
50,000 times the mass. What's the diameter?
At the heart of every large galaxy resides a supermassive black hole, and astroboffins have found the smallest one yet – about 340 million light years away. "In a sense, it's a teeny supermassive black hole," said Elena Gallo, assistant professor of astronomy at the University of Michigan College of Literature, Science, and …
"50,000 times the mass. What's the diameter?"
The boffins don't say (their paper is here).
However, what we can do is calculate the Schwarzschild radius of the phenomenon, which is 147km.
That means if you take enough material that's 50,000 times the mass of our Sun (9.942175 x 10^34 kg in total) and compress it right down to an object with a radius of 147km, the escape velocity from its surface is the speed of light.
C.
How do you arrive at a figure of 443m? That's tiny for a black hole of that mass. I banged together the following PHP to check it:
----------
//Calculate the Schwartzschild radius of an object of 50,000 solar masses
$msun = 1.989 * pow(10.0, 30.0);
echo sprintf("%1.4f", Schwartzschild(50000.0 * $msun))." metres";
function Schwartzschild($mkg){
$g = 6.67384 * pow(10.0, -11.0);
$c = 299792458.0;
return ((2.0*$g*$mkg)/($c * $c));
}
----------
This gave me a figure of 147,696,147.6723 metres or roughly 150,000 km which is about right for a supermassive black hole. Given that the very page you linked in the article gives the Schwartzschild radius of the Sun as 3 km, your figure seems way off.
If my working is wrong, I'd appreciate you showing me where. Otherwise you might want to revisit your own working and update the article?
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"perhaps it's supposed to be 443 times the sun's `black hole radius'?"
The formula for calculating the Schwarschild Radius is:
R = 2*G*M / c²
So that means the radius (R) is proportional to the mass (M). So a mass of 50,000 suns will have a radius 50,000 times as big as for one sun.
Here's some Python code:
>>> G = 6.67384e-11 # Gravitational constant (m³ / kg s²)
>>> c = c=299792458.0 # Speed of light (m/s)
>>> M = M=1.989e30 # Solar mass (kg)
>>>
>>> def schwarz(m):
... return 2*G*m/(c*c)
...
>>> schwarz(M)
2953.922953446948
>>> schwarz(50000*M)
147696147.67234743
So R for one solar mass is just under 3 km, and for 50,000 solar masses is almost 148 million km (approx 0.001 AU).
Thank you, thank you, thank you!
I thought I was the last of the band of people that winces when they see 'five times lighter' (instead of a fifth the weight) or 'three time smaller', etc, etc.
I had given up complaining about it thinking I was now the anomaly
I feel little less lonely today.
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In theory if they do exist they would evaporate almost instantly.
A 10-million ton black hole, which would have a radius well under a nanometer and thus qualify as 'microscopic,' would evaporate in 2 million years*. Particle and photon production during the evaporation process would average about 3 terawatts. Admittedly, it'd get a bit more exciting** in the last minutes of evaporation because evaporation rate increases inversely with mass.
*Which can be instant on certain astrophysical time scales.
**As in, "Oh, God, oh, God, we're all going to die."
It sounds an unlikely scenario but, depending on exactly how small it was, the 'evaporation radiation' may be enough to clear the local area of gas - at least far enough out to keep anything left nearby in orbit. In that case it can continue evaporating, the rising radiation output will drive the gas further away and the blackhole will evaporate to non-existence.
It seems more likely, to me, that the radiation level will only succeed in reducing the infall but not to the point that the blackhole evaporates faster than it munches down gas. The black hole grows until it clears the local gas cloud or the radiation released by the infalling mass drives everything else away. maybe.
"Would it survive long enough in the cloud to grow into something self sustaining?"
As the AC noted, it depends on the evaporation rate and whether that cleans out local space. A sub-nanometer black hole blasting out some terawatts is not conducive to accretion disk. However, Hawking radiation drops inversely with mass. Double the mass, half the power. A black hole as massive as Ceres will only emit a few hundred watts (or 30-ish watts, if I misremembered Ceres' mass), which
1) means it takes a very long time to evaporate 1 kilogram of its respectable mass, E=MC^2 meaning there's a lot of Joules in each kilogram
2) probably won't disrupt mass inflows
It depends on the characteristics you want to give your hypothetical microscopic black hole.