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Using $M87^*$ data from EHT observation (mass, temp of the surrounding accreting disk) and approximating area of EH by euclidean geometry $4\pi r^2)$, error comes around of order one or two.

One gets following result: $T_{M87^*}\approx0K$ and since surrounding temperature is $6\times10^9K$ therefore by Boltzmann formulae of blackbody radiation, $P_{rad}\approx7.3\times10^{31}W/m^2$, Area of EH $\approx4.7\times10^{27}$ therefore luminosity $L_{haw}\approx 3.4\times10^{66}\mathrm{ergs}/s$ which is ginormously large compared to X-ray luminosity $L_{0.5-7\mathrm{keV}}\approx6\times10^{43}$ which is only for a small range of energy of photon still there is a difference of order $23$ so it should make the Hawking radiation detectable.

Two things I am dubious of are

  1. The accretion happens in a planar region so the BH is not completely surrounded by temperature gradient of $10^9K$.
  2. The luminosity of Hawking radiation is calculated by including all the wavelengths of the spectrum whereas the second quantity is for only a selected portion of spectrum.

Edit: By $T_{M87^*}$ I meant the Hawking temperature of the black hole which is, neglecting the spin contribution $\frac{1}{4GM}\approx0$ and this black hole behaves like a black body of temp $\frac{1}{4GM}$ placed in the surrounding of the accreting disk.

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    $\begingroup$ The outside temperature is irrelevant, you have to use the Hawking temperature of the black hole. $\endgroup$ – Javier Sep 2 at 17:29
  • $\begingroup$ @Javier I don't think outside temperature is irrelevant since it's like a blackbody placed inside a room with the temp of blackbody given by Hawking temp and surrounding temp given by the accreting disk temp. $\endgroup$ – aitfel Sep 2 at 17:48
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I made a mistake we couldn't observe any Hawking radiation since the flux has to be from accreting disk to EH since the temp gradient is from accreting disk to BH. This case is analogous to a cold blackbody placed inside a hot room

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