Why couldn't they take pictures of a closer black hole? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?Do all of our discoveries of black holes in nature depend on the validity of GR?How will the super massive black hole affect our galaxy?Is there a horizontal event horizon on a spinning black hole?Are black holes in a binary system with white holes, and are they both wormholes?How do we know the stars orbiting Sgr A* are orbiting a supermassive black hole and not just the center of mass of the Milky Way galaxy?Why do galaxies have a super massive black hole at their center?Could dark matter consist of the supermassive black holes at the centers of galaxies?Are there super-super black holes in the center of super clusters?Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?First Black Hole Picture Takeaways
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Why couldn't they take pictures of a closer black hole?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?Do all of our discoveries of black holes in nature depend on the validity of GR?How will the super massive black hole affect our galaxy?Is there a horizontal event horizon on a spinning black hole?Are black holes in a binary system with white holes, and are they both wormholes?How do we know the stars orbiting Sgr A* are orbiting a supermassive black hole and not just the center of mass of the Milky Way galaxy?Why do galaxies have a super massive black hole at their center?Could dark matter consist of the supermassive black holes at the centers of galaxies?Are there super-super black holes in the center of super clusters?Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?First Black Hole Picture Takeaways
$begingroup$
The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 times 10^20$km) from our milky way.
Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?
My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.
black-holes
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
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PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 times 10^20$km) from our milky way.
Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?
My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.
black-holes
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
$begingroup$
"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
$endgroup$
– Federico
Apr 11 at 14:21
$begingroup$
See also this question: physics.stackexchange.com/questions/471792/…
$endgroup$
– Maxter
Apr 11 at 17:31
add a comment |
$begingroup$
The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 times 10^20$km) from our milky way.
Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?
My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.
black-holes
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 times 10^20$km) from our milky way.
Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?
My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.
black-holes
black-holes
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
asked Apr 11 at 8:58
PyRsquaredPyRsquared
18615
18615
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
PyRsquared is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
$begingroup$
"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
$endgroup$
– Federico
Apr 11 at 14:21
$begingroup$
See also this question: physics.stackexchange.com/questions/471792/…
$endgroup$
– Maxter
Apr 11 at 17:31
add a comment |
$begingroup$
I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
$begingroup$
"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
$endgroup$
– Federico
Apr 11 at 14:21
$begingroup$
See also this question: physics.stackexchange.com/questions/471792/…
$endgroup$
– Maxter
Apr 11 at 17:31
$begingroup$
I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
$begingroup$
I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
$begingroup$
"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
$endgroup$
– Federico
Apr 11 at 14:21
$begingroup$
"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
$endgroup$
– Federico
Apr 11 at 14:21
$begingroup$
See also this question: physics.stackexchange.com/questions/471792/…
$endgroup$
– Maxter
Apr 11 at 17:31
$begingroup$
See also this question: physics.stackexchange.com/questions/471792/…
$endgroup$
– Maxter
Apr 11 at 17:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.
My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.
Sgr A* : $M/d sim 4times 10^6/8 = 5times 10^5 M_odot$/kpc;
M87 : $ $ $M/d sim 6times 10^9/16times 10^3 = 3.8times 10^5 M_odot$/kpc.
Your suggested alternatives.
Andromeda : $M/d sim 2times 10^8/8times 10^2 = 2.5times 10^5 M_odot$/kpc;
Triangulum: doesn't have a known central supermassive black hole?
So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
$endgroup$
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
add a comment |
$begingroup$
Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
answered Apr 11 at 21:15
IsterIster
4816
$endgroup$
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.
My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.
Sgr A* : $M/d sim 4times 10^6/8 = 5times 10^5 M_odot$/kpc;
M87 : $ $ $M/d sim 6times 10^9/16times 10^3 = 3.8times 10^5 M_odot$/kpc.
Your suggested alternatives.
Andromeda : $M/d sim 2times 10^8/8times 10^2 = 2.5times 10^5 M_odot$/kpc;
Triangulum: doesn't have a known central supermassive black hole?
So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
$endgroup$
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
add a comment |
$begingroup$
The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.
My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.
Sgr A* : $M/d sim 4times 10^6/8 = 5times 10^5 M_odot$/kpc;
M87 : $ $ $M/d sim 6times 10^9/16times 10^3 = 3.8times 10^5 M_odot$/kpc.
Your suggested alternatives.
Andromeda : $M/d sim 2times 10^8/8times 10^2 = 2.5times 10^5 M_odot$/kpc;
Triangulum: doesn't have a known central supermassive black hole?
So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
$endgroup$
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
add a comment |
$begingroup$
The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.
My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.
Sgr A* : $M/d sim 4times 10^6/8 = 5times 10^5 M_odot$/kpc;
M87 : $ $ $M/d sim 6times 10^9/16times 10^3 = 3.8times 10^5 M_odot$/kpc.
Your suggested alternatives.
Andromeda : $M/d sim 2times 10^8/8times 10^2 = 2.5times 10^5 M_odot$/kpc;
Triangulum: doesn't have a known central supermassive black hole?
So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
$endgroup$
The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.
My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.
Sgr A* : $M/d sim 4times 10^6/8 = 5times 10^5 M_odot$/kpc;
M87 : $ $ $M/d sim 6times 10^9/16times 10^3 = 3.8times 10^5 M_odot$/kpc.
Your suggested alternatives.
Andromeda : $M/d sim 2times 10^8/8times 10^2 = 2.5times 10^5 M_odot$/kpc;
Triangulum: doesn't have a known central supermassive black hole?
So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
edited Apr 11 at 15:45
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
answered Apr 11 at 10:20
Rob JeffriesRob Jeffries
71.4k7152248
71.4k7152248
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
add a comment |
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
1
1
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
$begingroup$
Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake
$endgroup$
– PyRsquared
Apr 11 at 12:39
2
2
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
$begingroup$
"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.
$endgroup$
– Acccumulation
Apr 11 at 15:43
1
1
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
$begingroup$
Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.
$endgroup$
– J.G.
Apr 11 at 21:13
2
2
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
$begingroup$
@J.G. Except for gravitational wave and neutrino astronomers. Lol.
$endgroup$
– Rob Jeffries
Apr 11 at 22:29
1
1
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
$endgroup$
– J.G.
Apr 11 at 22:30
$begingroup$
@RobJeffries Yes, I strictly meant electromagnetic view.
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– J.G.
Apr 11 at 22:30
add a comment |
$begingroup$
Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
answered Apr 11 at 21:15
IsterIster
4816
$endgroup$
add a comment |
$begingroup$
Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
answered Apr 11 at 21:15
IsterIster
4816
$endgroup$
add a comment |
$begingroup$
Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
answered Apr 11 at 21:15
IsterIster
4816
$endgroup$
Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
answered Apr 11 at 21:15
IsterIster
4816
answered Apr 11 at 21:15
IsterIster
4816
answered Apr 11 at 21:15
IsterIster
4816
answered Apr 11 at 21:15
IsterIster
4816
4816
add a comment |
add a comment |
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$begingroup$
I'd think perhaps it has got to do with them being smaller, but they may as well not have as active accretion disks. I don't know, i'm just guessing.
$endgroup$
– DakkVader
Apr 11 at 9:38
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"Duplicate" on Astronomy: astronomy.stackexchange.com/q/30339/7209 and see also: astronomy.stackexchange.com/q/30313/7209
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– Federico
Apr 11 at 14:21
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See also this question: physics.stackexchange.com/questions/471792/…
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– Maxter
Apr 11 at 17:31