<p>From now on <span class="math-container">$r_s$</span> represents <span class="math-container">$r_{star}$</span>, <span class="math-container">$r_o$</span> represents <span class="math-container">$r_{orbit}$</span>, <span class="math-container">$r_p$</span> represents <span class="math-container">$r_{planet}$</span>.</p>
<p>By trigonometry, the orbital inclination <span class="math-container">$i$</span> must be smaller than <span class="math-container">$\sin (\frac{r_s-r_p}{r_o})$</span> (draw a triangle on your paper, it will be very easy to see)</p>
<p>By small angle approximation, it gives that <span class="math-container">$i_{max} = \frac{r_s-r_p}{r_o}$</span></p>
<p>Note we need to multiply <span class="math-container">$i_{max}$</span> by <span class="math-container">$2$</span> because the orbit plane can be both &quot;above&quot; or &quot;below&quot; a plane for zero inclination.</p>
<p>You can think of a strip on the celestial sphere with a width of <span class="math-container">$2i_{max}$</span>, the area covered in the strip is where an observer is able to see the transit.</p>
<p>This strip has an <a href="https://en.wikipedia.org/wiki/Solid_angle" rel="nofollow noreferrer">solid angle</a> of:</p>
<p><span class="math-container">$\Omega = 2\pi \sin (2i_{max}) \approx 4\pi i_{max}$</span> (integrated over azimuthal angle and polar angle)</p>
<p>The entire sphere has a solid angle of <span class="math-container">$4\pi$</span>, thus the probability of the star's transit is:</p>
<p><span class="math-container">$P = \frac {\Omega}{4\pi} = \frac{4\pi i_{max}}{4\pi} = \frac{r_s-r_p}{r_o}$</span></p>
<p>When the planet is transiting through the sun, the angle it travels is <span class="math-container">$\theta = 2\sin^{-1} \frac{r_s}{r_o} \approx 2 \frac{r_s}{r_o}$</span>, so at any given moment, the probability of the planet transiting is <span class="math-container">$P^{'} = \frac{\theta}{2\pi}$</span></p>
<p>I don't know how to add the observation time into this problem, let someone else fix it.</p>
<p>If you want to see the mathematical proof for the solid angle equation, leave a comment and I will add it in there. Hope is helps ;)</p>

CC BY-SA 4.0
2024-07-04T03:10:31.710Z
Maybe this is just a question of terminology, but the orbital inclination should have nothing to do with the physical sizes of the planet and star (or the size of the orbit, for that matter).
HDE 226868
CC BY-SA 4.0
2024-07-04T03:49:58.227Z
@HDE 226868, You misunderstood me, I am using the size of the planets to figure out the maximum orbital inclination, the orbital inclination does not have anything to do with the size of the planets etc, but in this situation, the maximum orbital inclination is related to the size of the planet and size of the star and size of the orbit.
Polaris5744
CC BY-SA 4.0
2024-08-04T07:48:20.013Z
Yeah, I forgot that part, I will edit my answer but I am not familiar with probability.
Polaris5744
CC BY-SA 4.0
2024-07-04T02:51:43.950Z
2024-08-04T07:53:01.620Z
2024-08-04T07:53:01.620Z
Polaris5744
<p>An exoplanet is in a <em>circular</em> orbit around its star.</p>
<p>The radius of the star is <span class="math-container">$r_{star}$</span></p>
<p>The radius of the exoplanet is <span class="math-container">$r_{planet}$</span></p>
<p>The radius of the orbit is <span class="math-container">$r_{orbit}$</span>.</p>
<p>The orbital period of the planet is <span class="math-container">$T$</span></p>
<p>What is the probability that, from the perspective of a distant observer who is viewing the system from a random angle (in 3D, alternatively and equivalently, the orbital plane is randomly oriented), the (full) disk of the exoplanet occludes the disk of its star during an observation duration <span class="math-container">$t$</span> (randomly offset)?</p>
<p>Is there an analytic solution for this problem?</p>

CC BY-SA 4.0
2023-01-21T11:45:10.187Z
I started to answer but then realised what you are asking is more complex than I first thought... I think there is an analytic solution, though mostly I believe it is Monte-Carlo integrated to take into account observing cadence and signal to noise too.
ProfRob
<mathematics><transits-other-than-sun>
Probability of exoplanet transit
2080
# Probability of exoplanet transit

(Asked by: username_0 on 2023-01-20)

An exoplanet is in a *circular* orbit around its star.

The radius of the star is $r\_{star}$

The radius of the exoplanet is $r\_{planet}$

The radius of the orbit is $r\_{orbit}$.

The orbital period of the planet is $T$

What is the probability that, from the perspective of a distant observer who is viewing the system from a random angle (in 3D, alternatively and equivalently, the orbital plane is randomly oriented), the (full) disk of the exoplanet occludes the disk of its star during an observation duration $t$ (randomly offset)?

Is there an analytic solution for this problem?

**Tags:** mathematics, transits-other-than-sun
> **Comments:**
> **username_1**: I started to answer but then realised what you are asking is more complex than I first thought... I think there is an analytic solution, though mostly I believe it is Monte-Carlo integrated to take into account observing cadence and signal to noise too.  

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## 1 answers

### Answer by username_2 on 2024-07-04. Score: 1

From now on $r\_s$ represents $r\_{star}$, $r\_o$ represents $r\_{orbit}$, $r\_p$ represents $r\_{planet}$.

By trigonometry, the orbital inclination $i$ must be smaller than $\sin (\frac{r\_s-r\_p}{r\_o})$ (draw a triangle on your paper, it will be very easy to see)

By small angle approximation, it gives that $i\_{max} = \frac{r\_s-r\_p}{r\_o}$

Note we need to multiply $i\_{max}$ by $2$ because the orbit plane can be both "above" or "below" a plane for zero inclination.

You can think of a strip on the celestial sphere with a width of $2i\_{max}$, the area covered in the strip is where an observer is able to see the transit.

This strip has an [solid angle](https://en.wikipedia.org/wiki/Solid_angle) of:

$\Omega = 2\pi \sin (2i\_{max}) \approx 4\pi i\_{max}$ (integrated over azimuthal angle and polar angle)

The entire sphere has a solid angle of $4\pi$, thus the probability of the star's transit is:

$P = \frac {\Omega}{4\pi} = \frac{4\pi i\_{max}}{4\pi} = \frac{r\_s-r\_p}{r\_o}$

When the planet is transiting through the sun, the angle it travels is $\theta = 2\sin^{-1} \frac{r\_s}{r\_o} \approx 2 \frac{r\_s}{r\_o}$, so at any given moment, the probability of the planet transiting is $P^{'} = \frac{\theta}{2\pi}$

I don't know how to add the observation time into this problem, let someone else fix it.

If you want to see the mathematical proof for the solid angle equation, leave a comment and I will add it in there. Hope is helps ;)

> **Comments:**
> **username_3**: Maybe this is just a question of terminology, but the orbital inclination should have nothing to do with the physical sizes of the planet and star (or the size of the orbit, for that matter).  
> **username_2**: @HDE 226868, You misunderstood me, I am using the size of the planets to figure out the maximum orbital inclination, the orbital inclination does not have anything to do with the size of the planets etc, but in this situation, the maximum orbital inclination is related to the size of the planet and size of the star and size of the orbit.  
> **username_2**: Yeah, I forgot that part, I will edit my answer but I am not familiar with probability.  

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