A four foot diameter black hole destroys our Sun

  • #1
eggchess
9
5
Homework Statement: I am a writer completing a science fiction novel involving a four foot diameter black hole (with approx 1.5 times the mass of Saturn) In my novel, the black hole is being drawn toward our much more massive sun. I assume the black hole would begin to consume plasma/energy from the sun, slowly destroying it, and that this process would go on until the sun is completely spent. About how long might that process take? I wouldn't know how to solve such a problem. My background isn't physics. Any help you can lend would be appreciated. My thanks.
Relevant Equations: No idea.

I really don't know.
 
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  • #3
Very helpful! Thank you PeroK
 
  • #4
A rather relevant question is: How was the black hole captured in the first place? Based on just a cursory estimate where the black hole just absorbs a column of its own diameter when passing through the Sun, it will absorb of the order of ##10^{12}## kg in one pass. This is negligible in comparison to both the black hole's mass as well as to the Sun itself so the black hole won't really slow down enough to be captured. It will just continue out on the other side again with essentially the same velocity that it entered. Even if the capture estimate is off by several orders of magnitude, the conclusion remains the same.
 
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  • #5
PeroK said:

That article unfortunately just says
In an article published on the arXiv website, they presented the results of their calculations.
and does not give the actual reference to the arXiv post. This makes it difficult to verify what the paper actually says and what the ingoing assumptions are.
 
  • #6
Orodruin said:
A rather relevant question is: How was the black hole captured in the first place? Based on just a cursory estimate where the black hole just absorbs a column of its own diameter when passing through the Sun, it will absorb of the order of ##10^{12}## kg in one pass. This is negligible in comparison to both the black hole's mass as well as to the Sun itself so the black hole won't really slow down enough to be captured. It will just continue out on the other side again with essentially the same velocity that it entered. Even if the capture estimate is off by several orders of magnitude, the conclusion remains the same.
This is startling to me. I would never have considered such an event possible. Thank you very much.
 
  • #7
You mighty read Beford's Artifact for a similar story.

To add to @Orodruin's comment, if launched from Earth, it would only make a pass through the sun every six months. It would take ~1018 years to "eat" the sun. That's a billion times longer than the sun will last.

If anything - it will make the sun last infinitesmally longer. It will act as a source of convection, which will, on average add fuel to the core.

The biggest risk is to the Earth - if launched from the Earth, it will return to - or at least near - the Earth.
 
  • #8
Vanadium 50 said:
You mighty read Beford's Artifact for a similar story.

To add to @Orodruin's comment, if launched from Earth, it would only make a pass through the sun every six months. It would take ~1018 years to "eat" the sun. That's a billion times longer than the sun will last.

If anything - it will make the sun last infinitesmally longer. It will act as a source of convection, which will, on average add fuel to the core.

The biggest risk is to the Earth - if launched from the Earth, it will return to - or at least near - the Earth.
More likely it would come from outside of the solar system, meaning it will just pass through the Sun once, pick up its ##10^{12}## kg of mass, and then leave the solar system again. Even passing through the Sun once is extremely unlikely though.
 
  • #9
Wouldn't capturing that mass lead to an equal release of energy?
It's about the mass the sun converts to energy every four minutes.
How long would the transit last?
 
  • #10
haruspex said:
How long would the transit last?
Assuming the black hole has a parabolic trajectory (slowest speed for an object free falling from outside the solar system) that by an amazing coincidence passes right through center of the sun (thus really a rectilinear parabolic trajectory) then it should take approximately 37.5 minute to pass through (diameter of sun divided by orbital escape speed at sun surface, ignoring speed changes when inside sun). Pretty much any variation on initial conditions would yield a faster transit time so 37.5 minute is an approximate upper limit.

Later, regarding chance of hitting the Sun dead center: even if the black hole was deliberately aimed (by aliens) to hit the sun from outside our Solar system, I would expect its perihelion distance to end up being Rayligh distributed thus making a gracing pass much more likely. Could be interesting to calculate what velocity accuracy would be needed in order to have, say, a 90% chance of hitting the Sun from, say, the Oort cloud.
 
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  • #11
haruspex said:
Wouldn't capturing that mass lead to an equal release of energy?
Why? The mass simply gets absorbed into the black hole.

As mentioned earlier, you will get some amount of additional convection, but at the level of one pass my gut feeling is that it would be hardly noticable.
 
  • #12
Orodruin said:
Why? The mass simply gets absorbed into the black hole.

As mentioned earlier, you will get some amount of additional convection, but at the level of one pass my gut feeling is that it would be hardly noticable.
According to https://web.stanford.edu/~wilkinsd/docs/posters/BlackHolesPoster.pdf,
"Accretion onto a black hole is the most efficient process for emitting energy from matter in the Universe, releasing up to 40% of the rest mass energy of the material falling in."
 
  • #13
That is about stellar mass or even supermassive black holes. This is a much much smaller black hole passing through a star in a rather short time. I doubt there is sufficient time to accumulate a significant accretion disc.

I may be wrong, but even if it did release that kind of energy it would not be an energy release far beyond the typical solar conversion. And even if it was a couple of orders of magnitude bigger than the solar conversion for a short time, it would not matter much. Most of that energy would be released in the denser regions of the Sun. It takes 170000 years before the energy produced in the core diffuses out to the surface. During this time, the Sun converts another ##10^{22}## kg of mass to heat energy.
 
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  • #14
Where it comes from is a plot point. Was it manufactured here? Or out there. (It's also too big to be a PBH, unless its heavier than the average one by a factor of several thousands)

The light currently being scattered in the sun will have a path out behind the BH. However, its 4 feet wide, so the net effect is small. Still, if you were stationed right behind it, that could be a problem for you.
 
  • #15
Vanadium 50 said:
However, its 4 feet wide
... and will therefore be filled in almost instantaneously. Long before the passage has reached the center.
 
  • #16
It doesn't have to reach the center - the trapped light is distributed throughout the sun. The photosphere is essentially the spot where the light stops being trapped.

The solar constant at the surface is ~60 MW/m2. Until the hole fills in the trapped light will have a luminosity a trillion times larger. This is insignificant on the scale of the sun (a part per million), but I'd still not want to be standing right behind it.

How long do you think it will take for the hole to close behind the BH? A second?
 
  • #17
Vanadium 50 said:
How long do you think it will take for the hole to close behind the BH? A second?
I would imagine significantly less than a second. Already the gravity of the passing black hole will throw matter that was not accreted into the back path. It will close almost instantaneously.
 
  • #18
How long does a tube of vacuum last when the surrounding plasma pressure is upwards of ##26\cdot 10^{15}## Pa?
 
  • #19
The pressures and densities at the core will certainly cause it to fill faster than the surface.

I'm trying to figure out a good way to calculate this.

At the surface, the temperature is about 2.5 eV. At this temperature, the electrons are going fast enough to traverse the hole in a microsecond. But this would cause the charge to build up and pull the electrons back.

So maybe I should use the proton mass instead. That's more like 55 microseconds.

They will be fighting photon pressure, which will equilibrate event faster - a few nanoseconds. But once this happens it can be ignored, because there is equal pressure inside and outside the tube. So we only really need to worry about the matter.

So, I'd guess we're talking maybe half a millisecond. So I agree, "well under a second", but wouldn't be shocked if the real answer were 10x bigger or 10x smaller.
 
  • #20
Vanadium 50 said:
But this would cause the charge to build up and pull the electrons back.
If the BH was aimed by aliens it will likely also be charged enough to allow them to contain and move it around using an electric field. Not sure if that has any significant impact for how much mass it can gulp up while passing a star, but at least some parts of the plasma would get an extra acceleration towards or away from it.
 
  • #21
Except that a charged hole will neutralize as it passes through the charged solar wind.
 
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  • #22
Vanadium 50 said:
So, I'd guess we're talking maybe half a millisecond. So I agree, "well under a second", but wouldn't be shocked if the real answer were 10x bigger or 10x smaller.
I wouldn’t be shocked if the amount of mass gulped up was a factor 1000 larger than what I estimated. The point is that it still wouldn’t matter much for the Sun.

You have also ignored the motion of the solar plasma due to the passage of the black hole. This will induce velocities on the scale of the black hole speed due to the slingshot. There may be some local heating and additional slowing of the black hole due to this, but it remains negligible on the whole.
 
  • #23
My sincere thanks to those that have discussed this. Great help for an English and Computer Science major. You've brought up things I never would have considered. What a wonderful forum where less trained science-lovers like myself can go to get thoroughly thought out answers. Again, Thank You.
 
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  • #24
A particle in free fall will be captured by a black hole if the impact parameter is smaller than ~R_s * c/v up to some numerical prefactor. Assuming v = 1000 km/s as typical velocity inside the Sun we might capture a column with a radius of 0.6 m * 300, or ~90,000 times the mass of the direct cylinder. It will slow a v=1000 km/s black hole with 1 Saturn mass by dv=0.15 mm/s. To capture a black hole from such a small velocity change, its initial approach velocity needs to be below ##\sqrt{( (v+dv)^2-v^2 } \approx 0.017\, km/s## which is not very likely. It could be captured by interactions with Jupiter or Saturn, however, and then slowly lose more energy from repeated passes through the Sun.

I think my 0.15 mm/s is an underestimate for the real velocity change, however. In addition to the direct absorption of mass we also have induced drag - there will be a higher mass concentration behind the black hole compared to in front of it.

On human timescales it's still a very slow process - probably millions of years.
 
  • #25
mfb said:
It could be captured by interactions with Jupiter or Saturn, however, and then slowly lose more energy from repeated passes through the Sun.
I would like to see a probability estimate of being captured by Jupiter or Saturn and ending up in an orbit that actually passes through the Sun …
 
  • #26
Small. A single fly-by can do it but you need to be lucky with the parameters.
But that also applies to the probability of such a black hole flying through the Solar System in the first place.
 

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