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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?













2












$begingroup$


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.



Mars, ASA, ESA, and STScI



above: Cropped from Source NASA, ESA, and STScI



Mars, NASA/Hubble



above: Cropped from Source NASA/Hubble










share|improve this question











$endgroup$











  • $begingroup$
    First, you need to understand that magnification is not what you want to ask about, but angular resolution.
    $endgroup$
    – AtmosphericPrisonEscape
    2 hours ago















2












$begingroup$


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.



Mars, ASA, ESA, and STScI



above: Cropped from Source NASA, ESA, and STScI



Mars, NASA/Hubble



above: Cropped from Source NASA/Hubble










share|improve this question











$endgroup$











  • $begingroup$
    First, you need to understand that magnification is not what you want to ask about, but angular resolution.
    $endgroup$
    – AtmosphericPrisonEscape
    2 hours ago













2












2








2





$begingroup$


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.



Mars, ASA, ESA, and STScI



above: Cropped from Source NASA, ESA, and STScI



Mars, NASA/Hubble



above: Cropped from Source NASA/Hubble










share|improve this question











$endgroup$




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.



Mars, ASA, ESA, and STScI



above: Cropped from Source NASA, ESA, and STScI



Mars, NASA/Hubble



above: Cropped from Source NASA/Hubble







telescope mars hubble-telescope angular-resolution






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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
















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    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


















1












$begingroup$

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.



NPOI



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.






share|improve this answer











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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|improve this answer









    $endgroup$












    • $begingroup$
      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















    1












    $begingroup$

    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.






    share|improve this answer









    $endgroup$












    • $begingroup$
      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













    1












    1








    1





    $begingroup$

    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.






    share|improve this answer









    $endgroup$



    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.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    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
















    • $begingroup$
      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











    1












    $begingroup$

    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.



    NPOI



    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.






    share|improve this answer











    $endgroup$

















      1












      $begingroup$

      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.



      NPOI



      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.






      share|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        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.



        NPOI



        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.






        share|improve this answer











        $endgroup$



        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.



        NPOI



        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.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 16 mins ago

























        answered 42 mins ago









        Florin AndreiFlorin Andrei

        13k12945




        13k12945



























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