The Hawking wattage of a black hole

[marcus]

Assume the surface area of the event horizon of an ordinary black hole is 10⁵⁰,
ordinary meaning uncharged and nonrotating.
The hole glows with Hawking radiation----what is the radiant power?


I gave the event horizon area in natural units but it's easy to convert to square meters if you wish it that way.
The area 10⁵⁰ is 2.6 x 10^-20 square meters, or 2.6 square angstroms.

In natural units (c=G=hbar=k=1) the radiant power of a BH
with area A is simply

1/960A

So having the area 10⁵⁰ means that the luminosity is essentially 10^-53. To which of the four wattage figures given in the poll is this equivalent?

 

[kyle_soule]

For some reason I'm getting 1.0110686E-103 for the radiant power...is this right?

 

[marcus]

Originally posted by kyle_soule
For some reason I'm getting 1.0110686E-103 for the radiant power...is this right?

Could you have squared the area by mistake?

the area is E50 in Planck units
the formula for the overall power is 1/960A

so you should take reciprocal of 960E50
and it comes to about 1.04E-53 in units of Planck power

Planck power (c⁵/G) is an awesome amount of power.
It could supply the mass-energy of a whole galaxy in a few days.
You can easily calculate what it is in watts, if you look up the metric system values for G and c. It is around 3.6E52 watts---you can easily check this.

WAIT! Kyle you have calculated the radiant power PER UNIT AREA approximately correctly! The total power from the whole ball (namely 1.04E-53) divided by the area of the ball (namely E50) is indeed 1.04E-103.

Bravo Kyle! And thank you for responding.