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I just finished watching PBS Spacetime on primordial black holes, and the idea is that just after the big bang there could have formed 'small' black holes that would not have not hawking-evaporated till now.

As I understand for small mass (asteroid/planet weight) black hole normal gravity physics would apply, so they should just orbit bigger objects like planets and stars by now, which implies that some should also be here in the Solar system.

If there would be bunch of asteroid weight black holes somewhere within lets say Oort cloud or even asteroid belt would we have means of detecting them? What if they would be orbiting around the earth?

Could small (asteroid weight) primordial black hole be detected in solar system?

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Asteroids vary enormously in their weight, but let's take a mass of $10^{15}$kg as a starting point. This would have a Schwarzschild radius of:

$$ r_s = \frac{2GM}{c^2} \approx 1.5 \times 10^{-12}\,\text{m} $$

and a Hawking temperature of:

$$ T = \frac{\hbar c^3}{8\pi G m k_b} \approx 1.6 \times 10^8 \, \text{K} $$

To get the power radiated by the black hole we use the Stefan-Boltzmann law:

$$ P = A\,J = 4\pi r_s^2 \sigma T^4 \approx 1150 \, \text{W} $$

And finally the peak wavelength of the radiation would be given by the Wien displacement law:

$$ \lambda_\text{max} = \frac{b}{T} \approx 0.018 \, \text{nm} $$

If we're trying to detect the black hole the mass doesn't help because it would behave the same as the countless other medium mass asteroids in the asteroid belt, so the question is whether we could detect it by the X-rays it emitted. And I must confess that I have no idea how sensitive the current generation of X-ray telescopes is. However we could work out the number of photons per unit area that would be received by a satellite orbiting the Earth. The photon energy is:

$$ E = \frac{hc}{\lambda} $$

And if we divide our power of $1150$W by this we get about $10^{17}$ photons per second emitted. The asteroid belt is around $3$ AU from the Sun, so the closest approach to Earth would be about $2$ AU. Dividing our photon flux by the area of a sphere with a radius of $2$ AU gives us the photons per metre per second at the Earth:

$$ N = \frac{10^{17}}{4 \pi (2 \text{AU})^2} \approx 3.7 \times 10^{-7} $$

Which is a bit disappointing really. It's hard to see any satellite detecting a source that only emits one photon per square meter every 2.7 million seconds. I think we'd have to say that we have little chance of detecting the black hole.

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    $\begingroup$ or smaller mass black holes than the one discussed here , there [are thesis studies which give detectable signals. tigerprints.clemson.edu/cgi/… In the link they are talking of an 8.2x10^10 kg black hole. and signals detectable at the galactic level. I would think that excess of electron positrons from a specific direction in the Oort cloud might be a signal. the smaller mass of the black hole the higher the black body temperature afer all $\endgroup$ – anna v Oct 9 '17 at 8:49
  • $\begingroup$ Hi John, what about one circling the Earth. Lets say 100km from detector, with formula you used should be (10^17)/(4*pi*(100000)^2) = 795774.715459 photons per square meter... Is it safe to assume that there is no primordial black hole orbiting the earth? Or otherwise it would have been detected... $\endgroup$ – Matas Vaitkevicius Oct 9 '17 at 9:13
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    $\begingroup$ @MatasVaitkevicius: yes, while I'm not sure what the sensitivity of our X-ray detectors is I'm sure that would be plenty intense enough to detect. $\endgroup$ – John Rennie Oct 9 '17 at 9:16

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