# What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits)

Star number 12644769 from the Kepler Input Catalog was identified as an eclipsing binary with a 41-day period, from the detection of its mutual eclipses (9). Eclipses occur because the orbital plane of the stars is oriented nearly edge- on as viewed from Earth. During primary eclipses, the larger star, denoted “A,” is partially eclipsed by the smaller star “B,” and the system flux de- clines by about 13%

Here's the thing though: out of all possible edge-on configurations, there are far more configurations where a planet can never be in the position to be edge-on, rather than configurations where a planet could potentially be in the position to be edge on. (I'd guess that it happens in fewer than one in several hundred cases)

So why are we able to observe so many transits?

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Because there are so many planets out there!

Transits can only be detected if the planetary orbit is near the line-of-sight (LOS) between the observer and the star. This requires that the planet's orbital pole be within an angle of $d_*/a$ (part 1 of the figure below) measured from the center of the star and perpendicular to the LOS, where $d_{*}$ is the stellar diameter (= 0.0093 AU for the Sun) and $a$ is the planet's orbital radius.

This is possible for all $2\pi$ angles about the LOS, i.e., for a total of $4\pi d_*/2a$ steradians of pole positions on the celestial sphere (part 2 of figure).

Thus the geometric probability for seeing a transit for any random planetary orbit is simply $d_*/2a$ (part 3 of figure) (Borucki and Summers, 1984, Koch and Borucki, 1996).

For the Earth and Venus this is 0.47% and 0.65% respectively (see above Table). Because grazing transits are not easily detected, those with a duration less than half of a central transit are ignored. Since a chord equal to half the diameter is at a distance of 0.866 of the radius from the center of a circle, the usable transits account for 86.6% of the total. If other planetary systems are similar to our solar system in that they also contain two Earth-size planets in inner orbits, and since the orbits are not co-planar to within $2d_*/D$, the probabilities can be added. Thus, approximately $0.011 \times 0.866$ $= 1\%$ of the solar-like stars with planets should show Earth-size transits.

That's pretty freaking amazing! Kepler has been up there for a short while, and has a possible list of nearly 2000 planets just looking at about 150,000 stars for only a couple of years! So if only 1% statistically transit, that would mean that just randomly 1500 systems would have the correct orientation (given the results to date, that makes sense). And given that about 7500 stars were eliminated from consideration due to being variable of one sort or another... I think it would be pretty safe to say that pretty much every star out there has at least some sort of planetary body around it.

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I played with this a while back in the sense that I tried to find "interesting" stars that would see Earth transit the sun. I used the "SIMBAD" website to select stars close enough to the ecliptic (40 of about 40000 within 200LY). Keep in mind this is earth-size at 1AU. I picked my top five that I thought had the best chance of having intelligent life. (These were 1)HD27732, 2)HD95980, 3)HD53532, 4)HD115153 and 5)HD20477). I would welcome editing of these results. – Jack R. Woods Sep 14 '15 at 22:56