Smallest Black Hole to 'heat' a Gas Giant

So I want a large power source to heat my new real estate on the Jovian moons. I create a micro black hole (MBH), maybe a few kg, maybe more and drop it into Jupiter. I expect the following to happen [...] the MBH absorbs mass and emits Hawking radiation, Mass falling towards the MBH will heat up due to the high pressure near the MBH [...]

Or so I assumed in this question. However, it was pointed out to me in comments that such a small black hole is really, really small and that it emits enough Hawking radiation to effectivly push away all mass, meaning no accretion will happen and the MBH will evaporate.

Hawking radiation and resulting pressure will be lower with higher MBH mass, but that also means at some point Hawking radiation won't heat my gas giant - I still want a small MBH.

What is the smallest MBH that will, when shot into a gas giant, accrete mass fast enough for it to not evaporate?

I think this means that radiation pressure from Hawking radiation should be smaller than surrounding pressure, we may help by shooting the MBH to get some impact pressure in front.

It is entirely possible that the resulting MBH is too large to heat the gas giant via Hawking radiation.

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• It could just heat it through the accretion disk it would develop. – Andrew Dodds May 10 '17 at 12:28
• Yes, but it would still evaporate (and evaporate faster with lost mass). – mart May 10 '17 at 12:41
• You are better off dropping a metric tonne of muons into Jupiter and handwave away the short lifetime of a muon. Heating Jupiter using Muon Catalyzed Fusion. – Aron May 11 '17 at 2:44

First off, I'd like to plug JoeKissling's answer here, which I used as a basis for mine.

Hawking radiation emitted from a black hole acts as blackbody radiation, emitted equally from the surface area of the event horizon. Based on Wikipedia's 'crude analytic estimate', the equivalent temperature $T$ is given by

$$\frac{\hbar c^3}{8\pi GMk_{B}} \approx \frac{1.227\times10^{23} \text{ kg}}{M} \text{ K}.$$

Multiplying this factor to the fourth power by the emissivity $\epsilon = 1$ and the constant of proportionality $\sigma = 5.670373\times10^{-8} \text{W m}^{-2}\text{K}^{-4}$ gives us the power output per unit area ($j^*$), or radiant emittance, in accordance with the Stefan-Boltzman law.

\begin{align}5.670373\times10^{-8} \text{W m}^{-2}\text{K}^{-4} \cdot \left(\frac{1.227\times10^{23} \text{ kg}}{M} \text{ K}\right)^4 = 1.285\times10^{85} \text{ W m}^{-2} \cdot \frac{\text{kg}^4}{M^4}.\end{align}

So basically, we take the mass of the black hole, raise it to the fourth power, then divide that enormous constant by the result. This will give us power output in Watts per square meter.

Radiation pressure is function of the EM radiation flux ($E_f$) and the speed of light. It is also a function of the incident angle of the flux. We will assume a spherical event horizon, where the pressure of the gas giant is acting in the negative direction from the radiation pressure, thus making the incident angle $\alpha = 0$. Radiation pressure, plugging in our previous equation, is

\begin{align}P &= \frac{E_f}{c}\cos\alpha \\ &= \frac{1.285\times10^{85} \text{ W m}^{-2} \cdot \frac{\text{kg}^4}{M^4}}{2.998\times10^{8}\text{ m/s}}\cos0 \\&= 4.287\times10^{76} \text{ Pa} \cdot \frac{\text{kg}^4}{M^4}\end{align}

Pressure at the center of a Gas Giant

Since the temperature and pressure conditions in Earth's core are not well known, the same applies a million fold for other planets. So we can only take some best guess assumptions about Jupiter's core. Militzer et al., 2008 estimate 100-1000 GPa for Jupiter's core, while Wilson and Militzer, 2012 use 40 Mbar = 400 GPa. Incidentally, this guy Burkhard Militzer has a writing credit on about half the papers I can find on Jupiter, so lets take his word for it and use 400 GPa.

To solve for our minimum size black hole, set 400 GPa equal to our radiation pressure equation above.

\begin{align}4\times10^{11} \text{ Pa} &= 4.287\times10^{76} \text{ Pa} \cdot \frac{\text{kg}^4}{M^4} \\ \frac{M^4}{\text{kg}^4} &= 1.072\times10^{65} \\ M &= 1.809\times10^{16} \text{ kg} \end{align}

So there you have it. Your black hole must be roughly the mass of a 10 km radius asteroid.

• Have fun moving this thing around. – Draco18s May 10 '17 at 14:06
• I notice that Wikipedia gives the "Stefan–Boltzmann–Schwarzschild–Hawking power law" for pressure from Hawking radiation as having an $M^{-2}$ dependence, not $M^{-4}$. Which one is correct? – HDE 226868 May 10 '17 at 14:08
• @HDE226868 That is for total power emission. Since surface area depends on M$^2$, that explains the difference between total power and power flux (per unit area) – kingledion May 10 '17 at 14:11
• @kingledion Right, I was being dumb. Thanks. – HDE 226868 May 10 '17 at 14:11
• @mart I did find an error in the calculation. I was off by 2 orders of magnitude in the final answer. – kingledion May 10 '17 at 14:31

I went about this a bit differently than kingledion, and got a different answer (off by $\sim6$ orders of magnitude!). The difference is that I assumed that there would be accretion no matter what the mass of the black hole is; this is incorrect because accretion would probably be prevented because of pressure by Hawking radiation. I'll keep this answer here for posterity, and also to show that even if you ignore kingledion's key pressure assumption, there's still an even lower limit - and thus his solution still works. A black hole of $\sim10^{16}\text{ kg}$ would certainly be able to heat the gas giant.

The power emitted by a black hole from Hawking radiation is $$P=\frac{\hbar c^6}{15360\pi G^2M^2}=-c^2\dot{M}_H$$ where $\dot{M}_H$ is the change in mass of the black hole from Hawking radiation. Let's assume that the black hole also accretes mass; the equation for Bondi accretion should give us a good estimate: $$\dot{M}_B\simeq \frac{\pi\rho G^2M^2}{c_s^3}$$ where $\rho$ is the density and $c_s$ is the speed of sound. The central density of Jupiter is roughly $5\text{ g cm}^{-3}$, or $5000\text{ kg m}^{-3}$. We can find $c_s$ as $$c_s=\sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus - about $125\text{ GPa}$. This gives us $c_s\simeq5000\text{ m/s}$. We then set $$\dot{M}_H+\dot{M}_B=0$$ and solve $$\frac{\hbar c^4}{15360\pi G^2M^2}=\frac{\pi\rho G^2M^2}{c_s^3}\to M=\left[\frac{\hbar c^4c_s^3}{15360\pi^2\rho G^4}\right]^{1/4}$$ Plugging things in, we have $$M=\left[\frac{\hbar c^4(5000\text{ m/s})^3}{15360\pi^2(5000\text{ kg m}^{-3})G^4}\right]^{1/4}=5.158\times10^{10}\text{ kg}$$ This is, as I said, off from kingledion's result by a factor of one million.

There are only two things that could really be varied - the bulk modulus and the density. If we move the other factors out, we see that $$M=7.295\times10^8\text{ kg}^{5/4}\text{ s}^{3/4}\text{ m}^{-3/2}\left(\frac{K^{3/2}}{\rho^{5/2}}\right)^{1/4}$$ Even raising $K$ by an order of magnitude and lowering $\rho$ by an order of magnitude only multiplies our result by $10$.

This demonstrates the power of radiation pressure! It raises the lower limit by six orders of magnitude, which is pretty incredible. Be careful of what physical assumptions you make.