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The significance is this: it's very likely that there are a lot more such planets a lot nearer to us. This is because Kepler can only detect planets which lie

The significance is this: it's very likely that there are a lot more such planets a lot nearer to us. This is because Kepler can only detect planets which lie in the same plane as earth and the star so the planet occults the star. Assuming random distribution, there's probably thousand fold amount of off plane planets. There are 259 stars within about 30 light years. Communication could be conceivable with such distances... http://joy.chara.gsu.edu/RECONS/TOP100.posted.htm http://joy.chara.gsu.edu/RECONS/TOP100.posted.htm

jrlocke 11y ago
I believe you meant occludes rather than occults
- huehehue 11y ago
  Most of this discussion is beyond me, so I won't speculate on his intent, but occultation is a real thing too. https://en.wikipedia.org/wiki/Occultation https://en.wikipedia.org/wiki/Occultation
- rfreytag 11y ago
  https://en.wiktionary.org/wiki/occult#English https://en.wiktionary.org/wiki/occult#English The verb form is used thus: "The earth occults the moon during a lunar eclipse."
toomuchtodo 11y ago
> This is because Kepler can only detect planets which lie in the same plane as earth and the star so the planet occults the star. What would be necessary for a more comprehensive survey?
- antognini 11y ago
  If the planet is not transiting there are two other ways to detect it: radial velocity measurements, and direct imaging. In the radial velocity technique you measure the wobble of the star due to the gravitational pull of the planet. This also requires that the planet be somewhat edge-on, but the requirement isn't nearly as strict as with the transit method. The difficulty is that Earth size planets have very little mass, and so impart a very small wobble to the star, on the order of only ~10 cm / s. In the second method, you just try to block as much light from the star as you can and go look for the reflected light of the planet. This is much, much more difficult and so far has only been done for very large, distant planets.
  - phantarch 11y ago
    What kind of resolution is possible with direct imaging at 1400 ly away? I would imagine that it's probably milliseconds of arc within the field of vision of any telescope, so how close could we actually look at the planet?
    - antognini 11y ago
      With adaptive optics it's possible to achieve resolutions down to ~30 milliarcseconds. At 1400 ly that would subtend a distance of 13 AU, so you wouldn't be able to find Earth-like planets. But even if it's theoretically possible to resolve 13 AU separations at 1400 ly, in practice the light from the star is so much brighter than the planet that it bleeds all over the image and completely washes out the reflected light of the star. To do direct imaging you really need to look at systems near the Earth, maybe a few tens of ly away. Then you can separate the planet from the star by enough that you can hope to mask the light from the star and still be able to detect the planet. To get some idea of the techniques required to do this, take a look here: http://astrobites.org/2013/06/09/lowest-mass-exoplanet-discovered-via-direct-imaging/ http://astrobites.org/2013/06/09/lowest-mass-exoplanet-disco...
- InclinedPlane 11y ago
  Likely something facilitating direct imaging, which would require canceling out almost all of the light from the parent star (diminishing it by a factor of at least 10^6:1). While there are some techniques that could facilitate that (such as enormous star shades coupled with space telescopes) they wouldn't enable the sort of shotgun survey approach that Kepler has been able to do, but it would enable us to survey and extensively study stars one at a time for planets.
darkmighty 11y ago
So what's the current fraction of stars that have habitable zone planets? (i.e. how many stars were surveyed?) That would get us an estimate of how many of those 259 might contain Earth-like planets: we use the fraction of orbital those planets that Kepler detects to estimate the total fraction of planets per star, and multiply that by 259.
- spiralganglion 11y ago
  From a paper on the Kepler project[1]: "Kepler has observed over 156,000 stars simultaneously and near continuously to search for planets that periodically pass in front of their host star (transit)" From the article: "The confirmation of Kepler-452b brings the total number of confirmed planets to 1,030." — and I have verified that this refers specifically to planets confirmed by Kepler, not just planets in total. From Wikipedia[2]: "four, including Kepler-296f, were less than 2 1/2 the size of Earth and were in habitable zones where surface temperatures are suitable for liquid water." These figures give us an approximate sense of the number of planets that occult their host star, as a proportion of the number of stars surveyed. TRIGGER WARNING: SKETCHY BACK-OF-THE-ENVELOPE MATH 259 stars within 30 light years 1030 / 156000 = 0.00660256410256 (fraction of Kepler-surveyed stars possessing Kepler-confirmed planets that transit) 4 / 1030 = 0.00388349514563 (fraction of significantly Earth-like Kepler-surveyed planets) 259 * 0.00660256410256 * 0.00388349514563 = 0.00664102564102 or 0.66% (chance of a transiting Earth-like within 30 light years) And that's not even factoring in planets that do not transit their host star. [1] http://ntrs.nasa.gov/search.jsp?R=20100030619 http://ntrs.nasa.gov/search.jsp?R=20100030619 [2] https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_using_the_Kepler_spacecraft https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_...
  - darkmighty 11y ago
    Yup, but as others said it might be better since Kepler doesn't detect all orbital planes. So it would be 0.0066 * 1/(fraction of orbital planes) -- but this boost is probably less than an order of magnitude? So still probably no luck for planets reachable within a lifetime eh :P
flashman 11y ago
> Assuming random distribution, there's probably thousand fold amount of off plane planets. For a star like ours with a single Earth-size planet at 1AU, the odds of observing a transit are 0.47%. There is a page devoted to this question which gives some idea of the variables involved: http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/character.html http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Keple...
- Gravityloss 11y ago
  Thanks, that's an excellent response! Assuming a constant density of earth like stars, a density of 200 earth likes in a 1500 light year sphere would mean one earth like in 250 light year sphere. Assuming a disk instead, it would be about 100 light years. Don't know how it relates to the local star neighbourhood... https://en.wikipedia.org/wiki/Orion_Arm https://en.wikipedia.org/wiki/Orion_Arm Of course, sampling from this small amount of data is pretty prone to problems...
lovemenot 11y ago
>> Assuming random distribution (of plane of planets), Has there been any research on this? I have no actual insight, but I'd guess the plane of the galaxy would affect plane of stellar rotation and perhaps also planetary disk alignment.
- Nogwater 11y ago
  I've asked that before too and been told that they are randomly distributed. Here's a link! http://curious.astro.cornell.edu/236-are-the-planes-of-solar-systems-aligned-with-the-plane-of-the-galaxy-intermediate http://curious.astro.cornell.edu/236-are-the-planes-of-solar...
  - mohaine 11y ago
    Thank you. I've been wondering about that for years but never got around to looking it up.
- deleted 11y ago
  [deleted]
- japhyr 11y ago
  The milky way galaxy is about 200,000 light years in diameter, but only about 200 light years thick. Given this thickness, I imagine you'd need to look beyond the 200-light-year range to notice a significant influence of the plane of the galaxy on the distribution of stars (and planets).
  - lovemenot 11y ago
    This solar system orbits the galactic center every 250 million years or so; making us a young adult of around 18 Galactic years old, living in a suburb of quite a busy galaxy. Every eighteen year-old suburbanite[1] has already been pretty well socialised by local interactions in the context of a global system of local interactions, performed massively in parallel. [1] metaphors mixed and units of measure utterly messed-around with
ginmugen 11y ago
>There are 259 stars within about 30 light years. Communication could be conceivable with such distances... Wouldn't that mean a minimum of 60 years to get a reply to any message?
- markburns 11y ago
  Terrible latency but still possible to conceive communicating. Also 60 years would be the upper bound for those stars, unless you are suggesting crafting a reply takes a significant amount of time. Which it may. No need to rush a reply with that time lag.
jgord 11y ago
so we would only be able to detect those [non cross planar with earth ] planets if their orbit occluded another background star [ not their own sun ] .. wonder how rare that is.
deleted 11y ago
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jrlocke 11y ago

I believe you meant occludes rather than occults

huehehue 11y ago
Most of this discussion is beyond me, so I won't speculate on his intent, but occultation is a real thing too. https://en.wikipedia.org/wiki/Occultation https://en.wikipedia.org/wiki/Occultation
rfreytag 11y ago
https://en.wiktionary.org/wiki/occult#English https://en.wiktionary.org/wiki/occult#English The verb form is used thus: "The earth occults the moon during a lunar eclipse."

toomuchtodo 11y ago

> This is because Kepler can only detect planets which lie in the same plane as earth and the star so the planet occults the star. What would be necessary for a more comprehensive survey?

antognini 11y ago
If the planet is not transiting there are two other ways to detect it: radial velocity measurements, and direct imaging. In the radial velocity technique you measure the wobble of the star due to the gravitational pull of the planet. This also requires that the planet be somewhat edge-on, but the requirement isn't nearly as strict as with the transit method. The difficulty is that Earth size planets have very little mass, and so impart a very small wobble to the star, on the order of only ~10 cm / s. In the second method, you just try to block as much light from the star as you can and go look for the reflected light of the planet. This is much, much more difficult and so far has only been done for very large, distant planets.
- phantarch 11y ago
  What kind of resolution is possible with direct imaging at 1400 ly away? I would imagine that it's probably milliseconds of arc within the field of vision of any telescope, so how close could we actually look at the planet?
  - antognini 11y ago
    With adaptive optics it's possible to achieve resolutions down to ~30 milliarcseconds. At 1400 ly that would subtend a distance of 13 AU, so you wouldn't be able to find Earth-like planets. But even if it's theoretically possible to resolve 13 AU separations at 1400 ly, in practice the light from the star is so much brighter than the planet that it bleeds all over the image and completely washes out the reflected light of the star. To do direct imaging you really need to look at systems near the Earth, maybe a few tens of ly away. Then you can separate the planet from the star by enough that you can hope to mask the light from the star and still be able to detect the planet. To get some idea of the techniques required to do this, take a look here: http://astrobites.org/2013/06/09/lowest-mass-exoplanet-discovered-via-direct-imaging/ http://astrobites.org/2013/06/09/lowest-mass-exoplanet-disco...
InclinedPlane 11y ago
Likely something facilitating direct imaging, which would require canceling out almost all of the light from the parent star (diminishing it by a factor of at least 10^6:1). While there are some techniques that could facilitate that (such as enormous star shades coupled with space telescopes) they wouldn't enable the sort of shotgun survey approach that Kepler has been able to do, but it would enable us to survey and extensively study stars one at a time for planets.

darkmighty 11y ago

So what's the current fraction of stars that have habitable zone planets? (i.e. how many stars were surveyed?) That would get us an estimate of how many of those 259 might contain Earth-like planets: we use the fraction of orbital those planets that Kepler detects to estimate the total fraction of planets per star, and multiply that by 259.

spiralganglion 11y ago
From a paper on the Kepler project[1]: "Kepler has observed over 156,000 stars simultaneously and near continuously to search for planets that periodically pass in front of their host star (transit)" From the article: "The confirmation of Kepler-452b brings the total number of confirmed planets to 1,030." — and I have verified that this refers specifically to planets confirmed by Kepler, not just planets in total. From Wikipedia[2]: "four, including Kepler-296f, were less than 2 1/2 the size of Earth and were in habitable zones where surface temperatures are suitable for liquid water." These figures give us an approximate sense of the number of planets that occult their host star, as a proportion of the number of stars surveyed. TRIGGER WARNING: SKETCHY BACK-OF-THE-ENVELOPE MATH 259 stars within 30 light years 1030 / 156000 = 0.00660256410256 (fraction of Kepler-surveyed stars possessing Kepler-confirmed planets that transit) 4 / 1030 = 0.00388349514563 (fraction of significantly Earth-like Kepler-surveyed planets) 259 * 0.00660256410256 * 0.00388349514563 = 0.00664102564102 or 0.66% (chance of a transiting Earth-like within 30 light years) And that's not even factoring in planets that do not transit their host star. [1] http://ntrs.nasa.gov/search.jsp?R=20100030619 http://ntrs.nasa.gov/search.jsp?R=20100030619 [2] https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_using_the_Kepler_spacecraft https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_...
- darkmighty 11y ago
  Yup, but as others said it might be better since Kepler doesn't detect all orbital planes. So it would be 0.0066 * 1/(fraction of orbital planes) -- but this boost is probably less than an order of magnitude? So still probably no luck for planets reachable within a lifetime eh :P

flashman 11y ago

> Assuming random distribution, there's probably thousand fold amount of off plane planets. For a star like ours with a single Earth-size planet at 1AU, the odds of observing a transit are 0.47%. There is a page devoted to this question which gives some idea of the variables involved: http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/character.html http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Keple...

Gravityloss 11y ago
Thanks, that's an excellent response! Assuming a constant density of earth like stars, a density of 200 earth likes in a 1500 light year sphere would mean one earth like in 250 light year sphere. Assuming a disk instead, it would be about 100 light years. Don't know how it relates to the local star neighbourhood... https://en.wikipedia.org/wiki/Orion_Arm https://en.wikipedia.org/wiki/Orion_Arm Of course, sampling from this small amount of data is pretty prone to problems...

lovemenot 11y ago

>> Assuming random distribution (of plane of planets), Has there been any research on this? I have no actual insight, but I'd guess the plane of the galaxy would affect plane of stellar rotation and perhaps also planetary disk alignment.

Nogwater 11y ago
I've asked that before too and been told that they are randomly distributed. Here's a link! http://curious.astro.cornell.edu/236-are-the-planes-of-solar-systems-aligned-with-the-plane-of-the-galaxy-intermediate http://curious.astro.cornell.edu/236-are-the-planes-of-solar...
- mohaine 11y ago
  Thank you. I've been wondering about that for years but never got around to looking it up.
deleted 11y ago
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japhyr 11y ago
The milky way galaxy is about 200,000 light years in diameter, but only about 200 light years thick. Given this thickness, I imagine you'd need to look beyond the 200-light-year range to notice a significant influence of the plane of the galaxy on the distribution of stars (and planets).
- lovemenot 11y ago
  This solar system orbits the galactic center every 250 million years or so; making us a young adult of around 18 Galactic years old, living in a suburb of quite a busy galaxy. Every eighteen year-old suburbanite[1] has already been pretty well socialised by local interactions in the context of a global system of local interactions, performed massively in parallel. [1] metaphors mixed and units of measure utterly messed-around with

ginmugen 11y ago

>There are 259 stars within about 30 light years. Communication could be conceivable with such distances... Wouldn't that mean a minimum of 60 years to get a reply to any message?

markburns 11y ago
Terrible latency but still possible to conceive communicating. Also 60 years would be the upper bound for those stars, unless you are suggesting crafting a reply takes a significant amount of time. Which it may. No need to rush a reply with that time lag.

jgord 11y ago

so we would only be able to detect those [non cross planar with earth ] planets if their orbit occluded another background star [ not their own sun ] .. wonder how rare that is.

deleted 11y ago

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