I'm trying to calculate the probability of transit of an exoplanet in orbit around a star of diameter $d$, at a distance of $4d$ from its centre. Here's what I did:

enter image description here

enter image description here

In this figure, we are only concerned about the right hemisphere. The idea I have used is that for a transit to be visible for an observer situated directly to the right, the orbit must extend to within $l$ of the right extreme of this sphere.

To calculate $l$, I applied basic trigonometry:

$$ \frac{\theta}{2}=tan^{-1}\frac{1}{8},$$

and $\theta$ came out to be roughly $0.2$ radians.

$$l = 4d * \theta \approx 0.8 \ d.$$

Now the probability is simply given by $\large\frac{0.8 \ d}{4 \pi d}$ and this comes out to be about $\large\frac{1}{5\pi}$.

Where have I gone wrong, because the answer to this problem is supposed to be $\large\frac{1}{8\pi}$?

Even after approximating $l$ to be nearly $d$, the answer I get is $\large\frac{1}{4\pi}$ and not $\large\frac{1}{8\pi}$.

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    $\begingroup$ Please edit the image to make it smaller, I don't know how to do that. $\endgroup$ Commented Jan 13, 2022 at 8:17
  • $\begingroup$ Your two images seem to disagree about the definition of d. In the upper it is l = d. In the lower it is $l = 4d\cdot \theta / (2\pi) \approx 2d$. Is the lower sketch is supposed to be perpendicular to the top view? $\endgroup$ Commented Jan 13, 2022 at 8:31
  • $\begingroup$ @planetmaker Thanks for pointing it out. I have edited it. $\endgroup$ Commented Jan 13, 2022 at 9:10
  • $\begingroup$ @AmbicaGovind I've made some edits, using a trick to display the images smaller and adjusted the MathJax formatting. Please have a look and feel free to edit further. $\endgroup$
    – uhoh
    Commented Jan 13, 2022 at 10:05
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    $\begingroup$ I haven't throught this through. However there is a probability for a central passage to happen right now (orbit inclination of 0°), but the probability of this has to be multiplied with the probability of the orbits inclination being such that it transits at all due to inclination (I recon that could be $\theta$ in the lower sketch). I don't see this being taken care of in the calculation. $\endgroup$ Commented Jan 13, 2022 at 10:43

2 Answers 2


Before I tackle this problem, I will assume that the size of the planet is negligible compared to the size of the star, so I won't take into account partial transits. I will also assume circular orbits. I will also assume uniform random probability distribution of orbital revolution axes on the unit sphere.

Suppose the orbital revolution axis of the planet is pointed directly at us, the observer. Then, we would be perfectly perpendicular to the orbital plane and there would be no transit. Instead, from our perspective, the planet's orbit would be perfectly circular at a distance of $4d$ from the star.

Only when the orbital revolution axis (along r in the below image) was some angle $\theta$ from our line of site (along h), would the planetary orbit cross the star from our perspective. The revolution axes where no transit exists could be projected onto a unit sphere as a spherical cap, with area $A = 2\pi (1-\cos{\theta})$. The area of a unit hemisphere is $A_h=2\pi$, so the fractional area of no transits (which is the probability of no transits) is simply $P(\sim t)=A/A_h=1-\cos{\theta}$. Image credit wikipedia

enter image description here

The angle $\theta$ is simply the complementary angle of the maximum transit angle, so $\theta=\cos^{-1}{(1/8)}$. So $P(\sim t)=1-1/8$, and the probability of a transit is then $P(t) = 1- P(\sim t)=\frac{1}{8}$.

Note: The OP claims the answer should instead be $P(t) = \frac{1}{8\pi}$. Please let me know if you find my missing $\pi$!

  • $\begingroup$ "The angle θ is simply the complementary angle of the maximum transit angle,..." is difficult to follow. The "maximum transit", if it is defined as the transit with longest time, would it not be with θ =90°? in your convention for θ. In this case the non-transit cap is defined by that max θ that results in the minimum transit time (the planet just touches the star disk w/o entering it). Did I miss something? $\endgroup$
    – Ng Ph
    Commented Jan 17, 2022 at 19:29
  • $\begingroup$ @NgPh Sorry I was late with this reply. In the figure in your answer, my $\theta$ is the same as $i$, Since cosine is the Adjacent over Hypotenuse of a right triangle, $\cos{\theta} = 1/8$. The maximum transit angle is the (complementary) unlabeled interior angle. $\endgroup$
    – Connor Garcia
    Commented Jan 18, 2022 at 16:34

This is an amusing (and tough) math quiz.

We are confronted with 3 numerical results and various mathematical reasonings (of which one is not given):

    1. the OP's calculation gives $Pt=0.8/4\pi$ (about 6.4%).
    1. the OP’s announced correct result $Pt=1/8\pi$ (about 4%), w/o demonstration or source.
    1. @Connor Garcia calculation gives $Pt=1/8$ (12.5%).
    1. The tutorial linked by @ProfRob, Slide 7, gives $Pt=R/a$=(d/2)/(4d)=1/8

Definition of Pt

Given an Observer(O) at a far distance from a Star (S) of radius R, having a much smaller Exoplanet orbiting at distance $a$, the probability that the Observer can detect the Exoplanet by the transit method is denoted Pt. Note that the duration of transit, the luminosity of the Star, the observation time and the sensitivity of the detector are not considered. It is purely a geometrical exercise.

For example, when the orbital plane contains the line between the Observer and the Star (O-S), the transit time is maximum. We can call these orbits « maximum transit orbits » (for that Observer position). When the orbital plane is such that the exoplanet just touches the disk of the Star, viewed from the Observer, the transit time is minimal (near zero). We call these « minimum transit orbits ». Both types of orbits count equally as detectable in the calculation of Pt. When O-S coincides with the axis of rotation, the Observer will never see a transit.

The reasonings

The reasoning in (3) computes (1-Pt) as the ratio of the areas of two spherical caps: the "non-transit cap" and the hemisphere cap facing the observer, both on a unit sphere.

The reasoning in (1) computes Pt as the ratio of arc lengths: the arc of maximum transit and the half circumference, both of a circle of radius a (=4d).

The reasoning in (4) computes Pt as the probability Prob{abs(cos(i)) < $R/a$}, $i$ being the angle that the orbital plane makes with the perpendicular to O-S (see Fig).

enter image description here source

Walking through the reasoning in (3)

Take a perfect sphere and make the following « experiment » (I owe this intuitive and nice reasoning to @ConnorGacia):

  • pick a random point on the sphere,

  • paint in blue a cap of half-cone angle $\theta$ centered at the chosen point. enter image description here

  • Paint also the « opposite » cap (180° symmetric).

  • Let the sphere roll on a bumpy surface so that, when it comes to rest, any point on the sphere has equal opportunity of being the rest point of contact with the surface. The chance that the sphere comes at rest at a point inside any of the two blue areas is equal to the ratio of one cap area to the area the sphere, multiplied by 2. This is the same as taking area of one cap and divide it by the area of the hemisphere. When the sphere has a radius r=1, the area of each blue cap is $2\pi(1-cos \theta)$ and the hemisphere has an area equals to $2\pi$. Hence the probability of not coming to rest in the blue area is $cos \theta$.

The connection to our Pt is as follows :

Call the angle between the line O-S and the axis of orbit rotation, $\delta$. Call $\theta$ that value of $\delta$ resulting in minimum transit orbits (in the direction of the given Observer). If we increase $\delta$ from 0° (the orbit is orthogonal to the O-S), then the orbit becomes a transiting one (from the Observer’s point of view), when and only when $\delta$ crosses $\theta$.

For any given R and a, it turns out that the value of $\theta$ is related only to R and a in the following way $cos \theta = sin (90°- \theta) = R/a$.

For a given orbit, consider a (any) sphere centered at the star and draw on it the two blue caps, each cap is centered on the orbit axis of rotation. Choosing a random direction for O-S is akin to rolling that sphere on a bumpy surface. The Observer can detect the planet if the O-S does not intersect any of the 2 blue caps.

Dissecting and comparing (1) and (4)

Both (1) and (4) reasonings are based on a reduction to a « 2D » geometry. Both make use, more or less, of the same angles.

In (1), it is reasoned that, when we increase the inclination angle (called $\theta/2$ in the figure), from 0° (maximum transit), the orbit leaves the transit zone when $sin(inclination)=R/a$.

In (4), it is reasoned that when we reduce the angle $i$ (see Fig) from 90° (maximum transit), the orbit leaves the transit zone when $i$ reaches the value that satisfies $cos i =R/a$.

They are both correct up to this point, but diverges on a (very subtle) mathematical point. In (1), it is assumed obvious that the inclination angle is uniformly distributed, whereas in (4) it is $\cos i$ that is uniformly distributed. We can suspect that the calculation of (4) is correct because the result of (4) matches that of (3). But why is it that saying that the inclination is uniformly distributed is flawed ?

Another ball rolling experiment

I will resort again to the intuitive (and nice) « experiment » with the rolling sphere above (again, I am indebted to @ConnorGarcia for this). This time we do not paint the sphere. Instead, once an orbit is chosen, we identify the Equator and the Poles and we draw the « latitude» markers, from Equator to Poles. We roll the ball again on our bumpy surface and when it comes to rest, we note the latitude of the resting point. This latitude is the inclination angle of the orbit with respect to the (random) observation line O-S, which is the outcome of our experiment run. If we repeat several runs of this, we can observe that we get more inclination below 45° than inclination above 45°. Another test is to count how many times we get a rest at, or close to any Pole, and how many time we get, at or close to the Equator. We can observe that we get more often closer to the Equator than to the Pole. This shows that the random inclination angle is not uniformly distributed.

The reasoning in (1) however is based on this false evidence and therefore its result is incorrect.

If you like this answer, please upvote @Connor Garcia too.

  • $\begingroup$ Related: astronomy.stackexchange.com/questions/35471/… $\endgroup$
    – Connor Garcia
    Commented Jan 18, 2022 at 16:38
  • $\begingroup$ I think a key takeaway from this answer is that the probability distribution of orbital inclinations should not be treated as uniform random from 0 to 180. Instead, one should consider the distribution of orbital axes on the unit sphere. $\endgroup$
    – Connor Garcia
    Commented Jan 18, 2022 at 16:53
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    $\begingroup$ @Connor Garcia, neatly formulated. Nevertheless, as much as I am convinced I have identified where (1) is wrong, I am still unable to find an intuitive way to answer the "why this is wrong?". Especially when the 2 functions $1-cos \theta$ and $2\theta/\pi$ agree at the two extremes of $\theta=0$ and $\theta=\pi/2$. Need more coffee ! $\endgroup$
    – Ng Ph
    Commented Jan 18, 2022 at 18:30
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    $\begingroup$ @Connor Garcia, I think you are taking me closer to a "AHA! moment". I may be able to connect the dots to the related question (your 1st comment post). But I haven't fully got the connection with this problem, yet. Please, do not give another hint before a couple of days. I like math quiz! $\endgroup$
    – Ng Ph
    Commented Jan 19, 2022 at 12:02
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    $\begingroup$ @Connor Garcia, AHA! Please check my edits. $\endgroup$
    – Ng Ph
    Commented Jan 20, 2022 at 21:18

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