How many black holes can fit on the head of a pin?

Black holes are regions of space from which nothing can escape. If you consider a spherical object of mass M and radius R and set the escape velocity from the object to be cc, the speed of light, you can determine a relationship between R and M, R=2GM/c2R=2 GM/c^2, where GG is Newton's gravitational constant. This radius is called the Schwarzschild radius, denoted by RsR_s. If mass M is concentrated into a region with a radius smaller than RsR_s then you have a black hole, and if not, there is no black hole.

From the above relation you can determine the minimum mass of a black hole, as roughly speaking the Schwarzschild radius must be larger or equal to the Compton wavelength - the minimum size of the region in which an object at rest can be localized.

Find the minimum mass of a black hole in μg\mu g.

Finally, a bonus thing to think about. What does this result mean for the masses of the particles that we see in nature?

Details and assumptions

  • The value of the gravitational constant is G=6.67×1011 m3/kg s2G=6.67 \times 10^{-11} \text{ m}^3\text{/kg s}^2.
  • The speed of light is c=3×108 m/sc=3 \times 10^8\text{ m/s}.
  • Planck's constant is h=6.63×1034 kgm2/sh=6.63 \times 10^{-34} \text{ kgm}^2\text{/s}.
  • 1 μg=106 g=109 kg 1 ~\mu g = 10^{-6} \text{ g} = 10^{-9} \text{ kg}

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