What detail can Hubble see on Mars?Why can't I see Mars clearly?What is the faintest magnitude a telescope can see?what can I see with this telescope?What can we expect to see with a telescope with a 70mm aperture and a 10mm eyepiece?Why can’t I see any detail on planets?What magnification is required to see detail on Mars / other local bodies?what can i see using celestron nexstar 127SLTCan't see any detail on planets with my new Celestron NexStar 130 SLT Computerized TelescopeCan we see the color of nebulae?If we had the right technology could we see a distant star in detail?
Copper as an adjective to refer to something made of copper
The selling of the sheep
Installing Debian 10, upgrade to stable later?
What does the copyright in a dissertation protect exactly?
What detail can Hubble see on Mars?
As a GM, is it bad form to ask for a moment to think when improvising?
While drilling into kitchen wall, hit a wire - any advice?
Is there precedent or are there procedures for a US president refusing to concede to an electoral defeat?
What happens if I accidentally leave an app running and click "Install Now" in Software Updater?
What is the thing used to help pouring liquids called?
Has the United States ever had a non-Christian President?
Which "exotic salt" can lower water's freezing point by –70 °C?
What's the 2-minute timer on mobile Deutsche Bahn tickets?
How to speed up large double sums in a table?
How to preserve a rare version of a book?
If I no longer control a permanent when it would be sacrificed, is it still sacrificed?
How to say something covers all the view up to the horizon line?
Can you figure out this language?
Rock, Paper, Scissors, Lizard, Spock Tournament
Emergency stop in plain TeX, pdfTeX, XeTeX and LuaTeX?
Python 3 - simple temperature program version 1.3
Why increasing of the temperature of the objects like wood, paper etc. doesn't fire them?
What is more safe for browsing the web: PC or smartphone?
Endgame puzzle: How to avoid stalemate and win?
What detail can Hubble see on Mars?
Why can't I see Mars clearly?What is the faintest magnitude a telescope can see?what can I see with this telescope?What can we expect to see with a telescope with a 70mm aperture and a 10mm eyepiece?Why can’t I see any detail on planets?What magnification is required to see detail on Mars / other local bodies?what can i see using celestron nexstar 127SLTCan't see any detail on planets with my new Celestron NexStar 130 SLT Computerized TelescopeCan we see the color of nebulae?If 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
2 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 4 mins ago
uhoh
7,96522275
7,96522275
asked 3 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
2 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
2 hours ago
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
2 hours ago
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
2 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
21 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 |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "514"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fastronomy.stackexchange.com%2fquestions%2f31775%2fwhat-detail-can-hubble-see-on-mars%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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
21 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
21 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
21 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
21 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
21 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
21 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 16 mins ago
answered 42 mins ago
Florin AndreiFlorin Andrei
13k12945
13k12945
add a comment |
add a comment |
Thanks for contributing an answer to Astronomy Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fastronomy.stackexchange.com%2fquestions%2f31775%2fwhat-detail-can-hubble-see-on-mars%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
First, you need to understand that magnification is not what you want to ask about, but angular resolution.
$endgroup$
– AtmosphericPrisonEscape
2 hours ago