# What exactly would happen if a black hole was introduced into the sun?

A ship powered by a black hole of a few hundred thousand tons (say, under $6 \times10^8$ kg, which would have a lifetime of several years and a power output of a few hundred petawatts) crashes (malfunctions or is piloted deliberately, it doesn't matter) into the sun. The heat destroys the hull and exposes the black hole to the interior of the sun. What precisely would happen and on what kind of timescales?

Obviously, the sun would fall into the black hole, heating up and giving out huge amounts of radiation in the process. But how much radiation? How long would it be before effects were noticeable from orbit (eg from the Earth), and what would the effects be? How long would it take for the sun to be completely consumed? How would the radiation given off vary (roughly) over this time period? What would happen to the solar system throughout this process?

I'm looking for a timeline with (rough) details of the process and how it would develop. This question is similar but the only answer with any details only calculates time to failure of the sun, and the answerer mentions that he seems to have made a mistake in his calculation, so it isn't a duplicate because it doesn't give the information I'm looking for.

• How is it obvious? The Earth has a mass of ~$6 \times 10^{24}$ kg. The black hole would be torn apart by the sun which wouldn't even register such a puny mass in its gravity well. – nzaman Mar 21 '17 at 13:50
• @nzaman By definition you can't "tear apart" a black hole. Nothing can leave the event horizon. – Tim B Mar 21 '17 at 13:51
• I recently proposed this sun/hole maneuver as a downvote hungry method of destroying a black hole on an idea here. I asked exactly this question on the astronomy stack. astronomy.stackexchange.com/questions/20457/… The answer: black hole always wins. – Willk Mar 21 '17 at 13:56
• @TimB: Hawking radiation does. Symmetry would imply a sufficiently large mass should be able to speed up the process. The only reason black holes exist is that they're the biggest bullies on their playground. An BH 16 orders of magnitude less than the Earth going against the Sun? It won't last – nzaman Mar 21 '17 at 14:09
• This is answered over in Physics.SE. "Throwing a micro black hole into the sun: does it collapse into a black hole or does it result in a supernova?" – Schwern Mar 21 '17 at 22:51

Using your black hole from the actual size would be much smaller than a proton so it would have difficulty accreting mass because its effective cross section is very small. It may even only be able to absorb neutrinos, electrons, and gamma rays. Also, its overall gravity would still be very weak. It weighs as much as a building and you don't see people being drawn to buildings; at least until you get very very close to the singularity itself. Then there is the outpouring of Hawking Radiation as it evaporates which would certainly make it very difficult for mass to get close to it and would likely destructively interfere with any light trying to get in.

The black hole could probably fall all the way through the sun because its outpouring of radiation would clear a path for it. If it somehow got caught in the core of the star it may be unable to accrete mass for reasons previously mentioned, except by catching neutrinos. In the end, I suspect nothing much would happen.

Math to support my claims and edited for clarity

The black hole in question would be tiny, meaning that its Event horizon or Schwarzschild Radius is small. Knowing its mass we can calculate its size using this. The equation is:

$R_s = \frac{2MG}{c^2}$

Where:

$R_s$ is the Schwarzschild Radius

$M$ is the mass of the black Hole

$G$ is the universal gravitational constant

Plugging it all in:

$\frac{2\times6*10^{8}kg\times6.67*10^{-11}m^3 kg^{-1} s^{-2}}{(3*10^8m/s)^2}$

Gives a radius of

$R_s=8.91\times10^{-19} m$

For comparison the radius of a proton is around $8.5\times10^{-16}$. So its rate of accretion, if any at all would be very low.

Because of its low mass (relatively), its gravity won't be very strong at all. Using Newton's Law of Gravitation:

$F=\frac{GM_1M_2}{r^2}$

Dividing by $M_2$ so we can get the acceleration due to gravity.

$a_g = \frac{GM_{1}}{r^2}$

Now we plug in the mass of the black hole and a few distances 10, 1, .1, .01, .001 meters to see what the acceleration due to gravity would be.

At 10 m the acceleration is $4\times10^{-4}m/s^2$

At 1 m the acceleration is $4\times10^{-2}m/s^2$

At .1 m the acceleration is $4m/s^2$

At .01 m the acceleration is $4\times10^{2}m/s^2$

At .001 m the acceleration is $4\times10^{4}m/s^2$

So even if it was a meter away from you would probably not notice it was there at all. Reaching out to it would end badly for you, but its sphere of influence is rather small.

Now there is the outpouring of radiation from the tiny singularity that would keep all matter far away from it because of the pressure the radiation exerts. First, we need to calculate the power being radiated from the black hole using the Stefan–Boltzmann–Schwarzschild–Hawking power law (really that's its name)

$P_b=\frac{\hbar c^6}{15360\pi G^2 M^2}$

where $\hbar$ is the Reduced Plank Constant

Plugging in our values we get a power output of

$P_b=9.89\times10^{14}$ watts

Now knowing the power output we can calculate the pressure exerted by the radiation using the planer Radiation Pressure Equation with the assumption of being normal to the surface we get:

$P_{rad}=\frac{E_f}{c}$

Where:

$E_f$ is energy flux in $w/m^2$

$c$ is the speed of light

$P_{rad}$ is pressure exerted by the radiation

In order to see if the out flowing of radiation would be enough to keep matter away even if the black hole passed through the core of the star, we are going to solve for the distance that the radiation pressure is equal to the pressure in the core of the sun. If that distance is less than the radius of the event horizon then matter will fall into the black hole, if it is larger then no matter will fall in. I am also assuming that the radiation is emitted from the black hole evenly in all direction, which may not be the case if the black hole has a large charge or is spinning rapidly. So we will solve:

$P_{sun} = P_{rad}$

Expanding

$P_{sun}=\frac{E_f}{c}$

Expanding a little more

$P_{sun}=\frac{\frac{P_b}{4\pi r^2}}{c}$

Where:

$P_{sun} =2.4*10^{16} Pa$

Plugging in our values and solving for $r$ we get:

$r=3.3\times10^{-6}m$

Meaning that the pressure from outflow of radiation will be equal to the pressure from the core of the star at that distance, which is much greater than the Schwarzschild radius. Therefore no matter will even be able to reach the singularity. I also suspect that the heating resulting from the outpouring of radiation would cause some expansion, but given the overall size of the sun it would be insignificant and would still find some kind of equilibrium.

• Yes: the black hole would establish some kind of orbit through the sun's path the same way any object would, except it would fling itself through the sun without being destroyed a few times. It wouldn't just stop when it hits the sun's center, it would probably be flung straight through the sun, and then head into a very, very unstable orbit. – SpaceMouse Mar 21 '17 at 14:42
• Cf. a similar fictional scenario in Larry Niven's short story The Hole Man where instead of the Sun it is Mars that gets a nano-sized black hole. IIRC the main character guesstimates that nothing will happen for a long time. – pablodf76 Mar 21 '17 at 14:44
• It would even probably end oscillating, falling through the sun several times (Or hundreds of times) until it evaporated or somehow managed to experience drag. I think a nanoscale back hole would be much more dangerous. It's orders of magnitude larger (mass and actual size) and has less hawking radiation pouring out of it. That's a weird thing to say: nanoscale and orders of magnitude larger. – Joe Kissling Mar 21 '17 at 14:47
• Will add an edit to see if the radiation pressure from the black hole is enough to really keep the matter away from it in the core of the sun. – Joe Kissling Mar 21 '17 at 14:50
• Thank you - excellent point about the size and radiation pressure! But I am a little puzzled by a discrepancy in your calculation compared to values given in the paper (arxiv.org/pdf/0908.1803.pdf) mentioned in the link in the question. The authors used 6.06e8kg, and got very similar answers to you for radius, but their power calculation is 1.6e17 W - three orders of magnitude above yours - any idea why? – Tharaib Mar 22 '17 at 4:59