<p>The answer comes from the number of stars examined by each method.
Kepler in the first part of its mission examined 150,000 stars. After the extended mission it has examined 503,506 <a href="https://en.wikipedia.org/wiki/Kepler_space_telescope" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Kepler_space_telescope</a>.
Kepler stared at one patch of sky at a time measuring the brightness of many 10's of thousands of stars at a time. The new TESS satellite mission will examine about 200,000 stars.</p>

<p>The radial velocity method has to take repeated measurements of each candidate star then move onto the next. For bright stars the exposure might be 2 minute, while fainter stars 10 minutes or so ... The HARPS instrument uses the majority of available nights  on a 4m class telescope. The HARPS initial catalog of candidates was 376 <a href="https://phys.org/news/2011-09-exoplanets-harps.htmlstars" rel="nofollow noreferrer">https://phys.org/news/2011-09-exoplanets-harps.htmlstars</a>. This has been expanded and changed over the years. There are several other major radial velocity searches. Their lists overlap so all-in-all they are checking 5,000 stars A personal estimate).</p>

<p>These 2 techniques are fairly comparable as they are both most sensitive to close in planets. So the difference in the number of plants found is because transit surveys examine more than enough stars to overcome the low probability of detecting a planet transiting it host star.</p>

<p>The other techniques for finding exoplanets favor finding different sorts of planets. For example microlensing tends to find high mass planets about the distance of Jupiter from its star. Of the millions of stars check for microlensing roughly 3,000 (now, far less in early searches) show microlensing in a year, of these only 10's have the signatures of planets. It typically takes a year or so to model each event ... so relatively few planets are found.</p>

<ul>
<li>Method           number of stars examined    number of planets found</li>
<li>Transits          500,000+                     3126 </li>
<li>Radial velocity     5,000                       778</li>
<li>Microlensing         100's estimate              84</li>
<li>Direct imaging       100 guess                   47</li>
<li>Astrometry            10 guess                    1</li>
</ul>

<p>see <a href="https://exoplanets.nasa.gov/alien-worlds/ways-to-find-a-planet/" rel="nofollow noreferrer">https://exoplanets.nasa.gov/alien-worlds/ways-to-find-a-planet/</a> 
the estimates of the number of stars are my estimates. </p>

CC BY-SA 4.0
2019-11-06T01:37:03.010Z
2019-11-06T01:37:03.010Z
TazAstroSpacial
<p>Because there are so many planets out there!</p>

<p><a href="http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/character.html" rel="noreferrer">There just happens to be an entire web page dedicated to calculating that answer</a>.</p>

<blockquote>
  <p>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.</p>
  
  <p>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).</p>
  
  <p>Thus the geometric probability for seeing a transit for any random planetary orbit is simply $d_*/2a$ (part 3 of figure) (<a href="http://dx.doi.org/10.1016/0019-1035%2884%2990102-7" rel="noreferrer">Borucki and Summers, 1984</a>, <a href="http://kepler.nasa.gov/files/mws/koch.circumstellar1.pdf" rel="noreferrer">Koch and Borucki, 1996</a>).</p>
  
  <p><img src="https://i.sstatic.net/2KYOW.gif" alt="Diagram"></p>
  
  <p>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.</p>
</blockquote>

<p>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 (<a href="http://en.wikipedia.org/wiki/Kepler_%28spacecraft%29#Mission_results_to_date" rel="noreferrer">given the results to date, that makes sense</a>). 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.</p>

CC BY-SA 3.0
2015-09-14T22:56:14.433Z
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
CC BY-SA 4.0
2021-04-20T02:40:57.577Z
The first link seems broken, but the text occurs [here](https://www.nasa.gov/kepler/overview/abouttransits).
Keith McClary
CC BY-SA 3.0
2013-09-26T02:04:38.347Z
2015-05-03T20:15:25.530Z
2015-05-03T20:15:25.530Z
Larian LeQuella
<blockquote>
  <p>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). <strong>Eclipses occur because the orbital plane of the
  stars is oriented nearly edge- on as viewed from Earth</strong>. 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%</p>
</blockquote>

<p>From <a href="http://www.sciencemag.org/content/333/6049/1602">http://www.sciencemag.org/content/333/6049/1602</a></p>

<p>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)</p>

<p>So why are we able to observe so many transits? </p>

<observational-astronomy><exoplanet><eclipsing-star-systems>
What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits)
Larian LeQuella
# What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits)

(Asked by: username_0 on 2013-09-26)

> 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%

From <http://www.sciencemag.org/content/333/6049/1602>

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?

**Tags:** observational-astronomy, exoplanet, eclipsing-star-systems


---

## 2 answers

### Answer by username_0 on 2013-09-26. Score: 11 | Accepted answer

Because there are so many planets out there!

[There just happens to be an entire web page dedicated to calculating that answer](http://certificate.ulo.ucl.ac.uk/modules/year_one/NASA_Kepler/character.html).

> 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](http://dx.doi.org/10.1016/0019-1035%2884%2990102-7), [Koch and Borucki, 1996](http://kepler.nasa.gov/files/mws/koch.circumstellar1.pdf)).
>
> ![Diagram](https://i.sstatic.net/2KYOW.gif)
>
> 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](http://en.wikipedia.org/wiki/Kepler_%28spacecraft%29#Mission_results_to_date)). 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.

> **Comments:**
> **username_2**: 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.  
> **username_3**: The first link seems broken, but the text occurs [here](https://www.nasa.gov/kepler/overview/abouttransits).  

---

### Answer by username_1 on 2019-11-06. Score: 1

The answer comes from the number of stars examined by each method.
Kepler in the first part of its mission examined 150,000 stars. After the extended mission it has examined 503,506 <https://en.wikipedia.org/wiki/Kepler_space_telescope>.
Kepler stared at one patch of sky at a time measuring the brightness of many 10's of thousands of stars at a time. The new TESS satellite mission will examine about 200,000 stars.

The radial velocity method has to take repeated measurements of each candidate star then move onto the next. For bright stars the exposure might be 2 minute, while fainter stars 10 minutes or so ... The HARPS instrument uses the majority of available nights on a 4m class telescope. The HARPS initial catalog of candidates was 376 <https://phys.org/news/2011-09-exoplanets-harps.htmlstars>. This has been expanded and changed over the years. There are several other major radial velocity searches. Their lists overlap so all-in-all they are checking 5,000 stars A personal estimate).

These 2 techniques are fairly comparable as they are both most sensitive to close in planets. So the difference in the number of plants found is because transit surveys examine more than enough stars to overcome the low probability of detecting a planet transiting it host star.

The other techniques for finding exoplanets favor finding different sorts of planets. For example microlensing tends to find high mass planets about the distance of Jupiter from its star. Of the millions of stars check for microlensing roughly 3,000 (now, far less in early searches) show microlensing in a year, of these only 10's have the signatures of planets. It typically takes a year or so to model each event ... so relatively few planets are found.

* Method number of stars examined number of planets found
* Transits 500,000+ 3126
* Radial velocity 5,000 778
* Microlensing 100's estimate 84
* Direct imaging 100 guess 47
* Astrometry 10 guess 1

see <https://exoplanets.nasa.gov/alien-worlds/ways-to-find-a-planet/>
the estimates of the number of stars are my estimates.


---

