Orion1
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In order for the black hole to evaporate it must have a temperature greater
than that of the present-day black-body radiation of the Universe.
Cosmic microwave background radiation temperature:
T_u = 2.725 \; \text{K}
Hawking radiation temperature:
T_H = \frac{\hbar c^3}{8 \pi G M k_B}
Hawking radiation temperature is greater than or equal to cosmic microwave background radiation temperature:
T_H \geq T_u
\frac{\hbar c^3}{8 \pi G M k_B} \geq T_u
Hawking total black hole mass:
M_H \leq \frac{\hbar c^3}{8 \pi G k_B T_u} \leq 1.226 \cdot 10^{23} \; \text{kg}
\boxed{M_H \leq 1.226 \cdot 10^{23} \; \text{kg}}
Earth total mass:
M_{\oplus} = 5.9722 \cdot 10^{24} \; \text{kg}
\frac{M_H}{M_{\oplus}} = 0.007539 = 0.754 \; \text{%}
According to reference 4 - p. 2, eq. 1:
\frac{M_H}{M_{\oplus}} = 0.8 \; \text{%}
Are these equations correct?
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Reference:
Cosmic microwave background radiation - temperature - Wikipedia
Hawking radiation - black hole evaporation - Wikipedia
Earth mass - Wikipedia
The Last Eight Minutes Of A Primordial Black Hole - Joseph Kapusta