<p>Consider a circular orbit and approximate the planet as a point.</p>

<p>The distance from Earth to a star is so large compared to the diameter of the star, which we'll call d, that we approximate the lines joining the two poles of the star with the observer as parallel to each other and to the x-axis that we choose. Thanks to symmetry, we choose our coordinates so that the plane of the planet's orbit passes through the tangent plane of the celestial sphere at the star in two places on the y-axis. By symmetry again we ignore half the cases for rotation and also ignore the direction of the orbit and only have to consider how the orbit is inclined at between 0 and pi/2 radians from the x-axis (Edit for clarity: in the XZ plane). Call that angle a.</p>

<p>When the radius r of the planet's orbit is such that d/r = sin(a), the planet will just barely touch the limb of the star when it is closest to the Earth. If sin(a) > d/r it will never happen and if sin(a) &lt; d/r there will be a transit of finite length. Since there is no preference for the value of a, we conclude that the probability of the planet transiting must be $$P(sin(a) &lt; d/r) = P{ \left( a &lt; {arcsin \left( { d \over r } \right)  } \right) } = { 2 arcsin \left( { d \over r } \right) \over \pi } $$</p>

<p>You will notice that for a star-grazing planet, the probability of a transit approaches certainty. Overall probability will depend on the probability distribution of semi-major axes and possibly eccentricity which I ignored here.</p>

CC BY-SA 3.0
2016-11-26T14:22:17.963Z
2016-11-26T14:22:17.963Z
user25972
<p>do most or only a minority of all planetary systems to show transits?</p>

<p>I feel like its minority due to the random orientation of their orbits?</p>

<p>But the thing is I seem to find conflicting information.</p>

<p>Could someone explain which answer is correct because now I am confused?</p>

CC BY-SA 3.0
2016-11-24T07:39:21.307Z
What source tells you that the majority of planetary systems exhibit transits?
ProfRob
CC BY-SA 3.0
2016-11-24T07:49:12.307Z
I just realized that were referring to the fact that the majority of discovered planets were discovered by detecting transits
aa11
CC BY-SA 3.0
2016-11-24T11:42:34.953Z
I was going to vote to close but I think there is a useful question in there. Maybe you mean something like "What fraction of planets do our detection methods miss?" I don't know the answer to that it isn't trivial. But you might also mean something simpler like "What fraction of known exoplanets are transiting?"
Warrick
CC BY-SA 3.0
2016-11-24T14:26:50.250Z
My intended question was What fraction of all exoplanets are transiting  from our line of sight? But its clear that it is very very small due to the randomness of the orbits.
aa11
<planet><fundamental-astronomy><observational-astronomy>
does a minority of all planetary systems show transits?
aa11
# does a minority of all planetary systems show transits?

(Asked by: username_0 on 2016-11-24)

do most or only a minority of all planetary systems to show transits?

I feel like its minority due to the random orientation of their orbits?

But the thing is I seem to find conflicting information.

Could someone explain which answer is correct because now I am confused?

**Tags:** planet, fundamental-astronomy, observational-astronomy

> **Comments:**
> **username_1**: What source tells you that the majority of planetary systems exhibit transits?  
> **username_0**: I just realized that were referring to the fact that the majority of discovered planets were discovered by detecting transits  
> **username_2**: I was going to vote to close but I think there is a useful question in there. Maybe you mean something like "What fraction of planets do our detection methods miss?" I don't know the answer to that it isn't trivial. But you might also mean something simpler like "What fraction of known exoplanets are transiting?"  
> **username_0**: My intended question was What fraction of all exoplanets are transiting  from our line of sight? But its clear that it is very very small due to the randomness of the orbits.  

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## 1 answers

### Answer by username_3 on 2016-11-26. Score: 1

Consider a circular orbit and approximate the planet as a point.

The distance from Earth to a star is so large compared to the diameter of the star, which we'll call d, that we approximate the lines joining the two poles of the star with the observer as parallel to each other and to the x-axis that we choose. Thanks to symmetry, we choose our coordinates so that the plane of the planet's orbit passes through the tangent plane of the celestial sphere at the star in two places on the y-axis. By symmetry again we ignore half the cases for rotation and also ignore the direction of the orbit and only have to consider how the orbit is inclined at between 0 and pi/2 radians from the x-axis (Edit for clarity: in the XZ plane). Call that angle a.

When the radius r of the planet's orbit is such that d/r = sin(a), the planet will just barely touch the limb of the star when it is closest to the Earth. If sin(a) > d/r it will never happen and if sin(a) < d/r there will be a transit of finite length. Since there is no preference for the value of a, we conclude that the probability of the planet transiting must be $$P(sin(a) < d/r) = P{ \left( a < {arcsin \left( { d \over r } \right) } \right) } = { 2 arcsin \left( { d \over r } \right) \over \pi } $$

You will notice that for a star-grazing planet, the probability of a transit approaches certainty. Overall probability will depend on the probability distribution of semi-major axes and possibly eccentricity which I ignored here.


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