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tdvance
2006-Oct-02, 03:23 PM
I wonder if this would work--all math is approximate (e.g. I assume linearity where things are only approximately linear) but this gives ballpark estimates.

A sun-sized star 30 LY away with a planet 1 AU from it is occulted by a 50-mile diameter asteroid 4 AU away.

Ok--our sun is 8 light minutes away and appears 30 arcminutes in diameter, so the 30LY star's disk appears 30*60/(30*365.25*24*60/8)=0.0009126 arcseconds in diameter. Since an AU is about 93 million miles, the asteroid's disk appears to be about arcsin(50/(93000000*4))*180/pi*3600 or about 0.02772 arcseconds in diameter, about 30x as big. 1 AU at 30 LY is about 10 arcseconds because a parsec is 3.26LY so 1 arcsecond is an AU (or is it 2 AU? well, same order of magnitude) viewed from 3.26LY and we pretend this is a linear scaling. Thus, the asteroid's disk is 1/300 the diameter of the orbit of the planet. Conclusion--such a planet should be visible in, say, the Hubble telescope!

Now, by adjusting asteroid sizes, distances, and star sizes, distances, and planet distances, we should come up with lots of opportunities for finding extrasolar planets this way. This will probably be the minority of asteroid occultations, since I know we time them to determine sizes and shapes of asteroids! But if we could predict a suitable one in advance, aim Hubble, and photograph it, we might just detect an extrasolar earth!

Todd

01101001
2006-Oct-02, 03:30 PM
Thus, the asteroid's disk is 1/300 the diameter of the orbit of the planet. Conclusion--such a planet should be visible in, say, the Hubble telescope!
That was a sudden leap to a conclusion. How long of an exposure we can do for this dim, indirectly illuminated planet?

Rob1ooo1oo
2006-Oct-02, 03:35 PM
As soon as I saw this I just thought "exposure time" :rolleyes:

antoniseb
2006-Oct-02, 03:41 PM
This is similar to an idea that was a UT story several weeks ago. The idea for that was to construct a giant sun-shade that would be millions of miles away from a large telescope. In your idea, you get ad hoc lucky chances to see these planets. Not realy a bad idea, but it won't produce huge repeatable numbers of observations.

kzb
2006-Oct-02, 04:39 PM
Actually I seem to remember some schemes with balloons in orbit. All you need is a giant black balloon that you can steer around a bit and an orbiting telescope, the balloon doesn't need to have much mass in space. I don't remember "millions" of miles distance been involved, it was more like several hundred to detect an Earth size planet around the nearer stars. The balloon diameter was several hundred meters. I have not done the maths to check it.

Romanus
2006-Oct-02, 10:21 PM
I think the idea is workable in theory, but not in practice.

I've crunched a few numbers myself, and come up with the following, which assumes that our asteroid has a semimajor axis of 4 AUs, and that our target planet is as bright as Earth, and 1 AU from its primary.

An asteroid with an average distance of 4 AUs would have an orbital period of about 8 years. This works out to a heliocentric motion of 45 degrees per year, or ~443 arcseconds per day. This in turn gives the asteroid a sidereal motion of about 18 arcseconds per hour. If the planet is only ten arcseconds from its primary, we'll only have about a half-hour to image it, even if Earth's own motion buys us a few minutes.

It gets worse. Earth's "absolute magnitude"--its brightness if it were viewed from 1 AU with 100% illumination--is about -3.86, according to the Earth fact sheet. At 30 l.y., the planet would be about 1.9 million AUs distant. Square that in accordance with the inverse square law, take the log and divide by 0.4, and you should end up with a difference of ~31.4 magnitudes. This means that at the very best, the planet would be only +27.5; furthermore, since it would only be half-illuminated, we could probably tack another full magnitude onto that, giving us +28.5. Although that's within Hubble's capacity, I seriously doubt that even that magnificent instrument could pull it off with only a half-hour exposure.

StupendousMan
2006-Oct-02, 11:11 PM
I think the idea is workable in theory, but not in practice.

I've crunched a few numbers myself, and come up with the following, which assumes that our asteroid has a semimajor axis of 4 AUs, and that our target planet is as bright as Earth, and 1 AU from its primary.

An asteroid with an average distance of 4 AUs would have an orbital period of about 8 years. This works out to a heliocentric motion of 45 degrees per year, or ~443 arcseconds per day. This in turn gives the asteroid a sidereal motion of about 18 arcseconds per hour. If the planet is only ten arcseconds from its primary, we'll only have about a half-hour to image it, even if Earth's own motion buys us a few minutes.


It's worse that this -- a LOT worse. The asteroid will cover the star for only a few seconds, not for a half hour.

I know -- I observe asteroid occultations. In fact, there's one tonight: (25) Phocaea occults an eighth mag star for a maximum of about 6.3 seconds.

Romanus
2006-Oct-03, 02:25 AM
^
Wow--I didn't even think about that. Makes perfect sense.

Geez...

astromark
2006-Oct-03, 06:28 AM
The observance of occultations is a very useful tool for the accurate plotting of the path of asteroids. As stated they are brief and often involve low magnitude stars. By very careful time keeping and recording the information from many observation points, a image can be compiled of the tracked objects path. If a spike was to be detected in the spectrographic image. It would not necessarily follow the star has a planet. It could be a much smaller debris orbiting the asteroid.

hxPer
2006-Oct-03, 08:03 AM
Hi Tdvance:

It's good that you're thinking creatively about this. In addition to the very short occultation timescale and the extremely long integration time (you would have to integrate *far* longer than the occultation lasts, so unless you have a nearly perfect coronagraph you're kinda screwed) there's another, more practical problem that you're faced with even if the occultation lasted for 5 hours.

You stated that the seeing disk of the solar analog would be ~0.001 arcseconds. If the simple scaling is correct then, yes, this would be true. However, the light from the star is not spread over 0.001 arcseconds. Assuming, for the sake of argument, infinitesimally small pixels on your CCD or IR detector, the actual size of the star's disk in a (again hypothetical) diffraction limited case would be omega = 0.25 lambda/D where lambda is in microns and D is the telescope diameter in meters and omega is in arcseconds. You want, at bare minimum, for light from the star and that from the planet to land on different pixels. You also want the asteroids disk to fully occult the star. From the latter you need an angular resolution of < 0.027. You also want to observe the planet where it's putting out most of its flux. Assuming MS solar analog the equilibrium temperature at 1 AU is T ~280K*(L/Lsun)^0.25*(r/1AU)^-.5 or ~ 280 ~ 300K. This peaks in the infrared. Taking the most favorable stance that 2-3microns is ok, this works out to a telescope diameter of 28 meters. For comparison, the Keck telescopes are 10 meters across & JWST is not going to be anywhere near 28 meters in size. It's going to be a very long time before we put a 20-30m class telescope in space.

So let's say you want to do this from the ground: good luck. Even with the best seeing (near the VLT site) you're not going to do much better than 0.1" seeing (and even then you're going to need one hell of a wavefront sensor). At the MMT a few nights ago they got ~0.5" seeing and that was considered good! You can work around this a little bit by doing interferometric observations (e.g. with the Keck interferometer), but then you run into the pixel size issue as another logical constraint. I don't know of any IR detector that has <0.01" pixels. Usually their size is on the order of the theoretical performance limit of the telescope (usually 0.1"-1"/pixel).

Good idea, though ;). The Coronagraph design is one of the possible designs for TPF.

StupendousMan
2006-Oct-03, 01:02 PM
Hi Tdvance:
However, the light from the star is not spread over 0.001 arcseconds.


After the light enters a telescope, this statement is correct. But before the light enters a telescope, this statement is not correct. If an asteroid of angular diameter 0.010 arcsec (as seen from the Earth) passes in front of a star of angular size 0.001 arcsec (as seen from the Earth), then the light from the star will be blocked (aside from the very very small fraction which diffracts around the asteroid, given these relative sizes).

So, if only asteroids would move very slowly across the sky, and go exactly where we wished, they might allow us to do a better job of finding planets orbiting nearby stars.

tdvance
2006-Oct-03, 05:29 PM
I didn't think of the exposure time, my mistake--just the fact that the Hubble can "see down to mag. 30" (or something close to that, 29 maybe). I should have realized that was with several-hours' exposure!

I do believe the quantum-mechanical and seeing problems are not a problem when the star is occulted by a large, distant asteroid. That is why simply putting a disk in an earth-bound telescope won't help.

The balloon or other artificial object suggested above might be the best way if we can get it to work.

Todd

hxPer
2006-Oct-04, 03:22 AM
After the light enters a telescope, this statement is correct. But before the light enters a telescope, this statement is not correct. If an asteroid of angular diameter 0.010 arcsec (as seen from the Earth) passes in front of a star of angular size 0.001 arcsec (as seen from the Earth), then the light from the star will be blocked (aside from the very very small fraction which diffracts around the asteroid, given these relative sizes).

So, if only asteroids would move very slowly across the sky, and go exactly where we wished, they might allow us to do a better job of finding planets orbiting nearby stars.

I know, I was using that as an illustration that the angular size of a star from his calculations is not what you would actually see if you looked at the star ( I was writing very late so perhaps I didn't make this point clear). Even if angular resolution weren't a problem (which it is), a very,very small fraction diffracting around the asteroid hurts you given the very small fractional luminosity of the planet compared to the star (hence the statement about a 'near perfect' coronograph). Thus I think you would be hard-pressed to get a 5sigma detection out of a 'planet' (instead of statistical noise or diffracted light). The point I was making is that to get a convincing detection you're still going to have to have the planet and the star (being occulted) landing on different pixels, and the quoted 0.027 " separation must be the separation between the planet and where the occultation is. If you can't resolve things on a 0.027" scale you're not going to be able to detect the planet. The calculation of the telescope size comes from that. In reality you probably need a pixel of background count value in between the planet and the (occulted) star for anyone to believe you which exacerbates the problem.

hxPer
2006-Oct-04, 03:34 AM
I didn't think of the exposure time, my mistake--just the fact that the Hubble can "see down to mag. 30" (or something close to that, 29 maybe). I should have realized that was with several-hours' exposure!

I do believe the quantum-mechanical and seeing problems are not a problem when the star is occulted by a large, distant asteroid. That is why simply putting a disk in an earth-bound telescope won't help.

The balloon or other artificial object suggested above might be the best way if we can get it to work.

Todd

It's not quite so much seeing as angular resolution: seeing just limits your angular resolution if you don't have a good AO system. The two problems with directly detecting extra-solar terrestrial planets are the high level of angular resolution you need to be able to resolve such a planet (you did the angular separation of the planet wrong, the angle gets smaller than 1" as you move past 1pc, not larger) and the extremely high-level contrast you need to be sensitive to between the star and the planet (>10^8 contrast in brightness). What you're proposing is basically using an asteroid as a coronagraph, instead of making one yourself. The occultation helps to resolve the latter problem (though it's unlikely to be good enough) but it doesn't help the former problem since you still need the angular resolution to distinguish between light from a planet and a non-planetary body. Even good coronagraphs don't fully block out the light from a star: for example see the image of the AU Mic disk http://images.google.com/imgres?imgurl=http://www.ifa.hawaii.edu/info/press-releases/aumic-B%2BW.jpg&imgrefurl=http://www.ifa.hawaii.edu/info/press-releases/AU_Mic_images.html&h=1890&w=1365&sz=141&hl=en&start=7&tbnid=FEEUD6Ij1M91oM:&tbnh=150&tbnw=108&prev=/images%3Fq%3DAU%2BMic%26svnum%3D10%26hl%3Den%26lr% 3D