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Assuming some magic handwavium caused the sun to go supernova, how long would it take to reach Earth and wipe out the planet? I know it would be eight minutes until the light reached us, but how fast would the physical explosion travel?

ArtOfCode
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Kyle
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    As you suspect, physically speaking there's no mechanism that could cause the Sun to supernova. Not enough mass there. If you magically replaced our Sun with the smallest supernova star though, that might work. – Jim2B May 30 '15 at 23:01
  • You might be better off asking on [astronomy.se] how fast the ejected material travels in a supernova explosion. (Assuming that's known in the first place.) That and the Earth's distance from the sun (150Gm) will pretty much give you the answer. – user May 30 '15 at 23:06
  • This might be answerable if you pick a different star-and-planet system, specifying the mass of both objects, the luminosity of the star, orbital radius, etc. Note, though, that many supergiants have radii greater than Earth's orbital radius, so you'd have to change some parameters. But what @Jim2B said trumps all. Oh, and as another note, by "it", I assume you mean the ejecta? Don't assume that it would completely destroy the planet. – HDE 226868 May 30 '15 at 23:06
  • Working off of @MichaelKjörling's suggestion: No post yet exists on Astronomy, nor Physics, so you could ask on one of those two, preferable Astronomy. Chris White's answer here gives a speed of ~10,000 km/s, which is reasonable. You can do the math here, but keep in mind that, once again, stars that are massive enough to undergo core collapse supernova are most likely already large enough to reach earth's orbit. Also, this could be of use. – HDE 226868 May 30 '15 at 23:12
  • If you need handwaving to make the sun go supernova, you must handwave an answer. The sun is not massive enough for a core collapse, and does not have any circumstances for any other type. Since the physical details of this exceptional phenomenon is not known, we can't give an answer. – JDługosz Jun 01 '15 at 06:35
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    As far as anyone present is concerned, about 8 minutes 19 seconds. – Travis Christian Jun 01 '15 at 15:25

2 Answers2

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Using the physicists rule of thumb that "However big you think supernovae are, they're bigger than that."

A supernova detonated 1 AU from you is $ 9 $ orders of magnitude brighter than a hydrogen bomb detonated against your eyeball.

Here's a nice video of a Type 1a supernova explosion.

Why wait?
The Earth's destruction doesn't need to wait for the "physical explosion". Just the electromagnetic radiation from the supernova will do the job handily.

How much time do we have?
From the Physics stack exchange, I find:

30 seconds from the radiation front reaching Earth until the Earth has absorbed enough energy to vaporize.

How much time, really?
In response to some comments, I dug up a little research from "What If". Even if you were on the side of the Earth away from the explosion, you wouldn't get the extra 30 s (or more) it took for the supernova to vaporize and disperse the Earth. It turns out that at 1 AU, the supernova puts out a lethal dose of neutrinos and the Earth provides no protection.

Since neutrinos travel at >0.999976c, you really only get less than 0.012 seconds before you received a lethal dose. I estimate the dosage at ~23 grays of radiation. Which On this table indicates the following:

Immediate disorientation and coma will result, onset is within seconds to minutes.

Prognosis: Certain death

But the neutrino front does NOT possess enough energy to destroy the Earth, just enough to sterilize it.

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Jim2B
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    Actually, it'll take a bit longer, since the vaporized rock/iron will form an ablative shield for the unvaporized remnant. Getting the numbers right would be an interesting exercise in high-energy hydrodynamics. Got to throw in neutrino interactions, too. What's the opacity of vaporized silica/nickel-iron? – WhatRoughBeast May 31 '15 at 00:55
  • I think you're right. The stuff closest the the explosion simply couldn't get out the way of the streaming radiation fast enough. It'd act as a sort of shield for a little while. However, I still wouldn't want to be living on that planet ;) – Jim2B May 31 '15 at 01:04
  • But the neutrinos kill us in 5 s or less anyway :( – Jim2B May 31 '15 at 01:37
  • That's not what xkcd https://what-if.xkcd.com/73/ says ;) – WhatRoughBeast May 31 '15 at 02:48
  • Correct, xkcd stated that you'd get a lethal dose. I merely applied the math to figure out how far behind the radiation front the neutrinos were. My guestimate was they traveled at 99% c. This was wrong. I'll correct my answer. But the other bit was I gave the time it took to get a lethal dose of neutrinos, I don't know how fast you'd die from it though. – Jim2B May 31 '15 at 02:53
  • I found the answer and included it. It looks like you'd still die from the vaporization of the planet before the radiation damage from the neutrinos. – Jim2B May 31 '15 at 03:03
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    Neutrinos travel at the speed of light, for all practical purposes. Certainly within .01%, so the neutrinos would hit only (at most) a few milliseconds after the radiation. Actually, the neutrinos might well precede the radiation, since they aren't slowed by propagation through the outer layers of the sun the way radiation is. And, actually, when I mentioned neutrinos I was thinking about thermal effects, which I'm pretty sure would be negligible. – WhatRoughBeast May 31 '15 at 03:14
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    From the xkcd: "Core collapse supernovae happen to giant stars, so if you observed a supernova from that distance, you'd probably be inside the outer layers of the star that created it." (emphasis mine). All of these calculations are based on a star much larger than the Sun. – 2012rcampion May 31 '15 at 16:22
  • Even travelling at 0.999...c, neutrinos would still take a little more than 8 minutes 20 seconds to reach us from the Sun. To what does your 0.012s refer? – ArtOfCode May 31 '15 at 20:57
  • The difference in time from when the em wave front hits to when the neutrino wave front hits. But as @WhatRoughBeast mentioned, it is possible that the neutrinos hit first because light interacts with the matter and is slowed on its way out, while the neutrinos do not. So you still might drop dead from radiation poisoning before the light wave front hits. – Jim2B May 31 '15 at 21:26
  • Also note that the vaporized, sunward surface layers of the Earth will block and absorb incoming radiation, heating and expanding so fast that the rest of the Earth will be shattered and thrown outward at a significant fraction of the speed of light, instead of being completely vaporized. – 2012rcampion May 31 '15 at 22:26
  • @Jim2B - No, the total dose would be on the order of 20S, which would hardly drop you in your tracks. – WhatRoughBeast May 31 '15 at 22:52
  • You might want to incorporate http://physics.stackexchange.com/questions/136425/how-did-all-of-the-heavy-elements-on-earth-get-here into this answer. – HDE 226868 Jun 12 '15 at 00:54
  • No mention at all of the 8m 20s required for the explosion to reach Earth. – RonJohn Oct 28 '17 at 10:28
  • @RonJohn, it was assumed. The debate was regarding whether the EM radiation wave front or neutrino wave front would arrive and kill you faster. Both would take the 8m 20s +/- us once liberated from the exploding star. The debate was regarding how long for the EM radiation to get through the photosphere (might be up to weeks) vs what your expected lifespan would be after receiving the lethal dose of neutrinos (I do not know what a 20 Sievart dosage means for expected lifespan). – Jim2B Dec 25 '17 at 19:08
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Which type of supernova? There are several:

Type Ia

A white dwarf is a type of compact star composed of electron-degenerate matter, and is the endpoint in the evolution of most stars in the universe (including the Sun). No fusion occurs in a white dwarf: its luminosity comes entirely from stored thermal energy.

If a large ($1.3-1.4~M_\odot$) white dwarf occurs in a binary system and accretes matter from its companion, it can eventually become heavy enough to begin carbon and oxygen fusion. The extreme density of the star means that the fusion 'flame front' takes a little more than one second to propagate through the star, releasing a huge amount of energy as it does so: enough to unbind the star, resulting in a supernova.

The mass at which a type Ia supernova occurs is pretty much an invariant. This makes type Ia supernovae useful as standard candles, and also means that we can quote (relativly) precise statistics on their products:

  • Ejecta: $1.4~M_\odot~@~6\%~c$
  • Energy release: $<2\cdot 10^{44}~\text{J}$
  • Luminosity: $\approx 5\cdot 10^{9}~L_\odot$

So if the Earth somehow stuck around until after the Sun's red giant phase, and the mass of the Sun was increased around 40%, the ejecta would take around:

$$ \frac{1~\text{AU}}{6\%~c}\approx 2~\text{hours}~20~\text{minutes} $$

to reach us. The supernova doesn't reach full luminosity immediately, since most of its energy is trapped inside the dense, opaque, expanding outer layers. It takes around two weeks for the star to become optically thin, when peak luminosity is reached. However, the initial luminosity is still high enough to vaporize the sunward side of the Earth, and to sterilize the night side of the Earth via reflected Moonlight (if you're close to a full Moon).

Type II/Ib/Ic

These are core collapse supernova (the type discussed in the What-If article mentioned many times in the comments), and only happen to stars large enough to burn silicon, producing iron in their cores. The lower limit is around $8~M_\odot$, making it impossible for a core-collapse to occur in the Sun unless two things happen:

  • You increase the mass of the Sun by 40% to the Chandrasekhar limit

  • You convert the entire (increased) mass of the Sun into iron.

This would result in some sort of weird fully-stripped core-collapse supernova which could not occur in nature. Usually core-collapse supernovae blow off most of their mass, leaving a neutron star remnant behind. In this case, some of the core would be expelled, and I don't know if a remnant would form. In either case, we'd be hit with an Earth-vaporizing dose of gamma rays from the unobscured photodisintegration before the fatal dose of neutrino radiation reached us microseconds later.

2012rcampion
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  • For type Ia supernovas, you make it sound like 2e44 J isn't much by using that less-than sign... – user May 31 '15 at 20:03
  • @MichaelKjörling Writing $1-2\cdot 10^{44}$ didn't look as nice... but yeah, it's small compared to the $10^{46}~\text{J}$ released by a(n ordinary) type II. – 2012rcampion May 31 '15 at 20:31
  • It's a little more than the 8e31 J needed to vaporize the Earth, which we've done here a few times. :) – user May 31 '15 at 20:37
  • @MichaelKjörling Assuming the energy distribution is isotropic, it's around 1000 times more than you need =) – 2012rcampion May 31 '15 at 20:39
  • Wolfram Alpha. I should have guessed. – user May 31 '15 at 20:40
  • MathJax's \cdot is really intended for unit denotation rather than numeric multiplication - I've edited your answer to reflect this. – ArtOfCode May 31 '15 at 21:00
  • @ArtOfCode I usually use \cdot to denote a "tighter" multiplication than with \times, which is always reserved for multiplying two quantities. For example, $2\times 10^3\times 4\times 10^5$ just looks silly, but $2\cdot 10^3\times 4\cdot 10^5$ makes it clear that we're multiplying two numbers in scientific notation. Units use \cdot because they're part of the 'number unit' too; otherwise something like $2\times 10^3~\text{J}\cdot\text{s}$ looks to me like $2\times\left(10^3~\text{J}\cdot\text{s}\right)$. I would write $2\cdot10^3~\text{J}\cdot\text{s}$. – 2012rcampion May 31 '15 at 21:09
  • @ArtOfCode Since your edit was purely stylistic, I've reverted it to be in line with the formatting of the rest of my answers. – 2012rcampion May 31 '15 at 21:11
  • @ArtOfCode I also always use $\cdot$ when handwriting for the slight speed benefit, so it seems more natural to me than $\times$. – 2012rcampion May 31 '15 at 21:13
  • @ArtOfCode Also, if you're going to be a stickler, \cdot is not MathJax: MathJax is a rendering library, which in this case is rendering the $\LaTeX$ notation \cdot/\times. $\LaTeX$ is intended purely for typesetting, not for automated formula input (although it is sometimes used for that purpose), so all of its symbols have whatever meaning you ascribe to them. – 2012rcampion May 31 '15 at 21:17
  • @ArtOfCode If you want a particular style of formatting to be used, then I suggest you propose creating a style guide on meta (but beware: the community may end up with a compromise style that neither of us like). – 2012rcampion May 31 '15 at 21:20
  • Damn committees, :P – Jim2B Jun 01 '15 at 03:21
  • I liked this answer. Why so few upvotes? – Jim2B Jun 01 '15 at 03:23
  • @Jim2B My guess is that 1) The question already had an accepted answer, and 2) my statements to the effect of, "this can't happen." – 2012rcampion Jun 01 '15 at 03:25