
#1
Dec411, 12:59 PM

P: 21

Ok so first I know that this equation was presented by Stephen Hawking to describe to Temperature of a black hole:
T = hc^{2} / 16∏^{2}GMk so I did the calculations and got that the temperature of a black hole with the mass of our sun would be ≈ .57° Is this right? Is this the right equation? Here is what I used for the variables can you guys check if these are correct?: g = 6.67*10^11 h = 62606956*10^34 k = 1.3806583*10^23 and then just the mass of our sun and the speed of light Thanks 



#2
Dec411, 01:13 PM

PF Gold
P: 184

I'm getting closer to 10^{16} °K, so a Black Hole Sun would evaporate very slowly, unlike the Soundgarden video.




#3
Dec511, 01:52 AM

Sci Advisor
P: 4,721

Looks like you missed a factor of the speed of light in your equation. Should be:
[tex]T = {h c^3 \over 16\pi^2 GMk}[/tex] Anyway, the easiest way to calculate these things is to just plug them into Google. The Google calculator knows about units, fundamental constants, and a lot of common values, so you can simply type in: h*c^3/(16*pi^2*G*(mass of sun)*k) ...to Google, and it will give you the right result (about 10^8 K). Oh, and there's also a nifty calculator for all of the values related to a black hole: http://xaonon.dyndns.org/hawking/ 



#4
Dec511, 04:40 AM

Sci Advisor
PF Gold
P: 9,186

Black Hole Temperature
Shouldn't that be [tex]\hbar c^3 / 8 \pi GMk[/tex] or did I miss something?




#5
Dec511, 04:57 AM

Sci Advisor
P: 4,721





#6
Dec511, 06:31 AM

Sci Advisor
PF Gold
P: 9,186

My error, I am so accustomed to hbar I overlooked the obvious equivalence.




#7
Dec1311, 10:21 AM

P: 21

Thanks for your help guys! And thanks Chalnoth for the calculator and link :)



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