A question about Kepler

[robnissen]

From SDC today:

"Kepler . . . [will] sift through . . . millions of targets for 100,000 pre-selected candidate stars that may have Earth-sized planets around them. Those target stars sit between 600 and 3,000 light-years from Earth."

Why, oh why, is Keppler not also looking for stars closer to earth within say 100 light years. If planets are found in the habitable zone, it seems like closer planets would be better than farther, so that we could like for signs of life (O2 in the atmosphere, etc.). I doubt we can do much study at 3K lys. Now going out to 3K is fine, but why would the lower value be limited to 600 lys. Is there something about Kepler that doesn't allow it to look for closer stars?

 

[MeteorWayne]

There may not be 100,000 stars within 100 ly.

 

[Mee_n_Mac]

There may not be 100,000 stars within 100 ly.

I was thinking that perhaps in the sector of the sky where Kepler was looking there weren't any (or very few) stars closer than 600 ly. I can't think of a reason it wouldn't image "close" stars if they were there and in it's FOV.

 

[xXTheOneRavenXx]

There may not be 100,000 stars within 100 ly.

I was thinking that perhaps in the sector of the sky where Kepler was looking there weren't any (or very few) stars closer than 600 ly. I can't think of a reason it wouldn't image "close" stars if they were there and in it's FOV.

Then it posses the question why they wouldn't select a different target area that is closer. I'm sure Kepler is capable of adjusting it's target area. It does seem odd that they would look for planets at such a distance if there are closer prospects, to which we could potentially gather more detail of those discovered planets.

 

[MeteorWayne]

I haven't had a chance to follow up on this yet, I'd suggest visiting the NASA Kepler website for why they chose the place they did. Obviously, the chose the plane of the galaxy so that the stellar density was high enough, but not toward the galactic center where there would be too many stars and too much dust.
MW

 

[emudude]

They wanted to get as many stars in a single field as possible. That just happened to be the region which best suited them. I know how you feel though, if they do find something then it will be all the more difficult for us to learn anything about it if it's so far away...let alone get close to it anytime soon; to get to Alpha Centauri (one of the closest stars to our own), using the best theoretical technology we have (nuclear pulse propulsion - banned internationally because it involves a rocket which sets off small nuclear explosions as a method of propulsion, and a problem with launch could be disastrous), it would take something like 40 years and it is only a few light years away...oh, and that's just to get there, to decelerate you can tack on another 40.

 

[RS_Russell]

[##

 

[erioladastra]

You also don't want anything to bright in the field of view or it will saturate the display, causing a significant loss of data. That helps rule ou most nearby stars.

 

[thnkrx]

There is something on the order of 35,000 - 50,000 stars within about 160 lightyears (50 parsecs) of earth, depending on which stellar distribution model pans out (this gets into detailed counts of very faint M and L class dwarf stars).

Kepler, if I remember right from the 'Kepler Input Catalog' info, is supposed to check out stars brighter than visual magnitude (or is it photometric magnitude?) 15, (give or take a few decimal digits - the sources for magnitudes, particularly those of very faint stars, tend to give different values). The area scanned by Kepler (again, if memory serves) works out to something on the order of one seventh of the night sky - apparently just about everything north of Dec +45?, with a large swath extending down to about Dec -20, mostly from RA 19? (counting up) over to around RA 1 or 2?

With these parameters, Kepler should nail something on the order of 90-95% of the stars within 160 lightyears within that field, so...35,000 - 50,000 stars within 160 lightyears divided by 7 comes to 5000 - 7200 stars, give or take. I seem to recollect that there are something on the order of 170,000 stars in Keplers target area....so that works out to something on the order of 3-5% of the total. (Within 100 lightyears...that percentage probably drops to around 1%).

That said, there are several thousand Kepler stars marked out for special investigation (mostly giant stars of various sorts for some reason). There was a (overly) long selection process for the various stars and area to be surveyed by Kepler.

I am not sure, but I believe that the COROT satelite is looking at a different part of the sky than Kepler; the preliminary claims being the discovery of hundreds of extrasolar planets.

 

[MeteorWayne]

Your estimate of the area surveyed by Kepler is probably an order of magnitude or more too high. It's probably closer one sevetyth rather than one seventh...and is probably less than that. It's a rather small area near Cygnus, much smaller than the average comstellation. I'll try and find the actual area is square degrees.

 

[thnkrx]

Made me go track down my source from the catalog:

170,000 stars for the first year; 100,000 stars after that.

Area is 177 square degrees, centered RA 19 22 40 by Dec +44 30. Now...at this point I'm not entirely sure what they are describing.

If it is a 'block' of 177 1 degree squares total, then we are talking about something on the order of 13.5 degrees each way, total width just under 1 hour.

However, the plot maps projections show a sort of funky hour glass type shape which is what you'd get if you projected a much larger area on a sphere. In that case...177 / 2 = 88.5. Take that 'up' from Dec +44 30 and it goes right over the pole and down the other side. Likewise, take it 'down', and it covers things down to about Dec -44. Hour wise...from around RA 14 over to RA 00. That is a big area. (This is what I used in my previous post).

If it is a 'ragged' block of 177 square degrees...that could account for the shape of the projection. In that case, the area surveyed could extend across a couple of hours RA wise (RA 18 to 21?) by a roughly comparable distance in terms of Dec.

There is also a list of several thousand stars targeted for detailed examination; if I understand the material correctly, many of these stars are outside the target area.

 

[MeteorWayne]

If I'm doing this right (which I haven't had time to examine in detail) the sky area is at least 360 x 360 degrees = 129600 degrees. If they are covering 177 sq degrees, that is about 0.1366% of the sky or 1/732 of the visible sky.
MW

 

[halcyondays]

If I'm doing this right (which I haven't had time to examine in detail) the sky area is at least 360 x 360 degrees = 129600 degrees.......

I don't think it works like that. I am not a mathematician, but I think it's something to do with the fact that the sky (or celestial sphere) is not square. Dim memory cells from books a couple of decades back tell me that there are about 40,000 square degrees in the sky. I am sure someone with the requisite skills will come on here and explain further.

 

[centsworth_II]

Dim memory cells from books a couple of decades back tell me that there are about 40,000 square degrees in the sky.

That's what Wikipedia says:
"The number of square degrees in a whole sphere is 4π(180 / π)2 = 129600 / π or approximately 41,253 deg²."http://en.wikipedia.org/wiki/Square_degree

And Bad Astronomy:
"4 x pi x 57.3^2=41253 square degrees."http://www.badastronomy.com/bitesize/bigsky.html

 

[thnkrx]

Hmmm...accepting the above last two posts, and assuming the Kepler info refered to 177 suare degrees, then...

...the area surveyed by Kepler comes to around 0.4% of the total sphere of the night sky. Assuming that the figure of 35,000 - 50,000 stars within approx 160 light years holds up...and we have 140 - 200 stars within that distance.


In terms of stellar density, it is the southern hemisphere that has the most stars. The 'belt' between Dec -50 and Dec -20 has more stars in it than the entire northern hemisphere. So...if the idea is to present the maximum number of targets for Kepler, why aim it at the northern hemisphere?

 

[MeteorWayne]

IIRC, if you look in the Kepler thread, it is monitoring 100 thousand stars. I'll have to go back and check.

Yes:

The Kepler instrument is a specially designed 0.95-meter diameter telescope called a photometer or light meter. It has a very large field of view for an astronomical telescope — 105 square degrees, which is comparable to the area of your hand held at arm's length. The fields of view of most telescopes are less than one square degree. Kepler needs that large a field—105 square degrees—in order to observe the necessary large number of stars. It stares at the same star field for the entire mission and continuously and simultaneously monitors the brightnesses of more than 100,000 stars for the life of the mission—3.5 years

From the FAQ:

Why did you choose the area around Cygnus?

We need a region of the sky that is both rich in stars, and one where the Sun does not get in the way throughout the entire orbit of the Kepler spacecraft. Cygnus is far enough north of the plane of Earth' orbit (the ecliptic) that the Sun will not encroach on Kepler's view, yet is in a very star-rich part of our Milky Way galaxy.


BTW, here is the Kepler thread in Missions and Launches:

viewtopic.php?f=6&t=740

On page 3 is an image of the Kepler Field, as well as the first light images.

Here is the Kepler home page:

http://www.kepler.arc.nasa.gov/

And the NASA Kepler Home Page:

http://www.nasa.gov/mission_pages/keple ... index.html

Latest News Release:

2009 May 1. Mission Manager Update - Kepler's calibration data collection is drawing to a close. Several hundred data sets have been acquired to characterize and map the optical and noise performance of the telescope and the electronics for the focal plane array (the area where light is focused). The data sets are now being analyzed on the ground. Optimally shaped "windows" of pixels will be defined for each of the more than 100,000 target stars and a table of these pixels uploaded to the spacecraft. These are the pixels that will ultimately help the science team find planets -- the pixels will be downlinked to Earth and used to construct light curves, or measurements of brightness over time, for each star.

After science observations begin, the data analysis "pipeline" at the Science Operations Center at NASA's Ames Research Center in Moffett Field, Calif., will process the light curves to identify "threshold crossing events," which is the first step in identifying potential transiting planets. Various tests will be applied to these events to weed out false indications. Once confidence is built for candidate transits, observations by ground-based telescopes will be performed to further rule out phenomena that can masquerade as transiting planets.

2009 May 01 14:00 UTC - Distance to Kepler: 5,382,000 km; 3,344,000 mi; 0.036 AU; 14.00 times the distance to the Moon.