Basic idea: Primordial black hole (2.5% of Sun's mass) directly approaches Sun at fairly rapid pace (300 km/s).

Consequently, six days after passing Earth's orbit, it will collide with the Sun; accounting for additional gravitational acceleration as it approaches, the collision will result in many orders of magnitude more kinetic energy (~1E+40 J) dissipating into heat than the Sun radiates every second (3.828E+26 J), i.e. equivalent to about one million years of solar radiation.

How long would it take for this to make itself felt?

Will there be an instantaneous, GRB-like "flash" that scours Earth's surface of life? Or will it be concentrated in a few coronal mass ejections? Is luminosity going to build up over a period of minutes, hours, days, weeks, maybe even years? I assume most of the heat will be dissipated deep within the Sun, and it is going to need time to work its way up to the surface. When is it going to reach its peak?

Could at least people in deep bunkers survive that "flash"? Or will it basically scour away the surface of the planet.

(Idea is that the black hole is detected in advance and there is a race to build an Orion Drive powered colony ship. The ship lifts off just in time. Will anybody in ground control survive that flash? If so, how long before the world warms to such an extent that they die anyway? I would like to have people on Earth survive the initial flash and for as long as possible thereafter as scientifically plausible... but they do need to die in the end).

Thanks in advance.

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    $\begingroup$ If the people on Earth doesn't survive, then what are the odds that the people aboard a hastily-built spacecraft will survive? Detecting a non-radiating object, which is moving at 300 km/s relative to us, sufficiently far out to give enough time to even launch an existing spacecraft is going to be quite the stretch. While it isn't exactly the same thing, you might find my question How far from the Sun could we detect an alien spacecraft similar to the Voyagers? to be of interest. $\endgroup$
    – user
    Commented Feb 27, 2019 at 21:28
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    $\begingroup$ I'm also not sure what your "0.025M" is supposed to mean. It's probably possible to work backwards through your impact energy calculations to figure that out, but instead of having us do that, could you please just Edit to clarify that part? (I also get the feeling that you're assuming an inelastic collision; something tells me that's not an accurate approximation here.) $\endgroup$
    – user
    Commented Feb 27, 2019 at 21:31
  • $\begingroup$ The spacecraft will - presumably - be making its way away from the Sun as these events go down. Object detection - realistically, it probably won't be, but can be for purposes of scifi through - some jargon about gravitational lensing; perhaps it hits an asteroid on its way in; or just plain luck. Inelastic collision - I suppose it will yo-yo a bit around there, but this object will fall within Sun's escape velocity if it approaches closer than 3 million km (as it will - otherwise, it will just pass by and nothing particularly interesting will happen). $\endgroup$
    – ak7
    Commented Feb 27, 2019 at 21:39
  • $\begingroup$ Black hole cannot reasonably collide with Sun in a way this question would make sense. If there still is " Earth's surface" to be scorched, then the black hole was probably small enough to just fly thorough Sun. $\endgroup$
    – Mołot
    Commented Feb 27, 2019 at 22:53
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    $\begingroup$ Also I see a mention of the planet warming up but in an accurate scenario where the sun is engulfed by a hypervelocitic black hole the heat source would be gone and the planet would actually freeze leaving people to find artificial sources of heat, unless of course the planet is caught in the gravity of the black hole. That would change alot $\endgroup$ Commented Feb 28, 2019 at 23:46

2 Answers 2


Well, what you are missing is the fact of how tiny such a black hole is. Wikipedia says, a black hole with a weight of 10 suns is only 30km in diameter.

Also, you need to add the sun's escape velocity to the equation, which is 617.7km/s. So the black hole will hit the surface of the sun with roughly 686.7km/s.

The effect is, that the black hole will basically just punch away a cylinder of sun mass with a diameter of 30km, going all the way through the sun's core, and leaving on the other side. You need to realize that this cylinder of mass is much lighter than the black hole that travels through it. Thus, the black hole will not loose enough speed to be captured within the sun, and it won't be able to eat the sun from within.

The whole process is actually quite like a bullet hitting a box of marshmallows. The marshmallow box may be much more heavy than the bullet, but still the bullet will pass through relatively unhindered. And, just like the bullet is much more dense than the marshmallows, the black hole is much more dense than the sun. So, just like the bullet simply does not encounter enough marshmallows on its path through the box to stop it, the black hole does not encounter enough sun plasma on its path through the sun to hinder its progress significantly.

Detour: Estimation of slowdown

The sun's core has a density of $\rho\approx150\frac{g}{cm^3}$. For simplicities sake, let's assume that the entire mass of the sun ($m_s=1.99\cdot 10^{30}kg$) is confined within a sphere of this density. This results in a sphere with a radius of $r_s=146819km$ (much too small, the sun has a radius of $696342km$, but we want an upper bound on the effect on the black hole).

The stellar mass that is on the path of the black hole of radius $r_{bh}=15km$ would be

$$m_{collision}=2r_s\cdot\pi r_{bh}^2\cdot\rho$$ $$=207560185km^3\cdot\rho$$ $$=3.11\cdot10^{22}kg$$ $$=0.0000000157m_s$$

Setting this into relation with the mass of your black hole ($m_{bh}=0.025m_s$), we can compute the speed that results from the fully inelastic collision between that mass and the black hole:

$$v_{bh}\cdot m_{bh}=v_{out}\cdot (m_{bh} + m_{collision})$$ $$\Leftrightarrow v_{out}=v_{bh}\frac{m_{bh}}{m_{bh} + m_{collision}}$$ $$=686.7\frac{km}{s}\cdot\frac{0.025m_s}{0.0250000157m_s}$$ $$\approx686.7\frac{km}{s}$$

Looks like my analogy with the bullet and the marshmallows was totally off. It's more like an anti-tank bullet going through a box of fluffy cotton wool...

And that's even with the assumption that the sun were core-only, the real mass distribution would lead to even less of an effect...

The point is, the stellar mass that the black hole interacts with is just way too small for any appreciable effect.

Of course, you must expect some hard gamma radiation when the black hole enters the sun, and some more when it leaves on the other side. But those ray bursts will be tiny in comparison to the 150 million kilometers, so I doubt that they will be strong enough to be catastrophic. I may be mistaken on that one, though, as I can't do the math on this.

So, sorry, the apocalypse won't happen that way...

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    $\begingroup$ I agree whole heartedly with most of your answer except the conclusion. Stellar dynamics is a voodoo science, and even though the impact may be small we can’t really say for sure what the end result of removing that column (with all the magnetic and material shockwaves it would entail) would be. Might be nothing happens. Might be that thirty years from now a coronal mass ejection swats mercury from the sky. Might cause long term climate effects that only our grandchildren will care about. Nobody knows! $\endgroup$
    – Joe Bloggs
    Commented Feb 28, 2019 at 0:15
  • $\begingroup$ Thanks for the comment. I agree with what you wrote. However, 617.7km/s is the escape velocity at the Sun's surface. My understanding is that if the BH merely grazes the surface of the Sun, it will indeed fly off. But if it goes any deeper into the photosphere, it will become gravitationally bound (elliptic orbit?); and if it goes even deeper, then that orbit becomes tighter and tighter the closer to the center of gravity it hits. $\endgroup$
    – ak7
    Commented Feb 28, 2019 at 7:09
  • $\begingroup$ While a black hole going through the Sun will indeed be like a bullet going through a ball of marshmallows, but in this analogy, the bullet falls into orbit within said ball of marshmallows. I assume that this will slow said bullet over time? Since the core of the Sun in particular is really quite dense. What I wonder about is how long this would take - minutes; days; months; years? $\endgroup$
    – ak7
    Commented Feb 28, 2019 at 7:15
  • $\begingroup$ @ak7 Good point about the slowdown, I didn't fully consider this. I'm positive that, even when the black hole hits the sun straight on, it will still go through relatively unhampered. And once it enters the sun it will continue to accelerate at first. However, the question is whether it will leave with more or less than those 617.7km/s. I'll need to do the math on that, which I can't do right now. I'll try to come back to this question with the answer on the slowdown. $\endgroup$ Commented Feb 28, 2019 at 8:04
  • $\begingroup$ The cylinder would be much thinner than 30 km, since the mass of the black hole as specified by OP is only 1/40 of that needed for a 30 km diameter black hole... $\endgroup$
    – user
    Commented Feb 28, 2019 at 19:25

If a black hole "collides" with your sun not much energy is dispersed. Maybe a little escapes but, beyond the event horizon, nothing gets out of the black hole, and your planet is now orbiting a black blob of sheer gravity and probably falling inward slowly, but yes people could survive this.

People don't seem to think this is helpful, so I will provide some of my knowledge that might be useful. Due to the reaction to my answer I assume you already are set on having an explosion of some sort. It might help to know that nothing (aside from things affected by quantum occurrences) can travel faster than light, approx 830,000 miles per second. Not an exact, but I do know earth is 8 light minutes from the sun on average so the people on the planet would have at least 8 minutes to find a solution

And not to rain on this parade but I must say that something of any less mass of the sun really shouldn't be considered a black hole. But this is for abother discussion.

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    $\begingroup$ Thanks. That would be great, but what we have here is a very heavy body (2.5% of Sun's mass) that goes from hypervelocity to stationary within a short period of time. Where else could that energy go? $\endgroup$
    – ak7
    Commented Feb 27, 2019 at 21:46
  • $\begingroup$ First of all why would it stop. There is no known object that is evenough nominally close to being able to exert any sort of force on a black hole. As modern science is concerned everything bows to these things. Realistically if it directly touched the star it would swallow it instantly and continue cruising through space $\endgroup$ Commented Feb 28, 2019 at 22:15
  • $\begingroup$ @ak7 second that is not a possible mass estimation for a black hole. I encourage you to research black holes and by all means I am open to receive anything you find that makes this possible. I will do some research to and am sorry if I'm wrong $\endgroup$ Commented Feb 28, 2019 at 22:35

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