What detail can Hubble see on Mars?Inability of Hubble to clearly resolve nearby celestial objectsWhat “revolutionary” discoveries were made by Hubble TelescopeHubble HST scheduling algorithmClaim that 30-m class telescopes will have resolution far superior to Hubble: true?who invented the Hubble telescopewhat can I see with this telescope?Why can’t I see any detail on planets?What magnification is required to see detail on Mars / other local bodies?Can't see any detail on planets with my new Celestron NexStar 130 SLT Computerized TelescopeIf we had the right technology could we see a distant star in detail?
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What detail can Hubble see on Mars?
Inability of Hubble to clearly resolve nearby celestial objectsWhat “revolutionary” discoveries were made by Hubble TelescopeHubble HST scheduling algorithmClaim that 30-m class telescopes will have resolution far superior to Hubble: true?who invented the Hubble telescopewhat can I see with this telescope?Why can’t I see any detail on planets?What magnification is required to see detail on Mars / other local bodies?Can't see any detail on planets with my new Celestron NexStar 130 SLT Computerized TelescopeIf we had the right technology could we see a distant star in detail?
$begingroup$
I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.
The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.
Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?
From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.
above: Cropped from Source NASA, ESA, and STScI
above: Cropped from Source NASA/Hubble
telescope mars hubble-telescope angular-resolution
$endgroup$
add a comment |
$begingroup$
I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.
The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.
Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?
From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.
above: Cropped from Source NASA, ESA, and STScI
above: Cropped from Source NASA/Hubble
telescope mars hubble-telescope angular-resolution
$endgroup$
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago
add a comment |
$begingroup$
I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.
The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.
Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?
From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.
above: Cropped from Source NASA, ESA, and STScI
above: Cropped from Source NASA/Hubble
telescope mars hubble-telescope angular-resolution
$endgroup$
I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.
The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.
Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?
From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.
above: Cropped from Source NASA, ESA, and STScI
above: Cropped from Source NASA/Hubble
telescope mars hubble-telescope angular-resolution
telescope mars hubble-telescope angular-resolution
edited 33 mins ago
uhoh
7,96522275
7,96522275
asked 4 hours ago
Eric J.Eric J.
1745
1745
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago
add a comment |
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Hubble space telescope has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about $10^-7$ radians. Mars' closest distance to Earth is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast. Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.
$endgroup$
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
add a comment |
$begingroup$
Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.
Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.
Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.
The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but it varies all the time).
Let's apply the tangent of 0.04 arcsec at 55 mil km:
https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000
It's 10 km. It would only be able to see the major geographical features.
To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are places several km apart and are coupled optically to function as a single huge mirror.
This would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.
Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.
All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.
Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the surface, the ground provides that rigidity. In space you'd have to... I dunno, use space magic.
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The Hubble space telescope has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about $10^-7$ radians. Mars' closest distance to Earth is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast. Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.
$endgroup$
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
add a comment |
$begingroup$
The Hubble space telescope has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about $10^-7$ radians. Mars' closest distance to Earth is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast. Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.
$endgroup$
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
add a comment |
$begingroup$
The Hubble space telescope has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about $10^-7$ radians. Mars' closest distance to Earth is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast. Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.
$endgroup$
The Hubble space telescope has a 2.4m mirror and is pretty much diffraction limited, so at near-UV wavelengths of say 240nm it has an angular resolution of about $10^-7$ radians. Mars' closest distance to Earth is about 54.6 million km, so the theoretical minimal resolution is between 5 ad 6 km. So large cities might be visible if they have lots of contrast. Since at closest approach an observer on Mars is looking straight at the night side of the Earth, a well lit city might be easy to spot, on the other hand, they are also looking more or less straight at the Sun, which might give some problems. Still there will be times when the distance is only 70 or 80 million km and the angle from the Sun is more manageable, so a resolution of 10km in the UV, 20 in visible light is credible.
answered 3 hours ago
Steve LintonSteve Linton
2,9391323
2,9391323
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
add a comment |
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
$begingroup$
What's not clear to me is whether the minimum Earth-Mars distance coincides with the transit of Earth over the solar disc, seen from Mars. Since the orbits are not exactly coplanar, the coincidence is not guaranteed - at least that's my intuition. I could be wrong.
$endgroup$
– Florin Andrei
49 mins ago
add a comment |
$begingroup$
Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.
Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.
Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.
The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but it varies all the time).
Let's apply the tangent of 0.04 arcsec at 55 mil km:
https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000
It's 10 km. It would only be able to see the major geographical features.
To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are places several km apart and are coupled optically to function as a single huge mirror.
This would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.
Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.
All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.
Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the surface, the ground provides that rigidity. In space you'd have to... I dunno, use space magic.
$endgroup$
add a comment |
$begingroup$
Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.
Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.
Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.
The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but it varies all the time).
Let's apply the tangent of 0.04 arcsec at 55 mil km:
https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000
It's 10 km. It would only be able to see the major geographical features.
To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are places several km apart and are coupled optically to function as a single huge mirror.
This would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.
Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.
All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.
Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the surface, the ground provides that rigidity. In space you'd have to... I dunno, use space magic.
$endgroup$
add a comment |
$begingroup$
Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.
Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.
Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.
The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but it varies all the time).
Let's apply the tangent of 0.04 arcsec at 55 mil km:
https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000
It's 10 km. It would only be able to see the major geographical features.
To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are places several km apart and are coupled optically to function as a single huge mirror.
This would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.
Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.
All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.
Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the surface, the ground provides that rigidity. In space you'd have to... I dunno, use space magic.
$endgroup$
Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.
Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.
Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.
The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but it varies all the time).
Let's apply the tangent of 0.04 arcsec at 55 mil km:
https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000
It's 10 km. It would only be able to see the major geographical features.
To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are places several km apart and are coupled optically to function as a single huge mirror.
This would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.
Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.
All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.
Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the surface, the ground provides that rigidity. In space you'd have to... I dunno, use space magic.
edited 45 mins ago
answered 1 hour ago
Florin AndreiFlorin Andrei
13k12945
13k12945
add a comment |
add a comment |
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$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
3 hours ago