6 ms·

Someone want to do the math for me? How many arc-angles per pixel in this image? And then, with those arc-angles, what sized object could you resolve (as larg

Someone want to do the math for me? How many arc-angles per pixel in this image? And then, with those arc-angles, what sized object could you resolve (as larger than one pixel) on the moon?

teraflop 6y ago
Not sure that "per pixel" is the most meaningful metric, because in astronomy, resolution is typically limited by telescope optics and atmospheric conditions, rather than pixel count. But if I'm reading the paper correctly ([1], table 2) the observed angular resolution -- the FWHM, i.e. the "blurriness" of an assumed point source -- was on the order of 0.05 arcseconds. Given the distance from the Earth to the Moon (about 384,000 km), that would correspond to a resolution of about a hundred meters. [1]: https://www.eso.org/public/archives/releases/sciencepapers/eso2011/eso2011a.pdf https://www.eso.org/public/archives/releases/sciencepapers/e...
mturmon 6y ago
> ESO ... has taken the first direct image of a planetary system around a star like our Sun, located about 300 light-years away... > The two gas giants orbit their host star at distances of 160 and about 320 times the Earth-Sun distance. For our purposes, a parsec [pc] is 3 ly, so the host star is at 100 pc. The inner planet is at 160 AU, which we'll round to 100 AU. So the angular separation is 100 AU / 100 pc = 1 AU/pc = 1 arcsec, and their resolution must exceed this. And then we have this handy chart: https://astronomy.stackexchange.com/questions/20695/how-big-is-one-arcsecond-at-various-distances https://astronomy.stackexchange.com/questions/20695/how-big-... This illustrates that the key issue is not resolution itself. It's contrast, at this high resolution. Because the planet must be distinguished from the host star. I don't know what the contrast difference here is. For Earth-like exoplanets (smaller targets at 1AU), the contrast difference is 10^10. That is, for every 10 billion photons from the host star, you get one reflected from the exoplanet. [edited to add: using the link provided by @teraflop, Table 1, column 3 seems to show a contrast of about 10-12 in magnitude units, which is 10^4 to 10^5 in physical units like photons]
- raducu 6y ago
  300 AU? Isn't that exeedingly far? Like 10 times farther away from Pluto? Would we even be able to detect such a planet if it orbited the Sun?
  - vl 6y ago
    In the article they say that these planets are young and hot and they detected them in infrared by blocking light from the host star using special device. We would easily detect such super-massive hot planet at 300 AU in the solar system, but there is no reason for them to exist here since solar system is much older.
    - raducu 6y ago
      I was thinking that in out case hot or cold is not the main issue, but the area where to look. If you look at a star 300 light years away, you just have to search a couple of pixels away from the star. But if you searched for such a planet at 300 AU from out Sun, you would have a massive amount of space to search throug -- like a massive cilindrical wall of space around the Sun(if it was not exactly on our plane around the Sun). But in anycase, I was wondering if we had such a planet in our solar system, but obviously cold by now, could we detect it?
      - riidom 6y ago
        If it has a gravitational influence on other planets, then yes, same way as the discovery of Neptune. https://en.wikipedia.org/wiki/Discovery_of_Neptune https://en.wikipedia.org/wiki/Discovery_of_Neptune Now, 300 AU is far away, but then, 14 times the mass of jupiter is heavy. But I'm not knowledged enough to do the math here. On a 2nd thought, maybe not. How long would a year be for a planet that far outside? Maybe several hundred years? We'd need to be in the right window of time to spectate such effects in first place.
        cgriswald 6y ago
        To be equivalent to the gravitational impact of Neptune the mass would have to be about 70,000 times the mass of Neptune. This is more equivalent to trying to find Planet Nine [0], which, although was also “discovered” through gravitational effects, those effects are much subtler and the planet might not exist and if it does, hasn’t been found. [0] - https://en.m.wikipedia.org/wiki/Planet_Nine https://en.m.wikipedia.org/wiki/Planet_Nine
s1artibartfast 6y ago
For your first question: Tan^-1[(160/2 Au)/(300 lightyears)] = 2.4×10^-4° for the distance between the inner planet and the star. You can divide this by the pixel length. With respect to your second question: Resolution is not tied to do pixel size for this type of measurement. You can resolve an object of any size if it is emitting enough photons.
- VikingCoder 6y ago
  I meant resolve as in "determine what it is," like, how big would the letters have to be on the moon in order for you to read them.
  - s1artibartfast 6y ago
    So then resolve that two objects are distinct.

Someone want to do the math for me? How many arc-angles per pixel in this image? And then, with those arc-angles, what sized object could you resolve (as larger than one pixel) on the moon?

teraflop 6y ago

Not sure that "per pixel" is the most meaningful metric, because in astronomy, resolution is typically limited by telescope optics and atmospheric conditions, rather than pixel count. But if I'm reading the paper correctly ([1], table 2) the observed angular resolution -- the FWHM, i.e. the "blurriness" of an assumed point source -- was on the order of 0.05 arcseconds. Given the distance from the Earth to the Moon (about 384,000 km), that would correspond to a resolution of about a hundred meters. [1]: https://www.eso.org/public/archives/releases/sciencepapers/eso2011/eso2011a.pdf https://www.eso.org/public/archives/releases/sciencepapers/e...

mturmon 6y ago

> ESO ... has taken the first direct image of a planetary system around a star like our Sun, located about 300 light-years away... > The two gas giants orbit their host star at distances of 160 and about 320 times the Earth-Sun distance. For our purposes, a parsec [pc] is 3 ly, so the host star is at 100 pc. The inner planet is at 160 AU, which we'll round to 100 AU. So the angular separation is 100 AU / 100 pc = 1 AU/pc = 1 arcsec, and their resolution must exceed this. And then we have this handy chart: https://astronomy.stackexchange.com/questions/20695/how-big-is-one-arcsecond-at-various-distances https://astronomy.stackexchange.com/questions/20695/how-big-... This illustrates that the key issue is not resolution itself. It's contrast, at this high resolution. Because the planet must be distinguished from the host star. I don't know what the contrast difference here is. For Earth-like exoplanets (smaller targets at 1AU), the contrast difference is 10^10. That is, for every 10 billion photons from the host star, you get one reflected from the exoplanet. [edited to add: using the link provided by @teraflop, Table 1, column 3 seems to show a contrast of about 10-12 in magnitude units, which is 10^4 to 10^5 in physical units like photons]

raducu 6y ago
300 AU? Isn't that exeedingly far? Like 10 times farther away from Pluto? Would we even be able to detect such a planet if it orbited the Sun?
- vl 6y ago
  In the article they say that these planets are young and hot and they detected them in infrared by blocking light from the host star using special device. We would easily detect such super-massive hot planet at 300 AU in the solar system, but there is no reason for them to exist here since solar system is much older.
  - raducu 6y ago
    I was thinking that in out case hot or cold is not the main issue, but the area where to look. If you look at a star 300 light years away, you just have to search a couple of pixels away from the star. But if you searched for such a planet at 300 AU from out Sun, you would have a massive amount of space to search throug -- like a massive cilindrical wall of space around the Sun(if it was not exactly on our plane around the Sun). But in anycase, I was wondering if we had such a planet in our solar system, but obviously cold by now, could we detect it?
    - riidom 6y ago
      If it has a gravitational influence on other planets, then yes, same way as the discovery of Neptune. https://en.wikipedia.org/wiki/Discovery_of_Neptune https://en.wikipedia.org/wiki/Discovery_of_Neptune Now, 300 AU is far away, but then, 14 times the mass of jupiter is heavy. But I'm not knowledged enough to do the math here. On a 2nd thought, maybe not. How long would a year be for a planet that far outside? Maybe several hundred years? We'd need to be in the right window of time to spectate such effects in first place.
      - cgriswald 6y ago
        To be equivalent to the gravitational impact of Neptune the mass would have to be about 70,000 times the mass of Neptune. This is more equivalent to trying to find Planet Nine [0], which, although was also “discovered” through gravitational effects, those effects are much subtler and the planet might not exist and if it does, hasn’t been found. [0] - https://en.m.wikipedia.org/wiki/Planet_Nine https://en.m.wikipedia.org/wiki/Planet_Nine

s1artibartfast 6y ago

For your first question: Tan^-1[(160/2 Au)/(300 lightyears)] = 2.4×10^-4° for the distance between the inner planet and the star. You can divide this by the pixel length. With respect to your second question: Resolution is not tied to do pixel size for this type of measurement. You can resolve an object of any size if it is emitting enough photons.

VikingCoder 6y ago
I meant resolve as in "determine what it is," like, how big would the letters have to be on the moon in order for you to read them.
- s1artibartfast 6y ago
  So then resolve that two objects are distinct.