There are 259 stars within about 30 light years. Communication could be conceivable with such distances...
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.
http://curious.astro.cornell.edu/236-are-the-planes-of-solar...
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).
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
Yes, I know, it's still from Earth, but I do think it's a weird way of looking at dinosaur fossils.
More other-side-of-the-galaxy dinosaurs:
http://www.enchantedlearning.com/subjects/dinosaurs/mesozoic...
- What did you see out there? What horrors and wonders hide on the other side of space? Tell me, what did you see?
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_Keple...
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
Of course, sampling from this small amount of data is pretty prone to problems...
"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 [2] https://en.wikipedia.org/wiki/List_of_exoplanets_discovered_...
Assume the plane needs to be within 1 degree from the line joining the star and kepler. I would guess it probably less. It would be a function of size of the planet, size of the star, radius of the orbit and distance of star from kepler.
That would give about 179 planes that we cannot see, which is two orders of magnitude, but the probability calculation won't be that simple.
A naive calculation for Earth would be Theta ~= 2 * (Earth diameter) / (Mean distance to Sun), which is ~2*10^-4. Since most planets found by kepler were much larger than earth, it seems 10^-2 is a reasonable guesstimate.
Assuming you mean 1 degree in either direction, or 2/360 = 1/180, this is off by about a factor of pi/2 ~= 1.57. The correct answer is sin(pi/360) ~= pi/360 ~= 1/115. Still two orders of magnitude, but slightly better. The reason is that the degrees are not equally likely (small degrees, that you want, are more likely, and large degrees less likely).
Here's the calculation: Consider the unit vector normal to the equator of the orbit, using say the right-hand rule. This vector is uniformly distributed over the unit sphere (surface area 4pi). You want the vector to lie within +-1 degree of the rim of the sphere. Call the region of such vectors R; the probability of a good planet is P = area(R)/(4pi).
R is an annulus going around the equator and up/down by 1 degree. In area, this is close to a rectangle with width 2pi (circumference of sphere) and height 2pi/360 (i.e., 2 degrees), so P ~= (pi^2/90) / (4pi) = pi/360.
The area of R (an equatorial annulus) can be computed exactly using calculus---or looked up on Wikipedia [1]: area(R) = 4pi - 2(2pi(1-sin(pi/360))) = 4pi sin(pi/360). Thus, P = sin(pi/360).
[1] https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surfa...
What would be necessary for a more comprehensive survey?
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.
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-disco...
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.
Wouldn't that mean a minimum of 60 years to get a reply to any message?
The verb form is used thus: "The earth occults the moon during a lunar eclipse."