there is some land for enterprising developers, although since the "sun" is a neutron star, things might be a bit cold and dark.

  • $\begingroup$ Isn't this just what Samuel's answer says? $\endgroup$
    – HDE 226868
    Commented Jun 13, 2015 at 22:24
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    $\begingroup$ It appears this answer doesn't conform to the hard-science tag. Zero sources cited (let alone referenced). $\endgroup$
    – Samuel
    Commented Jun 14, 2015 at 2:51

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