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Fraser
2006-Jul-05, 07:04 PM
Space telescopes designed to observe distant planets need to be powerful, but they also need some method of blocking the light from the parent star, which completely washes out any dimmer objects orbiting it. A strategy from CU-Boulder professor Webster Cash would use a large, daisy-shaped space shield to block the light from the star. A space telescope trailing the shade by thousands of kilometres would then be able to see much fainter objects surrounding the star.

Read the full blog entry (http://www.universetoday.com/2006/07/05/sky-shade-could-reveal-planets/)

antoniseb
2006-Jul-05, 07:17 PM
That's pretty cool (and inexpensive, and low-tech compared to the interferometers).

George
2006-Jul-05, 11:01 PM
I would guess a spherical shape would be better, put perhaps not as practical?

atlaurie
2006-Jul-06, 12:16 AM
Sky Shade sounds like a brilliant idea and may be inexpensive as antoniseb says - but would it be possible, cheaper and easier to use the same method by positioning the space telescope to use an existing (preferably dark) body such as a moon or asteroid as the shade?

AlfaCentavra
2006-Jul-06, 03:32 AM
It is possible to see planets in 2006! Place a Hubble Space Telescope in point L2 and use “starshade”.It it will be cheap.

George
2006-Jul-06, 03:59 AM
Wow. Welcome to you tandem two. :clap:

Hi Atlaurie. I would guess the shade would be much easier to manuever and hold still relative to the scope, compared to existing bodies. The shade would need to be very close to round, since stars are round, and asteroids aren't that round. Moons are active orbiting something, so aren't that still.

Hi AlfaCentavra. The Hubble is not powerful enough to get much of a planetary image. L4 and L5 are more stable, though L2 is stable for about three weeks, I think.

[edit: oops, I originally said L3 and L4]

murphyme
2006-Jul-06, 01:28 PM
Hi All,

Is there a reason the shade has to be so large and so far away from the telescope? Could a smaller, rigid, shade attatched to the telescope achieve the same effect?

Michael.

GBendt
2006-Jul-06, 03:06 PM
The basic concept of the telescope looks promising. However, it may be difficult to keep the distant shade exactly centered on the line of sight, and in proper orientation between the telescope and the star that is being searched for planets.

Further, if the shade´s diameter is too large, it may block out not only the light from the star, but also the light from a planet. Diffraction effects caused by the shade´s edge may also hamper the shade´s starlight blocking performance.

A star, although it seems to be just a spot in the heavens, is in fact a very very large object, emitting light from a really vast surface area. If the light of the star falls on the shade, the shade will cast a shadow. The sharpness of the edge of that shadow is dependent on the distance of the shade from the shadow and the diameter of the star. If you observe at a location in that shadow where you are within the unsharp edge of that shadow, you will be hit be light from the star that you don´t want. How large is the "area of unsharp edge" compared to the "dark" shadow?

If we look at solar eclipse pictures taken by satellites, we see that the diameter of the shadow cast by the moon is much wider than the diameter of the small center of that shadow, from where a total eclipse of the sun can be seen. The area of "unsharp edge" in that shadow makes up more than 99%of the entire shadow area.
It is evident that a very precise alignment of shade, telescope and star is absolutely crucial for a successful planet detection by this design. But it can be done!

Regards,

Günther

antoniseb
2006-Jul-06, 03:20 PM
Could a smaller, rigid, shade attatched to the telescope achieve the same effect?

Part of the issue is that the diameter of the telescope has to be smaller as an angle when seen from the shade than the angular separation between the star and the planet. If you had precise enough positioning, you could work with a shade only slightly bigger in diameter than the telescope.

ioresult
2006-Jul-06, 09:44 PM
I did a little calculation with the example cited in the article. Assuming a 2m mirror for the telescope with a 50m diameter shade at 25 000 km, the shade would hide a 100 million km radius circle at 11 light years distance. The partially shaded circle would be 108 million km. So with such a technique, an earth-like planet orbiting at 150 million km from a sun-like star at 11 light years distance would be revealed in all its glory!

The partial shade would get bigger with a bigger mirror, of course. For example, for a 4m mirror, the "penumbra" would extend from 95 to 112 million km from the star. It still would reveal planets in an earth-like orbit. It would even reveal a planet in a venus-like orbit (108M km), although part of the telescope's mirror wouldn't receive the planet's light. It wouldn't be a problem if we were willing to shift the shade to one side a little.

Anyway, that's a great idea and I hope it gets realized. Or at least tested.

AlfaCentavra
2006-Jul-06, 09:52 PM
Hi AlfaCentavra. The Hubble is not powerful enough to get much of a planetary image. L4 and L5 are more stable, though L2 is stable for about three weeks, I think.

[edit: oops, I originally said L3 and L4]

Hi All,

"Darwin" is a proposed European Space Agency (ESA) mission designed to directly detect Earth-like planets orbiting nearby stars, and search for evidence of life on these planets. "Darwin" are going to use 3 meter telescopes (all 3 scopes).Earlier, mission planned to use 2 meter telescopes.
The Hubble has 2.4 meter. I am assured, it is enough to see even terrestrial planets.

George
2006-Jul-06, 11:27 PM
Hi All,

"Darwin" is a proposed European Space Agency (ESA) mission designed to directly detect Earth-like planets orbiting nearby stars, and search for evidence of life on these planets. "Darwin" are going to use 3 meter telescopes (all 3 scopes).Earlier, mission planned to use 2 meter telescopes.
I sure hope they do it. It will have the resolving power of a 100 meter diameter telescope as the spacing is 50m each and is an interferometer.


The Hubble has 2.4 meter. I am assured, it is enough to see even terrestrial planets. It might detect a large, bright planet nearby, but it would not reveal any detail. I hope I'm wrong, but I doubt it.

Grand_Lunar
2006-Jul-07, 02:09 AM
This sounds like a great idea.

I hope funding for it is forthcoming.

AlfaCentavra
2006-Jul-07, 03:35 PM
It might detect a large, bright planet nearby, but it would not reveal any detail. I hope I'm wrong, but I doubt it.

I did a little calculation. From Earth, apparent magnitudes of Mars has -2.8. Between the Earth and Mars there is a distance of 57000000 km. At 11 light years distance apparent magnitudes of Mars will be 28. Faintest objects observable in visible light with Hubble Space Telescope has apparent magnitude of 30.
That is, terrestrial planets can be observed. It would not reveal any detail, of course.

filrabat
2006-Jul-07, 04:11 PM
Welcome to BAUT AlfaCentavia

If you have any questions and answers, or even some non-astronomy related topics, there are other forums for you to enjoy yourself on, too. <grins>

While the distance-magnitude calculations are fresh on your mind, would you mind doing some more calculations - namely, ones showing how far away one must get for a planet to be Mag +30 -- for each planet in our system?

George
2006-Jul-07, 05:58 PM
I did a little calculation. From Earth, apparent magnitudes of Mars has -2.8. Between the Earth and Mars there is a distance of 57000000 km. At 11 light years distance apparent magnitudes of Mars will be 28. Faintest objects observable in visible light with Hubble Space Telescope has apparent magnitude of 30.
That is, terrestrial planets can be observed. It would not reveal any detail, of course.
Yes, that seems correct ( I get 28.5). But what will it see? The resolving power of the Hubble is much too limited to see it as anything other than a point source of light. Can the Hubble be useful in cases like this? Probably not.

It is worth mentioning that your value assumes a brightness value based on what is reflected off Mars. Mars has a low albedo, 15%, and is small compared to all the other known exoplanets. This should make most of the known exoplanets appear a couple of magnitudes brighter. Using a brighter star would also increase the brightness.


While the distance-magnitude calculations are fresh on your mind, would you mind doing some more calculations - namely, ones showing how far away one must get for a planet to be Mag +30 -- for each planet in our system?

It would be double the distance for Mars. This would be true for Jupiter since Mars and Jupiter are essentially equal when they are at their brightest.

Interestingly, using the sun as your light source, you don't have to go all that far from the sun to reach a 30 mag. level. Move the planets 1500 a.u. (0.024 lyr) outward and the Hubble would not be able to see them. The Oort Cloud may extend beyond 1 light year from the Sun. No Jupiter-sized objects are expected out there, but if they are, we wouldn't be able to see them with the Hubble.

Astrowannabe
2006-Jul-21, 02:54 AM
Hello everyone. I'm an astrophysics student here at CU Boulder, and this past semester I had an astronomy class with Professor Webster Cash. He was understandably excited about the New Worlds Observer and gave us updates on it quite a bit in class. I also went into his office and spoke with him about it once, so perhaps I can shed some light onto the discussion here.

First of all, anyone who wants to read his proposal can find it here: http://casa.colorado.edu/~wcash/Planets/new_worlds.pdf

You'll need a pdf viewer to read it.

First of all, the shape of the shade is most deffiently NOT spherical. While I don't pretend to fully understand all of it, I know the shape has something to do with the light from the star interfering with itself in the proper way to cancel out the starlight, leaving only light from any planets getting through. He has a picture of it in his presentation, as well as a fair amount of math for you guys to chew over. The shape must be very precise, which is where a large chunk of the 400 million dollars is going.

So since just a spherical body won't do, that rules out moons/asteroids as well.

Also, even though getting a nice, beautiful picture of an extra-solar planet that you could hang on your bedroom wall would be very nice, it's not required. Even seeing an extra-solar planet as just a point of light, one or two pixels on a computer screen we can do some very real science with it. As long as we get some light, ANY light, from the planet we can run it through a spectrograph to determine the chemical make-up of the planets atmosphere.

One intriguing possibility of that is that the atmosphere of Earth (and we assume anyplace with living things) is actually not stable. If you seal the Earths atmosphere in a jar and leave it on a shelf for a few million years, it's no longer made up of the same compounds. The air will very slowly degrade. What keeps out air the way it is is that life slowly and constantly refreshes the air. For example, left on it's own the oxygen in our air would slowly get reabsorbed back into the rocks. So if we found a planet with an atmosphere that is "unstable", then it's a good possibility that life could be there. Not 100% though, since many geological events can do similar things.

At any rate, read through his presentation and it should clear up a lot of the questions you guys have. I personally have high hopes for this mission!