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Minimum mass of a black hole 
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#1
Dec904, 04:47 PM

P: 501

Knowing that the escape velocity of the Sun is 617 km/s and that it's mass is 1.989 e30 kg, and the speed of light: 299 792 km/s, tell me if this calculation is right. I'm trying to calculate the minimum mass of a black hole. And we know that light cannot escape a black hole so...
Comparing to the sun: 299 792/617 km/s = approx. 485.88, so now if a mass of 1.989 e30 kg has an escape velocity of 617 km/s then: 1.989 e30 kg x 485.88 = 9.66 e32 kg (a pretty much exact mass of an object that can barely emit light.) So this would mean that a black hole can have a minimum mass of about 1 e33 kg??? 100000000000000000000000000000000 kg is some heavy stuff lol. So are my calculations right? Thnx 


#2
Dec904, 05:14 PM

P: n/a

I was under the impression that black holes had to exceed a certain density, not mass. I am not sure of the exact reasons but I am sure someone else will have a better explination.



#3
Dec904, 05:19 PM

P: 1,116

Balck holes are singularities. They are very dense and will not allow anything to esape them as their gravity it so large. I was under the impression (because of all of the talk about time travel) that it was simply like a hole in spacetime or a continous curving of it.
About the mass though I am not sure if it actually has any or not. I believe it must do but if it did then it would get bigger and the universe would be getting emptier and emptier. I don't know. The Bob (2004 ©) 


#4
Dec904, 05:22 PM

P: 1,116

Minimum mass of a black hole
The Bob (2004 ©) 


#5
Dec904, 05:30 PM

P: 501

485.88 is how many times more the escape velocity of an object of a mass like the sun would have to be to have an escape velocity of the speed of light 299792 km/s



#6
Dec1004, 01:46 AM

P: 989

[tex]n_1 = \frac{c}{v_{\odot}} = \sqrt{ \frac{r_{\odot}}{2GM_{\odot}}} c[/tex] [tex]n_1 = \sqrt{ \frac{r_{\odot}}{2GM_{\odot}}} c[/tex] [tex]n_1 = 485.436[/tex] [tex]v_{\odot} = 6.176*10^5 m*s^{1}[/tex]  Solar escape velocity Solving for M,n for 'constant' solar radius: [tex]M_{BH} = n_1^2 M_{\odot} = \frac{r_{\odot}c^2}{2G}[/tex] [tex]\boxed{M_{BH} = 4.687*10^{35} kg}[/tex] 


#7
Dec1004, 03:04 AM

Sci Advisor
PF Gold
P: 9,360

The current theoretical limit for minimum size of a black hole is the plank mass  about 21.77 micrograms. That yields a schwarzchild radius of a planck length. Why would you think minimum black hole mass has anything to do with a solar mass? In your calculations, you treat the sun as stationary. In the presence of a mass as imposing as that you propose, the sun would have about as much chance of winning that gravitational tugowar as a flea against a locomotive.



#8
Apr1408, 01:37 PM

P: 3

I think the proper question that needs to be asked is: What is the theoretical minimum size that a black hole can be that can sustain itself?
The problem is Hawking Radiation, the smaller the black hole the more Hawking Radiation it releases, and the quicker the black hole dissipates. So, at some point the black hole is going to be releasing so much energy out that it will collapse, explode, dance a merry jig (I'm not entirely sure about what will happen). I heard once that the theoretical minimum is somewhere around the mass of Mount Everest, but I could be wrong. 


#9
Apr1408, 02:38 PM

P: 145

if you consider a black hole in a theoretical universe that is really completely "empty" (no radiation, no matter), then it would continously loose energy (and therefore mass) by the Hawking Radiation you described. for a big black hole this might take a very long time, but in the end it would vanish. on the other hand, in our "real" universe, we always have the cosmic background radiation (2.7K I think). so if the black hole is so big that the temperature of its Hawking Radiation is colder than 2.7K, the black hole would absorb more radiation than it emits, gaining energy/mass and growing bigger. furthermore, our "real" universe is not completely empty of matter but contains some "dust". so the black hole might also survive if it could suck in this cosmic dust faster than it looses energy/mass due to Hawking Radiation, even if its so small that Hawking Radiation is hotter than 2.7 K 


#10
Apr1408, 06:27 PM

P: 3

Oberst Villa
You are right about the Cosmic Background Radiation being too hot for most black holes to evaporate due to Hawking Radiation, but there might be black holes that are small enough that they have heated up beyond 2.7k, so called primordial black holes which were created right after the big bang that are small enough that they do emit more energy than they absorb and could evaporate within the current lifetime of the universe. They are theoretical at this point, but, even if they do not exist, eventually the CBR will drop to a point below the temperature of all black holes and they will all start evaporating. It will just take a really, really long time before any black hole will be small enough to worry about reaching the evaporation point. 


#11
Apr1408, 07:24 PM

P: 456




#12
Apr1508, 01:19 PM

P: 145




#13
Apr1508, 01:33 PM

P: 145

but when we talk about velocities that approach the speed of light we have to use Relativistic calculations. in this case, because we are talking about a gravity field, those of the General Theory of Relativity. unfortunately, doing this calculations is way beyond my own (current) capabilities, but I just wanted to "warn" you. 


#14
May708, 04:58 PM

P: 989

Schwarzschild black hole blackbody radiation temperature: (Hawking radiation temperature) [tex]T_s = \frac{\hbar c^3}{8 \pi G M_s k_b}[/tex] [tex]k_b[/tex]  Boltzmann constant Cosmic microwave background radiation black body spectrum temperature: [tex]T_u = 2.725 \; \text{K}[/tex]  Universe temperature Black Hole blackbody radiation temperature equivalent to Cosmic microwave background radiation temperature: [tex]\boxed{T_s = T_u}[/tex] Minimum mass black hole that can existentially exist naturally in known Universe in absentia of quantum black holes: [tex]\boxed{M_s = \frac{\hbar c^3}{8 \pi G k_b T_{u}}}[/tex] Theoretical minimum black hole mass: [tex]\boxed{M_s = 4.502 \cdot 10^{22} \; \text{kg}}[/tex] Schwarzschild radius: [tex]r_s = \frac{2G M_s}{c^2}[/tex] [tex]\boxed{r_s = 6.687 \cdot 10^{5} \; \text{m}}[/tex] Reference: Hawking radiation  Wikipedia Cosmic microwave background radiation  Wikipedia Boltzmann constant  Wikipedia 


#15
May708, 05:34 PM

P: 989

Integration by substitution: [tex]M_s = \frac{r_s c^2}{2G} = \frac{\hbar c^3}{8 \pi G k_b T_u}[/tex] Theoretical minimum Schwarzschild radius: [tex]\boxed{r_s = \frac{\hbar c}{4 \pi k_b T_u}}[/tex] [tex]\boxed{r_s = 6.687 \cdot 10^{5} \; \text{m}}[/tex] Reference: Schwarzschild radius  Wikipedia 


#16
May908, 06:49 AM

P: 49

i think that Krusty the clown isright, but id also look up on google: chandransaker limit. That qwill give you your answer!



#17
May1008, 05:43 AM

P: 989

Chandrasekhar limit maximum mass: [tex]M_c = \frac{\omega_3^0 \sqrt{3\pi}}{2}\left ( \frac{\hbar c}{G}\right )^{3/2}\frac{1}{m_H^2}\; \; \; \; \; \; \mu_e = 1[/tex] Black Hole minimum mass: [tex]\boxed{M_c = 1.1401 \cdot 10^{31} \; \text{kg}}[/tex] Black Hole minimum mass: [tex]\boxed{M_c = 5.7318 \cdot M_{\odot}}[/tex] Schwarzschild radius: [tex]r_s = \frac{2G M_s}{c^2}[/tex] Schwarzschild mass is equivalent to Chandrasekhar mass: [tex]\boxed{M_s = M_c}[/tex] Integration by substitution: [tex] M_s = \frac{r_s c^2}{2G} = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left( \frac{\hbar c}{G}\right )^{3/2}\frac{1}{m_H^2}[/tex] [tex]r_s = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left( \frac{2G}{c^2} \right) \left( \frac{\hbar c}{G}\right )^{3/2}\frac{1}{m_H^2}[/tex] [tex]\boxed{r_s = 16.933 \; \text{km}}[/tex] key: [tex]m_H[/tex]  Hydrogen mass [tex]\omega_3^0[/tex]  LaneEmden equation solution constant The LaneEmden equation solution for the [tex]\omega_3^0[/tex] constant proof, has not yet been published on Wikipedia. Reference: LaneEmden_equation  Wikipedia Chandrasekhar limit  Wikipedia 


#18
May1008, 06:23 AM

P: 989

SchwarzschildChandrasekhar radius: [tex]r_s = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left( \frac{2G}{c^2} \right) \left( \frac{\hbar c}{G}\right )^{3/2} \frac{1}{m_H^2} = \frac{\omega_3^0 \hbar}{c} \sqrt{ \frac{3 \pi \hbar c}{G} } \frac{1}{m_H^2}[/tex] SchwarzschildChandrasekhar radius: (minimum black hole radius) [tex]\boxed{r_s = \frac{\omega_3^0 \hbar}{c} \sqrt{ \frac{3 \pi \hbar c}{G} } \frac{1}{m_H^2}}[/tex] [tex]\boxed{r_s = 16.933 \; \text{km}}[/tex] 


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