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How much mass would the Sun lose to a primordial black hole that (initially) has 10 Earth masses, passing through or close to the center of the Sun at solar escape velocity? How massive would the Sun end up and how massive would the black hole become?

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    $\begingroup$ Note that the question assumes the black hole goes through the Sun's polar regions so that it doesn't meddle with the planet's orbits. $\endgroup$
    – Greenhorn
    Dec 3 '20 at 13:39
  • $\begingroup$ @fasterthanlight It doesn't necessarily have to be the center of the Sun, it doesn't have to go through the Sun's core (but it can). I wonder about multiple options, but in the core there's the most mass of course. $\endgroup$
    – Greenhorn
    Dec 3 '20 at 13:53
  • $\begingroup$ If the black hole just goes through the edge of the sun and grazes the interior, the result would be much more different than if the black hole passed through the core $\endgroup$ Dec 3 '20 at 13:57
  • $\begingroup$ @fasterthanlight Right, this is what I mean by above comment. $\endgroup$
    – Greenhorn
    Dec 3 '20 at 13:58
  • $\begingroup$ I think a rough estimate could be made with a column through the sun, 7 inches in diameter, plus an accretion disk maybe 3 radius distant. Beyond that the heat generated might equalize the infalling gravity. It becomes tricky to estimate because the area around the black hole would radiate enough energy that it would push a lot of the sun out of the way and perhaps not eat less than that, but I'm only guessing. With a black hole that small passing through, I think the sun would lose a negligible amount of mass. If it got stuck inside, that might be different. $\endgroup$
    – userLTK
    Dec 3 '20 at 14:26
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The answer is that there is a limited rate at which matter that can be crammed into a black hole of this size. That rate is small enough that the black hole will traverse through the Sun hardly picking up any mass (when expressed as a fraction of its original mass).

Details:

You could use the Bondi-Hoyle accretion rate: $$\dot{M} = \frac{dM}{dt} = \pi \rho \frac{G^2 M^2}{c_s^3}\ ,$$ where $\rho$ is the density of the solar material and $c_s$ is its sound speed.

The biggest values will occur at the centre because the density is greatest at the centre and increases more rapidly than the cube of the sound speed. Using rough values of say, $c_s \sim 500$ km/s and a density of $10^5$ kg/m$^3$, we get, for $M=6\times 10^{25}$ kg, $$ \dot{M} = 4\times 10^{19} \left(\frac{\rho}{10^3\ {\rm kg\ m}^{-3}}\right)\left( \frac{c_s}{300\ {\rm km\ s}^{-1}}\right)^{-3}\ {\rm kg\ s}^{-1} $$

The black hole would be travelling at around 600 km/s, so would traverse the central 10% of the Sun in around 100s (the accretion rate gets orders of magnitude lower in the outer parts of the Sun, so can be neglected), so might increase its mass by $4\times 10^{21}$ kg (or 0.007%). This would be insufficient to slow it appreciably and it would shoot out of the other side and disappear into space again. Note that a larger black hole would pick up more mass as a fraction of its original mass because $\dot{M} \propto M^2$.

All this assumes that the accretion can occur at the Bondi-Hoyle rate. This is very unlikely for two reasons - the first is turbulence which reduces the accretion efficiency. The second is radiation pressure. There will be some sort of super-heated region immediately around the black hole (just imagine trying to compress $4 \times 10^{19}$ kg of material into a 20 cm diameter hole every, the event horizon diameter for a 10 Earth-mass black hole, every second) and the radiation from this region will exert an outward pressure on the surrounding gas. The limiting Eddington accretion rate, where the radiation pressure from the hot gas immediately around the black hole, exerts enough pressure to halt further infalling gas, is given by $$ \dot{M} \leq \frac{4 \pi G M m_p}{\epsilon c \sigma_T}\ ,$$ where $m_p$ is the proton mass and $\sigma_T = 6.6\times 10^{-29}$ m$^2$ is the Thomson scattering cross-section for free electrons in the solar plasma (i.e. the effective area an electron presents to radiation). The $\epsilon$ parameter is the radiative efficiency of the infalling gas (i.e. what fraction of its kinetic energy is radiated away) and is usually assumed to be of order 0.1. Taken together, this gives a maximum accretion rate (for a ten Earth-mass black hole) of $$ \dot{M} = 4 \times 10^{10} \left(\frac{\epsilon}{0.1}\right)^{-1}\ {\rm kg\ s}^{-1}$$

Whether the Eddington limit scales down to micro black holes and dense environments like this, I am unsure. But it seems clear that the accretion rate will be way lower than the Bondi-Hoyle rate. I conclude that the Bondi-Hoyle accretion rate is probably a vast over-estimate, and that the black hole would shoot through the Sun and leave on the other side, having hardly accreted any mass (as a fraction of its own mass).

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  • $\begingroup$ You may hate non-metric units but that's no reason not to convert for English-speaking folks (who're probably a multitude on SE). $\endgroup$
    – Greenhorn
    Dec 3 '20 at 18:05
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    $\begingroup$ @Greenhorn: American units are useful for relating to US citizens who aren't sure if 30C is hot or cold or if 5km is a long walk. But we are so far away from human experience that "sigma=100 octillionths of a square inch" isn't helpful. $\endgroup$ Dec 3 '20 at 22:15
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    $\begingroup$ @Greenhorn Do you think that anyone who has difficulty with unit conversion would be able to get anything else of substance from this answer (aside from the TLDR at the end)? $\endgroup$ Dec 3 '20 at 22:21
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    $\begingroup$ @userLTK The gravitational pull is of course included as part of the Eddington limit. Any accretion requires a gravitational pull. If it helps, just imagine trying to cram $10^{17}$ kg of gas into a 1cm hole every second, and how hot that is going to get. $\endgroup$
    – ProfRob
    Dec 4 '20 at 9:33
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    $\begingroup$ @GreenHorn not all English-speakers use non-metric units. Indeed, if I were a linguistic purist I could argue that all Englsh-speakers are taught metric in school and it's only speakers of strange North American dialects who aren't :-). $\endgroup$ Dec 4 '20 at 10:25

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