Black hole collides with Sun: How rapidly is energy released?

Question

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.

Answer 1

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.

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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.

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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...

Answer 2

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.