4 ms·
Iirc a black hole with the mass of an asteroid (whose event horizon would be very small indeed) would have a hawking temperature smaller than the cosmic microwa
by ithkuil 2y ago
Iirc a black hole with the mass of an asteroid (whose event horizon would be very small indeed) would have a hawking temperature smaller than the cosmic microwave radiation so there wouldn't be any net evaporation.
- Ancalagon 2y agoHmm what mass of black hole is needed to rip your arm off? Probably more than an asteroid I’m guessing?
- rcxdude 2y agoProbably a fairly significant multiple of earth's mass. What multiple of your arm's weight would be required to pull it off? (and tidal forces are generally less than the absolute gravitational pull)
- ithkuil 2y agoThe heavier the black hole, the lower the hawking temperature. Only the tiniest black holes would have hawking temperature high enough to evaporate given the presence of the CMB
- Ancalagon 2y agoI asked o1-preview a couple times, it gave me: *Answer:* Approximately 27 trillion kilograms—the black hole would need about 2.7 × 10¹³ kg of mass to exert tidal forces strong enough at 1 meter to tear off an arm. - This radius is much smaller than an atomic nucleus, so at a distance of 1 meter, the person is far outside the event horizon. No, the black hole with a mass of approximately \( 2.7 \times 10^{13} \) kilograms would not evaporate quickly due to Hawking radiation. Even though this mass is relatively small compared to stellar black holes, such a black hole would have an extremely long evaporation time—*on the order of \( 5 \times 10^{16} \) years, which is about 50 quadrillion years*. This duration is vastly longer than the current age of the universe, which is approximately \( 13.8 \times 10^{9} \) years (13.8 billion years).