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

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