RE: [Utah-astronomy] Can't quite figure this one
Did a little quick math. Assume you wish to block a star the size of our sun at a distance of 10 light years. That star subtends an angle of approximately .0188 arcseconds. For every doubling of distance, that angle is halved. Assuming your block is a distance of 10 meters from your scope, it will require a diameter of .4555 microns. Of course that only works if your scope mirror is a point, since the occluding disk has an umbra and a penumbra. I don't know the distance required to make the shadow negigible, but thankfully the disk can get bigger as that distance increases. Of course, the difficulty of keeping that mechanical system in alignment is substantial. But let's assume you've cleared that technical hurdle. At the 10 lightyear distance, an earth-sized planet covers a whopping .000173 arcseconds. Pretty good scope, eh? I'm guessing you're still a lot better off with a great big interferometer and a good CCD system. You can throw away the bright data (the star) and focus your attention on whatever's left over. As I said before, interesting issue. Maybe the real optical guys can chime in on this. [Utah-astronomy] Can't quite figure this one
It shouldn't matter that the blocking piece isn't a point. If covers the star, you'd think it would then allow a view of the planet to show up. Yet I don't think the star really would be a point -- but why not? thanks, Joe
Pretty small, but maybe that's doable. Why does it have to be 10 meters from the scope? Why not 100 meters? It would be difficult, but it still seems easier than building a fleet of telescopes linked together. I seem to remember reading that the difficulty isn't in picking up light from the planet, it's in finding that light when it's swamped by light from the star. -- Joe
Did a little quick math. Assume you wish to block a star the size of our sun at a distance of 10 light years. That star subtends an angle of approximately .0188 arcseconds. For every doubling of distance, that angle is halved. Assuming your block is a distance of 10 meters from your scope, it will require a diameter of .4555 microns. Of course that only works if your scope mirror is a point, since the occluding disk has an umbra and a penumbra. I don't know the distance required to make the shadow negigible, but thankfully the disk can get bigger as that distance increases. Of course, the difficulty of keeping that mechanical system in alignment is substantial.
But let's assume you've cleared that technical hurdle. At the 10 lightyear distance, an earth-sized planet covers a whopping .000173 arcseconds. Pretty good scope, eh?
I'm guessing you're still a lot better off with a great big interferometer and a good CCD system. You can throw away the bright data (the star) and focus your attention on whatever's left over.
As I said before, interesting issue. Maybe the real optical guys can chime in on this.
[Utah-astronomy] Can't quite figure this one
It shouldn't matter that the blocking piece isn't a point. If covers the star, you'd think it would then allow a view of the planet to show up. Yet I don't think the star really would be a point -- but why not? thanks, Joe
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Joe Bauman -
Michael Carnes