Why not take a picture of a closer black hole? The 2019 Stack Overflow Developer Survey...
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Why not take a picture of a closer black hole?
The 2019 Stack Overflow Developer Survey Results Are InWhy didn't the Event Horizon Telescope team mention Sagittarius A*?Can A Black Hole Exist?Star versus Black HoleCan things move faster than light inside the event horizon of a black hole?What conditions would lead to this event around the black hole in the Pictor A galaxy?Why do “they” portray colliding black holes like that?Black Hole growthWhat will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)?Black hole, escape velocity, going up?Shouldn't we not be able to see some black holes?Is this a black hole?
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There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?
black-hole supermassive-black-hole event-horizon-telescope
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There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?
black-hole supermassive-black-hole event-horizon-telescope
New contributor
$endgroup$
add a comment |
$begingroup$
There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?
black-hole supermassive-black-hole event-horizon-telescope
New contributor
$endgroup$
There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?
black-hole supermassive-black-hole event-horizon-telescope
black-hole supermassive-black-hole event-horizon-telescope
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MorganMorgan
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I was surprised too when I first heard they were trying to image M87's black hole.
The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*.
A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
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add a comment |
$begingroup$
There are a few criteria necessary to see a black hole with the event-horizons telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 15 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, it’s distance is made up for by sheer scale.
$endgroup$
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The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I was surprised too when I first heard they were trying to image M87's black hole.
The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*.
A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
$endgroup$
add a comment |
$begingroup$
I was surprised too when I first heard they were trying to image M87's black hole.
The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*.
A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
$endgroup$
add a comment |
$begingroup$
I was surprised too when I first heard they were trying to image M87's black hole.
The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*.
A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
$endgroup$
I was surprised too when I first heard they were trying to image M87's black hole.
The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*.
A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
edited 25 mins ago
answered 55 mins ago
IngolifsIngolifs
1,5721619
1,5721619
add a comment |
add a comment |
$begingroup$
There are a few criteria necessary to see a black hole with the event-horizons telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 15 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, it’s distance is made up for by sheer scale.
$endgroup$
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
add a comment |
$begingroup$
There are a few criteria necessary to see a black hole with the event-horizons telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 15 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, it’s distance is made up for by sheer scale.
$endgroup$
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
add a comment |
$begingroup$
There are a few criteria necessary to see a black hole with the event-horizons telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 15 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, it’s distance is made up for by sheer scale.
$endgroup$
There are a few criteria necessary to see a black hole with the event-horizons telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 15 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, it’s distance is made up for by sheer scale.
answered 24 mins ago
cmscms
2164
2164
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
add a comment |
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
$begingroup$
The distance is about 50 million ly, I thought.
$endgroup$
– Ingolifs
7 mins ago
add a comment |
Morgan is a new contributor. Be nice, and check out our Code of Conduct.
Morgan is a new contributor. Be nice, and check out our Code of Conduct.
Morgan is a new contributor. Be nice, and check out our Code of Conduct.
Morgan is a new contributor. Be nice, and check out our Code of Conduct.
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